Communications Script 2013 En

Lectures script

Introduction to Communications

Summer semester 2013

Prof. Dr.-Ing. Dr. h.c. Gerhard P. FettweisTechnical University of DresdenFaculty of Electrical Technology

Vodafone ChairMobile Communications Systems

D-01062 Dresden

Status: February 12, 2013Version: 2.1.8

CONTENTS 3

Contents

1 Preface 7

2 Goal of the Lecture 8

2.1 Systems for Information Transmission . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Transmission Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.5 Signal Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Signal Theory 12

3.1 Sine Signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2 Generalized Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2.1 Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.2.2 Integration of Generalized Functions . . . . . . . . . . . . . . . . . . . . . 14

3.2.3 Differentiation of Generalized Functions . . . . . . . . . . . . . . . . . . . 15

3.2.4 Sifting Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2.5 Calculation with Dimensional Parameters . . . . . . . . . . . . . . . . . . 15

3.3 Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Linear Time Invariant System 22

4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.2 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3 Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.4 Convolution as the Basic Operation in Communications . . . . . . . . . . . . . . . 24

4.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.4.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.4.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.5 Consideration in Time and Frequency . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.5.1 Fourier Transformation of Rect Function . . . . . . . . . . . . . . . . . . . 27

4.5.2 Zeit- und Frequenzbegrenzung . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.5.3 Raised-Cosine-Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.6 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30


4 CONTENTS

5 Bandpass Signal 32

5.1 Up and Down Conversion of Real Baseband Signals . . . . . . . . . . . . . . . . . 32

5.2 Up and Down Conversion of Complex Baseband Signal . . . . . . . . . . . . . . . 37

5.3 General Bandpass Signal – Equivalent Low-Pass Signal . . . . . . . . . . . . . . . 41

6 Analog Modulation 42

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6.2 Modulation Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 Amplitude Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3.1 Description in Time Domain . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3.2 Description in Frequency Domain . . . . . . . . . . . . . . . . . . . . . . . 44

6.3.3 Power Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.3.4 AM-Modulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.3.5 AM-Demodulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.3.6 Different Types of Amplitude Modulation . . . . . . . . . . . . . . . . . . 48

6.4 Phase and Frequency Modulation (Angle Modulation) . . . . . . . . . . . . . . . . 49



6.4.3 Power Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.4.4 FM-Modulator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6.4.5 FM-Demodulator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.5 Comparison between Amplitude and Angle Modulation . . . . . . . . . . . . . . . 56

7 Analog-Digital-Conversion 57

7.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.2 Time Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.2.1 Dirac-Comb-Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

7.2.2 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

7.2.3 Sampling Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.2.4 Signal Reconstruction – Digital-Analog-Conversion . . . . . . . . . . . . . 62

7.3 Value Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

7.3.1 Quantization Characteristic Curve . . . . . . . . . . . . . . . . . . . . . . 65

7.3.2 Quantization Error Signal . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7.3.3 Signal-to-Noise Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

7.3.4 Oversampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

7.3.5 Error Modelling of AD Converter . . . . . . . . . . . . . . . . . . . . . . . 69

Script ”Introduction to Communications”

CONTENTS 5

8 Digital Modulation 70

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

8.2 Types of Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

8.3 BPSK – Binary Phase Shift Keying . . . . . . . . . . . . . . . . . . . . . . . . . . 71


8.3.2 Sender . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

8.3.3 Phase Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74


8.3.5 Receiver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

8.3.6 Data Transmission over Noisy Channel . . . . . . . . . . . . . . . . . . . . 79

8.3.7 Calculation of Bit Error Ratio . . . . . . . . . . . . . . . . . . . . . . . . . 81

8.4 QPSK – Quaternary Phase Shift Keying . . . . . . . . . . . . . . . . . . . . . . . 85

8.5 Further ASK/PSK Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

8.6 Frequency Shift Keying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

References 88

A Formulas 90

A.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

A.2 Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

A.3 Notes on Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

B Appendix 94

B.1 Analytical Signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

C Excercises 96

C.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

C.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

D Exams 147

D.1 Exam SS 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

D.2 Solutions: Exam SS 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

D.3 Exam WS 2006/2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

D.4 Solutions: Exam WS 2006/2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

D.5 Exam SS 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172


D.7 Exam WS 2007/2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185


6 CONTENTS


D.9 Exam SS 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198


D.11 Exam WS 2008/2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213


D.13 Exam SS 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228


D.15 Exam WS 2009/2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244


D.17 Exam SS 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258


D.19 Exam WS 2010/2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272


D.21 Exam SS 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289


D.23 Exam WS 2011/2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304


D.25 Exam SS 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320


D.27 Exam WS 2012/2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337



7

1 Preface

The script contains the lectures for Communications course given in the Summer semester. ThisEnglish version of the script is made in order to help the foreign, German language non-speakingstudents, to follow the lectures. The content of the script doesn’t however differ from the one inGerman which is normally used. The same holds for the contents of the lectures.

The first version of the German script was created in 1996 by a student Michael Hosemann andfrom 1997 to 1999 was revised by Dipl. Ing. Achim Nahler. The second complete revision tookplace from 1999 to 2001 by Dipl. Ing. Matthias Henker. Additional revisions were done by Dipl.Ing. Denis Petrovic, Dipl. Ing. M.Sc. Peter Zillmann, Dipl. Ing. Andreas Frotzscher, Dipl. Ing.Stefan Krone, Dipl. Ing. Fabian Diehm and Dipl. Ing Jan Dohl.

This English version of the script was translated from German by the M.Sc student Jafar SadequeAdnan (e-mail: [email protected]) under the supervision of Dr.-Ing. Denis Petrovic.

The script is now managed by Dipl.-Ing. Bjorn Almeroth (Telefon: (463) 41040, e-mail:[email protected]).

Please report any errors or mistakes to improve the script.

Books for self study

• Books on signal theory and LTI systems: [Hof98], [Fli91], [WS93], [Fet96]

• Books on communications: [Kam96], [Luk95], [Pro95], [Cou93]


8 2 GOAL OF THE LECTURE

2 Goal of the Lecture

The task of communications consists of carrying information from a sender to a receiver. Thecommunications technology can be roughly divided into two large areas, the

• transmission techniques and the

• switching techniques.

In this lecture the emphasis will be on the problems of transmission techniques. Since this areais also very heterogeneous, only the most important and elementary things will be dealt here.Examples for applications of communications-engineering are :

• Audio broadcast and television

– analog: AM-Radio (Medium wave), FM-Radio (USW),

– digital: DAB (digital audio broadcasting), DVB (digital video broadcasting),

• Telephone

– Fixed network,

– Cellular network.

2.1 Systems for Information Transmission

One can describe the information transmission systems by a block diagram as in Fig. 2.1.

lecture contents

SourceSource-

coder

Channel-

coder

Modulator

Multiplexer

ChannelInterference

(noise,fading)

Demodulator

Demultiplexer

Channel-

decoder

Source-

decoderSink

Channel-source-adaptationMulti-medium

utilization

Redundancyreduction(Data

compression)

Increasingredundancy(higher errorrobustness)

Figure 2.1: General transmission system in communications


2.2 Concept 9

2.2 Concept

The building blocks source, source coder, channel coder, modulator and multiplexer are summa-rized under the term sender. Accordingly the modules demultiplexer, demodulator, error correctionelements, source decoders and sink belong to the receiver. Senders and receivers can be both sta-tionary (e.g. television station) and can be mobile (e.g. Handy). They are however always powerlimited. The channel, as transmitting medium, is bandwidth limited. The transmitted informationis influenced by noise, amplitude oscillations (fading, caused by drift and shadowing), interferenceappearances, time dispersion (delay spread, caused by multi-path propagation) and frequency dis-persion (Doppler spread, caused by movement of sender, receiver and/or scatterer/reflectors etc.).

2.3 Transmission Medium

The selection of the transmitting medium depends greatly on the demands imposed on the trans-mission channel (e.g. concerning frequency range or signal bandwidth, in addition, regarding theconcurrent number of users). Possible transmitting mediums are

• ”‘Twisted Pair”’ (twisted copper cable),

– e.g. Telephone cable (link terminal)

• coax cable,

– e.g. Antenna cable, TV cable

• wave guide,

– e.g. antenna feed with high frequencies (GHz range)

• Optical fibre,

– e.g. transmission with very high data rates

• Radio channel.

– e.g. cellular, radio and television broadcast

From the point of radio communications, one can differentiate between indoor and outdoor appli-cations. A typical example of indoor applications could be a WLAN (wireless local area network)in one Building. Outdoor applications are e.g. cellular radio networks widely spread throughoutthe country, at present GSM-900, DCS-1800 (D1, D2, Eplus and E2 network) and in the futurealso UMTS. Also the frequency or wavelength ranges can be differentiated, begin from the well-known MW and USW ranges up to the millimeter wave bands and far infrared ranges of opticalcommunications technology.


10 2 GOAL OF THE LECTURE

2.4 Properties

In the following, we name few characteristics of the communication systems.

Simplex/Duplex A distinction criterion is whether a system can operate in simplex or duplexmode. Simplex mode means that messages are transferred only into one direction (e.g. broad-cast), while in the duplex operation the information can be transferred in both directions(e.g. Telephone).

Single-Cast/Multi-Cast There are Single-Cast-System (Telephone: 1 source, 1 receiver) andMulti-Cast-System (Broadcast: 1 source with many receivers)

Packet/circuit-switching Another way to distinguish between systems is based on switching,circuit-switched (e.g. ”‘the good old” ’ telephone) or package-switched (e.g. data communi-cation in the Internet–IP-protocol).

2.5 Signal Level

Often signals with large power differences are given. Typical values for signal power P lie between1μW and 1kW. That corresponds to a difference of 109. For this reason a logarithmic scale isadvantageous. One example of such a scale is the dBm scale, where the power level LP is normalizedon Pref = 1 mW, as follows

P = power(s(t)) (2.1)

LP = 10 log10P

Pref

dBm

LP = 10 lgP

Pref

dBm mit Pref = 1 mW (2.2)

abs. Power in [mW] 0.1 0.5 1 2 4 8 10 100 1000

rel. Power in [dBm] -10 -3 0 3 6 9 10 20 30

Table 2.1: Absolute power vs. dBm-level

• In Tab. 2.1 some absolute powers and their corresponding dBm values are given.

• 2 W-Handy (D-Net): Pmax = 2 W, equivalent representation as level LPmax =10 lg 2W

1mWdBm = 10 lg(2 · 103)dBm = (10 lg(2) + 10 lg(103))dBm = 33dBm. In GSM-

Standard it is specified that the level at the base station has to be minimum -102 dBm, i.e.it can result in a level difference up to 135 dBm respectively.

• 0.8 W-Handy (E-Net): ≡ 29dBm. In E-Net signals are transmitted with less power than inD-Net. In addition at frequency range where this system operates (1.8 GHz) attenuationis much stronger. Therefore in E-network a larger number of base stations is necessary fortransmission in contrast with the D-network, which is a drawback regarding to infrastructurecosts.


2.5 Signal Level 11

• Also signal voltages can be given in the logarithmic scale. As reference voltage 0.775 Vis usually used and it corresponds 0 dBu (Choice of reference voltage: Which voltageis necessary, in order to develop a power of 1 mW at a standard resistance of 600Ω?= 0.7752V2/600Ω = 1mW). The selection of another standard resistance or reference voltagewill shift the dB-scale accordingly.

Likewise system amplifications can be specified as equivalent to level. Indicating e.g. x and yrespectively as input and output of a system, yields the levels as follows,

gP =Power(y)

Power(x)resp. gA =

Amplitude(y)

Amplitude(x)(2.3)

− Power amplification (gain) − Amplitude amplification

LgP = 10 lg(gP ) dB LgA = 20 lg(gA) dB (2.4)

In passive systems attenuation is often given instead of amplification.

aP =Power(x)

Power(y)resp. aA =

Amplitude(x)

Amplitude(y)(2.5)

aP =1

gPaA =

1

gA(2.6)

− Power attenuation − Amplitude attenuation

LaP = 10 lg(aP ) dB LaA = 20 lg(aA) dB (2.7)

LaP = −LgP LaA = −LgA (2.8)

Often the relation between user and interfering signals is of special interest. The ratio

SNR = 10 lg

(Signal power

Noise power

)dB (2.9)

gives the signal-to-noise-ratio.

Please note,

• that level specification in dB always defines ratios between two powers or amplitudes (e.g.amplification factor, signal-to-noise ratio)

• that level specification defines always absolute performances or voltages in dBm (Reference:power Pref = 1 mW), dbW (Reference: power Pref = 1 W), dBu (Reference: power Uref =0.775 V)

• that while calculating with levels the followings hold:

dB± dB = dB (2.10)

dBm± dB = dBm (2.11)

dBm− dBm = dB (2.12)

dBm + dBm = not defined (2.13)

(2.14)


12 3 SIGNAL THEORY

3 Signal Theory

3.1 Sine Signal

Many electrotechnical and also communications engineering problems can be modelled by thefollowing differential equations.

U = LQ(t) +1

CQ(t) = 0 frictionless vibration equation (3.1)

U = LQ(t) +RQ(t) +1

CQ(t) = 0 damped oscillation equation (3.2)

Q(t) = i(t) current flow in the resonant circuit (3.3)

Assuming that Q(t) has the form Q(t) = Q0 eλt+ϕ the nontrivial solution of a frictionless vibration

equation isQ(t) = Q0 e

± j(ω0t+ϕ) (3.4)

with ω0 =√1/(LC). Only real part of the solution has practical interest here, which is as well a

solution of the differential equation.

Q(t) = Re{e± j(ω0t+ϕ)

}= cos(ω0t+ ϕ) (3.5)

Since the solution of the damped oscillation equation also consists of sine functions (generalharmonic functions), these functions are used as basic signals in the electrical and communicationstechnology 1.

In physics, one usually works with the rotative frequency ω. In communications, one uses usuallythe frequency f for practical reasons, so that with the substitution f = ω

2πresults in the expression

Q(t) = Q0 cos(2πf0t+ ϕ).

3.2 Generalized Functions

A very important criterion for the classification of signals is the characteristic of absolute integra-bility. A Signal s(t) is absolutely integrable, when∫ ∞

−∞|s(t)| d t < ms <∞ (3.6)

Many interesting signals are however not absolutely integrable, as for example the sine signals.In order to examine and also to describe these signals, the term generalized functions is intro-duced [BS96]. In mathematics this theory is known as distribution theory [GZZZ95]. Only a shortintroduction relevant to communications technology is given here.

A generalized function s(t) can be understood as (see exercise 4) a limit of the series

{sn(t)} = {s0(t), s1(t), s2(t), . . .} (3.7)

with

s(t) = limn→∞

sn(t) (3.8)

1Sometimes in the script functions are called signals, in order to clarify situations (even if this is not completelycorrect).


3.2 Generalized Functions 13

3.2.1 Dirac Delta Function

In physics, usually the mass m = m0 of a body is assigned to a point �x = �x0 in a space R3 space

(center of gravity of the body), i.e. the entire mass is concentrated at one point. The resultingmass density ρ(x) possesses the following characteristics

ρ(�x) =

{0 for �x �= �x0

∞ for �x = �x0(3.9)

∫R3

ρ(�x) d �x = m0 (3.10)

i.e. in many cases a function is looked for which fulfills the following demands

δ(x) =

{0 for x �= 0

∞ for x = 0(3.11)

∫ ∞

−∞δ(x) d x = 1 (3.12)

A quotation from a mathematics paperback [BS96]

There exists no classical function y = δ(x) with characteristic like (3.11) and (3.12).. . . Nevertheless the physicists had worked for an approximation of the dirac deltafunction, and finally it was successfully introduced by physicist Paul Dirac in 1930.Experience from the history of mathematics shows that successful formalcalculations can always be strictly justified in a suitable formulation . . . .

Also in electrical engineering, the Dirac pulse δ (the Dirac delta function is often defined shortly)possesses an important meaning.

The Dirac impulse δ(x) is an ”‘arbitrarily”’ narrow signal with unit area, which cause a definedeffect at the output of a system. Not the concrete process of the signal δ(x), but its energy content(area) is crucial.

The Dirac impulse δ(x) can be approximated as in Fig.1 3.1 by a series of functions.

The following conditions must hold:

1. Asymptotic (limit) value

δ(x) = limn→∞

δn(x) =

{0 for x �= 0

∞ for x = 0(3.13)

2. Unit area ∫ ∞

−∞δn(x) d x = 1 (3.14)

3. Symmetryδn(x) = δn(−x) (3.15)

The Dirac pulse is a Model. In practice, it is not possible to generate a voltage waveform e.g.u(t) = δ(t). The Dirac impulse can be symbolically plotted as in Fig. 3.2.


14 3 SIGNAL THEORY

δn(x)

n

− 12n

12n

x

a) δn(x) = n rect(nx)

δn(x)

n

− 1n

1n

x

b) δn(x) = n triang(nx)

δn(x)

n/π

− 12n

12n

x

c) δn(x) =nπ

11+(nx)2

Figure 3.1: Dirac impulse approximation by a series of functions (n > 0)

f(x)

1

0 x

a) f(x) = δ(x)

f(x)

A0

0 x0 x

b) f(x) = A0 δ(x− x0)

Figure 3.2: Graphical representation of Dirac impulse

3.2.2 Integration of Generalized Functions

Based on a generalized function definition, it is sensible to define the integral over the limit valueof the function sequence ∫

s(x) d x =

{∫sn(x) d x

}= lim

n→∞

∫sn(x) d x (3.16)

Example: Integration of Dirac pulse∫ x

−∞δ(χ) dχ = lim

n→∞

∫ x

−∞δn(χ) dχ (3.17)

with e.g. (n > 0)

δn(x) :=n

π

1

1 + (nx)2(3.18)

becomes ∫ x

−∞δ(χ) dχ = lim

n→∞1

πarctannχ

∣∣∣x−∞

(3.19)

=

⎧⎪⎪⎨⎪⎪⎩0 for x < 0

12

for x = 0 (here δn(x) is symm.)

1 for x > 0

(3.20)


3.2 Generalized Functions 15

= σ(x) (3.21)

The function σ(x) is often called step function or unit step (see Fig. 3.3). This function is also a

σn(x)

1

x

Figure 3.3: Step function σ(x)

generalized function and can be, for example,presented as (see exercise 4).

σ(x) = limn→∞

σn(x) = limn→∞

1

πarctannx+

1

2(3.22)

3.2.3 Differentiation of Generalized Functions

The differentiation of generalized functions, also analogous to the integration, can be defined as

d

d xs(x) = lim

n→∞d

d xsn(x) (3.23)

The derivative of the step function provides again the Dirac impulse: ddxσ(x) = δ(x) (see exer-

cise 4).

3.2.4 Sifting Property

A very interesting characteristic of the Dirac pulse is the so-called shifting property (see exercise 4).∫ ∞

−∞s(x) δ(x− x0) d x = lim

n→∞

∫ ∞

−∞s(x) δn(x− x0) d x (3.24)

= s(x0) (3.25)

3.2.5 Calculation with Dimensional Parameters

In the preceding paragraph, x was always used as argument of the Dirac pulse or step function.There x was assumed as dimensionless. In communications, however,one generally operates withdimensional parameters like time t or frequency f .

In such cases one needs to make sure whether such dimensional measures are valid as argumentsof functions, then e.g. sin(5) or rect(−1) should be defined, instead of sin(5 s) or rect(−1 Hz).Here such measures must be standardized.

Possible substitutions are e.g. x → t/T or x → f/F . In many books t and f are worked out asdimensionless measures, i.e. t and f are standardized measures with T = 1 s or F = 1 Hz, even ifare referred explicitly.


16 3 SIGNAL THEORY

In this script, the following x should be treated as dimensionless; t and f are on the other handdimensional parameters representing respectively time and frequency.

For the Dirac pulse in time therefore the following definition yields

δ(t) = limT→+0

1

Trect

(t

T

)(3.26)

= limT→+0

1

Ttriang

(t

T

)(3.27)

= limT→+0

1

T

1

π(1 +

(tT

)2) (3.28)

By definition here applies∫ ∞−∞ δ(t) d t = 1 or

∫ ∞−∞ δ(f) d f = 1 and accordingly the dimension from

δ(t) 1 s−1 and from δ(f) follows 1 s = 1 Hz−1.

3.3 Fourier Transformation

3.3.1 Definition

The idea of the Fourier transformation is based on the fact that a signal s(t) can be represented asa superposition of the harmonic functions A ej 2πft where A is amplitude and f is frequency. TheFourier transformation is treated in detail in the lectures ”‘Systemtheorie” ’ and ”‘Mathematik” ’.Therefore here only once the definition is recalled and insisted on some important characteristicsfor communications [BS96].

The Fourier transformation is defined in engineering as 2 (see also Appendix A.2)

S(f) =

∫ ∞

−∞s(t) e− j 2πft d t = F {s(t)} (3.29)

A time domain function s(t) can be transformed into a function S(f) in complex or frequencydomain. The back or inverse Fourier transformation is defined as

s(t) =

∫ ∞

−∞S(f) ej 2πft d f = F−1 {S(f)} (3.30)

3.3.2 Properties

The relationship between the time domain function and corresponding frequency domain functionobtained using the Fourier transformation (shortly: FT) is one-to-one. FT pair is symbolicallydenoted as s(t) S(f). Usually functions in time domain are denoted with lowercase letterswhile uppercase letters usually correspond functions in frequency domain.

The integrals (3.29, 3.30) are not always convergent, e.g. the transformation of the important sinesignals does not succeed in the context of classical functions, so that the terms of the generalizedfunctions and distributions must also be used here.

On the other hand, if the signal s(t) is absolutely integrable (see Section. 3.2), then the Fourierintegral is also convergent and the Fourier transform S(f) exists and converges for f → ±∞towards 0. Due to the symmetry of the FT this applies also in the reverse case.

2In Mathematics, as well as in Physics sometimes slightly deviated definitions are used.


3.3 Fourier Transformation 17

Reversibility It appliess(t) = F−1 {F {s(t)}} (3.31)

Proof:

s(t) =

∫ ∞

f=−∞

∫ ∞

t′=−∞s(t′) e− j 2πft′ d t′ ej 2πft d f (3.32)

=

∫ ∞

t′=−∞s(t′)

∫ ∞

f=−∞ej 2πf(t−t′) d f d t′ (3.33)

=

∫ ∞

t′=−∞s(t′)δ(t− t′) d t′ (3.34)

= s(t) (3.35)

Due to symmetry of FT S(f) = F {F−1 {S(f)}} is also valid.

Linearity It applies in general∑i

ai si(t)∑i

ai Si(f) (3.36)

i.e. signals can be decomposed into ”‘component-signals”’ to get the result more easily.

Proof: ∫ ∞

−∞

∑i

ai si(t) e− j 2πft d t =

∑i

ai

∫ ∞

−∞si(t) e

− j 2πft d t (3.37)

=∑i

ai Si(f) (3.38)

FT od Conjugate Complex Signals It states,

s∗(t) S∗(−f) (3.39)

Proof:

F {s∗(t)} =

∫ ∞

−∞s∗(t) e− j 2πft d t =

∫ ∞

−∞s∗(t) ej 2π(−f)t d t (3.40)

=

[∫ ∞

−∞s(t) e− j 2π(−f)t d t︸︷︷︸

S(−f)

]∗

= S∗(−f) (3.41)

By substituting t′ = −t one gets a very similar relationship

s∗(−t) S∗(f) (3.42)

Time Reversal With substituting t′ = −t in the Fourier integral one gets the interrelation

s(−t) S(−f) (3.43)


18 3 SIGNAL THEORY

Symmetry The integrals of Fourier and inverse Fourier transform differ only by the sign in theexponent of the exponential function. It states therefore the duality principle, i.e. the Fourier andinverse Fourier transform are exchangeable to their signs. The relation, which applies e.g. in timedomain, can also be transferred to frequency domain.

S(t) s(−f) (3.44)

Proof:

s(t) =

∫ ∞

−∞S(f) ej 2πft d f (3.45)

Substituting f = t and by Fourier transform yields,∫ ∞

−∞S(t) e− j 2πft d t =

∫ ∞

−∞S(t) ej 2π(−f)t d t = s(−f) (3.46)

FT of Real Signals For a real signal s(t), s(t) = s∗(t) and therefore (see FT conjugate complexsignals)

S(f) = S∗(−f) (3.47)

From this we get,

Re{S(f)}+ j Im{S(f)} = Re{S(−f)} − j Im{S(−f)} (3.48)

|S(f)| ej arg(S(f)) = |S(−f)| e− j arg(S(−f)) (3.49)

X(f)

f

Re{X(f)}

Im{X(f)}

Figure 3.4: Real- and imaginary part of the Fourier transform S(f) of a real signal s(t)

The following consideration is somewhat more illustrative

S(f) =

∫ ∞

−∞s(t) e− j 2πft d t (3.50)

=

∫ ∞

−∞s(t) (cos 2πft− j sin 2πft) d t (3.51)

As s(t) is pure real-valued, we can conclude the following

Re{S(f)} =

∫ ∞

−∞s(t) cos 2πft d t – even function in f , as cos-term is even in f

Im{S(f)} = −∫ ∞

−∞s(t) sin 2πft d t – odd function in f , as sin-term is also odd in f

Both considerations show that both real part of the spectrum Re{(S(f)} and the amplitudespectrum |S(f)| are even functions; and both imaginary part of the spectrum Im{S(f)} andphase − arg(S(f)) are odd functions. These characteristics are demonstrated in Fig. 3.4.



FT of Pure Imaginary Signals For pure imaginary signals s(t), s(t) = −s∗(t) and so (see FTof pure real signals)

S(f) = −S∗(−f) (3.52)

Re{S(f)}+ j Im{S(f)} = −Re{S(−f)}+ j Im{S(−f)} (3.53)

|S(f)| ej arg(S(f)) = |S(−f)| e− j(arg(S(−f))+π) (3.54)

The following consideration is somewhat more descriptive

S(f) =

∫ ∞

−∞s(t) e− j 2πft d t (3.55)

=

∫ ∞

−∞s(t) (cos 2πft− j sin 2πft) d t (3.56)

Since s(t) is pure imaginary, we have,

Re{S(f)} =

∫ ∞

−∞s(t) sin 2πft d t – odd function in f , as sin-term is also odd in f

Im{S(f)} =

∫ ∞

−∞s(t) cos 2πft d t – even function in f , as cos-term is also even in f

Here imaginary part of the spectrum Im{S(f)} and the amplitude spectrum |S(f)| are evenfunctions, as well as real part of the spectrum Re{(S(f)} and phase difference − arg(S(f)) areodd functions.

Shifting Theorem – FT of Time Delayed Signals A shift of the signal s(t) in time domaincorresponds to a linear, frequency dependent, phase shift of the spectrum S(f)

s(t− t0) e− j 2πft0 S(f) (3.57)

Proof (Substitution t′ = t− t0):

F {s(t− t0)} =

∫ ∞

−∞s(t− t0) e

− j 2πft d t =

∫ ∞

−∞s(t′) e− j 2πf(t′+t0) d t′ (3.58)

= e− j 2πft0

∫ ∞

−∞s(t′) e− j 2πft′ d t′ = e− j 2πft0 S(f) (3.59)

Shifting Theorem – FT of Frequency Shifted Signals Due to the symmetry of FT, reverseis also valid (see also modulation)

ej 2πf0t s(t) S(f − f0) (3.60)

Proof:

S(f − f0) =

∫ ∞

−∞s(t) e− j 2π(f−f0)t d t =

∫ ∞

−∞

(s(t) ej 2πf0t

)e− j 2πft d t = F {

s(t) ej 2πf0t}

(3.61)


20 3 SIGNAL THEORY

Scaling Property If a signal is stretched or squeezed in time, the spectrum in frequency axisis squeezed (stretched) and also the reverse holds.

s(at)1

|a| S(f

a

), a �= 0 (3.62)

Proof (Substitution t′ = at and distinction of cases a > 0, a < 0):∫ ∞

−∞s(at) e− j 2πft d t =

1

|a|∫ ∞

−∞s(t′) e− j 2πf t′

a d t′ =1

|a|S(f

a

)(3.63)

The fact which lies behind this prediction is that fast changing of time signals possess a broadspectrum and slowly changing signals have a narrow spectrum.

FT of Dirac Impulse The Fourier transformation of Dirac pulse results directly from its siftingcharacteristic:

F {δ(t)} =

∫ ∞

−∞δ(t) e− j 2πft d t = e− j 2πf ·0 = 1 (3.64)

The Dirac pulse possesses thus an infinitely widened constant spectrum. This characteristic followsdirectly as regarding to the limiting value of previously treated time elongation or scaling property.

Due to the symmetry of the FT and the shifting theorem, the following holds:

1 δ(f) δ(t) 1 (3.65)

ej 2πf0t δ(f − f0) δ(t− t0) e− j 2πft0 (3.66)

FT of Sine Signals The extremely important basic signals cos 2πf0t and sin 2πf0t are notabsolutely integrable, as∫ ∞

−∞| cos 2πf0t| d t = ∞ or

∫ ∞

−∞| sin 2πf0t| d t = ∞ (3.67)

It is however well known that one can write

cos 2πf0t =1

2

(ej 2πf0t +e− j 2πf0t

)(3.68)

sin 2πf0t =1

2 j

(ej 2πf0t − e− j 2πf0t

)(3.69)

Thus one obtains the following transformation rules (see Fig. 3.5)

cos 2πf0t1

2

(δ(f + f0) + δ(f − f0)

)(3.70)

sin 2πf0t1

2 j

(δ(f − f0)− δ(f + f0)

)=

1

2j(δ(f + f0)− δ(f − f0)

)(3.71)

With the help of inverse transform these rules can be easily checked.



Re{S(f)}(Im{S(f)} = 0)

A2

f−f0 f0

s(t) = A cos 2πf0t

Im{S(f)}(Re{S(f)} = 0)

A2

f−f0 f0

s(t) = A sin 2πf0t

Figure 3.5: Spectrum of cosine and sine function

Convolution Theorem Multiplication of two signals in time domain corresponds to a convo-lution of the corresponding spectrum in frequency domain. Due to the symmetry of the FT, thereverse case is also true.

s1(t) · s2(t) S1(f) ∗ S2(f) (3.72)

s1(t) ∗ s2(t) S1(f) · S2(f) (3.73)

Proof: see eqn. (4.8-4.13)

Modulation The multiplication of a signal s(t) with a sine signal corresponds to a shift of thecorresponding spectrum S(f) in frequency domain.

s(t) ej 2πf0t S(f − f0) (3.74)

s(t) cos 2πf0t1

2(S(f − f0) + S(f + f0)) (3.75)

s(t) sin 2πf0t1

2 j(S(f − f0)− S(f + f0)) (3.76)

The proof follows with the help of the relations between sine signal and convolution theory

s(t) ej 2πf0t S(f) ∗ δ(f − f0) (3.77)

as well as the sifting property of Dirac pulse (see Fig. 3.2.4, convolution with respect to f , seealso notes for convolution Fig. A.3)

S(f) ∗ δ(f − f0) =

∫ ∞

−∞S(f − ν) δ(ν − f0) d ν (3.78)

= S(f − f0) (3.79)

The correspondenceS(f) ∗ δ(f − f0) = S(f − f0) (3.80)

will appear again and again in the script and would be very helpful to understand it.


22 4 LINEAR TIME INVARIANT SYSTEM

4 Linear Time Invariant System

4.1 Definition

A system H with input x and output y is given as in Fig. 4.1.

Hx y

Figure 4.1: System H

At first some terms are defined here to describe the system:

Causality A system is causal, if for any point in time t0 where t0 < t the system reaction y(t0)is independent of the value of x(t) after t0. The system reaction is determined exclusivelyby past values of the input signal. (cause-effect principle, see also exercise 11).

Time invariance A system is time invariant if the transfer function is time-independent, i.e. ifx(t) → y(t), it follows that any time shifting of τ of the input signal causes the same timeshifting of output signal: x(t− τ) → y(t− τ).

Linearity A system is linear if it holds x1(t) → y1(t) and x2(t) → y2(t) holds also x1(t)+x2(t) →y1(t) + y2(t) and a · x1(t) → a · y1(t) (a ∈ R).

Source free system A system is source free if from x(t) ≡ 0 follows y(t) ≡ 0. Linear systemsare always source free.

In the sequel, the systems fulfilling the characteristics linearity and time invariance will bestudied. Such systems are called (linear time invariant systems), shortly LTI-systems. LTI-systemsdo not have to be causal although practically realizable systems are always causal.

4.2 Transfer Function

Let the LTI system H be excited with a harmonic (complex) input signal x(t) = ej 2πf0t with afrequency of f0. In the stationary state the output signal is yf0(t). So,

ej 2πf0t → yf0(t) (4.1)

Because of time invariance property

ej 2πf0(t−τ) → yf0(t− τ) (4.2)

and from linearitye− j 2πf0τ︸︷︷︸

a

· ej 2πf0t → e− j 2πf0τ︸︷︷︸a

·yf0(t) (4.3)


4.2 Transfer Function 23

At time t = 0

e− j 2πf0τ → yf0(0) · e− j 2πf0τ (4.4)

and with substituting t = −τej 2πf0t → yf0(0) · ej 2πf0t (4.5)

Using Fourier transform one gets,

δ(f − f0) → yf0(0) · δ(f − f0) (4.6)

Therefore, a LTI system with a sinusoidal excitation with frequency f0 always generates a sinu-soidal output signal with the same frequency f0 (Properties of LTI-systems [Fli91]). This outputoscillation is however weighted with a complex factor yf0(0), i.e. the input signal is altered only inamplitude and phase. Assigning the weighted factor yf0(0) to all frequencies f = f0 over f givesthe transfer function of the LTI system H in frequency domain. The transfer function is usuallydenoted as H(f). So H(f0) = yf0(0) with −∞ < f0 <∞ (see Fig. 4.2).

|H(f)|0.8

−fg 0 fg f

|X(f)|

|Y (f)| = |H(f) ·X(f)|

−fg 0 fg f

Figure 4.2: Amplitude spectrum of the transfer function |H(f)|, as well as that of input andoutput signal |X(f)| and |Y (f)|

Let X(f) be the Fourier transformation of x(t), i.e. the input signal x(t) can be represented assuperposition of infinite number of sine waves. Exploiting linearity the relation between input andoutput signal of an LTI system is as follows

Y (f) = H(f) ·X(f) (4.7)

The individual spectral lines of X(f) produce weighted spectrum Y (f). A LTI system cannotproduce sine signals with frequencies which are not contained in the input signal.

The function H(f)

• is a complex weighting function of the input signal spectrum,

• indicates the transfer function of the LTI system H,

• can be separated into amplitude response (amplitude |H(f)|) and phase response(phase− argH(f)).



4.3 Impulse Response

Consideration in time domain gives

y(t) = F−1 {H(f)X(f)} (4.8)

=

∫ ∞

−∞

∫ ∞

−∞h(θ) e− j 2πfθ d θ

∫ ∞

−∞x(τ) e− j 2πfτ d τ ej 2πft d f (4.9)

=

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞h(θ)x(τ) ej 2πf(t−(θ+τ)) d τ d θ d f (4.10)

by substitution ϑ = θ + τ follows

=

∫ ∞

−∞

∫ ∞

−∞h(ϑ− τ)x(τ) d τ

∫ ∞

−∞ej 2πf(t−ϑ) d f dϑ (4.11)

and by using the relation e− j 2πfτ δ(t− τ) (see Tab. A.2) follows

=

∫ ∞

−∞

∫ ∞

−∞h(ϑ− τ) x(τ) δ(t− ϑ) d τ dϑ (4.12)

And with sifting property finally yields

y(t) =

∫ ∞

−∞h(t− τ) x(τ) d τ = (h ∗ x)(t) (4.13)

The signal h(t) is the inverse Fourier transform of the transfer function H(f) and is named asimpulse”-response, because the system H generates exactly the impulse response y(t) = h(t)at the output when the excitation is Dirac pulse x(t) = δ(t).

y(t) =

∫ ∞

−∞h(t− τ)x(τ) d τ (4.14)

=

∫ ∞

−∞h(t− τ)δ(τ) d τ (4.15)

= h(t) (4.16)

As a summery, one can say that a linear and an time-invariant (LTI) system can be completelydescribed by its impulse response h(t) or its transfer function H(f) = F {h(t)}.

4.4 Convolution as the Basic Operation in Communications

4.4.1 Definition

In eqn. (4.13) the relation between input and output signal of a LTI system H is obtained as

y(t) =

∫ ∞

−∞h(t− τ) x(τ) d τ (4.17)


4.4 Convolution as the Basic Operation in Communications 25

= (h ∗ x)(t) (4.18)

This integral is known as convolution integral or briefly convolution and possesses an very im-portant meaning in system theory and communications technology (see also the annotation inthe Appendix A.3). The output signal y(t) results from the convolution of input signal x(t) withimpulse response h(t) of the system H.

The transfer characteristic of a LTI system may occur thereby alternatively in frequency or timedomain. Both representations are equivalent.

Y (f) = H(f) ·X(f) (4.19)

y(t) = (h ∗ x)(t) = h(t) ∗ x(t) =∫ ∞

−∞h(t− τ) x(τ) d τ (4.20)

4.4.2 Properties

The convolution is

• commutative: g ∗ h = h ∗ g,• associative: f ∗ (g ∗ h) = (f ∗ g) ∗ h,• distributive: f ∗ (g + h) = f ∗ g + f ∗ h and

• from g ∗ h = 0 gives g = 0 or h = 0

These characteristics can be proved by simple substitution of the integration variables, e.g. com-mutativity ∫ ∞

−∞x1(t− τ)x2(τ) d τ = −

∫ −∞

∞x1(τ

′)x2(t− τ ′) d τ ′ with τ ′ = t− τ (4.21)

=

∫ ∞

−∞x1(τ

′)x2(t− τ ′) d τ ′ (4.22)

=

∫ ∞

−∞x1(τ)x2(t− τ) d τ withτ := τ ′ (4.23)

4.4.3 Example

Another characteristic of convolution will be given with an example of convolution of two rectan-gular functions. (see Fig. 4.3 and Exercise 12).

y(t) = rect(t/T )︸︷︷︸x1(t)

∗ rect(t/T )︸︷︷︸x2(t)

(4.24)

=

∫ ∞

−∞rect(τ/T ) rect((t− τ)/T ) d τ (4.25)



x1(τ), x2(t− τ)

1

−T2

0 T2

τ

t− T2

t t+ T2

⇒x1x2

y(t) = x1(t) ∗ x2(t)

T

−T 0 T t

a) 1. case: t < −T ⇒ y(t) = 0

x1(τ), x2(t− τ)

1

−T2

0 T2

τ

t− T2

t t+ T2

⇒x1x2

y(t) = x1(t) ∗ x2(t)

T

−T 0 T t

b) 2. case: −T ≤ t < 0 ⇒ y(t) = T + t

x1(τ), x2(t− τ)

1

−T2

0 T2

τ

t− T2

t t+ T2

⇒x1 x2

y(t) = x1(t) ∗ x2(t)

T

−T 0 T t

c) 3. case: 0 ≤ t < T ⇒ y(t) = T − t

x1(τ), x2(t− τ)

1

−T2

0 T2

τ

t− T2

t t+ T2

⇒x1 x2

y(t) = x1(t) ∗ x2(t)

T

−T 0 T t

d) 4. case: T ≤ t ⇒ y(t) = 0

Figure 4.3: Convolution of two rectangular functions rect(t/T ) ∗ rect(t/T )

=

∫ T2

−T2

rect((t− τ)/T ) d τ =

∫ T2

−T2

rect((τ − t)/T ) d τ (4.26)


4.5 Consideration in Time and Frequency 27

=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

0 for t < −T∫ t+T/2

−T/2d t = T + t for − T ≤ t < 0∫ T/2

t−T/2d t = T − t for 0 ≤ t < T

0 for T ≤ t

(4.27)

The convolution of two signals y(t) = (x1 ∗ x2)(t) causes a signal spreading in time. If x1(t) isdifferent from zero only in the interval T1 and the signal x2(t) is different from zero in interval T2,then the output signal y(t) is in the interval T1 + T2 different from zero.

Further examples can be found in Exercise 11, 13, 14 and 15.

4.5 Consideration in Time and Frequency

In this paragraph relations between the time and frequency response of signals will be explored.

4.5.1 Fourier Transformation of Rect Function

The Fourier transformation of the rect function s(t) = rect(t/T ) yields

S(f) =

∫ ∞

−∞rect(t/T ) e− j 2πft d t (4.28)

=

∫ T2

−T2

e− j 2πft d t =1

− j 2πf

(e− jπfT − ejπfT

)(4.29)

and using the relation sinx = 12 j(ejx− e− jx)

=sin(πfT )

πf= T

sin(πfT )

πfT(4.30)

= T si(πfT ) (4.31)

The function si(x) = sin(x)/x is called sinc function (see Fig. 4.4 and also definition in Ap-pendix A.1).

si(t/T )

1

−π π t/T

∼ 1t

Figure 4.4: Sinc function si(t/T )



4.5.2 Limitation in Time and Frequency 3

Most signals are time limited, just because of the fact that a signal generator is first switched onand later switched off; therefore the output signal during switch-off is zero, so is time limited. Thequestion is whether signals can be produced to beboth time and frequency limited. Let us proceedwith a time limited signal.

Let s(t) = 0 for |t| > T/2. It follows then,

s(t) = rect(t/T ) · x(t) (4.32)

⇒ S(f) = T si(πfT ) ∗X(f) (4.33)

Convolution of X(f) with sinc function si(πfT ) causes a spectral dispersion. Since si(f) possessesa level decay proportional to 1/f , S(f) represents an infinitely expanded spectrum.

The impulse response or transfer function of a LTI system can be either time or frequency limited,but never can be both at the same time4. This means that a time limited signal possesses aninfinitely widened spectrum and the reverse also holds.

|S(f)|

−fg 0 fg f

Figure 4.5: Quasi band limited signal

The spectrum S(f) of a time limited signal may however tend to zero for |f | → ∞, but itnever becomes zero exactly. Therefore the following picture results in practice (see Fig. 4.5). Thespectrum S(f) of the signal is lost starting from a certain critical frequency fg in noise, is thussmaller than the noise level. This noise floor results for example from thermal noise of elements.Such signals are called quasi bandlimited. In practice these quasi bandlimited signals are treatedlike bandlimited signals.

Time domain Frequency domain

finite signal duration ⇒ infinitely expandedspectrum

infinitely expandedsignal

⇐ bandlimited spectrum

3see duration-bandwidth product [Kam96] and applications of band limited non-periodic functions [Hof98]4The remarks made above are not as proof, rather serve as practical illustration of facts.


4.5 Consideration in Time and Frequency 29

4.5.3 Raised-Cosine-Filter

It is desirable to have signals which are both time and frequency limited. In the previous paragraphit was however stated that this is not possible.

An ideal low-pass filter has the transfer function H(f) = rect(f/2fg), where fg is the criticalfrequency. The impulse response h(t) is obtained as a result of the inverse Fourier transformationof the H(f) which gives sinc function h(t) = 2fg si(2πfgt). This impulse response is neithercausal nor time limited. By ”‘shifting”’ and ”‘chopping”’ causality and time limitation can beachieved 5. The sinc function however has still the crucial drawback that its envelope decay is onlyproportional with 1/t which is not sufficient for many practical applications.

For this reason is Raised Cosine Filter Hrc used in many applications (see Fig. 4.6). The transferfunction and impulse response are given in the following equations.

Hrc(f) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 for |f |

fg< 1− α

cos2( π4α( ffg

− (1− α))) for 1− α ≤ |f |fg< 1 + α

0 for 1 + α ≤ |f |fg

(4.34)

hrc(t) = 2fg si(2πfgt)cos(2παfgt)

1− (4αfgt)2(4.35)

This filter is likewise frequency-limited, has however in contrast to the ideal low-pass a ”‘smooth”’,cosine-shaped transition between pass and stop band. In a limit, when α → 0 this filter is an ideallow pass filter.

Stronger decaying of impulse response hrc(t) can be achieved by these less abrupt transitionbetween pass and stop band. The steepness at which the envelope declines is a function of α (forα = 0 with 1/t and for α = 1 with 1/t3). For α > 0 this decay is strong enough to achieve asufficient ”‘delimitation”’ in time and frequency for most applications.

The following example demonstrates how causality and time limit can be achieved by ”‘shifting”’and ”‘chopping”’

h(t) = hrc(t− 5Tg) rect

(t− 5Tg10Tg

)with Tg = 1/fg (4.36)

=

{hrc(t− 5Tg) for 0 < t < 10Tg ⇒ thereby causal

0 otherwise ⇒ thereby time limited(4.37)

In signal processing such ”‘chopping”’ is called windowing and allows the description of signalswhich can be observed in a limited time interval. However, for example, the calculation of theFourier transform is practically impossible as it requires an integration over the time period−∞ < t < ∞. A measurement of a signal can take place only in a finite period. Expected effectsregarding windowing are treated in Exercise 6.

5In order to not to distort signals unnecessarily with filtering (detection made more difficult), the linear phasing isparticularly required in communications technology for filters. This leads also to equal requirement of symmetricalimpulse response of the filter. Since only causal systems are realizable, this impulse response must be shiftedaccordingly in time.



H(f)

1

−fg 0 fg f

h(t)

2fg

− 12fg

12fg

t

a) Transfer function H(f) and impulse response h(t) of ideal low pass filters

Hrc(f)

1

−fg 0 fg f

hrc(t)

2fg

− 12fg

12fg

t

b) Transfer function Hrc(f) and impulse response hrc(t) of Raised Cosine Filters (α = 0.3)

Figure 4.6: Comparison between ideal low pass and Raised Cosine Filter

4.6 Central Limit Theorem

The central limit theorem states that the distribution function Fn of the sum of n independentrandom variables Xi converges for n → ∞ towards a normal distribution. There exist a set oftheories, which under all different assumptions state about the convergence [BHPT95], [BS96]. Aquite general form of the central limit theorem is Ljapunov’s theorem, which is described herebriefly.

Theory of Ljapunov

Conditions:

Xi − independent,

not necessarily identically distributed

μi = E{Xi} = finite

σ2i = E{(Xi − μi)

2} finite > 0

limn→∞

Bn

σ= 0 with Bn = 3

√√√√ n∑i=1

bi

and bi = E[|Xi − μi|3

]and σ =

√√√√ n∑i=1

σ2i

If

Z =1

σ

n∑i=1

(Xi − μi) (4.38)


4.6 Central Limit Theorem 31

belongs to normalized random variable X =∑n

i=1Xi, then:

limn→∞

Fn(z) = Φ(z, 0, 1) =1√2π

∫ z

−∞e−

t2

2 d t (4.39)

I.e. the distribution function Fn(z) converges asymptotically towards the standardized normaldistribution Φ(z, μ, σ)|μ=0, σ=1 = Φ(z, 0, 1). With

Z =1√∑ni=1 σ

2i

(X −

n∑i=1

μi

)(4.40)

follows

X =

√√√√ n∑i=1

σ2i Z +

n∑i=1

μi (4.41)

The random variable X =∑n

i=1Xi is likewise asymptotically normalized with E[X] =∑n

i=1 μi

and D2[X] =∑n

i=1 σ2i .

The interpretation of the central limit theorem exists as follows: If a random vari-able can be understood as a sum of large number of independent random variables, of whicheach supplies only an insignificant contribution to the total sum, then this random variable isapproximately normal distributed[BS79].


32 5 BANDPASS SIGNAL

5 Bandpass Signal

5.1 Up and Down Conversion of Real Baseband Signals

When signals are defined as low pass or baseband signal s(t), the following relation applies for thespectra

S(f) =

{any for |f | ≤ fg

0 for |f | > fg(5.1)

The direct transmission of baseband signals, as described in the Article 6.1, is not meaningful.Baseband signals are therefore shifted to a higher frequency band by multiplication with a carriersignal sc(t) = cos 2πfct. Such a signal operation is generally called mixing and particularly hereas upward mixing. The baseband signal should be regarded for the time being as real valued. Asa result a bandpass signal is obtained (see Fig. 5.1)

sBP (t) = Re{s(t) ej 2πfct

}(5.2)

= s(t) cos 2πfct (5.3)

⇒ SBP (f) = S(f) ∗(1

2δ(f + fc) +

1

2δ(f − fc)

)(5.4)

=1

2

(S(f + fc) + S(f − fc)

)(5.5)

Since the baseband signal s(t) was assumed as real valued, the amplitude spectrum |S(f)| is aneven function (symmetric over f = 0) and thus the amplitude spectrum of the bandpass signal is”‘local”’ symmetric around f = ±fc.

|S(f)|A

−fg fg f

⇒|SBP (f)|

A

−fg,o −fc −fg,u fg,u fc fg,o f

Figure 5.1: Spectrum of a real baseband signal (left) and corresponding spectrum of the up-mixed bandpass signal with fg,u = fc − fg and fg,o = fc + fg (right)

So for bandpass signals

SBP (f) =

{any for fg,u ≤ |f | ≤ fg,o

0 for fg,u > |f | > fg,o(5.6)

Such a rectified bandpass signal can be radiated over an antenna. In the receiver, of course, oncemore the conversion of the signal to baseband is necessary. Consequently the bandpass signal issupplied again to a mixer, where it is multiplied by the same carrier signal, likewise in the sender.

r(t) = sBP (t) · cos 2πfct (5.7)


5.1 Up and Down Conversion of Real Baseband Signals 33

= s(t) cos 2πfct cos 2πfct (5.8)

=1

2s(t) +

1

2s(t) cos 2π2fct (5.9)

u(t) = r(t)∣∣∣|f |<fc

=1

2s(t) (5.10)

The received signal r(t) contains both the original baseband signal s(t) and two images at ±2fc.Therefore the baseband signal s(t) can be recovered by an ideal low pass filtering with a criticalfrequency of fc (see Section. 5.2, 5.3). These filtering is symbolized as ”‘

∣∣|f |<fc

”’.

|SBP (f)|A

−fg,o −fc −fg,u fg,u fc fg,o f⇓|R(f)|, |U(f)|

A

−2fc −fc −fg fg fc 2fc f

Low-pass

Figure 5.2: Spectrum of the received real bandpass signal (above) and pertinent spectrum ofthe signal r(t) at the output of the mixer or u(t) at the output of the low-pass(below).

Sender Receiver

× × LPs(t) sBP (t)r(t)

u(t) = 12s(t)

cos 2πfct cos 2πfct

Figure 5.3: Up and down mixture of a real baseband signal s(t) with ideal carrier synchroniza-tion

During a radio transmission senders and receivers are spatially separated from each other andone can not guarantee in advance that the oscillators run synchronously in phase for the upand downward mixing. Both oscillators operate independently and therefore there exists a smallfrequency offset Δf and an unknown initial phase difference ϕ0 between both carrier signals.

sc,S(t) = cos 2πfct carrier signal in the sender (5.11)

sc,E(t) = cos(2π(fc −Δf)t− ϕ0) carrier signal in the receiver (5.12)

= cos(2πfct− (2πΔft+ ϕ0)︸︷︷︸ϕ(t)

) (5.13)



The down-converted received signal u(t) results thereby in the case of non-ideal carrier synchro-nization to (see Fig. 5.4)

r(t) = sBP (t) · sc,E(t) (5.14)

= s(t) cos(2πfct) cos(2πfct− ϕ(t)) (5.15)

=1

2s(t) cos(−ϕ(t)) + 1

2s(t) cos(2π2fct− ϕ(t)) (5.16)

u(t) = r(t)∣∣∣|f |<fc

=1

2s(t) cosϕ(t) (5.17)

A time-varying s(t) factor cosϕ(t) is impressed to the baseband signal. Particularly a phase offset

Sender Receiver

× × LPs(t) sBP (t)r(t)

u(t) = 12s(t) cosϕ(t)

cos 2πfct cos(2πfct− ϕ(t))

Figure 5.4: Up and down mixture of a real low-pass signal s(t) with non-ideal carrier synchro-nization

of ϕ ≈ nπ/2 affects adversely, because then cosϕ ≈ 0 and u(t) goes to zero (buried in noise ). Thecarrier signal in the receiver must be synchronized thus to the phase of the carrier in the sender.

The simplest version is to add the carrier sc,S(t) of a certain level to the signal sBP (t) and thentransmit. In the receiver, the carrier can be regenerated with a very narrow bandpass filter andthen used for the down mixing. The disadvantage of this procedure is however the worsened powerbalance, because the transmitting power is not available only in the user signal sBP (t), but is ratherdivided in user and carrier signal. ⇒ It is implemented with analog modulation methods(Amplitude modulation – AM).

A further possibility is to insert certain known data (training sequences) in the baseband signals(t). Then try to detect these training sequences in the receiver and to estimate the frequencyoffset. It must be guaranteed before that the received signal u(t) cannot degrade to zero. ⇒ It isimplemented with digital modulation methods.

In addition the following consideration: The problem with a phase offset of ϕ ≈ nπ/2. If however,with a phase shift of π/2 is applied into the carrier cos(2πfct − ϕ(t) + π/2) before mixing, thenthe signal u(t) can be received error free despite the phase offset of ϕ ≈ nπ/2. This transposedcarrier corresponds to a sine carrier − sin(2πfct − ϕ(t)). So the bandpass signal sBP (t) can bedown converted with two carriers in parallel (see Fig. 5.5). So the signal down converted witha cosine carrier is known as (in phase component) uI(t) and the one down converted with sinuscarrier is the (quadrature phase component) uQ(t). Therefore the name I-Q down mixing comesfor this process 6.

After mixing the signals rI(t) and rQ(t) arise as

rI(t) = sBP (t) cos(2πfct− ϕ(t)) rQ(t) = −sBP (t) sin(2πfct− ϕ(t)) (5.18)

6Often also the term I-Q Demodulation or quadrature demodulation is used.


5.1 Up and Down Conversion of Real Baseband Signals 35

Sender

Receiver

s(t) ×

cos 2πfct

sBP (t)

×

cos(2πfct− ϕ(t))

rI(t)LP uI(t) =

12s(t) cosϕ(t)

×

− sin(2πfct− ϕ(t))

rQ(t)LP uQ(t) =

12s(t) sinϕ(t)

Figure 5.5: I-Q down mixing of the real bandpass signal sBP (t)

= s(t) cos(2πfct) cos(2πfct− ϕ(t)) = −s(t) cos(2πfct) sin(2πfct− ϕ(t))(5.19)

=1

2s(t)

(cos(−ϕ(t)) + cos(2π2fct− ϕ(t))

)= −1

2s(t)

(sin(−ϕ(t)) + sin(2π2fct− ϕ(t))

)(5.20)

The images at ±2fc are filtered here by a low-pass, and the received signals result uI(t) and uQ(t)in the baseband as

uI(t) = rI(t)∣∣∣|f |<fc

uQ(t) = rQ(t)∣∣∣|f |<fc

(5.21)

=1

2s(t) cosϕ(t) =

1

2s(t) sinϕ(t) (5.22)

One may interpret the inphase and quadrature component (both real) of the received signal asreal and imaginary part of a complex signal

r(t) = rI(t) + j rQ(t) (5.23)

= sBP (t) e− j(2πfct−ϕ(t)) (5.24)

=1

2s(t) ejϕ(t)

(1 + e− j 2π2fct

)(5.25)

and after low-pass filtering as complex baseband signal

u(t) = uI(t) + juQ(t) (5.26)

=1

2s(t)

(cosϕ(t) + j sinϕ(t)

)(5.27)

=1

2s(t) ejϕ(t) (5.28)

the transmitted baseband signal s(t) can be represented as pointers in the complex number plane.This pointer has an amplitude of s(t) and an argument of ϕ(t) (see Fig. 5.6). As s(t) is realvalued, it represents the real part of the axis in the coordinate system of the sender. The Iand Q-components uI(t) and uQ(t) stretch against the coordinate system of the receiver. The



Coordinate system

of the sender

Im {s(t)}

Re {s(t)}

uQ(t) = Im {u(t)}

uI(t) = Re {u(t)}

Coordinate system

of the receiverϕ(t)

Figure 5.6: Phase diagram of the baseband signal s(t)

phase difference between the carrier signals in the sender and receiver leads to a twist of the twocoordinate systems exactly around this phase difference ϕ(t). The synchronization of the receiverwith the sender means thus the harmonization two coordinate systems. During transmission ofdigital signals (see Chapter 8) a wrong synchronization may lead to a false ”‘perspective”’ overthe received data and finally to an erroneous conclusion. Thus e.g. with ϕ = π a transmitted +1would be interpreted as −1. During transmission of analog signals wrong synchronization leads toamplitude fluctuations in the demodulated received signal.

Here it becomes also clear why during asynchronous down mixture the multiplication with onlyone carrier cos(2πfct + ϕ(t)) is not sufficient. The received signal u(t) = 1

2s(t) ejϕ(t) would only

comprise the projection on the real axis uI(t) =12s(t) cosϕ(t) instead of the entire send signal. In

contrast by I-Q down conversion, always a constant signal power

Pu = |u|2(t) = 1

4s2(t) (5.29)

exists in the receiver.

From the complex received signal u(t) the phase difference ϕ(t) can be estimated by using a knowntraining sequence (in digital modulation). By a complex multiplication afterwards with e− j ϕ(t),the coordinate system in the receiver is ”’rotated back”’ and is then identical to the coordinatesystem of the sender with ideal carrier synchronization ϕ(t) = ϕ(t) (see Fig. 5.7).

Re{u(t) e− j ϕ(t)

}= Re {(uI(t) + juQ(t)) (cos ϕ(t)− j sin ϕ(t))} (5.30)

= uI(t) cos ϕ(t) + uQ(t) sin ϕ(t) (5.31)

=1

2s(t) cos(ϕ(t)− ϕ(t)︸︷︷︸

Δϕ(t)

) (5.32)


5.2 Up and Down Conversion of Complex Baseband Signal 37

For ideal carrier synchronization, i.e. an approximate error Δϕ(t) = 0, gives

Re{u(t) e− jϕ(t)

}=

1

2s(t) (5.33)

uI(t) =12s(t) cosϕ(t) ×

cosϕ(t)

uQ(t) =12s(t) sinϕ(t) ×

sinϕ(t)+ 1

2s(t)

Figure 5.7: Principle of the phase correction for real baseband signals in the receiver (Δϕ(t) =0)

5.2 Up and Down Conversion of Complex Baseband Signal

In general, the baseband signal s(t) should be regarded as complex valued. Whereas Re {s(t)} andIm {s(t)} indicate the quadrature components of the baseband signal s(t).

sI(t) = Re {s(t)} in phase component (5.34)

sQ(t) = Im {s(t)} quadrature phase component (5.35)

s(t) = sI(t) + j sQ(t) complex base band signal (5.36)

For the up mixing in this case gives (compare with (5.3))


}(5.37)

= Re {(sI(t) + j sQ(t)) · (cos 2πfct+ j sin 2πfct)} (5.38)

= sI(t) cos 2πfct− sQ(t) sin 2πfct (5.39)

This means that to each real bandpass signal one complex baseband signal can be assigned orequivalently two real valued baseband signals which can be independent and represent real andimaginary part of the complex baseband signal.

This process is called I-Q up conversion (see Fig. 5.9) 7.

As the signal sBP (t) is real valued, the spectrum SBP (f) is symmetrical about f = 0. Since forcomplex signals s(t) the symmetry of the corresponding amplitude spectrum |S(f)| generally does

7One can regard this up mixing also as a modulation procedure (quadrature amplitude modulation – QAM),i.e. both the quadrature components sI(t), sQ(t) are independent signal source. The modulation product sBP (t) isthen called QAM signal. (see Digital modulation methods).



not hold, the ”’local”’ symmetry of the amplitude spectrum of the BP signal around f = ±fc alsodoes not hold. (see Fig. 5.8).

SBP (f) = SI(f) ∗(1

2δ(f + fc) +

1

2δ(f − fc)

)− SQ(f) ∗

(1

2 jδ(f − fc)− 1

2 jδ(f + fc)

)(5.40)

=1

2

(SI(f + fc)− jSQ(f + fc) + SI(f − fc) + jSQ(f − fc)

)(5.41)

With SI(f) + jSQ(f) = S(f) and SI(f) − jSQ(f) = S∗I (−f) − jS∗

Q(−f) = S∗(−f) follows (seealso eqn. 3.47)

SBP (f) =1

2

(S∗(−(f + fc)) + S(f − fc)

)(5.42)

|S(f)|A

−fg fg f

|SBP (f)|A

−fg,o −fc −fg,u fg,u fc fg,o f

|R(f)|, |U(f)|A

−2fc −fc −fg fg fc 2fc f

Low-pass

Figure 5.8: Spectrum of the complex baseband signal s(t) (above), the real bandpass signalsBP (f) (middle) and the signal r(t) at output of the mixture or u(t) at output ofthe low-pass (below)

The real bandpass signal sBP (t) can be transmitted e.g. over a radio link. Then the I-Q down mix-ing back into the baseband (quadrature demodulation) takes place in the receiver. (see eqn. (5.24)).

r(t) = sBP (t) e− j(2πfct−ϕ(t)) (5.43)

= Re{s(t) ej 2πfct

}e− j(2πfct−ϕ(t)) (5.44)


5.2 Up and Down Conversion of Complex Baseband Signal 39

Sender Receiver

sI(t) ×

cos 2πfct

sQ(t) ×

− sin 2πfct+ sBP (t)

×

cos(2πfct− ϕ(t))

rI(t)LP uI(t)

×

− sin(2πfct− ϕ(t))

rQ(t)LP uQ(t)

Figure 5.9: I-Q up and down mixing of a complex baseband signal s(t) = sI(t)+j sQ(t) (Quadra-ture modulator and quadrature demodulator)

With Re {z} = 12(z + z∗) (see Exercise 1) gives

r(t) =1

2

(s(t) ej 2πfct +s∗(t) e− j 2πfct

)e− j(2πfct−ϕ(t)) (5.45)

=1

2

(s(t) ejϕ(t) +s∗(t) e− j 2π2fct ejϕ(t)

)(5.46)

Consideration of the I-Q down mixing as multiplication with a complex carrier signal e− j 2πfct−ϕ(t)

shows, with the help of corresponding fourier transformation, that the spectrum R(f) of thesignal results r(t) at the output of the mixer originates from a ”’left shift”’ of the spectrum ofthe bandpass signal sBP (t).(see Fig. 5.8). With the approximation ϕ(t) ≈ ϕ = const yields thespectrum

R(f) =1

2

(S(f) ejϕ +S∗(−(f + 2fc)) e

jϕ)

(5.47)

The transmitted baseband signal s(t)can be recovered by a low-pass filter with the critical fre-quency fc (see Section 5.1).

u(t) = r(t)∣∣|f |<fc

(5.48)

=1

2s(t) ejϕ(t) (5.49)

So the inphase and quadrature components are

uI(t) =1

2

(sI(t) cosϕ(t)− sQ(t) sinϕ(t)

)(5.50)

uQ(t) =1

2

(sQ(t) cosϕ(t) + sI(t) sinϕ(t)

)(5.51)

In Section 5.1 the problem of the carrier synchronization was discussed in details. Also the correc-tion of the phase mismatch ϕ(t) is necessary during the transmission of complex baseband signals,in order to interpret the transmitted information correctly. The phasor diagram in Fig. 5.10 shouldclarify this once more.

The carrier synchronization can also be implemented here in two ways. 1.) The carrier signal ofcertain level is added to the bandpass signal of the sender and can be regenerated in the receiver



Im {s(t)}

Re {s(t)}

Coordinate system

of the sender

uQ(t) = Im {u(t)}

uI(t) = Re {u(t)}

Coordinate system

of the receiverϕ(t)

Figure 5.10: Phasor diagram of the baseband signal s(t)

by a narrow bandpass filter, i.e. the down convertion follows synchronous in phase ⇒ analogmodulation. 8 2.) The down convertion is executed independent of the carrier signal of the sender(phase asynchronously) and the correction takes place additionally with estimation of the phasemisalignment ϕ(t) and multiplication of the received signal u(t) with e− j ϕ(t) (”‘rotating back”’ ofthe coordinate system /phasors) (see Fig. 5.11) ⇒ Digital Modulation.

u(t) e− j ϕ(t) =1

2s(t) ej(ϕ(t)−ϕ(t)) (5.52)

=1

2s(t) ejΔϕ(t) (5.53)

=1

2s(t) for Δϕ(t) = 0 (5.54)

So the baseband signals sI(t) and sQ(t) can be transferred in the same frequency band andseparated in the receiver again.

u(t) = 12s(t) ejϕ(t) ×

e− jϕ(t)

12s(t)

Figure 5.11: Principle of phase correction for complex baseband signals in the receiver (Δϕ(t) =0)

8This corresponds to an analog QAM, which is however not applicable in practice; since even a small phasemisalignment ϕ(t) leads to a crosstalk between both source signals. Instead one uses single-sideband modulationwith two independent sidebands – ISB, see also Section 6.3.6


5.3 General Bandpass Signal – Equivalent Low-Pass Signal 41

5.3 General Bandpass Signal – Equivalent Low-Pass Signal

In general sinusoidal carriers are used for transmitting information. These can be generally mod-ulated in amplitude A, frequency f and phase φ (see Section 6.2). A general bandpass signal canbe represented as

sBP (t) = A(t) cos(2πf(t)t+ φ(t)) (5.55)

the carrier cos(2πfct) can be rearranged into

= A(t) cos(2πfct+ 2πΔf(t)t+ φ(t)︸︷︷︸ϕ(t)

) (5.56)

= Re{A(t) ejϕ(t)︸︷︷︸

s(t)

ej 2πfct︸︷︷︸Carrier

}(5.57)

Since the carrier signal does not contain any information, it can be omitted while consideringsignal transmission. The (complex) low-pass signal s(t) which equivalently represents bandpasssignal sBP (t) is given by

s(t) = A(t) ejϕ(t) (5.58)

= A(t) cosϕ(t) + jA(t) sinϕ(t) (5.59)

= sI(t) + j sQ(t) (5.60)

This low-pass signal (baseband signal) represents the complex Envelope (complex envelope –amplitude and phase) of the real bandpass signal.

A bandpass system can be thus be represented exactly with an equivalent low-pass system. Com-puter simulation of the time-continuous systems requires sampling at high enough sampling fre-quencies (sampling theorem). For example, a time step of T � 1 ns is necessary for the simulationof a radio transmission according to GSM standard in the 900 MHz band. The resulting simulationtimes would be very large. Since the GSM bit rate (270,83 kbit/s) is low and the (radio) channelcharacteristics change slowly, the system in the baseband can be simulated with a much smallersampling rate than the original bandpass signal.


42 6 ANALOG MODULATION

6 Analog Modulation

6.1 Introduction

Modulation is understood as imposing a source signal on a carrier signal. The source signal canbe a low-pass signal or baseband signal. The goal is to transmit the source signal over a certainchannel. The channel is an abstract model for describing the transmitting medium, e.g. coax cable,LWL-cable, radio link, for wire or free space transfer of electromagnetic waves.

Source Mod. Channel Demod. Sink

Figure 6.1: Transmission of source signal over a channel

At the beginning it requires to clarify why should modulation be used. Principal reasons for thisare

• the adjustment of the source signal into the transmitting medium (a source signal with a dccomponent cannot be directly transferred e.g. over a radio link.),

• the possibility to transfer several independent source signals over the same medium(e.g.frequency multiplex - ”’Shifting”’ the source signal into the desired frequency range(e.g. USW-broadcasting, television channel); Time multiplex - temporal progressive rate ofseveral source signals (each source has, in each case, access to the channel for a certain periodof time )).

At the same time

• to increase the quality of the transfer (e.g. signal-to-noise ratio SNR)

• to minimize the transmitting power,

• to minimize the structural complexity of sender and recipient.

If the transmission has to be implemented over a radio link, it is to be noted that an effectiveradiation is possible if the wavelength of the signal lies in the order of magnitude of the antenna.

λ =c

for, λ/m =

300

f/MHz(6.1)

For example, an antenna of length 1 m can radiate signals in the frequency range 100 MHz-1 GHz.In principle, low-pass signals cannot be transmitted as radio signals.


6.2 Modulation Schemes 43

6.2 Modulation Schemes

As carriers, particularly application of sinusoidal signals is found in communications (conceivablehowever also different signal forms, e.g. pulse-type carriers are with pulse modulation methods).

The sinusoidal carrier signal generally has the form (see Fig. 5.3)

sc(t) = A0 cos(2πf0t+ ϕ0) (6.2)

and contains thereby three parameters A0, f0, ϕ0; which can be used for information transmission(see Fig. 6.2). One parameter (or perhaps more) can be changed by the modulation to be propor-tional to the source signal (modulation signal). The source signal is an analog signal correspondingto analog modulation and depends on selected parameters particularly of

Amplitude Modulation AM: A0 → A(s(t)),

Frequency Modulation FM: f0 → f(s(t)) and

Phase Modulation PM: ϕ0 → ϕ(s(t)).

Source Modulators(t)

sc(t) = A0 cos(2πf0t+ ϕ0)

sBP (t) = A(t) cosψ(t)

Figure 6.2: Principle of modulation with sinusoidal carrier

6.3 Amplitude Modulation

The source signal s(t) is imposed with the amplitude of carrier signal sc(t). At the beginningclassical double sideband amplitude modulation(DSB) with carrier will be regarded. This typeof modulation is found e.g. in the medium wave broadcast application. The source signal s(t) isassumed to be zero-mean with symmetrical amplitude distribution.

6.3.1 Description in Time Domain

The source signal s(t) together with an overlaid dc component A0 forms the modulated amplitudeA(t) of the carrier signal sc(t), is also called envelope of the carrier sc(t).

sBP (t) = (A0 + s(t)︸︷︷︸A(t)

) cos 2πfct = A0

(1 +

s(t)

A0

)cos 2πfct (6.3)

The ratiomax(|s(t)|)

A0

=smax

A0

= m = μ (6.4)

defines the modulation factor or modulation index and it may vary the measure of carrier ampli-tude. If the modulation index is selected as μ ≤ 1, (→ A0 − smax ≥ 0), it is guaranteed that the



information is contained in the amplitude of the carrier. The demodulation becomes thereby verysimply (see envelope detector, e.g. medium wave broadcast). As opposed to that, if the modulationindex is selected as μ > 1 (→ A0 − smax < 0), then the information is contained in the amplitudeand phase of the carrier. Where the baseband signal A(t) = A0 + s(t) has a zero crossover (signchange), a rapid phase change ±π in the modulation product sBP (t) appears (see Fig. 6.3). Duringthe demodulation, therefore also the carrier phase must be included (see synchronous demodulator,e.g. short wave amateur radio).

sBP (t)

t

A0 + smax

A0

A0 − smax

sBP (t)

t

A0 + smax

A0

A0 − smax

Figure 6.3: Amplitude modulated transmitting signal sBP (t) with μ = 0.84 (left) and μ = 1.65(right)

6.3.2 Description in Frequency Domain

The Fourier transformation of eqn. (6.3) yields

SBP (f) = F {sBP (t)} = F {A0 cos 2πfct + s(t) cos 2πfct} (6.5)

=A0

2

(δ(f + fc) + δ(f − fc)︸︷︷︸

Carrier

)+

1

2

(S(f + fc) + S(f − fc)︸︷︷︸

Mixer product

)(6.6)

In the case of double sideband AM, the spectrum of the bandpass signal arises as a result ofshifting the low-pass spectrum of the source signal by the carrier frequency ±fc, in addition tothe carrier spectral line (see Fig. 6.4).

Example: 1-Ton-Modulation. The Spectrum SBP (f) is shown in Fig. 6.5 (see also Exercise 21).

s(t) = Am cos 2πfmt (6.7)

sBP (t) = A0(1 + μ cos 2πfmt) cos 2πfct with μ =Am

A0

(6.8)

⇒ SBP (f) =A0

2

(δ(f + fc) + δ(f − fc)

)+μA0

4

(δ(f + (fc + fm))+

δ(f + (fc − fm)) + δ(f − (fc − fm)) + δ(f − (fc + fm)))

(6.9)


6.3 Amplitude Modulation 45

|S(f)|S0

−fg fg f

|SBP (f)|A0

2

S0

2

−fc fc f

B = 2fgupper

Sideband

lowerSideband

carrier

Figure 6.4: Spectrum S(f) of the source signal s(t) (above) and Spectrum SBP (f) of the am-plitude modulated Bandpass signal sBP (t) (below)

|S(f)|Am

2

−fm fm f

|SBP (f)|A0

2

mA0

4

−fc − fm −fc −fc + fm fc − fm fc fc + fm f

Figure 6.5: Spectra of the source and transmitting signals for a 1-Ton-Modulation (μ = 0.67)

6.3.3 Power Balance

The signal power of the transmitted signal sBP (t) is composed of the signal power of the twosidebands PSB and that of the carrier signal Pc together. The two sidebands (shifted LP-spectrumof the source signal) represent the information signal. The carrier signal however does not containinformation about the source signal. The total transmitted power is given as

PAM = Pc + PSB (6.10)

The efficiency ηAM is given as

ηAM =PSB

PAM

(6.11)



With help of the above example of 1-Ton modulation, the efficiency ηAM can be calculated as afunction of the modulation index μ.

PAM = 2 · A20

4+ 2 · 2 · μ

2A20

16=

A20

2︸︷︷︸Carrier power Pc

+μ2A2

0

4︸︷︷︸Sideband power PSB

(6.12)

=A2

0

2

(1 +

μ2

2

)(6.13)

⇒ ηAM =μ2

2

1 + μ2

2

(6.14)

Full rejection of the modulator can be achieved by μ = 1 only with an efficiency of ηAM = 33 %.(The modulation with a square wave signal provides the highest efficiency of ηAM = 50 %. Thiscase is only theoretical, where the square wave signal needs an infinitely high bandwidth.)

6.3.4 AM-Modulators

Fig. 6.6 shows the principle of an AM-Modulator. It is assumed that the source signal is normalizedto smax (⇒ s(t) ∈ [−1;+1]).

s(t) μ × + A0 sBP (t)

cos 2πfct

Figure 6.6: Principle of an AM-Modulators

6.3.5 AM-Demodulators

The recovery of the source signal s(t) from the modulated band-pass filter signal sBP (t) is calleddemodulation. With an inspection of AM in the frequency range, it is clear that the source signals(t) contained in the modulation product sBP (t) is not located in its original frequency but shiftedwith the carrier frequency. Thus a frequency conversion is necessary in the receiver in order tobring the source signal back into the baseband from the carrier frequency. This is always donewith the help of amplitude modulation with same carrier signal (see also Section 5.1). Thus inthe receiver for the demodulation of the received signal sBP (t) (distortion-free transmission it isassumed here) the carrier signal is necessary once more. With regard to double sideband ampli-tude modulation with carrier (μ < 1), the carrier signal with a sufficient amplitude is containedin the modulation product sBP (t) and can be used directly for demodulation (see Envelope de-modulator). In Section 6.3.6 further versions of AM are presented, in which the carrier signal isonly is partly (μ > 1) or not at all, contained in the modulation product. The carrier signal mustbe supplied externally to the demodulator (see Synchronous demodulator). The book ”‘AnalogeModulationsverfahren”’[Mau92] is recommended for interested readers.


6.3 Amplitude Modulation 47

Envelope Detector The envelope detector

• is the simplest circuit for the demodulation of an amplitude modulated carrier signal (seeFig. 6.7 and Exercise 21) and assumes that the source signal is contained in the envelope ofthe carrier wave exclusively (μ < 1),

• consists of a half-wave rectifier, a RC low-pass and a RC high-pass;

• is practically found in each LW-, MW-, SW-radio;

• forms together with an antenna and a Tank circuit to the transmitter selection, to thedetector receiver as the first and simplest radio;

• needs no external supply of the carrier signal;

• is an incoherent AM demodulator, where phase position of the received signal is insignificant.

sBP (t)u(t)

≈ s(t)

rectifier low pass high pass

Figure 6.7: Circuit diagram of an envelope demodulator (detector recipient)

The adjoining amplitude modulated signal is rectified by means of the diode (see Fig. 6.8 a)). Thefollowing RC low-pass reconstructs the approximate envelope of the carrier wave (see Fig. 6.8 b)).The RC high-pass removes the dc component. (see Fig. 6.8 c)).

uDi(t)

t

a) after half-wave rectification

uTP (t)

t

b) after LP filtering

u(t) ≈ s(t)

t

c) after HP filtering

Figure 6.8: Processing in different places at the envelope demodulator



Synchronous Demodulator The Synchronous demodulator

• in the comparison with envelope demodulator somewhat more complex,

• shows however smaller distortions of the demodulated Signal,

• needs smaller input voltages, is thus more sensitive,

• can also be used with over modulation, i.e.. μ > 1 (reduced carrier),

• needs the phase position of the carrier,

• multiplies the received signal by the phase of the carrier signal and mixes so that the sourcesignal gets back into the baseband (see Section 5.1),

• is also well-known as product demodulator.

Fig. 6.9 shows the principle of a synchronous demodulator. The carrier signal must be regenerated

Synchronous down mixer

sBP (t) × u(t) ≈ s(t)LP HP

narrow BP with fc PLL

Carrier retrieval

Figure 6.9: Principle of the Synchronous demodulator (with AM with reduced carrier)

and strengthened the received signal sBP (t) from phase position. This is done with the help ofa very narrow bandpass filter and PLL circuits (phase locked loop). The Low-pass filter followsthe mixer removes those images by ±2fc (see Section 5.1). The high-pass filter removes the dccomponent from the output signal.

Still there exist some further versions, e.g. quadrature demodulator for single-sideband AM.

6.3.6 Different Types of Amplitude Modulation

So far mainly double sideband is considered with carrier (μ < 1), like it is found e.g. in the mediumwave broadcast applications. However still some other versions of AM exist. Reasons to use otherAM methods are [Mau92]

• to improve power balance ⇒ by reducing or suppressing the carrier,

• need to reduce the bandwidth (in modulation with a real source signal, the upper and lowersideband contains the same information, see Section 5) ⇒ single sideband modulation.

Through these measures, however, reference values for frequency and phase of the carrier are lostwhich must be known during the demodulation.


6.4 Phase and Frequency Modulation (Angle Modulation) 49

The AM methods are differentiated with regard to modulated sidebands and carrier power.

• Sidebands

– Double sideband DSB

– Vestigial sideband VSB

– Single sideband SSB with the distinction whether upper or lower sideband is transferredupper sideband USB, lower sidebandLSB

– Independent sideband ISB

• Carrier

– with carrier

– with reduced carrier -RC

– with suppressed carrier -SC

6.4 Phase and Frequency Modulation (Angle Modulation)

If the source signal s(t) is impressed on the frequency of the carrier sc(t), one speaks of frequencymodulation FM. On the other hand, impressing the source signal on the phase of the carrieris phase modulation PM. In both cases, the frequency as well as the instantaneous phase of thecarrier is modified ⇒ modulation product. One can generally speaks therefore of angle modulationFM [Mau92].


The Modulation product has the general form

sFM(t) = A0 cosΦ(t) = A0 cosΦ(s(t)

)(6.15)

With the fundamental relationship (angular speed ω(t) results from the derivative of the angleφ(t) with respect to time t)

ω(t) = 2πf(t) =d

d tΦ(t) (6.16)

can be also written as

sFM(t) = A0 cos(∫ t

t0

2πf(τ) d τ + ϕ(t0))

with t0 < t (6.17)

Frequency modulation FM:

f(t) = fc + kFM s(t) with the frequency shift ΔF = kFM smax (6.18)

⇒ Φ(t) = 2πfct︸︷︷︸carrier

+2π kFM

∫ t

t0

s(τ) d τ︸︷︷︸mod. Phase

+ϕ(t0)− 2πfct0︸︷︷︸ϕ0

(6.19)



Thus the modulation product results to

sFM(t) = A0 cos(2πfct + 2π kFM

∫t

t0

s(τ ) d τ + ϕ0

)(6.20)

Phase modulation PM:

Φ(t) = 2πfct+ kPM s(t) + ϕ0 with the phase shift ΔΦ = kPMsmax (6.21)

⇒ f(t) = fc +kPM

2πs(t) with s(t) =

d

d ts(t) (6.22)

Thus the modulation product yields

sPM(t) = A0 cos(2πfct + kPM s(t) + ϕ0

)(6.23)

The modulation product sFM(t) or sPM(t) has a constant Amplitude. Both the frequency shift δFand phase shift δφ are usually given by international agreements (e.g. USW-broadcast: FM withΔF = 75 kHz). Those Amplification factors kFM , kPM must be selected accordingly.

Example: 1-Ton Modulation s(t) = Am cos 2πfmt

⇒ sFM(t) = A0 cos(2πfct+ 2π kFM

∫ t

t0=0

Am cos 2πfmτ d τ + ϕ0

)for 0 < t (6.24)

= A0 cos(2πfct+

2π kFMAm

2πfmsin 2πfmt+ ϕ0

)(6.25)

= A0 cos(2πfct+

ΔF

fmsin 2πfmt+ ϕ0

)(6.26)

⇒ sPM(t) = A0 cos(2πfct+ kPM Am cos 2πfmt+ ϕ0

)(6.27)

= A0 cos(2πfct+ΔΦcos 2πfmt+ ϕ0

)(6.28)

Here the common view is underlined again by FM and PM. Both procedures supply the samemodulation product with the 1-Ton modulation (see Fig. 6.10). Here frequency and phase shiftsare connected by the following context:

ΔF

fm= ΔΦ (6.29)

A distinction of FM and PM becomes only possible with modification of the frequency fm of thesource signal s(t). Despite this, both procedures differs in modulation gain, bandwidth, modulatorand demodulator circuits, noise behavior etc.. Whereupon is not to be further explained in thecontext of this lecture. Further readings may be continued in the lecture ⇒ Rundfunksysteme).


Frequency and phase modulation are non linear modulation method in contrast to AM.While the modulation product (bandpass signal) of the AM is obtained simply by ”’shifting”’the spectrum (low-pass) of the source signal around the frequency ±fc , the spectrum of themodulation product by angle modulation is not so easily definable. Fundamentally it can be said,however, that the necessary bandwidth of the modulation product sFM(t) is usually larger thanone of the modulating source signal (see Fig. 6.11) 9.

9Theoretically the bandwidth of the modulation product is yet infinite, but practically one can assume thespectrum outside of a certain bandwidth B in the disappeared noise.



sFM(t)

t

A0

−A0

Figure 6.10: 1-Ton angle modulated Signal

|S(f)|

−fg fg f

B = 2fg

|SFM(f)|

−fc fc f

B ≥ 2fg

Figure 6.11: Spectrum S(f) of the source signal s(t) (above) and Spectrum SBP (f) of the anglemodulated bandpass filter signal sFM (t) (below)

The bandwidth depends thereby on the rejection of the modulator (frequency shift ΔF or phaseshift ΔΦ). Sufficient approximation for practice represents the Carson rule, which is the worstcase consideration (1-Ton modulation), and is usually bigger than the bandwidth required.

FM: BFM = 2(ΔF + fg) (6.30)

PM: BPM = 2(Δφ+ 1)fg (6.31)

Here fg the critical frequency of the source signal. The USW broadcast results in the case of afrequency shift by ΔF = 75 kHz and a critical frequency of fg >= 15 kHz (mono transmitter), anecessary bandwidth of BFM = 180 kHz.

For the above example of the 1-Ton modulation the spectrum can be given still quite clearly. Sincefrequency and phase shifts over eqn. (6.29) are interconnected, the modulation product can begenerally indicated as (see eqn. (6.26) and (6.28))

sFM(t) = A0 cos(2πfct+ΔΦsin 2πfmt+ ϕ0

)(6.32)

The difficulties in the calculation of the spectrum lie in the fact that here trigonometric functionsemerge again as argument of trigonometric functions. With expansions in powers series derivefrom the following relation [SS88]

cos(α + x sin β) =∞∑

n=−∞Jn(x) cos(α + nβ) (6.33)



The modulation product can be now written as (with ϕ0 = 0)

sFM(t) = A0

∞∑n=−∞

Jn(ΔΦ) cos(2π(fc + n fm)t

)(6.34)

In addition the Bessel function Jn(ΔΦ) of first type and n-th order is indicated (see Fig. 6.12).

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

Phasenhub ΔΦ

J n(ΔΦ

)

J0(ΔΦ)

J1(ΔΦ)

J2(ΔΦ)J3(ΔΦ)

J4(ΔΦ)

Figure 6.12: Bessel function Jn(ΔΦ) first type, 0. to 4. Order, It applies additionally J−n(x) =(−1)n Jn(x)

The Fourier transformation from (6.34) yields the spectrum of the modulation product, out ofwhich exist many infinite spectral lines.

SFM(f) =A0

2

∞∑n=−∞

Jn(ΔΦ)(δ(f + (fc + n fm)) + δ(f − (fc + n fm))

)(6.35)

Fig. 6.13 shows the dependency of output spectrum with respect to phase shift ΔΦ (Frequencyshift ΔF = fmΔΦ).

6.4.3 Power Balance

Discussion is quite difficult with angle modulation from the point of view of power balance becausethe carrier contained in the modulation product is also to be contained in the information which



|SFM(f)|A0

2

−fc fc f

BCarson ΔΦ = 0.3

|SFM(f)|A0

2

−fc fc f

BCarsonΔΦ = 1

|SFM(f)|A0

2

−fc fc f

BCarsonΔΦ = 2.4

|SFM(f)|A0

2

−fc fc f

BCarsonΔΦ = 5

|SFM(f)|A0

2

−fc fc f

BCarsonΔΦ = 10

Figure 6.13: Spectrum SFM (f) of the Modulation product sFM (t) in 1-Ton Modulation withrespect to Phase shift ΔΦ



is transmitted. The total operating performance of the Modulation product /transmitting signalis constant due to the non-changing envelope.

PFM =A2

0

2(6.36)

The total output divides into the power of the sidebands and the carrier power. The allocationdepends on the selected rejection(ΔΦ, ΔF ).

For the 1-Ton modulation again very good numerical values can be indicated.

PFM =A2

0

2=A2

0

2

(J20(ΔΦ)︸︷︷︸

Carrier power

+ 2∞∑n=1

J2n(ΔΦ)︸︷︷︸Side band power

)(6.37)

With the help of the diagram in Fig. 6.12 two values can be read off immediately: Pc = PFM forΔΦ = 0 (unmodulated carrier) and Pc = 0 for ΔΦ ≈ 2.4.

6.4.4 FM-Modulator

There exists a large variety of modulators and will not be dealt here further (⇒ Lecture Rund-funksysteme). For FM modulators often voltage-controlled oscillators come to the application, e.g.control of a varactor diode in the tank circuit A) directly over the source signal ⇒ FM, b) overthe high-pass-filtered (differentiated) source signal ⇒ PM (see Fig. 6.14).

a)

s(t) VCO sFM(t)

b)

s(t) HPs(t)

VCO sPM(t)

Figure 6.14: FM modulator with the help of a voltage-controlled oscillator (VCO)

Voltage-controlled oscillators are usually named VCO briefly. The Oscillator frequency fV CO isdependent on the adjoining input voltage u(t) (here u(t) ∼ s(t) – FM or u(t) ∼ s(t) – PM, seeFig. 6.15).

fV CO

fc

−umax umax u(t)

Figure 6.15: Dependence of the oscillator frequency of a VCO on input voltage u(t)



6.4.5 FM-Demodulator

There exist tremendous variety of demodulators here too. Only one version is briefly pointed outhere. A possible FM demodulation is the FM to AM conversion with the following AM demodula-tion. A flank frequency discriminator (flank demodulator) executes such a conversion with the helpof an easily detuned tank circuit (see Fig. 6.16). The impedance Z(f) of a tank circuit is periodic

|Z(f)|

fc fr f

sFM(t) input signal

output signal

sAM(t)

Figure 6.16: Frequency discriminator characteristic of a tank circuit

whose maximum value occurs with resonant frequency fr. Beside of the resonant frequency frimpedance Z(f) decreases, which can be adjusted by damping of the resonant circuit. In a certainarea of the characteristic this is almost linear and can be used for the FM to AM conversion. Inaddition the FM signal can be fed into the resonant circuit. The varying frequency leads to animpedance modification, which again leads to a change in voltage. Therefore an amplitude modu-lated signal is present at the output, which can be demodulated like an AM signal (see envelopedemodulator in Section 6.3.4).

The FM to AM conversion can be usually produced by differentiation of the frequency modulatedsignal.

sFM(t) = A0 cos(2πfct+ 2π kFM

∫ t

t0

s(τ) d τ)

(6.38)

sAM(t) =d

d tsFM(t) (6.39)

= −A0

(2πfc + 2π kFMs(t)

)︸︷︷︸

A(t)

sin(2πfct+ 2π kFM

∫ t

t0

s(τ) d τ)

︸︷︷︸Carrier

(6.40)



6.5 Comparison between Amplitude and Angle Modulation

Properties of Amplitude Modulation – AM

• very easy method

• for μ < 1 the information of the source signal is exclusively contained in the envelope of themodulation product → demodulation becomes particularly simple (see envelope detector)

• for μ > 1 the information is also contained in the phase of the carrier and must be consideredwith the coherent demodulation → phase synchronization is necessary (see synchronousdemodulator)

• by double sideband modulation with (full) carrier (DSB) puts practically more than 70 %of the transmitted power in the carrier signal

• the level of demodulator output signal is linearly dependent on the input level. Thus it canresult strong amplitude oscillations when fluctuation occurs in the transmission path.

• the transmission amplifier must be linear in a large dynamic range

Properties of Frequency and Phase Modulation – FM, PM

• the modulation product possesses constant envelope

• therefore the transmission amplifier can be operated without power reserve

• quite noise insensitive(Modulation gain > 1)

• higher bandwidth required

• Narrow band FM (small frequency, phase shift) behaves like AM

Criteria AM FM / PM

Bandwidth SSB: B = fg PM: B = 2(ΔΦ + 1)fgDSB: B = 2fg FM: = 2(ΔF + fg)

(theoretically B → ∞)

Distortions no distortion from suppression increasing distortion with increas-ing phase shift above threshold

Requirement of atransmission sys-tem

linear amplitude frequency re-sponse required, amplifier withscope”-reserve

no linear amplitude frequency re-sponse required, amplifier can beoperated without power”-reserve

Application analog signal transmissionwith inferior quality, inferiortransmission”-bandwidth, simplerreceiver, carrier”-frequency technic

analog signal”-transmission withbetter quality


57

7 Analog-Digital-Conversion

7.1 Motivation

Signals around us are usually continuous both in value and time. Such signals are generally calledanalog signals. The direct processing of these analog signals takes place by means of analog systems(see lecture System Theory). Since however digital systems are less vulnerable against distortions,universally applicable, parameterized and reconfigurable, and often offer more possibilities forsignal manipulation, it is meaningful in many cases to implement the processing of analog signalsby means of digital systems. Therefore an Analog-Digital- or Digital-Analog-Conversion of thesignals is necessary. Such conversions are also needed when digital signals are transferred with thehelp of analog signals, e.g. over a radio interface. (see Fig. 7.1).

AnalogSignal

ADCDigital Systeme.g. μ-Processor

DACAnalogSignal

DigitalSignal

DACAnalog System

e.g. Radio channelADC

DigitalSignal

Source/Sink Source/Sink

Figure 7.1: Main applications of AD or DA converter

This paragraph is particularly concerned with possible problems of the analog-to-digital conver-sion. Hereunder it would clarify, how and which signal property is changed by such conversions,and whether reconstruction of analog signal is possible again. In regard with practical implemen-tations, error sources should be analyzed and considered for the development of converters.

A digital signal is characterized by discrete in time and discrete in value. In many applicationsdigital signals are attained from analog signals, by means of an analog-digital converter, wherecontinuous in time and value analog signal is converted into discrete in time and value (digital)signal. The discretization in time is called sampling, whereas the discretization in amplitude iscalled quantization.

The advantages of digital signal processing are in particular:

• The processing is independent of technology and temperature.

• The processing of signals in digital systems is usually simpler i.e., determination of signalmanipulations is realizable only with the help of digital signal processing.

• The signals can be easily stored in a memory.

7.2 Time Discretization

Time continuous signals s(t) are function of time, i.e., exactly one function value is assigned tothem at each point in time t. Time discrete signals {s(n)} can be thought as sequences, in which


58 7 ANALOG-DIGITAL-CONVERSION

exactly one value is assigned to each n. At the same time each n is assigned to a certain time pointt. The transformation s(t) → {s(n)}, which ideally takes place without any information loss, iscalled time discretization in time or sampling.

7.2.1 Dirac-Comb-Function

For a better description of sampling, a new function, the dirac-comb function XT (t) is introduced.This function consists of equally spaced time-shifted dirac pulses and is defined as follows.

XT (t) =∞∑

n=−∞δ(t− nT ) (7.1)

Since the dirac-comb function is a periodic function with period T , one can extend it into a Fourierseries (see Exercise 16)

XT (t) =∞∑

n=−∞

1

Tej 2π

nTt (7.2)

Its Fourier transformation can be obtained easily as

F {XT (t)} =1

T

∞∑n=−∞

δ(f − n

T

)(7.3)

=1

TX 1

T(f) (7.4)

So it is clear that the spectrum of a dirac-comb function in time domain corresponds again to thedirac-comb function in frequency domain.

XT (t)1

TX 1

T(f) (7.5)

7.2.2 Sampling

Sampling means ”‘taking”’ of function values (samples) of a time-continuous signal. The samplingvalues are combined into a sequence.

s(tn) → {s(n)} with ∀n∈Z tn < tn+1 (7.6)

The sampling can take place at any successive points in time, however only equidistant samplingis to be regarded here, since it possesses the most practical meaning.

s(nT ) → {s(n)} with T – sample period (7.7)

The sampling process can be modelled by masking (setting to zero) all function values s(t �= nT ).Since after masking of samples the area

∫s(t) d t tends to zero, then the (remaining) sampling

values must be weighted accordingly.

Starting point should be a real sampling or gating circuit with a gate open time of T0 (see Fig. 7.2).So the sampled signal can be described by


7.2 Time Discretization 59

s(t)

0 T 2T 3T 4T t

T0

T0

Tss(t) = s(t)

∞∑n=−∞

rect(

t−nTT0

)

Figure 7.2: Input signal s(t) and output signal ss(t) of the real sampler

ss(t) = s(t)∞∑

n=−∞

T

T0rect

(t− nT

T0

)with T0 < T (7.8)

The factor T/T0 is essential, so that the window function remains dimensionless (unit less) as wellas the area remains constant independently of sampling rate 1/T and gating time T0 constant. Ifone starts making the duration of the sampling pulse shorter and finally aims at T0 → 0, then themodel of ideal sampler becomes 10 (see Fig. 7.3 and 7.4)

ss(t) = s(t)∞∑

n=−∞Tδ(t− nT ) (7.9)

With the help of Dirac comb function introduced above the sampling can now be described as

s(t)

0 T 2T 3T 4T t

ss(t) = s(t)∞∑

n=−∞T δ(t− nT )

Figure 7.3: Input signal s(t) and output signal ss(t) of the ideal samplers (Dirac impulses canalso be drawn only as symbol, see also Fig. 3.2)

ss(t) = s(t)∞∑

n=−∞Tδ(t− nT ) = s(t)T XT (t) (7.10)

Here T defines the sample period and fs = 1/T is the sample rate.

10In older versions of the scripts the conventional way in the literature was taken, the factor T allowed in timedomain (it appeared therefore in the frequency domain as 1/T ). It yields ss(t) = s(t)

∑∞

n=−∞δ(t − nT ) (see also

the exam in 1998-1999). In principle therefore nothing changes, but working with dimension afflicted parameters,like time t and frequency f , is facilitated by the way they have been selected here. (see Article. 3.2.5). Since thesedimension afflicted parameters are worked out in the script consistently, it is only possible to compare s(t) andss(t) and later S(f) and Ss(f) together and to draw a diagram.



s(t) ×

T∞∑

n=−∞δ(t− nT )

ss(t)

Figure 7.4: Model of an ideal sampler

The sampled signal ss(t) is a time-continuous signal. The multiplication of the signal s(t) with thedirac comb function TShT (t) in time domain corresponds to the convolution of the spectra S(f)and F {T XT (t)}. Thus with eqn. (7.5)

ss(t) = s(t)T XT (t) (7.11)

⇒ Ss(f) = S(f) ∗X 1T(f) (7.12)

Considering the additional important characteristic (3.80) of the dirac delta function yields

Ss(f) = S(f) ∗X 1T(f) = S(f) ∗

[ ∞∑n=−∞

δ(f − n

T

)](7.13)

=∞∑

n=−∞S(f) ∗ δ

(f − n

T

)=

∞∑n=−∞

S(f − n

T

)(7.14)

By convolution of the original spectrum S(f) with the periodic spectrum of dirac comb function,the resulting sampled signal spectra is continued periodically with period fs = 1/T (compareto (7.14)).

Ss(f) =∞∑

n=−∞

S(f − nfs) with fs = 1/T (7.15)

Fig. 7.5 clarifies this duplication of the original spectrum. The signal s(t) was assumed as bandlimited low-pass signal with critical frequency fg. The obtained duplicates of the original spectrumare called images and they do not carry any further information.

Ss(f) =∞∑

n=−∞S(f − nfs)

−fs −fg fg fs f

S(f) images

Figure 7.5: Spectral representation of the sampled signal (fs = 1/T )

7.2.3 Sampling Theorem

In Fig. 7.5 the sampled signal was assumed as band limited low-pass signal. Now naturally themain questions that arise in digital signal processing and communications technology are whether



these hypothesis are necessary and which relation exists between the critical frequency fg and thesampling rate fs of the signal.

Many researchers dealt with these questions and published the well known sampling theorem indifferent occasions. The first publications were in mathematics: e.g. La Vallee Poussin (1908), E.T.Whittaker (1915). In the communications technology particularly two names should be quoted:V.A. Kotelnikov (1933) and C.E. Shannon (1949), whereas formerly the sampling theory by Shan-non found a broad publicity.

Sampling Theorem A band limited signal with fg can be reconstructed accurately from itssampled values with the help of sampling series

s(t) =∞∑

n=−∞s(nT ) si

(π

(t

T− n

))(7.16)

if during the sampling the condition

fs =1

T> 2fg (7.17)

is kept.

The sampling rate fs = 2 > fg is also known as Nyquist rate. From the sampling theory thefollowing conclusions can be drawn (see Exercise 17, 18):

• If the signal s(t) is band limited and if it is sampled under the condition specified in thesampling theory (7.17), then the signal is uniquely described by the (samples) and can bereconstructed accurately (7.16).

• If the signal s(t) is sampled with a rate which is smaller than the Nyquist rate, then s(t) cannot be sufficiently described by its samples s(n > T ) and reconstruction of such signal willproduce distortions (errors). This effect is graphically representable in the frequency domain(see Fig. 7.6). Due to the periodic continuation of the spectrum (f) with fs, it overlapswithin the space of Ss(f). Such overlaps is denoted as Aliasing.

• Since aliasing error cannot be cancelled, the mentioned condition (7.17) must be guaranteedfor sampling. This is done by an anti-aliasing filter before sampling (see Fig. 7.7). Theanti-aliasing filter is an analog low-pass filter with H(f) = 0 for f ≥ fs

2.

• A band limited signal s(t), which is sampled exactly with the Nyquist rate, can be accuratelyreconstructed only if the phase relation between s(t) and sampling points nT is known.

• If a signal is sampled with a rate fs which is larger than the Nyquist rate 2fg, oversamplingoccurs. By oversampling one does not gain additional information (oversampling howeverbrings advantages for technical circuiting – quantization noise, filter design).



Ss(f)

−fs −fg fg fs f

fs > 2fg

Ss(f)

−fs −fg fg fs f

fs < 2fg

Aliasing

Figure 7.6: Aliasing effect

Anti-Aliasing

FilterADCs′(t)

s(t)ss(t)

Figure 7.7: To avoid aliasing errors an so-called anti-aliasing filter (analog low-pass filter) hasto be provided before the sampler.

7.2.4 Signal Reconstruction – Digital-Analog-Conversion

In order to recover a time-continuous (Analog) signal from a time-discrete (Digital) signal, thesignal processing between the sampling points must be reconstructed. Particularly eqn. (7.17)have to be fulfilled so that the reconstruction takes place without any error.

Images arise during sampling (see eqn. (7.15)), which are reflections of the original spectrum. Forsignal reconstruction these images must be removed by suitable filtering. For that purpose anideal low pass (analog) filter with a critical frequency fg,TP = fs/2 is used (see Fig. 7.8 and 7.9).With the application of real filters (the ideal low pass is not realizable), the reconstructed signaldeviates srek(t) more or less strongly from the original signal s(t).

ideal

low passss(t) srek(t) = s(t)

Figure 7.8: The signal reconstruction takes place with an ideal low pass filter with a criticalfrequency fg,TP = fs/2 (often called as reconstructing filters)

The DA converter with an ideal low pass filter with critical frequency fg,TP = fs/2 should beconsidered.

S(f) = Ss(f) · rect(f

fs

)with fs =

1

T(7.18)



Ss(f) = F {ss(t)}

−fs −fg fg fs2

fs f

ideal reconstructing

low pass filter

⇓Srek(f) = S(f)

−fg fg f

Figure 7.9: Spectral representation of the reconstruction of time continuous signal from a time-discrete signal (fg,TP = fs/2, fs = 1/T )

= (S(f) ∗X 1T(f)) · rect(fT ) (7.19)

The Fourier transformation in time domain yields the sampling series specified in the samplingtheory (7.16)

s(t) =

(s(t)

∞∑n=−∞

T δ(t− nT )

)∗(1

Tsi

(πt

T

))(7.20)

=

( ∞∑n=−∞

s(t) δ(t− nT )

)∗ si

(πt

T

)(7.21)

=∞∑

n=−∞s(nT ) si

(π

(t

T− n

))(7.22)



7.3 Value Discretization

Since digital systems can process only time and value discrete signals, the samples s(nT ) mustbe discretized additionally in amplitude. This process is also known as quantization and is anonlinear operation. In contrast to the (ideal) time discretization/sampling, the value discretiza-tion/quantization cannot be reversed. Therefore after the signal reconstruction in the DA convertera (as small as possible) deviation from the original signal must be established. (see Fig. 7.10, 7.11).This deviation is called quantization error.

s(t) sq(t)

Figure 7.10: Quantization

0 0.2 0.4 0.6 0.8 1 1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Zeit t/T

Am

plitu

de

Figure 7.11: Exemplary quantization of a sine signal with pertinent quantization error.

The order of operations, sampling and quantization, is arbitrary, in practice is however first sam-pled and then quantized. This is done via a real gate switch with a gating time T0 > 0 (seeFig. 7.2), followed by a short time integration (RC element) to the averaging of the signal in theinterval T0 and finally quantization of the average value. The quantized sampling value is pickedout and passed to the subsequent system.


7.3 Value Discretization 65

During quantization, the infinite scope of the (value continuous) input signal is mapped in a finitescope. The input signal must possess limited ranges of values, thus (s(t) is usually a voltage).

Ulower ≤ s(t) ≤ Uupper (7.23)

In AD converter the range ΔU = Uupper − Ulower is divided into q intervals and to each value ofthe input signal in that particular interval is assigned to a quantization step.

s(t) → sq(t) (7.24)

Since the number of values of the quantized signal is finite, the quantized signal can be representedwith the help of a signal of alphabets. Usually binary numbers are used for it. In this case thenumber of necessary bits b are derived from the number q of quantization levels

b = � ld q� (7.25)

Therefore ld designates the logarithm dualis and n = �x� is the smallest integer number n withx ≤ n (e.g. 4 = �3.1�).

7.3.1 Quantization Characteristic Curve

The quantization characteristic curve determines the power distortions and thus the quality ofquantization in connection with s(t) the amplitude density distribution (often uniform distri-bution is assumed) of the input signal. Even though the Quantization is nonlinear, oftenit is distinguished between linear (uniform quantizer) and nonlinear Quantization (nonuniformquantizer). It indicates thereby only the characteristic of quantization (companding characteristicdiagram).

Linear quantization characteristic diagram The Quantization step Δs is constant

Δs =ΔU

q=

ΔU

2bwith q = 2b (7.26)

The output signal can take the following values (see Fig. 7.12):

sq(t) ∈{{±Δs

2· (1, 3, 5, . . . , q − 1)

} ⇒ mid-rise quantizer

{Δs · (0, 1, 2, . . . , q/2− 1); −Δs · (1, 2, 3, . . . , q/2)} ⇒ mid-tread quantizer(7.27)

Nonlinear quantization characteristic diagram With a linear quantization characteristic,the maximum relative error emax(t)/s(t) = Δs/(2s(t)) for smaller input signals is clearly biggerthan for larger input signals. Therefore it is sensible to quantize smaller signal values more preciselythan larger signal values (see Fig. 7.13).

Different nonlinear characteristics are used out of which at most A-law and μ-law characteristics.Since the handling of the nonlinear quantization characteristics goes beyond the framework of thislecture, only the linear quantization characteristic is going to be considered here.



sq

s

5Δs2

3Δs2

1Δs2

−1Δs2

−3Δs2

−5Δs2

sq

s

2Δs

Δs

−Δs

−2Δs

−3Δs

Figure 7.12: Linear mid-rise (left) and mid-tread (right) quantizer

sq

s

Figure 7.13: Quantizer with nonlinear quantization characteristics (non uniform quantizer)

7.3.2 Quantization Error Signal

The (quantization) error signal e(t) is defined as the deviation of the quantized signal sq(t) fromthe original signal s(t).

e(t) = s(t)− sq(t) (7.28)

If the quantizer is adequately well controlled and if the input signal s(t) is sufficiently variable intime, the error signal e(t) can be assumed as white noise with constant amplitude density in theinterval [−Δs/2; Δs/2] (see Fig. 7.14). Thus for linear quantization characteristic the quantizationerror power Pe can be specified as

Pe = E[e2] =

∫ Δs2

−Δs2

e21

Δsd e =

1

12(Δs)2 (7.29)



p(e)1Δs

−Δs2

Δs2

e

Figure 7.14: Assumed amplitude density function p(e) of the error signal

7.3.3 Signal-to-Noise Ratio

By signal-to-noise ratio SNR one understands the ratio of the user signal power and noise power.Considering the non-quantized input signal s(t) with average power Ps as user signal and theoriginating quantization error e(t) with average power Pe as noise signal, gives the signal-noise-ratio as

SNR = 10 lgPs

Pe

(7.30)

The average quantization error power is approximately known under certain conditions (see Ar-ticle. 7.3.2). The average input signal power however strongly depends on the dynamic range ofthe AD converter and the signal shape. Thus e.g. for a noise shaped signal as input (useful) signalwith uniform amplitude density over the entire (symmetrical around zero) dynamic range ΔU

Ps,N =

∫ ΔU2

−ΔU2

s21

ΔUd s =

1

12(ΔU)2 (7.31)

=1

12(2bΔs)2 (7.32)

and for a sine signal with the amplitude ΔU/2 (likewise full rejection)

Ps,S =1

2π

∫ 2π

0

(ΔU

2

)2

sin2 x d x =1

8(ΔU)2 (7.33)

=1

8(2bΔs)2 (7.34)

Now the signal-to-noise ratio can be given for both signals.

SNRN = 10 lg12(2bΔs)2

12(Δs)2= 20b lg 2 dB (7.35)

= 6.02b dB (7.36)

SNRS = 10 lg12(2bΔs)2

8(Δs)2= (10 lg 1.5 + 20b lg 2) dB (7.37)

= (1.76 + 6.02b) dB (7.38)

The dynamic range or the signal form changes the SNR only over a certain offset, however eachtime the SNR is increased by 6 dB independent of dynamic range and signal shape, if duringquantization 1 bit is more used.



7.3.4 Oversampling

Since the output signal of the AD converter is discrete in time, the entire quantization errorpower Pe must reside in the frequency range [−fs/2; fs/2]. Thus the power density spectrum ofthe error signal shown in Fig. 7.15 results on the assumption of a white quantization noise (seeArticle. 7.3.2). It is now to be demonstrated that oversampling, i.e. the sampling rate greater than

|N(f)|Pe

fs

−fs −fs2

fs2

fs f

images

Figure 7.15: Power density spectrum N(f) of the error signal

the least necessary Nyquist rate, implies an increased signal-to-noise ratio involved with the signalreconstruction. An (over sampling ratio) osr is introduced.

osr =fsfN

fN = 2fg – Nyquist rate (7.39)

An oversampling factor osr = 1 implies sampling with the Nyquist rate. If an oversampling factorosr > 1 is selected, then quantization noise power Pe distributes itself over a larger frequency range.A part of the quantization noise power is thereby in a frequency range outside of original signal

N(f)

0 fs/2 = fg f

Ps

Pe

Reconstructing

low pass filter

osr = 1SNR

N(f)

0 fg fs/2 f

Ps

Pe

Reconstructing

low pass filter

osr > 1SNR

Figure 7.16: Dependency of the spectral power densityN(f) of error signal on the oversamplingfactor osr

frequency range (0 ≤ f ≤ fg). For signal reconstruction, as mentioned in Article. 7.2.4, a low-passfilter with critical frequency fg is used. This low-pass filter filters a part of the quantization noisepower Pe (fg < f < fs/2) and thereby increases the signal-to-noise ratio (see Fig. 7.16).



The signal-to-noise ratio SNRRek after signal reconstruction increases thereby as follows

SNRRek = SNR + 10 lg osr (7.40)

Each doubling of sampling rate increases the signal-to-noise ratio SNRRek by 3 dB (it validateson the assumption that the quantization error remains white during oversampling).

As far an ideal low pass filter with Nyquist rate (osr = 1) is necessary for signal reconstruction,a filter having osr > 1 with finite edge steepness can also be used, which is technically realizablein contrast to an ideal low-pass filter. Therefore in practice, always osr > 1 is selected.

7.3.5 Error Modelling of AD Converter

By ”‘shaping”’ the quantization noise it is possible to reduce the quantization noise power den-sity in the frequency range of interest. A higher signal-to-noise ratio after reconstruction can beachieved despite of small quantization resolution of b.

N(f)

0 fg fs f

Ps

Pe

Reconstructing

low pass filterosr � 1

Figure 7.17: Spectral power density distribution of quantization error with the application ofa noise forming AD converter (ΣΔ-converter)


70 8 DIGITAL MODULATION

8 Digital Modulation

8.1 Introduction

By modulation, generally the suppression of a source signal on a carrier signal is understood. (seealso Section 6.1). In contrast to analog modulation, however, the source signal d{k} is a digitalsignal in case of digital modulation. Since digital signals represent only time-discrete sequences,they are not suitable for direct modulation with a time-continuous carrier, since otherwise themodulation product would be defined also only with appropriate cycle-time points kTs (Ts –symbol duration). Therefore the modulation must be viewed always in connection with the signalformation so that the process of the modulating signal corresponding to the modulation product isdefined between the points of cycle-time kTs. Since the digital source signal is discrete in valuewith M different values, the modulation product always presumes one of M different states of thecycle-time points kTs. The exact transition of one state to another is actually irrelevant for thedata transmission, however has substantial influences on

• the bandwidth requirement of the modulation product (send-signal),

• the resistance against interferences,

• the technical implementation.

8.2 Types of Modulations

If sinusoidal carriers are used, the parameters amplitude A0, frequency f0 and phase ϕ0 are alsoavailable for modulation (see Section 6.2). Since in first applications of the digital modulation,square pulses were used as modulating signals (hard shifting /switching of the selected parameters),the methods possesses the following titles (see Fig. 8.1)

Amplitude Shift Keying ASK: A0 → A(d{k})

Frequency Shift Keying FSK: f0 → f(d{k})

Phase Shift Keying PSK: ϕ0 → ϕ(d{k})

Three exists numerous variants on the basis of these methods, most of which hold independentnames and can be differentiated clearly. Distinguishing criterions are e.g.

• Number of states of the digital source signal

– 2 = 21 states (binary) e.g. BPSK – binary phase shift keying

– 4 = 22 states (quaternary) e.g. QPSK – quaternary phase shift keying

– 8 = 23 states e.g. 8-PSK

– 16 = 24, 64 = 26, 256 = 28 states e.g. 16-QAM, 64-QAM, 256-QAM quadratureamplitude modulation

• whether a real or a complex baseband signal is formed from the digital source signal


8.3 BPSK – Binary Phase Shift Keying 71

sBP (t)A0

0

−A0

t

a) ASK (special OOK)

sBP (t)A0

0

−A0

t

b) FSK (special 2-FSK)

sBP (t)A0

0

−A0

t

c) PSK (special BPSK)

Figure 8.1: Exemplary representation of the modulation of a sinusoidal carrier signal by meansof rectangular pulse (hard shifting /switching)

– real baseband signal e.g.OOK – on off keying (as alteration of the ASK, see Fig. 8.1 a)),BPSK (as alteration of the PSK, see Fig. 8.1 c))

– complex baseband signal e.g. QPSK, 8-PSK, 16-QAM

• Different variants of signal shaping

– rectangular baseband signal (hard switching) e.g. OOK

– half-wave sinus baseband signal (soft switching) e.g. MSK – minimum shift keying

– time mismatch between Inphasen and quadrature component of the complex basebandsignal e.g. O-QPSK – Offset QPSK, MSK

• modulation procedures with or without memory

– memory-free modulation e.g. BPSK, QPSK, 16-QAM

– modulation with memory e.g. CPFSK – continuous phase frequency shift keying,GMSK – gaussian minimum shift keying, DQPSK – differential QPSK

In the framework of this lecture a global picture to digital modulation cannot be presented. Insteadprinciples and important characteristics of the digital modulation are explained here using severalexamples.

To Interested readers the lecture ”‘Mobile Nachrichtensysteme I”’ is recommended for furtherdetails. Additional literatures are e.g. [GG98], [Cou93], [Mau85], [Kam96], [Kro91], [Pro95].

8.3 BPSK – Binary Phase Shift Keying

The binary phase shift keying BPSK is the simplest PSK-version and forms the basis for un-derstanding of the higher order PSK-versions (QPSK,8-PSK). The binary source signal d{k} isimpressed as the phase ϕ(t) on the sinusoidal carrier signal

sc = Re{A0 ej(2πfct+ϕ0)

}(8.1)



where

ϕ0 → ϕ(t) = 2πd{k}M

for (k − 1/2)Ts ≤ t < (k + 1/2)Ts (8.2)

=

{0 for d{k} = 0

π for d{k} = 1(8.3)

whereby Ts is the symbol duration and M is the number of states of digital source signal d{k},hereM = 2 (binary). At time points (k+1/2)Ts hard switching takes place between the two phasestates (see Fig. 8.2)11.


The modulation product can be described as

sBP (t) = Re

{∑k

A0 ejϕ(t) ej 2πfct

}= Re

{s(t) ej 2πfct

}(8.4)

By comparing eqn. (5.57) and (8.4) one recognizes that

s(t) = A0 ejϕ(t) (8.5)

= A0

∑k

ejπd{k} rect

(t− kTsTs

)(8.6)

= A0

∑k

b{k} rect

(t− kTsTs

)with b ∈ {−1, −1} (8.7)

represents the equivalent low-pass signal (baseband signal) of the modulation product (see Sec-tion 5.3 and Fig. 8.2). All considerations can focus at the baseband. One obtains the band-passsignal sBP (t) again by multiplying the equivalent baseband signal with the carrier and taking realpart of the product (Up mixing, see Section 5.1)12. In the frequency domain this corresponds to ashift in the spectrum of baseband signal about ±fc.

11It will be emerged later that this hard switching of a phase state is very uncomfortable in others regardingthe necessary bandwidth.One will try to arrange therefore these state swapping softly. One possibility is the phasetransition from ϕ(d{k}) to ϕ(d{k + 1}) as nonvolatile but to arrange e.g. linear. With the knowledge from theSection 6.4 it becomes however clear that such a phase change brings also a frequency change with itself. The resultwould be a special FSK-method (CPFSK – continuous phase frequency shift keying). With the PSK-procedurestherefore another way is chosen in order to economize bandwidth. The amplitude of the baseband signal (complexenvelope) is softly lowered at the phase saltuses, i.e. the phase saltuses are faded out. From this point of view onecan regard the PSK procedures as special ASK-methods (QAM) . Thence ASK- and PSK-methods are linear

Modulation methods. In contrary to analog modulation methods, only the FSK methods are nonlinear with thedigital modulation methods.

12This corresponds to a two-sideband amplitude modulation with suppressed carrier – DSB-SC (see Section 6.3.6)



sBP (t)A0

−A0

−Ts Ts 2Ts 3Ts 4Ts 5Ts t

s(t) = sI(t), sQ(t) = 0A0

−A0

−Ts Ts 2Ts 3Ts 4Ts 5Ts t

{0} {1} {0} {0} {1} {0} {1}

Figure 8.2: BPSK-modulated carrier signal (above) and pertinent Baseband signal (equivalentto low-pass signal, below), (the carrier frequency fc,with no special relation to thesymbol rate, must be fcTs ∈ R)

A digital data source delivers a time-discrete sequence d{k}, which are firstly mapped to symbolsb{k} which in this case have different amplitudes 13.

d{k} = 0 → b{k} = +1 (8.8)

d{k} = 1 → b{k} = −1 (8.9)

The problem is that these time-discrete values (data) need to be transferred over an analog channel.Due to this, amplitude values b{k} are weighted with delayed impulses hT (t− kTs) [Kam96]. Inorder to be able to consider the impulse hT (t) in usual way as impulse response of a time-continuousLTI system with dimension of 1/time, eqn. (8.7) is transformed into

s(t) = b(t) ∗ hT (t) (8.10)

withb(t) =

∑k

b{k}Ts δ(t− kTs) (8.11)

as output signal of an ideal DA-converter (see Section 7.2) and

hT (t) =A0

Tsrect

(t

Ts

)(8.12)

13Importantly the allocation of the symbols of the source signal to phase or amplitude of the carrier signal iscertain. This is arbitrary,however,must agree in sender and receiver. With high order modulations e.g. QPSK,8-PSK special allocations (Gray code) can be selected,in order to keep the bit error rate small.



as pulse shaper /transmission filters (for the time being square pulse shaper). This model permitsquite good and uniform mathematical description of sender and receiver, even if in practical im-plementations the transmission pulse shaping takes place usually digitally and the DA-conversionis only performed thereafter.

8.3.2 Sender

A block diagram of a BPSK-modulator can be derived from what is mentioned so far (see Fig. 8.3).

sBP (t) = Re

{∑k

b{k}︸︷︷︸Symbol

Ts hT (t− kTs)︸︷︷︸Send impulse

ej 2πfct︸︷︷︸carrier

}(8.13)

d{k} Symbol

allocation

b{k} ×

DAC

b(t)

∑k Ts δ(t− kTs)

hT (t)

pulse shapersend filter

s(t) = sI(t) ×

Mixer

cos 2πfct

sBP (t)

Figure 8.3: Circuit diagram of a BPSK-Modulator

8.3.3 Phase Diagram

Representation of baseband signal in complex plane has been found to be very suitable (parametricrepresentation with the time t as parameter). The time dependent progression of the basebandsignal depicts the complex envelope of the bandpass signal (modulated carrier). In the phasediagram, possible states of the baseband signal at time instances kTs are represented (see Fig. 8.4).Therefore, (see also Chapter 5)

s(t) = sI(t) + j sQ(t) (8.14)

⇒ s(kTs) = b{k}Ts hT (0) ={A0 ej 0 = +A0 for d{k} = 0

A0 ejπ = −A0 for d{k} = 1(8.15)

For both phase states 0 and π, the exponential function becomes real-valued and thus the basebandsignal s(t) in BPSK is also purely real-valued.

sQ(t) = 0 (8.16)

⇒ s(t) = sI(t) (8.17)



sQ(t)

sI(t)×× d{k} = {0}d{k} = {1}

A0−A0

Figure 8.4: Phase diagram of the baseband signal for BPSK


In analog modulation the considerations were made in the frequency domain by using Fouriertransformation of sent signals or of the modulating signal (amplitude density spectrum). This ishowever possible only if the time dependent signal itself is known or at least a typical time de-pendent signal in which case typical spectrum can be assumed. In digital modulation the questionthat arises is which the typical symbol sequences are. Since one may hardly indicate a typicalsymbol sequence as this is also code dependent, there are two possible ways.

• For calculating Fourier integrals over the time t it is necessary to integrate in the interval(−∞,∞). In practice signals are considered however only in a finite period and thereforesuch considerations do not give good insight. Rather we are interested in one period of e.g.where occurs 100 transmission symbols. (see Exercise 6 ⇒ Short time spectrum). During thisperiod certain data sequences are assumed, in order to examine the behavior of the receiver.For example, longer transmission of only one symbol (e.g. constant {1}) becomes a constantcomponent of the baseband signal and leads to loss of clock synchronization. A periodicsymbol sequence (constant {0}, {1}) thereagainst leads to a disturbing line spectrum of thesend signal. From these knowledge the goal for coding can be found avoiding such unsuitablesymbol sequences.

• As particularly favorable symbol sequences with frequently changing symbols has been found,whereby in the symbol sequences no regular patterns may appear for channel coding. Onetherefore assumes usually the symbol sequence, which can be transmitted, satisfies an un-correlated stochastic random process. In this case, nothing more can be worked out with theamplitude density spectrum (Fourier transform of send signal or baseband signal), since thisassumes the knowledge of the exact time dependent progression. Instead the power densityspectrum (Fourier transform of the autocorrelation function (AKF) of sent signal or base-band signal) is used . The power density spectrum of a signal describes how the signal poweris distributed as statistical average in the frequency domain. Its knowledge is importantin order to determine the necessary transmission bandwidth or to estimate the interferencescaused in data signal outside its assigned transmission bandwidth.

Here the average spectral power density (power density spectrum) will be considered on exampleof BPSK. It should be assumed that the symbol sequence d{k} is a binary, uncorrelated, random



process. Furthermore both symbols d{k} = {0} and d{k} = {1} are assumed equally probablewith the probability of P = 0.5 (symbol sequence d{k} = {1} is zero mean and both transmissionimpulses possess the same energy).

With this assumption the power density spectrum Ss,s(f) of the baseband signal s(t) can becalculated as

Ss,s(f) = F {ss,s(τ)} = F{∫ ∞

−∞s(t) s∗(t+ τ) d t︸︷︷︸AKF of s(t)

}(8.18)

Ss,s(f) = |b|2 Ts |HT (f)|2 (8.19)

i.e. the power density spectrum is proportional to the square of the transfer function of the pulseshaper hT (t).

For the case of a square-wave pulse shaper considered here (see Fig. 8.5)

Ss,s(f) = |b|2 Ts |F {hT (t)} |2 (8.20)

= A20 Ts

∣∣∣∣F{

1

Tsrect

(t

Ts

)}∣∣∣∣2 (8.21)

= A20 Ts | si(πfTs)|2 (8.22)

The power density spectrum Ss,s(f) of the BPSK-modulated bandpass signal sBP (t) =s(t) cos 2πfct (up-mixed baseband signal) arises simply as a result of weighted shifting ofthe power density spectrum Ss,s(f) of baseband signal s(t) around ±fc.Since so far not dealing with power density spectra, a short proof would put down here; the shiftingtheory of fourier transform can also be applied for power density spectra.

Ss,s(f) = F {ss,s(τ)} = F{∫ ∞

−∞sBP (t) s

∗BP (t+ τ) d t

}(8.23)

= F{∫ ∞

−∞s(t) cos(2πfct) s

∗(t+ τ) cos(2πfc(t+ τ)) d t

}(8.24)

Since sc(t) = cos(2πfct) is a determinate signal, it yields

Ss,s(f) = F{ss,s(τ) · lim

T→∞1

2T

∫ T

−T

cos(2πfct) cos(2πfc(t+ τ) d t)

}(8.25)

= F{1

2ss,s(τ) cos 2πfcτ

}(8.26)

and from hereon (see Fig. 8.5)

Ss,s(f) =1

4

(Ss,s(f + fc) + Ss,s(f − fc)

)(8.27)

The power density spectrum Ss,s(f) is thereby the square of the sinc function (see eqn. (8.22)).



Ss,s(f)

A20Ts

− 1Ts

1Ts

f

Ss,s(f)A2

0Ts

2

A20Ts

4

−fc fc f

Figure 8.5: Power density spectrum Ss,s(f) of the BPSK-modulated baseband signal (above)and Ss,s(f) the up-mixed bandpass signal sBP (t) = s(t) cos 2πfct (down)– withsquare-wave pulse shaper hT (t) = 1/Ts rect(t/Ts)

In Section 4.5 the slow damping behavior of the sinc function si(·) was addressed among otherthings. This leads to the fact that the sampled (not band-limited) BPSK has a very high bandwidthrequirement (see Fig. 8.7). Therefore in many applications a band limitation of the transmittedsignals is assumed. This band limitation can also take place equivalently in the baseband, asnarrow-band pulse-shaper filters hT (t) are used. This is also called base band signalling.

In practice root raised cosine filters are often used as pulse-shaper hT (t) (see Fig. 8.6 and 8.7).With smaller bandwidth requirement, the adjacent channel disturbances are smaller and a givenfrequency range can be occupied with a higher number of channels than to use square pulses.

hT (t) =1

Ts

4rt/Ts cos((1 + r)πt/Ts) + sin((1− r)πt/Ts)

πt/Ts(1− (4rt/Ts)2)(8.28)

hT (t)1−r+4r/π

Ts

Ts−Ts t

Figure 8.6: Root-Raised-Cosine-Impulse as baseband pulse shaper hT (t)



0 0.5 1 1.5 2 2.5 3 3.5−70

−60

−50

−40

−30

−20

−10

0

Leistungsdichtespektrum

normalized frequency f*Ts

leve

l in

dB

Figure 8.7: Comparison of the power density spectra Ss,s(f) when using square-wave impulse(dotted line) and time limited Root-Raised-Cosine-Impulse (solid line) as pulseshaper hT (t). (logarithmic scale of the ordinate)

8.3.5 Receiver

The receiver has the task to demodulate the received signal and afterwards make decisions withthe help of a detector, which symbols have been sent (see Fig. 8.8). During the detection processthe goal is to make as fewer wrong decisions as possible. Depending on the expected transfer ratios(channel model) and type of modulation, there exist many receiver concepts which with certaincomplexity guarantee fewer error rates. A receiver that achieves the theoretically lowest errorrate is the so called optimum receiver. Such receivers can often exceed the limit of technicalrealizability for certain channel models which requires the use of compromise solutions.

rBP (t) Demodulator Detector d{k}

Figure 8.8: Simplified representation of a receiver for digital data communication

Matched-Filter Receiver In the context of this lecture only the match filter receiver willbe briefly considered. For the AWGN-channel (see Section 8.3.6) this receiver 14 represents opti-mum receiver. The down-converted received signal is applied in parallel to several matched filters,

14An equivalent realization represents the correlation receiver.



whereby each of these filters is matched on exactly one transmission symbol pulse (therefore thename matched filter). The filters correlate the received signal with the corresponding pulses. Thedecision is made in favor of the symbol, whose matched filter shows the highest accordance (highestoutput amplitude).

In the case of BPSK both the transmission pulses for the symbols differ d{k} ∈ {0, 1} only bytheir sign, so that only one matched filter with

hR(t) = h∗T (−t) (8.29)

is needed15. An analog-digital converter follows with the symbol rate fs = 1/Ts after filtering.In the case of the BPSK detection symbol manipulation (bit by bit) is realized by a memorylessthreshold switch, i.e. upon the sign of the sampling value it is decided whether a {0} or {1} wassent. As however, it comes to wrong decisions, the following channel decoding must try to seekout and correct these erroneous bits again and again. Fig. 8.9 shows the circuit diagram of sucha matched filter receiver (with accepted ideal carrier and symbol synchronization).

rBP (t) ×

cos 2πfct

hR(t)

matchedfilter

ADC

fs =1Ts

≶ 0

Detector

d{k}

Mixer Filter AD-converter Decision maker

Figure 8.9: Principle diagram of a BPSK matched filter receiver

8.3.6 Data Transmission over Noisy Channel

The sent signal must bridge the spatial distance to the receiver. Different transmitting media canbe used thereby, e.g..

• metal conductors

– Two wire line

– Coaxial conductor

– Hollow conductor

• Optical fiber (LWL)

– Mono mode fiber

– Multi-mode fiber

• free space

– Radio link

15Such a filter maximizes the signal signal-to-noise ratio SNR with given receiving conditions (Eb/N0 at thereceiver input, see Section 8.3.7) to the sampling time kTs at the decision maker/detector input. Under the conditionof an AWGN-channel this SNR maximization is equivalent to the minimization of bit error rate BER.



– Mobile network channels

– Satellite route

Depending upon medium and length of the transmission path the sent signal is more or less stronglydistorted, attenuated and overlayed with disturbing signals. In order to achieve the higher datarate, the modest error rate and multiple use of a medium, the type of modulation, coding, carrierfrequency etc. must be adapted on the transmission path. Moreover it is inevitable to describe thetransmission path effectively. In communications the term channel is used for the transmissionpath as an abstract model. The channel model provides the description of the most importantcharacteristics of the transmission path.

AWGN Channel Here only very simple channel model is presented, the AWGN-channel (addi-tive white gaussian noise). It is characterized by its simplicity and easy mathematical treatment.

The received signal r(t) is understood as superposition of the undisturbed sent signals s(t) witha White Gaussian noise n(t) (see Fig. 8.10).

r(t) = s(t) + n(t) (8.30)

It is not important however whether the signals r(t) and s(t) are bandpass or baseband signals. Incase of assuming complex baseband signals the real and imaginary part of the signals are uncor-related with those of the complex noise signal; and thus the in-phase and quadrature componentof the baseband signals can be considered separately.

If we are interested only in the performance in the presence of additive white noise we can focusat the baseband (see Section 5.3), i.e. the modulation and demodulation with the carrier signal(see Section 5.1) can be ignored. The data transmission system is indicated as a block diagram inFig. 8.10.

d{k} → b{k} → b(t) hT (t)s(t)

Send filterpulse shaper

Canal

+

n(t)

r(t)hR(t)

rE(t)

Receiver filtermatched filter

ADCrE(kTs)

Detector d{k}

Figure 8.10: Blocks diagram of a data transmission system in Baseband

In case of the BPSK, the baseband signal s(t) = si(t) = s(t) is real valued. Assuming idealsynchronization we are interested in real part of the received signal r(t). From that the wholesystem can be analyzed only in real domain (see eqn. (8.30)).

The superimposed noise signal n(t) is assumed to be as a zero mean, stationary, white Gaussianstochastic process. All possible sources of noise ni(t) arising in the system are combined into onenoise source n(t) with noise power density of

Sn,n(f) =N0

2for −∞ < f <∞ (8.31)

The amplitude density distribution follows a normal distribution with μ = 0 (see Section 4.6)

fn(n) =1√2πσ

e−(n−μ)2

2σ2 (8.32)



8.3.7 Calculation of Bit Error Ratio

By symbol or bit error rate one understands the probability with which the detector decides awrong symbol or bit. In the case of BPSK, each transmitted symbol consists of only 1 bit, sothat symbol and bit error probability are identical. These error probabilities are random variableswhich specify the ratio of the number of wrong decisions to the number of all met decisions onaverage. The computation of the bit error rate BER will be presented here as an example ofBPSK assuming an AWGN-channel.

Also assuming ideal synchronization, the sampled values

rE(kTs) = sE(kTs) + nE(kTs) (8.33)

gained at the times t = kTs (see Fig. 8.10) are present at the entrance of the detector. Thesampling values sE(kTs) result from the receiver filtering and sampling of the baseband signal s(t)to the time-discrete convolution sum

sE(kTs) = s(t) ∗ hR(t)∣∣∣t=kTs

(8.34)

= Ts

∞∑n=−∞

b{n}h((k − n)Ts) (8.35)

the sender and receiver filter in this case are combined into a single filter 16 h(t) = hT (t) ∗hR(t). Inter-symbol interference freeness ISI is assumed here which is not to be dealt further(see Exercise 23). The Rectangular and Root-Raised-Cosine pulse shaper impulse response h(t)introduced in the script fulfills this ISI-freeness with the matched filter receiver. Sending of thesymbol d{k} = {0} is marked by d0 and the associated received value by s0, accordingly d1 ands1 for the sent symbol d{k} = {1}. It then holds

sE(kTs) = Ts h(0) b{k} (8.36)

=

{s0 for the event d0

s1 = −s0 for the event d1(8.37)

The noise signal n(t) is band-limited by the received filter hR(t) and thereto possesses a finitesignal power, which corresponds to the variance of the noise process.

σ2 =N0

2

∫ ∞

−∞|HR|2 d f (8.38)

The amplitude probability density function remains unaffected after filtering.

If the noise signal is n(kTs) = 0 (σ2 → 0), the received signal rE(kTs) adopts at the detectorentrance only the values s0 with a sent {0} or s1 with a sent {1}. If the noise signal power/varianceis larger than 0, the received value rE can deviate more or less from these two values. If one appliesa sufficiently large number of received values rE for the two events d0 and d1 (send a {0} or {1})

16For a matched filter receiver with hR(t) = h∗

T (−t) the total impulse response h(t) is identical with the (impulse)autocorrelation function of the sending filter hT (t).



in a histogram, thus results approximately the amplitude density distributions at the detectorentrance. (see Fig. 8.11)

frE |d0(rE|d0) =1√2πσ

e−(rE−s0)

2

2σ2 (8.39)

frE |d1(rE|d1) =1√2πσ

e−(rE−s1)

2

2σ2 (8.40)

s1 s0 rE

frE |d1(rE|d1)

frE |d0(rE|d0)small noise power

middle noise power

large noise power

Figure 8.11: Amplitude density distribution frE (rE) of the sampled values rE(kTs) at the en-trance of the detector for BPSK

The detector finds a decision rE(kTs), which can take the values d0 or d1 with knowledge of thereceived value. Four different decisions are possible (see Fig. 8.12).

P (d0) d0

sent symbol d{k}

P (d1) d1

d0

d{k}

d1

decided symbol

at the receiver

P (d{k} = d0 | d{k} = d0)

P (d{k} = d1 | d{k} = d1)

P (d{k} = d0 | d{k} = d1)

P (d{k} = d1 | d{k} = d0)

Figure 8.12: Model of distorted channels

Two cases of decision are wrong thereby:

• the detector decides a d{k} = d0, although d{k} = d1 was sent

• the detector decides a d{k} = d1, although d{k} = d0 was sent

The probability of a wrong decision is

BER = P (d{k} = d0 | d{k} = d1)P (d1) + P (d{k} = d1 | d{k} = d0)P (d0) (8.41)



where P (d0) and P (d1) are the a-priori-probabilities for the occurrence of the events d0 and d1,i.e. the probabilities for sending a {0} or {1}. The channel coding guarantees

P (d0) = P (d1) =1

2(8.42)

The conditioned probabilities for the wrong decisions result from the region under the appropriateamplitude density distribution within the range, where the detector decides wrongly. In the caseof BPSK, the detector makes its decision on basis of the sign of rE(kTs), i.e.,

P (d{k} = d0 | d{k} = d1) =

∫ ∞

0

frE |d1(rE|d1) d rE (8.43)

P (d{k} = d1 | d{k} = d0) =

∫ 0

−∞frE |d0(rE|d0) d rE (8.44)

With eqn. (8.39, 8.40), the relation s0 = −s1 > 0 and imposing the symmetry of the amplitudedensity distributions the error probability BER results to

BER =1

2

∫ 0

−∞

1√2πσ

e−(rE−s0)

2

2σ2 d rE +1

2

∫ ∞

0

1√2πσ

e−(rE−s1)

2

2σ2 d rE (8.45)

=

∫ 0

−∞

1√2πσ

e−(rE−s0)

2

2σ2 d rE =

∫ ∞

0

1√2πσ

e−(rE+s0)

2

2σ2 d rE (8.46)

These integrals can be alternatively reduced with help of the substitution x = rE−s0σ

or x = rE+s0√2σ

to the distribution function Φ(·) of standardized normal distribution or to the complementaryerror function erfc(·).

BER =1√2π

∫ − s0σ

−∞e−

x2

2 d x =1

2

2√π

∫ ∞

s0√2σ

e−x2

d x (8.47)

= Φ(−s0σ

)=

1

2erfc

(s0√2σ

)(8.48)

The bit error ratio thus exclusively depends on the ratio s0/σ at the detector entrance. This ratiocan also be indicated as signal-to-noise ratio. With s20 = s21 as power of the sampled data signaland σ2 as noise power, yields the SNR at detector entrance to

SNR =s20σ2

(8.49)

The bit error ratio can also be specified as

BER =1

2erfc

(√SNR

2

)(8.50)

Since the signal-to-noise ratio at the detector input depends on the receipt filter hR(t), the biterror ratio is rather interesting as a function of the signal-to-noise ratio at the receiver input.As suitable signal-to-noise measure proves the ratio Eb/N0, whereas Eb is the average energy of asingle character and N0/2 is the noise performance density at receiver input (see eqn. 8.31). With



matched filter receiver for a given Eb/N0 at the receiver input the signal-to-noise ratio SNR ismaximized at the detector entrance and it applies

SNR =Eb

N0/2=

|b|2T 2s

∫ ∞−∞ h2T (t) d t

N0/2(8.51)

So finally, for BPSK-modulation

BER =1

2erfc

(√Eb

N0

)(8.52)

Fig. 8.13 shows the dependence of bit error ratio with the ratio Eb/N0 for BPSK under thecondition of an AWGN-channel and a matched filter receiver

−15 −10 −5 0 5 10

BER

100

10−1

10−2

10−3

10−4

10 lg(Eb/N0) dB

Figure 8.13: Dependence of bit error ratio with Eb/N0 for BPSK


8.4 QPSK – Quaternary Phase Shift Keying 85

8.4 QPSK – Quaternary Phase Shift Keying

Same like BPSK unless the baseband signal is complex here.

8.5 Further ASK/PSK Methods

8-PSK, 16-QAM

perhaps also multi-valued QAM with real baseband signal

dealt briefly here only with phase diagram

0 2 4 6 8 10 12 14 16 18

BER

100

10−1

10−2

10−3

10−4

10−5

Eb/N0 in dB

BPSK,QPSK

8-QAM

8-PSK

16-QAM

16-PSK

Figure 8.14: Bit error ratio for different type of PSK or QAM modulations with respect tosignal-to-noise ratio Eb/N0



BPSK QPSK 8-PSKsQ

sI××

dmin

sQ

sI×

×

×

×dmin

sQ

sI×

×

×

××

×

×

×

dmin

8-QAM 16-PSK 16-QAMsQ

sI

×

×

×

××

×

×

×

dmin

sQ

sI×

×

×

××

×

×

×

×

×

×

×

×

×

×

×

dmin

sQ

sI

× × × ×× × × ×× × × ×× × × ×dmin

Figure 8.15: Phase diagram (due to closely related symbols, always arises in the case of samenoise power higher symbol or bit error rates )


8.6 Frequency Shift Keying 87

8.6 Frequency Shift Keying

auch nur kurz das Prinzip zeigen, vielleicht noch eine Sender- und Empfangerrealisierung vorstellen


88 REFERENCES

References

[BHPT95] O. Beyer, H. Hackel, V. Pieper, and J. Tiedge. Wahrscheinlichkeitsrechnung und mathe-matische Statistik. Mathematik fur Ingenieure und Naturwissenschaftler. B.G. TeubnerVerlagsgesellschaft, Leipzig, 7. edition, 1995. ISBN: 3-8154-2075-X.

[BS79] I. N. Bronstein and K. A. Semendjajew. Taschenbuch der Mathematik, volume 1.B.G. Teubner Verlagsgesellschaft, Leipzig, 24. edition, 1979. Lizenzausgabe fur denVerlag Harri Deutsch, Thun 1989.

[BS96] I. N. Bronstein and K. A. Semendjajew. Teubner-Taschenbuch der Mathematik, vol-ume 1. B.G. Teubner Verlagsgesellschaft, Leipzig, 1996. ISBN: 3-8154-2001-6.

[Cou93] Leon W. Couch II. Digital and Analog Communication Systems. Prentice-Hall Inter-national, Inc., 4th edition, 1993. ISBN: 0-13-571845-7.

[Fet96] Alfred Fettweis. Elemente nachrichtentechnischer Systeme. B.G. Teubner, Stuttgart,2. edition, 1996. ISBN: 3-519-16131-1.

[Fli91] Norbert Fliege. Systemtheorie. Informationstechnik. B.G. Teubner, Stuttgart, 1991.ISBN: 3-519-06140-6.

[GG98] Ian Glover and Peter Grant. Digital Communications. Prentice Hall, Europe, 1998.ISBN: 0-13-565391-6.

[GZZZ95] G. Grosche, V. Ziegler, D. Ziegler, and E. Zeidler, editors. Teubner-Taschenbuch derMathematik, volume 2. B.G. Teubner Verlagsgesellschaft, Leipzig, 7. edition, 1995.ISBN: 3-8154-2100-4.

[Hof98] Rudiger Hoffmann. Signalanalyse und -erkennung. Springer-Verlag, Berlin, Heidelberg,1998.

[Kam96] Karl Dirk Kammeyer. Nachrichtenubertragung. Informationstechnik. B.G. Teubner,Stuttgart, 2. edition, 1996. ISBN: 3-519-16142-7.

[Kro91] Kristian Kroschel. Datenubertragung. Springer-Verlag, Berlin, Heidelberg, 1991.ISBN: 3-540-53746-5.

[Luk95] Hans Dieter Luke. Signalubertragung – Grundlagen der digitalen und analogenNachrichtenubertragungssyteme. Springer Verlag Berlin, 6. edition, 1995. ISBN: 3-540-58753-5.

[Mau85] Rudolf Mausl. Digitale Modulationsverfahren. Dr. Alfred Huthig Verlag, Heidelberg,1985.

[Mau92] Rudolf Mausl. Analoge Modulationsverfahren. Dr. Alfred Huthig Verlag, Heidelberg,2. edition, 1992. ISBN: 3-7785-2130-6.

[Pro95] John G. Proakis. Digital Communications. McGraw-Hill International Editions, 3rdedition, 1995. ISBN: 0-07-113814-5.


REFERENCES 89

[SS88] Norbert Sieber and Hans-Jurgen Sebastian. Spezielle Funktionen. Number 12 in Math-ematik fur Ingenieure, Naturwissenschaftler, Okonomen und Landwirte. B.G. TeubnerVerlagsgesellschaft, Leipzig, 3. edition, 1988.

[WS93] Gerhard Wunsch and Helmut Schreiber. Analoge Systeme. Springer-Verlag, Berlin,Heidelberg, 1993.


90 A FORMULAS

A Formulas

A.1 Definitions

Sinc function one of the most frequently used function in communications and is defined as

si(x) =sin x

xsinc(x) =

sin πx

πx(A.1)

⇒ si(πx) = sinc(x) (A.2)

where si(·) is user in German language and sinc(·) is the American counterpart.

The rect function is also frequently used in the script and defined as follows

rect(x) =

⎧⎪⎪⎨⎪⎪⎩1 for − 1

2< x < 1

2

12

for |x| = 12

0 otherwise

(A.3)

The triangular function is used rather rarely and defined as follows here

triang(x) =

{1− |x| for − 1 ≤ x ≤ 1

0 otherwise(A.4)

A.2 Fourier Transformation

S(f) = F {s(t)} =

∫ ∞

−∞s(t) e− j 2πft d t Fourier transformation (A.5)

S(ω) =

∫ ∞

−∞s(t) e− jωt d t

s(t) = F−1 {S(f)} =

∫ ∞

−∞S(f) ej 2πft d f Inverse Fourier transformation (A.6)

=1

2π

∫ ∞

−∞S(ω) ejωt dω

The sufficient (however not necessarily) criteria for the convergence of Fourier integral is fromDirichlet-Jordan (absolute Integrable

∫ ∞−∞ |s(t)| d t < ∞, s(t) is decomposable in many finite

constant, monotonous subintervals (Due to the duality of FT this is also valid for S(f))) [BS79,BS96].

Some properties of Fourier Transformation:


A.2 Fourier Transformation 91

• For real signal s(t):

|S(f)| = |S(−f)| Amplitude spectrum is an even function (A.7)

− arg(S(f)) = arg(S(−f)) Phase response is an odd function (A.8)

Re {S(f)} = Re {S(−f)} Real part of the spectrum is an even function (A.9)

Im {S(f)} = −Im {S(−f)} Imaginary part of the spectrum is an odd function(A.10)

• Decomposition of real time signal in even and odd part:

s(t) = sg(t) + su(t) (A.11)

mit

s(t) S(f) (A.12)

sg(t) Re{S(f)} (A.13)

su(t) j Im{S(f)} (A.14)

• Parseval’s Equation: ∫ ∞

−∞|s(t)|2 d t =

∫ ∞

−∞|S(f)|2 d f (A.15)

• If the function s(t) is absolutely integrable in the interval (−∞,∞) , i.e.∫ ∞−∞ |s(t)| d t <∞,

then the function S(f) = F {s(t)} in −∞ < f <∞ is static and tends to zero for f = ±∞[BS79]. Due to symmetry of Fourier transformation (see also Tab. A.1) the opposite is alsovalid.

• Band limited signals have an infinite time expansion and time limited signals are bandunlimited (see also lecture Signalverarbeitung [Hof98] page 109f).

Important properties of dirac delta function:

• Sifting property ∫ ∞

−∞s(t) δ(t− t0) d t = s(t0) (A.16)∫ ∞

−∞S(f) δ(f − f0) d f = S(f0) (A.17)

• Adjournment property

s(t) ∗ δ(t− t0) = s(t− t0) (A.18)

S(f) ∗ δ(f − f0) = S(f − f0) (A.19)


92 A FORMULAS

Linearity∑i

aisi(t)∑i

aiSi(f)

Complex conjugate of time function s∗(t) S∗(−f)Complex conj. of frequency function s∗(−t) S∗(f)

Mirroring s(−t) S(−f)Symmetry S(t) s(−f)Time shifting s(t− t0) S(f) e− j 2πft0

Frequency shifting s(t) ej 2πf0t S(f − f0)

s(t) cos(2πf0t)12(S(f + f0) + S(f − f0))

s(t) sin(2πf0t)j2(S(f + f0)− S(f − f0))

Scaling s(at) 1|a|S(

fa)

Convolution in time domain s1(t) ∗ s2(t) S1(f) · S2(f)

Convolution in frequency domain s1(t) · s2(t) S1(f) ∗ S2(f)

Differentiation dn

d tns(t) (j 2πf)n S(f)

Integration∫ t

−∞ s(τ) d τ 1j 2πf

S(f) + 12S(0) δ(f)

Limiting value limn→∞

sn(t) = s(t) limn→∞

Sn(f) = S(f)

Table A.1: Properties of Fourier Transformation

δ(t) 1

δ(t− t0) e− j 2πft0

1 δ(f)

ej 2πf0t δ(f − f0)

cos(2πf0t)12(δ(f + f0) + δ(f − f0))

sin(2πf0t)j2(δ(f + f0)− δ(f − f0))

T XT (t) = T∞∑

n=−∞δ(t− nT ) X 1

T(f) =

∞∑n=−∞

δ(f − n

T

)Table A.2: Fourier transformation of generalized functions (with non-dimensional variables t

and f)


A.3 Notes on Convolution 93

rect(t) sinc(f) = si(πf)

sinc(t) = si(πt) rect(f)

triang(t) sinc2(f) = si2(πf)

jπt

sgn(f)

sgn(t) 1jπt

Table A.3: Some correspondence of Fourier transformation Fourier transformation of general-ized functions (with non-dimensional variables t and f)

A.3 Notes on Convolution

The connection between the two general complex functions f, g

(f ∗ g)(t) :=∫ ∞

−∞f(τ)g(t− τ) d τ (A.20)

is known as convolution. In (engineering) literature, as in this script, the notation is usually used asf(t)∗g(t) instead of (f ∗g)(t). This way of writing, however established, is not completely correct.The problem outcrops when notations occur like f(t) ∗ g(−t), f(at) ∗ g(t) or f(t) ∗ g(t − t0).Therefore it is always implicitly presupposed that the argument for the convolution is always t (intime domain) or f (in frequency domain).

f ∗ g = g ∗ f Commutativity

f ∗ (g ∗ h) = (f ∗ g) ∗ h Associativity

f ∗ (g + h) = f ∗ g + f ∗ h Distributivity

From f ∗ g = 0 follows f = 0 or g = 0.

Table A.4: Properties of convolution


94 B APPENDIX

B Appendix

under construction!

B.1 Analytical Signal

Since the transmitted signal sBP (t) has to be real, its Amplitude spectrum is an even functionwhile the phase spectrum is an odd function of frequencies (see Section. 3.3). From that (seeeqn. 3.47)

SBP (f) = S∗BP (−f) (B.1)

Therefore to describe the signal one needs knowledge of the spectrum for frequancies f ≥ 0. ”‘Theother half”’ can be omitted.

The dependence between the spectrum S(f) of the equivalent low pass signal s(t) (bandlimitedwith fg < fc) and spectrum from SBP (f) of the upconverted signal sBP (t) gives

F {sBP (t)} = F {Re

{s(t) ej 2πfct

}}(B.2)

Mit Re {z} = 12(z + z∗) (see Excercise 1) gives

F {sBP (t)} = F{1

2s(t) ej 2πfct +

1

2s∗(t) e− j 2πfct

}(B.3)

=1

2

(S(f − fc) + S∗(−(f + fc))

)(B.4)

As one can see, the spectrum SBP (f) is obtained through ”‘left-”’ and ”‘right-”’shift of the spec-trum S(f) or in other words S∗(−f) of the equivalent low-pass signal s(t).

If one compares this with (see also Fig. B.1)

F {s+(t)

}= F {

s(t) ej 2πfct}

(B.5)

= S(f − fc) (B.6)

it can be seen that exactly the ”‘left”’ shifted part of the spectrum is missing. Signal s+(t) iscalled analytical signal. Also holds

S+(f) = F {s+(t)

}= 0 fur f < 0 (B.7)

Signals sBP (t) and s+(t) are clearly dependent as follows.

sBP (t) = Re{s+(t)

}(B.8)

s+(t) = sBP (t) + j H{sBP (t)} (B.9)

Where H{sBP (t)} corresponds to the Hilbert transform of sBP (t).

To sum up the following holds

s(t) – complex baseband signal (B.10)

s+(t) = s(t) ej 2πfct – analytical signal (B.11)

sBP (t) = Re{s+(t)

}= Re

{s(t) ej 2πfct

}– real bandpass signal (B.12)

Use of the complex (low pass equivalent) representation of the signals makes the system analysismuch easier and because of this reason one often uses this way to analyse systems . Real bandpasssignals sBP (t) can that be easily obtained.


B.1 Analytical Signal 95

|S(f)| = |F {s(t)} |A

f−fg fg|S+(f)| = |S(f − fc)|

A

fc f|SBP (f)| =

∣∣12

(S(f − fc) + S∗(−(f + fc))

)∣∣A

−fc fc f

Figure B.1: Spectra of complex baseband signal, analytical signal and real bandpass signalfg < fc


96 C EXCERCISES

C Excercises

C.1 Problems

The exercises are created in order to help the students check their knowledge and for preparingexamination. In your own interest you should use the solutions which are also provided as achecklist for the solutions you have obtained and not to ”examine ” the solutions before evenreading the problems through. Most of the problems are solved as a part of the exercises on boardin parallel to the lectures.

Problems denoted with ∗ go beyond this lecture or are more difficult to solve.

Repetition

1. Each complex number z ∈ C has the following equivalent notations

z = x+ j y with x, y ∈ R (C.1)

x = Re{z} (C.2)

y = Im{z} (C.3)

z = r cosϕ+ j r sinϕ with r ∈ R ≥ 0, ϕ ∈ [0, 2π) (C.4)

r = |z| =√x2 + y2 (C.5)

ϕ = arg z =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

arctan yx

for x > 0 ∧ y ≥ 0

π2

for x = 0 ∧ y > 0

arctan yx+ π for x < 0

3π2

for x = 0 ∧ y < 0

arctan yx+ 2π for x > 0 ∧ y < 0

(C.6)

z = r ejϕ (C.7)

The following expression shows the relation between trigonometric and exponential functions

ejϕ = cosϕ+ j sinϕ (C.8)

Eqn. (C.8) is referred to as Euler’s Formula.

(a) Show with respect to eqn. (C.8) that the following addition theorems hold

cos(x± y) = cos x cos y ∓ sin x sin y (C.9)

sin x− sin y = 2 sinx− y

2cos

x+ y

2(C.10)

(b) Show that the following relations hold

z + z∗ = 2Re{z} (C.11)

z − z∗ = j 2Im{z} (C.12)

z · z∗ = |z|2 (C.13)

where z∗ stands for the conjugate complex of z.


C.1 Problems 97

2. A periodic function of time s(t), with a period T > 0 can be expanded into Furrier seriesFR{s(t)} as

s(t) = FR{s(t)} =a02

+∞∑k=1

ak cos(2πk

Tt)+ bk sin

(2πk

Tt)

(C.14)

with the following Fourier coefficients

ak =2

T

∫ T2

−T2

s(t) cos(2πk

Tt)d t k = 0, 1, 2, . . . (C.15)

bk =2

T

∫ T2

−T2

s(t) sin(2πk

Tt)d t k = 1, 2, . . . (C.16)

under the assumption that Fourier series (C.14) converges. The theory of Dirichlet [BS96]purveys a convergence criterion sufficient for many practical purposes. According to it, fora monotone function s(t) with period T which which is piecewise continuous applies thefollowing:

limn→∞

{a02

+n∑

k=1

ak cos(2πk

Tt) + bk sin(2π

k

Tt)

}=

{s(t) where s(t) is continuouss(t−0)+s(t+0)

2otherwise

(C.17)whereas s(ti − 0), s(ti + 0) represent correspondingly the left and right limit of s(t) at apoint of discontinuity ti and both limits must exist. According to equation (C.14) the periodictime function s(t) can be expanded into spectral allotments with the help of a Fourier seriesexpansion into an infinite sum of weighted sine and cosine components of different frequenciesfk =

kT. This type of process is called ”‘Harmonic analysis”’ or ”‘Spectral analysis”’. This has

great importance in the communications. However one often uses the (equivalent) complexform of the Fourier series:

s(t) = FR {s(t)} =∞∑

n=−∞cn e

j 2π nTt (C.18)

with the corresponding Fourier coefficients

cn = |cn| ejϕn =1

T

∫ T2

−T2

s(t) e− j 2π nTt d t (C.19)

Therefore the following relation applies between the coefficients cn and ak, bk:

c0 =a02

k = 0 (C.20)

ck =ak − j bk

2k = 1, 2, . . . (C.21)

c−k =ak + j bk

2k = 1, 2, . . . (C.22)


98 C EXCERCISES

One takes |cn| and ϕn = arg{cn} over the discrete frequency fn = nTto get the correspond-

ing amplitude and phase spectrum of the periodic time function s(t). This yields the socalled Line spectra, whereby the distance of two neighboring spectral lines comprises thefundamental frequency f1 =

1Tof s(t).

Further we consider the possibilities to obtain spectral representation for non-periodic func-tions. The Fourier series expansion, which can only be used for periodic functions, is notsuitable here as the signal is non periodic. However it provides a useful approach, if oneassumes that the non-periodic (absolutely integrable function in the interval (−∞,+∞))17

is periodically extended to a function s(t) which is periodic with an infinitesimal periodduration T [Hof98]. To carry out formally this limitation, putting eqn (C.19) in (C.18) gives:

s(t) =∞∑

n=−∞

(1

T

∫ T2

−T2

s(t) e− j 2π nTt d t

)ej 2π

nTt (C.23)

With the insertion of Δf = 1Tas distance of two neighboring spectral lines and the frequency

variables f = nT= nΔf results

s(t) =∞∑

n=−∞

(∫ T2

−T2

s(t) e− j 2πft d t

)ej 2πft Δf (C.24)

For T→∞, Δf would be infinitesimal small, i.e. neighboring spectral lines move infinitelyclose towards each other. By summing

∑∞n=−∞ . . . Δf in eqn (C.24) with the integral∫ ∞

−∞ . . . d f yields

s(t) =

∫ ∞

−∞

(∫ ∞

−∞s(t) e− j 2πft d t

)︸︷︷︸

S(f)

ej 2πft d f (C.25)

where the function S(f) is now a continuous succession of frequency f as spectrum or Fouriertransformation of the time function s(t). Non-periodic time functions therefore possess acontinuous spectrum.

The transition of time to frequency domain is named as Fourier transformation and thereverse process is inverse Fourier transformation. We may write as follows

S(f) = F {s(t)} =

∫ ∞

−∞s(t) e− j 2πft d t (C.26)

and

s(t) = F−1 {S(f)} =

∫ ∞

−∞S(f) ej 2πft d f (C.27)

Also in the case of non-periodic time function s(t) the amplitude and phase spectra can begiven as |S(f)| and ϕ(f) = arg{S(f)}.(a) Show considering the relation (C.20)-(C.22), that the Fourier series in (C.14) can be

transformed toa form as given by eqn (C.18). Use the Euler’s forms (C.8)

17This property of s(t) is assumed additionally for all the following considerations.


C.1 Problems 99

cos(ϕ) =ejϕ +e− jϕ

2(C.28)

sin(ϕ) =ejϕ − e− jϕ

2 j(C.29)

(b) Calculate the Fourier coefficients cn of the complex Fourier series for the real periodicrectangular signal s1(t) as presented in Fig. C.1. Sketch the corresponding amplitudespectrum for τ = T

4in frequency domain from −12

Tto 12

T.

s1(t)

s0

−T − τ2

0 τ2

T t

Figure C.1: periodic rectangular signal

(c) Calculate the spectrum S2(f) of the (real) rect impulse s2(t) in Fig. C.2. Sketch thecorresponding amplitude spectrum for τ = T

4in same frequency domain as in task (2b).

s2(t)

s0

− τ2

0 τ2

t

Figure C.2: Rect impulse

Basics

3. Figure C.3 shows an exemplary transmission system with given power-, gain-, andattenuation-levels. The path loss is given as LS = 20 log(100d) dB where d stands for thedistance between transmitter and receiver antenna (in meter).

Modulator � � + � Demodulator

Tx amplifier multistage Rx amplifier

PT = 12 dBm

LS

g1 = 23 dB g2 = 60 dBPN = −50 dBm

SNR ≥ 7 dB

Figure C.3: transmission system


100 C EXCERCISES

(a) What is the maximum distance dmax that ensures a minimum signal to noise ratio ofSNR = 7 dB at the input of the demodulator?

(b) Which power level PT (in dBm) is necessary in order to reach a signal to noise ratio ofSNR = 20 dB at a distance of d = 50 m?

Signal Theory and LTI Systems

4. Let {δn(t)} be a series of functions where n ∈ N, n > 0. An example graph is shown infig. C.4.

δn(x) =

⎧⎪⎪⎨⎪⎪⎩0 for 1

n< x < − 1

n

n4

for |x| = 1n

n2

for − 1n< x < 1

n

(C.30)

δn(x)n2

− 1n

1n

x

Figure C.4: The Dirac impulse δ(x) as the limit of certain known signals

(a) Show that the Dirac impulse δ(x) can be interpreted as the limit of the series {δn(x)}for n→ ∞.

δ(x) = limn→∞

δn(x) (C.31)

(b) Show that the following relation holds (which is called the sifting property of the Diracimpulse) ∫ ∞

−∞s(x)δ(x− x0) d x = s(x0) (C.32)

Use the result of subtask (4b). The signal s(x) is assumed to be continuous.

5. Fig. C.5 represents the spectrum S(f) (arg(S(f)) = 0). Calculate the corresponding timesignal s(t).

S(f)

a2

A2

−fc − fg −fc −fc + fg fc − fg fc fc + fg f

Figure C.5: Spectrum S(f)


C.1 Problems 101

6. The following short-time spectrum is given

ST (f) =

∫ T

0

s(t) e− j 2πft d t (C.33)

(a) Deduce a connection between the short-time spectrum ST (f) and the spectrum S(f)as a function of time duration T .

(b) Calculate the short-time spectrum ST (f) for the signal s(t) = cos(2πf0t) and sketch|ST (f)| with T = 1/f0, T = 10/f0 and T = 100/f0. Compare the spectrums.

7. Prove the Parseval’s Theorem∫ +∞

−∞|x(t)|2 d t =

∫ +∞

−∞|X(f)|2 d f (C.34)

8.∗ (a) Calculate the spectrum of the Gauss impulse

g(t) =1√2πσ2

e−t2

2σ2 (C.35)

(b) Calculate the area A =∞∫

−∞g(t) d t from g(t).

9.∗ The Hilbert transformation transforms a time function again on a time function. A veryimportant class of signals, so called analytic signals is defined as

xA(t) = x(t) + jxH(t) (C.36)

The Hilbert transformation is defined as

xH(t) = H{x(t)} = x(t) ∗ 1

πt(C.37)

The convolution in time domain corresponds multiplication in frequency domain (seeFig. C.6).

xH(t) = F−1

{F {x(t)} F

{1

πt

}}(C.38)

= F−1 {X(f) (− j) sgn(f)} (C.39)

(a) Given is a Signal x(t) = A0 cos(2πf0t). Calculate the Hilbert transformed xH(t) =H{x(t)}.

(b) Sketch the spectrum XA(f) = F {xA(t)} = F {x(t) + j xH(t)}. How do the spectrumsX(f) and XA(f) differ?

(c) Calculate the amplitude |xA(t)| = |x(t)+ jxH(t)|. Which feature can you conclude intofrom here?


102 C EXCERCISES

Figure C.6: Relation between Hilbert and Fourier transformation

10.∗ Any real-valued signal x(t) can be divided into a sum of an even (symmetrical) and an odd(skew-symmetrical) signal

x(t) = xg(t) + xu(t) (C.40)

where

xg(t) =1

2(x(t) + x(−t)) (C.41)

xu(t) =1

2(x(t)− x(−t)) (C.42)

(a) Show that, the following relation applies for real-valued signal x(t) = xg(t) + xu(t)(compare eqn. (A.11) – (A.14)).

X(f) = Xg(f) +Xu(f) (C.43)

mit

Xg(f) = Re{X(f)} (C.44)

Xu(f) = j Im{X(f)} (C.45)

Besides, if the signal x(t) is causal, i.e.

x(t) = 0 for t ≤ 0 (C.46)

so applies

xg(t) =

{12x(t) for t ≥ 0

12x(−t) for t < 0

(C.47)

⇒ xu(t) = xg(t) sgn(t) (C.48)

The real and imaginary part of the Fourier transformed causal signal are not independentof each other, however they are related to each other by the Hilbert transformation.

(b) Show that for real-valued causal signal x(t) = xg(t)+ xu(t) the following relation holds

Xg(f) = (− j)H{Xu(f)} (C.49)


C.1 Problems 103

11. A system is called strictly causal, if for each time instance t0, the output value y(t0) dependsonly on previous input values x(t), t < t0, and not on current or future input values. Showthat for LTI-systems this definition can be expressed equivalently, in the following way:

An LTI-system with its impulse response h(t) is called strictly causal if for all t ≤ 0, we haveh(t) = 0; otherwise, the system is noncausal.

h(t) = 0 for t ≤ 0 (C.50)

12. (a) Calculate the convolution product

triang(x) = rect(x) ∗ rect(x) (C.51)

and

y(x) = rect(x) ∗ rect(x) ∗ rect(x) = triang(x) ∗ rect(x) (C.52)

and sketch the behavior of the corresponding signal.

(b) Calculate and sketch the spectrum of the triangular function triang(t/T ).

13. Compute on the basis of Fig. C.10 on page 105 the impulse response of this low-pass filterfrom the depicted spectra H(f) = |H(f)|.Hint: You can represent H(f) as convolution product of two rectangular functions.

14. (a) An input signal x(t) with x(t) = 0 for 0 > t > T1 and the causal impulse response h(t)with h(t) = 0 for 0 > t > T2 are given, i.e. both signals are time limited (Signals withfinite duration). Show that the output signal y(t) = x(t) ∗ h(t) is also time limited, i.e.y(t) = 0 for 0 > t > T3, where T1, T2, T3 are positive real numbers. Calculate T3 inreliance of T1 and T2.

(b) Given is the non-causal LTI-System with the impulse response h(t) (see also Fig. C.7)

h(t) = rect

(t

T+ 1.5

)+ δ

(t

T− 6

)(C.53)

In which interval the non-causal input signal x(t) must be known in order to computethe initial value y(0)?

h(t)

tT

1

-2 -1 0 6Figure C.7: Impulse response h(t)

15. (a) Figure C.8 shows the impulse response h(t) of a certain LTI-system. What is thecorresponding mathematical operation performed by this system? What is the name ofsuch a system?


104 C EXCERCISES

h(t)ATs

t0 TS

Figure C.8: Impulse response h(t)

(b) This system is excited by a rectangular pulse x1(t) with the duration Ti.

h(t) =A

Tsrect

(t

TS− 1

2

)x(t) = rect

(t

Ti− 1

2

)(C.54)

Plot the output signal y1(t) = (x1 ∗ h)(t) for the following cases:

Ti < TS (C.55)

Ti = TS (C.56)

Ti > TS (C.57)

(c) The system is now excited with the signal x2(t) shown in figure C.9. Plot the outputsignal y2(t).

1

−1

x2(t)

t0 TS 2TS 3TS 4TS

Figure C.9: Given pulse stream x2(t)

Analog to Digital Conversion

16. Determine the spectrum (Fourier transform) of the following periodic pulse train

XT (t) =∞∑

n=−∞δ(t− nT ) (C.58)

17. A telephony signal can be interpreted as a low-pass signal with a cut-off frequency of fg =3.4 kHz.

(a) What is the minimum sampling frequency for perfect reconstruction of the signal(Nyquist rate)?


C.1 Problems 105

|H(f)|A

f−f2 −f1 0 f1 f2

Figure C.10: Non-ideal low-pass filter with a smooth transition between pass- and stop-band

(b) The sampled signal should be reconstructed by a non-ideal low-pass filter with a lowerstop band frequency f2 and a smooth transition between pass and stop-band (seeFig C.10). What is the minimum upper pass band frequency f1 and the minimumsampling frequency for perfect reconstruction of the signal?

18. Imagine you stroll through the forests with a DAT-recorder in order to tape birds singing.The sampling rate of the DAT-recorder is fs = 48 kHz. A model of the complete recordingand play-back setup is sketched in Figure C.11. The spectrum of the signals to be recordedis shown in figure C.12 (it should be noted that lower case indices refer to continuous-timesignals while upper-case indices refer to discrete-time i.e., sampled signals).

A/D-

Converter

DAT-

Recorder

D/A-

ConverterAmplifier

Microphone Loudspeaker

Figure C.11: Recording and play-back system

|Sv(f)|

f/kHz-10 -1 1 10

Figure C.12: Spectrum (absolute value) of singing birds

(a) Sketch the spectrum |SV (f)| of the recorded signal (after sampling) in the interval−96 kHz ≤ f ≤ 96 kHz.

(b) What are the necessary characteristics of the D/A-converter (filter characteristics, cut-off frequency) that ensure undistorted play-back?

(c) Unfortunately there is your neighbor walking around with his dog. He (your neighbor)plays an extremely loud dog whistle with a frequency of fh = 43 kHz. Sketch thespectrum |Sh(f)| (before sampling) and |SH(f)| (after sampling) of the tone producedby this very loud dog whistle in the interval −96 kHz ≤ f ≤ 96 kHz.

During play-back you realize that there is not much left of the birds singing when yourneighbor plays his dog whistle.


106 C EXCERCISES

(d) What is the reason for this? Sketch the spectrum |SA(f)| of the composite sampledsignal (birds singing and dog whistle) in the interval −48 kHz ≤ f ≤ 48 kHz. Describethe character of the distortion!

(e) Where would an analog filter have to be placed that ensures a proper play-back (perfectreconstruction) of the birds singing? What are the characteristics of such a filter (typeof filter, cut-off frequency)? Can a digital filter be used too (after sampling)?

(f) You regard such an analog filter as to costly. Moreover, you remember your party-affected ears that are completely deaf for all frequencies above 16 kHz. Therefore, youdecide to give a new dog whistle to your neighbor. Which frequency range (fh < 48 kHz)is allowed for this new whistle if you want to enjoy the birds singing when playing backthe recording?

|Sf (f)|

f/kHz-156 -146 146 156

Figure C.13: Spectrum (absolute value) of shrieking bats

(g) After all you want to record bat voices using the same equipment. The spectrum ofbat voices is sketched in Figure C.13. Is it principally possible to record and play backthese voices? If not, why is this not possible? If so, what would you have to change atthe setup with respect to (18b) and (18e) in order to enable perfect play-back? Sketchthe spectrum of the sampled signal |SF (f)| for a deeper understanding!

Bandpass Signals

19. Given is the sent signals(t) = Re

{u(t) ej 2πfct

}(C.59)

where u(t) = uI(t) + juQ(t) is an information bearing complex low-pass signal with a ban-dlimited spectrum [fTP,min, fTP,max] with fTP,min < 0 < fTP,max.

(a) Which condition must be fulfilled for the carrier frequency fc > 0 so that s(t) is abandpass signal in interval [fBP,min, fBP,max] with 0 < fBP,min < fBP,max ?

The signal s(t) is down-converted with a complex carrier signal of frequency fc on the receiverside, i.e. thus producing a signal as

r(t) = rI(t) + j rQ(t) = s(t) e− j 2πfct (C.60)

(b) Calculate the spectra R(f), RI(f) and RQ(f) as function of S(f).

(c) Calculate and sketch the spectra RI(f), RQ(f) and R(f) as function of U(f). Can thesignal u(t) be recovered as interference free from the signal r(t)? If so, which opera-tion /module is necessary for it?


C.1 Problems 107

20. Let s(t) = sI(t) + j sQ(t) be a complex baseband signal. The radio signal (real bandpasssignal) is generated by IQ-modulation


}(C.61)

Down conversion as well as low-pass filtering with fcut off = fc is performed at the receiver.Coherent (phase synchronous) reception is essential for many types of receivers which means

u(t) = uI(t) + juQ(t) = sBP (t) e− j 2πfct

∣∣∣|f |<fc

(C.62)

(a) Determine the in-phase and quadrature components uI(t) and uQ(t) as a function ofsI(t) and sQ(t). The phase offset between transmitter and receiver is ϕ0, i.e.

u(t) = sBP (t) e− j(2πfct+ϕ0)

∣∣∣|f |<fc

(C.63)

Which effect does a phase offset of ϕ0 = π (or ϕ0 = π/2) have?

Analog Modulation

21. A signal which is modelled by

s(t) = a1 cos 2πf1t+ a2 cos 2πf2t (C.64)

with f2 = 2f1 and a = a1 = a2, should be broadcasted by double sideband amplitudemodulation (DSB). The modulation product m(t) can be described as

m(t) = (s(t) + A) cos 2πfct (C.65)

where A is the amplitude of the carrier (sinusoid) and fc as the carrier frequency (herefc = 10f1).

(a) Give an inequality for the ratio aAthat keeps the modulation index μ = max(|s(t)|)

Asmaller

than 1.

(b) Sketch the spectrum of the modulated signal m(t) with A = 4a.

(c) Fig. C.14 shows a simplified envelope detector. The received signal is filtered to suppressnoise and adjacent channels, and rectified eventually giving the absolute value |m(t)|of the modulated signal. Calculate and sketch the spectrum of |m(t)|.Hints: The modulation index is always smaller than 1 (μ < 1). The following equationshold (Fourier series expansion):

for full wave rectifier: | cos 2πfct| = 2

π− 4

π

∞∑n=1

(−1)n

(2n− 1)(2n+ 1)cos 2π2nfct (C.66)


108 C EXCERCISES

BP LP HPm(t)

+ adjacent channels

m(t) |m(t)| s(t) + A s(t)

Figure C.14: Simplified envelope demodulator (idealized, neglecting noise and channel distor-tions)

22. Fig. C.15 shows a non-coherent envelope demodulator for the medium wave range(0.5 MHz < fc < 1.5 MHz). The received signal m(t) is down-converted to a con-stant intermediate frequency (here fIF < fc, see fig. C.16), before it is fed to an envelopedemodulator (see exercise 21).

The base band signals s1(t), s2(t) (with a cut off frequency of fg = 4.5 kHz) represent theprograms of two different radio stations. In this case the receive signal m(t) is

m(t) = m1(t) +m2(t) (C.67)

= (A1 + s1(t)) cos 2πfct+ (A2 + s2(t)) cos 2πfc,it (C.68)

The carrier frequencies of these two stations are fc and fc,i.

envelope demodulator

BP LP HP×m(t)

(+ noise)

cos 2πfM t

fIF s(t)

Figure C.15: Superhet

|M(f)|

ffc fM fc,i−fc−fM−fc,i

Figure C.16: Spectrum |M(f)| of the received signal at the input of the mixer

(a) Give a relation between the carrier frequency fc, the local oscillator frequency fM , andthe intermediate frequency fIF in the receiver.

(b) Show that after down-conversion the signals s1(t) at fc and s2(t) at fc,i can be super-imposing each other at the output of the mixer. This means that both stations willbe down-converted to the same frequency range f ∈ [fIF − fg, fIF + fg]. Therefore thedemodulation result in a mix of these two stations. Give the equations for the relationsbetween fc, fc,i, fIF and fM . The frequency fc,i is called image frequency. How can theseimage signals be suppressed?


C.1 Problems 109

(c) What is the minimum intermediate frequency fIF that ensures image rejection for allsignals of medium wave range? Assuming the same conditions what would be the IFfor the FM-frequency range (88 MHz < fc < 108 MHz)?

(d) Assuming an IF of fIF = 0.5 MHz, what is the necessary tuning range of fM?

(e) Which bandwidth is required for the band pass filters?

Digital Modulation

23.∗ The condition for (zero ISI (Inter symbol interference)) is known as 1. Nyquist criteriaand the received impulse h(t) (at output of the receiver filter, see Fig. 8.10 on page 80,h(t) = hT (t) ∗ hR(t)) must fulfils the following condition (in time domain):

h(t+ nT )∣∣t=0

=

{h(0) �= 0 for n = 0

0 for n �= 0(C.69)

where T is the symbol duration (see Fig. C.17). Show that the 1. Nyquist criterion in thefrequency domain corresponds the following condition

∞∑n=−∞

H(f − n

T

)= const. (C.70)

h(t)

h(0)

−T Tt

Figure C.17: 1. Nyquist condition in time domain

24. A bit sequence d{k} = {0, 1, 1, 0, 1, 0, 0, 1} for k = 0, 1, 2, . . . , 7 with a bit rate fbit = 1/T isto be transmitted in the baseband. For that purpose rect-shaped transmission impulses

hT (t) =A0

Trect

(t

T

)(C.71)

are used during bipolar transmission. Moreover symbol mapping of the form b{k} = 2d{k}−1applies, i.e.

d{k} = 1 → b{k} = +1 → +hT (t) to be sent (C.72)

d{k} = 0 → b{k} = −1 → −hT (t) to be sent (C.73)


110 C EXCERCISES

The send signal 18 s(t) has then the form

s(t) =7∑

k=0

b{k}T hT (t− kT ) (C.74)

= A0

7∑k=0

b{k} rect(t/T − k) (C.75)

= A0

7∑k=0

(2d{k} − 1) rect(t/T − k) (C.76)

(a) Sketch the send signal s(t) in the interval t ∈ [−2T, 9T ]. Emphasize the signal valuess(kT ), k ∈ [0, 7].

The channel will be treated for the time being as ideal, i.e. the received signal r(t) is givenas

r(t) = s(t) (C.77)

In the receiver the signal r(t) is filtered (receipt filters hR(t)), the filtered signal rE(t) =r(t) ∗ hR(t) is then sampled at time points kT and finally the sampling values are suppliedto a detector.

(b) Outline the block diagram of the entire data transmission system (transmitter, channeland receiver).

For the received filter hR(t) applies

hR(t) =1

Trect

(t

T

)(C.78)

(c) Sketch the filtered received signal rE(t) = r(t) ∗ hR(t) Def= sE(t) in the interval t ∈

[−2T, 9T ]. Also denote the sample values sE{k} = sE(kT ), k ∈ [0, 7].

(d) Compute the average power PsE of the sampled signal rE{k} = sE{k} at the entranceof the detector.

The channel is no more assumed as ideal. The received signal r(t) is to be interpreted assuperposition of the transmitted signals s(t) with a white noise signal n(t) the noise powerdensity N(f) = N0/2 for −∞ < f <∞.

r(t) = s(t) + n(t) (C.79)

rE(t) = sE(t) + nE(t) (C.80)

(e) Outline again the block diagram of the entire data transmission system (transmitter,channel and receiver).

(f) Compute the power density spectrum NE(f) of the filtered noise signal nE(t) at theoutput of the receipt filter hR(t).

NE(f) = |HR(f)|2N(f) (C.81)

18Such a signal is also designated as pulse amplitude modulated signal (PAM-Signal) (PAM –pulse amplitude

modulation).


C.1 Problems 111

(g) Compute the noise power PnEof nE(t) at the output of the receipt filter hR(t).

PnE=

∫ ∞

−∞NE(f) d f (C.82)

Hint: It applies ∫ ∞

−∞si2(ax) d x =

π

|a| (C.83)

Due to the assumed stationary noise signal n(t), the noise power PnEhas not been changed

through sampling. The noise power at the entrance of the detector is thereby also PnE.

(h) Compute the signal-to-noise ratio SNR at the entrance of the detector.

Hint: For simplification, both the transmitting and the receipt filters given here are indicatedas non-causal filters. If these filters are to be realized in an instrument, the impulse responseshT (t) and hR(t) must be delayed by T/2 in each case (see also Art. 4.5.3). From this a totaldelay of T , i.e. a sent bit d{k} thus is detected as d{k + 1} at the receiver one bit durationlater.

25. BPSK (binary phase shift keying) is used in the following transmission system. The proba-bility of the both transmit symbols d0, d1 are equal (P (d0) = P (d1) = 0.5). Assuming a idealnoise free transmission, the samples rE(kTs) = Re {rE(kTs)} at the input of the detectorare rE = s0 for the symbol d0 or rE = s1 for the symbol d1, respectively. Moreover it iss0 = −s1 > 0, Im {rE(kTs)} = 0.

(a) Plot the possible sample values of the received signal rE(kTs) in the complex plane(phase diagram at the detector input). Sketch the corresponding conditional probabilitydensity functions (pdf) of the amplitude frE |d0(rE|d0) and frE |d1(rE|d1) for the real partof the received signal rE(kTs) = Re {rE(kTs)}.

(b) Now a noise is superimposed on the received signal. Describe the differences of the newphase diagram at the receiver. Sketch again the corresponding conditional probabilitydensity function of the amplitude frE |d0(rE|d0) and frE |d1(rE|d1) for the case of additivewhite Gaussian noise (AWGN). Mark the area, which corresponds to the probability ofthe wrong decision P (d{k} = d0 | d{k} = d1) (conditional probability, that you detectd{k} = d0, even though d{k} = d1 was transmitted).

(c) The receiver is not properly synchronized to the transmitter now (non coherent downconversion). Assume a phase offset of ϕ = π

4. Sketch again the phase diagram of the

samples rE(kTs). Calculate the offset of the signal to noise ratio (SNR) for unchangednoise power.

(d) Assuming ideal phase synchronization again. The standard deviation of the noise signalis σ = 0.5 V. Calculate the signal amplitude |s0| = |s1| (in mV) corresponds to abit error rate of BER = 0.1. Take use of tables or math programs like MATLAB,Mathematica or Maple.


112 C EXCERCISES

C.2 Solutions

1. (a) For further computations the following properties are useful

cosϕ = cos−ϕ even function (C.84)

sinϕ = − sin−ϕ uneven function (C.85)

⇒ e− jϕ = cosϕ− j sinϕ (C.86)

cosϕ = Re{ejϕ} (C.87)

sinϕ = Im{ejϕ} (C.88)

Thus the first addition theorem can be checked

cos(x± y) = Re{ej(x±y)} = Re{ejx ej(±y)} (C.89)

= Re{(cosx+ j sin x)(cos y ± j sin y)} (C.90)

= Re{cos x cos y ∓ sin x sin y ± j cos x sin y + j sin x cos y} (C.91)

= cosx cos y ∓ sin x sin y (C.92)

and the second

sin x− sin y = Im

{ej(

x2+ y

2+x

2− y

2)− ej(

y2+x

2+ y

2−x

2)}

(C.93)

= Im

{ej

x+y2 ej

x−y2 − ej

x+y2 e− j x−y

2

}(C.94)

= Im

{ej

x+y2

(ej

x−y2 − e− j x−y

2

)}(C.95)

= Im

{(cos

x+ y

2+ j sin

x+ y

2

)2 j sin

x− y

2

}(C.96)

= 2 sinx− y

2cos

x+ y

2(C.97)

(b) These proofs are very simple

z + z∗ = x+ j y + x− j y = 2x (C.98)

= 2Re{z} (C.99)

and

z − z∗ = x+ j y − x+ j y = j 2y (C.100)

= j 2Im{z} (C.101)

also

z · z∗ = (x+ j y)(x− j y) (C.102)

= x2 − j xy + j xy + y2 (C.103)

= x2 + y2 (C.104)

= |z|2 (C.105)


C.2 Solutions 113

2. (a) By rearranging the eqn. (C.20)-(C.22) yields

a0 = 2c0 k = 0 (C.106)

ak = ck + c−k k = 1, 2, . . . (C.107)

bk = j(ck − c−k) k = 1, 2, . . . (C.108)

Using these relations in (C.14), then one gets

s(t) = FR{s(t)} = c0 +∞∑k=1

(ck + c−k) cos(2πk

Tt)+ j(ck − c−k) sin

(2πk

Tt)

(C.109)

and under consideration of the eqn. (C.28), (C.29) eventually,

s(t) = FR {s(t)} = c0 +∞∑k=1

(ck + c−k)ej 2π

kTt +e− j 2π k

Tt

2+ j(ck − c−k)

ej 2πkTt − e− j 2π k

Tt

2 j

(C.110)

= c0 +1

2

∞∑k=1

(ck e

j 2π kTt +ck e

− j 2π kTt +c−k e

j 2π kTt +c−k e

− j 2π kTt

+ ck ej 2π k

Tt −ck e− j 2π k

Tt −c−k e

j 2π kTt +c−k e

− j 2π kTt)(C.111)

= c0 +∞∑k=1

ck ej 2π k

Tt +c−k e

− j 2π kTt (C.112)

=∞∑

n=−∞cn e

j 2π nTt (C.113)

(b)

cn =1

T

∫ T2

−T2

s1(t) e− j 2π n

Tt d t (C.114)

=s0T

∫ τ2

− τ2

e− j 2π nTt d t (C.115)

=s0T

· 1

− j 2π nT

[e− j 2π n

Tt] τ

2

− τ2

(C.116)

=s0T

· 1

− j 2π nT

[e− j 2π n

Tτ2 − ej 2π

nT

τ2

](C.117)

=s0T

· 1πnT

· 1

2 j

[e+ j πnτ

T − e− j πnτT

]︸︷︷︸

sin(πnτT )

(C.118)

=s0τ

T· 1

πnτT

sin(πnτT

)(C.119)


114 C EXCERCISES

= s0τ

Tsi(πnτT

)(C.120)

For τ = T4

cn =s04

si(πn

4

)(C.121)

Fig. C.18 shows the corresponding amplitude spectrum.

|cn|s04

−12T

− 8T

− 4T

0 4T

8T

12T

f = nT

Figure C.18: Amplitude spectrum of a periodic rectangular signal

(c)

S2(f) =

∫ ∞

−∞s2(t) e

− j 2πft d t (C.122)

= s0

∫ τ2

− τ2

e− j 2πft d t (C.123)

The definite integral in (C.123) was calculated for the case f = nTin the solution for

subtask (2b) (compare (C.115)-(C.120)). By exploiting the result indicated there onefinally gets

S2(f) = s0τ si(πfτ) (C.124)

and with τ = T4

S2(f) =s04T si(πf

T

4) (C.125)

The corresponding amplitude spectrum is represented in Fig. C.19.

|S2(f)|s04T

−12T

− 8T

− 4T

0 4T

8T

12T

f

Figure C.19: Amplitude spectrum of the rect impulse

3. (a) Noise power at the input of the demodulator:

PN,dem = PN + g2 (C.126)


C.2 Solutions 115

= −50 dBm + 60 dB (C.127)

= 10 dBm (C.128)

Intelligence signal power at the input of the demodulator:

PT,dem = PT − LS + g1 + g2 (C.129)

= 12 dBm− 20 lg(100d) dB + 23 dBm + 60 dB (C.130)

= 95 dBm− 20 lg(100d) dB (C.131)

Signal-noise-ratio SNR at the input of the demodulator:

SNR = PT,dem − PN,dem (C.132)

= 95 dBm− 20 lg(100d) dB− 10 dBm ≥ 7 dB (C.133)

78 dB ≥ 20 lg(100d) (C.134)

d ≤ 103.9

100(C.135)

d ≤ 79.4 m (C.136)

(b) Solving for PT gives

SNR ≤ PT − LS + g1 + g2 − PN − g2 (C.137)

PT ≥ SNR + LS − g1 + PN (C.138)

≥ 20 dB + 20 lg(5000) dB− 23 dB− 50 dBm (C.139)

≥ 21 dBm (C.140)

≥ 126 mW (C.141)

4. (a) From definition

1.) δ(x) =

{+∞ for x = 0

0 for x �= 0(C.142)

2.)

∫ ∞

−∞δ(x) d x = 1 (C.143)

3.) δ(−x) = δ(x) (C.144)

(b) This integral converges to

limn→∞

∫ ∞

−∞s(x)δn(x− x0) d x = lim

n→∞

∫ x0+1n

x0− 1n

s(x)n

2d x (C.145)

= s(x0) limn→∞

∫ x0+1n

x0− 1n

n

2d x (C.146)

= s(x0) (C.147)


116 C EXCERCISES

(c) By integration over x one gets

σn(x) =

∫ x

−∞δn(χ) dχ (C.148)

=

⎧⎪⎪⎪⎨⎪⎪⎪⎩0 for x < − 1

n∫ x

− 1n

n2dχ for − 1

n≤ x < 1

n∫ 1n

− 1n

n2dχ for 1

n≤ x

(C.149)

=

⎧⎪⎪⎨⎪⎪⎩0 for x < − 1

n

n2x+ 1

2for − 1

n≤ x < 1

n

1 for 1n≤ x

(C.150)

The limit value giveslimn→∞

σn(x) = σ(x) (C.151)

with

σ(x) =

⎧⎪⎪⎨⎪⎪⎩0 for x < 0

12

for x = 0

1 for 0 < x

(C.152)

Fig. C.20 shows the function σn(x)

σn(x)

1

− 1n

1n

x

Figure C.20: Approximation of unit step function σ(x)

5. The computation of s(t) can take place explicitly using the inverse Fourier transformationwhich requires an enormous computation. This can be reduced under utilization of someproperties and correspondences of the Fourier transformation.

(a) Linearity ⇒ The task can be divided into subtasks (see Tab. A.1)

S(f) = S1(f) + S2(f) (C.153)

S1(f) =a

2(δ(f + fc) + δ(f − fc)) (C.154)

S2(f) =A

2

(rect

(f + fc2fg

)+ rect

(f − fc2fg

))(C.155)


C.2 Solutions 117

(b) Frequency shifting (Modulation, spectrum shifts towards ”‘left”’ and/or ”‘right”’) ⇒Bandpass signals can be analyzed as low-pass signal (see Tab. A.1)

s2(t) = F−1 {S2(f)} = A cos(2πfct) F−1

{rect

(f

2fg

)}(C.156)

(c) Scaling (time suppression frequency expansion, time expansion frequencysuppression) ⇒ the bandwidth can be normalized (see Tab. A.1)

S

(f

2fg

)2fgs(2fgt) (C.157)

(d) By FT (see Tab. A.2 and A.3)

s1(t) = F−1{a2(δ(f + fc) + δ(f − fc))

}= a cos 2πfct (C.158)

fur s1(t) und mit

rect

(f

2fg

)2fg si(2fgπt) (C.159)

wird s2(t) zu

s2(t) = A cos(2πfct)2fg si(2fgπt) (C.160)

The final result is

s(t) = s1(t) + s2(t) = (a+ 2Afg si 2πfgt) cos 2πfct (C.161)

6. Problem: The computation of the Fourier transformation S(f) assumes the knowledge ofthe progression of the function s(t) for −∞ < t < ∞. This is practically not possible, sothat only the process can be analyzed within the time window 0 ≤ t ≤ T for computation.⇒ Short-time spectrum ST (f)

Question: What is the effect of this so called windowing? How should this windowing takeplace, so that the deviation is as small as possible between S(f) and ST (f)?

(a) The computation short-time spectrum ST (f) can be considered also as Fourier trans-formation, where all unknown regions of the time signal s(t) be masked by a windowfunction w(t).

ST (f) =

∫ T

0

s(t) e− j 2πft d t =

∫ ∞

−∞s(t)w(t) e− j 2πft d t (C.162)

with (rect window function)

w(t) = rect

(t− T/2

T

)(C.163)


118 C EXCERCISES

By using convolution theorem

s(t)w(t) ST (f) = (S ∗W )(f) = S(f) ∗W (f) (see Art. A.3) (C.164)

I.e. the short-time spectrum ST (f) results from the convolution of the spectrum S(f)and the spectrum W (f) of the the window function w(t).

ST (f) = S(f) ∗ (T si(πfT ) e− jπfT

)(C.165)

Convolution with the sinc function (rect windowing) leads to a kind of ”‘Kantenver-schleifung”’ to the spectrum. The rect windowing is very simple in its application (com-putation of short-time spectrum), leads however to a large deviation between ST (f) andS(f). Therefore usually other window functions are used (Blackman, Kaiser, Hamming,Hanning etc.).

(b) Example for the effect of window length T

cos(2πf0t)1

2(δ(f + f0) + δ(f − f0)) (C.166)

⇒ ST (f) =T

2e− jπfT

(si(π(f + f0)T ) e

− jπf0T + si(π(f − f0)T ) ejπf0T

)(C.167)

Fig. C.21 shows the short-time spectrum ST (f) for different window lengths T . It

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

normierte Frequency f/f0

Am

plitu

de

a) |ST (f)| for T = 1/f0

−4 −3 −2 −1 0 1 2 3 40

1

2

3

4

5

6


Am

plitu

de

b) |ST (f)| for T = 10/f0

−4 −3 −2 −1 0 1 2 3 40

5

10

15

20

25

30

35

40

45

50


Am

plitu

de

c) |ST (f)| for T = 100/f0

Figure C.21: Amplitude the short-time spectrum |ST (f)|

becomes evident from it that with increasing window length T the deviation becomessmaller for Fourier transformation.

7. Starting from |x(t)|2 = x(t) · x∗(t), x∗(t) X∗(−f) and x1(t) · x2(t) X1(f) ∗ X2(f)one gets∫ ∞

−∞|x(t)|2 d t =

∫ ∞

−∞x(t) · x∗(t) e− j 2π0t d t (C.168)

= F {x(t) · x∗(t)}∣∣∣f=0

=[X(f) ∗X∗(−f)]∣∣∣

f=0(C.169)

=

∫ ∞

−∞X(u) ·X∗(−(f − u)) d u

∣∣∣f=0

(C.170)


C.2 Solutions 119

=

∫ ∞

−∞X(u) ·X∗(u) d u Subst. u = f, d u = d f (C.171)

=

∫ ∞

−∞X(f) ·X∗(f) d f =

∫ ∞

−∞|X(f)|2 d f (C.172)

8. (a) First the spectrum of the function g1(t) = e−t2 is computed:

G1(f) =

∫ ∞

−∞e−t2 e− j 2πft d t (C.173)

=

∫ ∞

−∞e−t2 cos 2πft d t− j

∫ ∞

−∞e−t2 sin 2πft d t (C.174)

Since both g1(t) is and cos t each are even function and sin t an odd function∫ ∞

−∞e−t2 sin 2πft d t = 0 (C.175)

then [BS79]

G1(f) = 2

∫ ∞

0

e−t2 cos 2πft d t (C.176)

=√π e−π2f2

(C.177)

Since

g(t) =1√2πσ2

e−t2

2σ2 (C.178)

=1√2πσ2

g1(t√2σ2

) (C.179)

the spectrum G(f) can be calculated with the help of the scaling property (see Tab. A.1)from G1(f).

G(f) =1√2πσ2

√2σ2 G1(

√2σ2f) (C.180)

=1√πG1(

√2σ2f) (C.181)

= e−2π2σ2f2

(C.182)

(b) The area A under g(t) (dc component) can be given as

A =

∫ ∞

−∞g(t) d t =

∫ ∞

−∞g(t) e− j 2π0t d t (C.183)

= G(f)∣∣f=0

= G(0) = 1 (C.184)

9. (a) In order to compute the Hilbert transformation, the detor over the fourier transforma-tion the usually is the simpler solution method. However both ways are to be demon-strated here.

xH(t) = H{x(t)} (C.185)


120 C EXCERCISES

= A0 cos 2πf0t ∗ 1

πt(C.186)

=A0

π

∫ ∞

−∞cos 2πf0(t− τ)

1

τd τ (C.187)

=A0

π

∫ ∞

−∞cos 2πf0t

cos 2πf0τ

τ︸︷︷︸uneven function ≡ 0

+sin 2πf0tsin 2πf0τ

τ︸︷︷︸even function

d τ (C.188)

= A0 sin 2πf0t2

π

∫ ∞

0

sin 2πf0τ

τ︸︷︷︸sine integral ≡ π

2

d τ (C.189)

= A0 sin 2πf0t (C.190)

bzw.

xH(t) = F−1 {F {x(t)} (− j) sgn(f)} (C.191)

= F−1 {F {A0 cos 2πf0t} (− j) sgn(f)} (C.192)

= F−1

{A0

1

2(δ(f + f0) + δ(f − f0)) (− j) sgn(f)

}(C.193)

= F−1

{A0

j

2(δ(f + f0)− δ(f − f0))

}(C.194)

= A0 sin 2πf0t (C.195)

(b) For computing the spectrum XA(f) we may use the result from the last subtask (seeFig. C.22)

XA(f) = F {xA(t)} = F {x(t) + jxH(t)} (C.196)

= X(f) + jXH(f) (C.197)

= A01

2(δ(f + f0) + δ(f − f0)) + A0

j

2j(δ(f + f0)− δ(f − f0)) (C.198)

=A0

2(δ(f + f0) + δ(f − f0))− A0

2(δ(f + f0)− δ(f − f0)) (C.199)

= A0 δ(f − f0) (C.200)

X(f)A0

f−f0 f0

XA(f)A0

f−f0 f0

Figure C.22: Spectrums of the signals x(t) and xA(t)

• the spectrum XA(f) is ”‘single-sided”’, but the signal xA(t) is complex,

• the spectrum X(f) is ”‘double-sided”’, but the signal x(t) is real.


C.2 Solutions 121

(c) The signal xA(t) possesses an constant amplitude (constant envelope).

|xA(t)| = |x(t) + j xH(t)| (C.201)

= |A0(cos 2πf0t+ j sin 2πf0t)| (C.202)

=√A2

0

(cos2 2πf0t+ sin2 2πf0t

)(C.203)

=

{A0 for A0 ≥ 0

−A0 for A0 < 0(C.204)

10. (a) For xg(t) applies

F {xg(t)} = F{1

2(x(t) + x(−t))

}(C.205)

Xg(f) =

∫ ∞

−∞

1

2(x(t) + x(−t)) e− j 2πft d t (C.206)

=

∫ ∞

−∞

x(t) + x(−t)2

cos 2πft︸︷︷︸even function

d t− j

∫ ∞

−∞

x(t) + x(−t)2

sin 2πft︸︷︷︸uneven function ≡ 0

d t (C.207)

=

∫ ∞

0

(x(t) + x(−t)) cos 2πft︸︷︷︸uneven function

d t (C.208)

=

∫ ∞

−∞x(t) cos 2πft d t (C.209)

= Re

{∫ ∞

−∞x(t) e− j 2πft d t

}(C.210)

= Re {F {x(t)}} (C.211)

and for xu(t) gilt

F {xu(t)} = F{1

2(x(t)− x(−t))

}(C.212)

Xu(f) =

∫ ∞

−∞

x(t)− x(−t)2

cos 2πft︸︷︷︸uneven function ≡ 0

d t− j

∫ ∞

−∞

x(t)− x(−t)2

sin 2πft︸︷︷︸even function

d t (C.213)

= − j

∫ ∞

0

(x(t)− x(−t)) sin 2πft︸︷︷︸uneven function

d t (C.214)

= − j

∫ ∞

−∞x(t) sin 2πft d t (C.215)

= j Im

{∫ ∞

−∞x(t) e− j 2πft d t

}(C.216)

= j Im {F {x(t)}} (C.217)

(b) With eqn. (C.48)

x(t) = xg(t) + xu(t) (C.218)


122 C EXCERCISES

= xg(t) + xg(t) sgn(t) (C.219)

since x(0) = 0 also applies the inversion

= xu(t) sgn(t) + xu(t) (C.220)

⇒ X(f) = Xu(f) ∗ F { sgn(t)}+Xu(f) (C.221)

mit X(f) = Xg(f) +Xu(f) folgt

Xg(f) = Xu(f) ∗ F { sgn(t)} (C.222)

= (− j)Xu(f) ∗ 1

πt(C.223)

= (− j)H{Xu(f)} (C.224)

11. By causality one generally understands the cause-effect principle, i.e. each effect excludesone or more timely back dated cause(n). An effect can however never occur before thecause takes place.

The way a regarded LTI system is indicated here is that an initial value y(t1) may dependonly on the past input value(s) x(t) with t < t1 and NOT on future or current values19. Ifthis condition is fulfilled, the system is called strictly causal, otherwise not strictly causal.

The signal y(t1) must be independent of x(t) for t ≥ t1.

y(t1) =

∫ ∞

−∞x(τ) h(t1 − τ) d τ (C.225)

=

∫ t1

−∞x(τ) h(t1 − τ) d τ︸︷︷︸

causal part

+

∫ ∞

t1

x(τ) h(t1 − τ) d τ︸︷︷︸anti-causal part

(C.226)

I.e. the second term must be equal to zero. Since for the signal x(t) no restrictions are to bemade, the condition is fastened with the impulse response.

0 =

∫ ∞

t1

x(τ) h(t1 − τ) d τ (C.227)

⇒ h(t1 − τ)∣∣τ≥t1

= 0 (C.228)

⇒ h(t)∣∣t≤0

= 0 (C.229)

12. (a) The triangular function triang(x) arises as a result of unique convolution of the rectfunction rect(x) with itself.

triang(x) = rect(x) ∗ rect(x) =

∫ ∞

−∞rect(χ) rect(x− χ) dχ (C.230)

To compute a distinction of cases is necessary

19Sometimes however dependence on the current input value is also permitted (strictly causal ⇒ causal)


C.2 Solutions 123

1. Case: x < −1 ∫ ∞

−∞rect(χ) rect(x− χ) dχ = 0 (C.231)

2. Case: −1 ≤ x < 0∫ ∞

−∞rect(χ) rect(x− χ) dχ =

∫ x+ 12

− 12

1 dχ = 1 + x (C.232)

3. Case: 0 ≤ x < 1 ∫ ∞

−∞rect(χ) rect(x− χ) dχ =

∫ 12

x− 12

1 dχ = 1− x (C.233)

4. Case: 1 < x ∫ ∞

−∞rect(χ) rect(x− χ) dχ = 0 (C.234)

Therewith one gets (see Fig. C.23)

triang(x) =

⎧⎪⎪⎨⎪⎪⎩1 + x for − 1 ≤ t < 0

1− x for 0 ≤ t < 1

0 otherwise

(C.235)

1

0 1−1

triang(x)

x

Figure C.23: The triangular function triang(x) = rect(x) ∗ rect(x)

To compute the convolution product proceeds in same manner

y(x) = triang(x) ∗ rect(x) (C.236)

=

∫ ∞

−∞triang(χ) rect(x− χ) dχ (C.237)

1. Case: x < −32 ∫ ∞

−∞triang(χ) rect(x− χ) dχ = 0 (C.238)

2. Case: −32≤ x < −1

2∫ ∞

−∞triang(χ) rect(x− χ) dχ =

∫ x+ 12

−1

(1 + χ) dχ =1

2x2 +

3

2x+

9

8(C.239)


124 C EXCERCISES

3. Case: −12≤ x < 1

2∫ ∞


∫ 0

x− 12

(1 + χ) dχ+

∫ x+ 12

0

(1− χ) dχ = −x2 + 3

4

(C.240)

4. Case: 12≤ x < 3

2∫ ∞


∫ 1

x− 12

(1− χ) dχ =1

2x2 − 3

2x+

9

8(C.241)

5. Case: 32≤ x ∫ ∞

−∞triang(χ) rect(x− χ) dχ = 0 (C.242)

So one gets (see Fig. C.24)

y(x) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

12x2 + 3

2x+ 9

8for − 3

2≤ x < −1

2

−x2 + 34

for − 12≤ x < 1

2

12x2 − 3

2x+ 9

8for 1

2≤ x < 3

2

0 otherwise

(C.243)

0 1 2−1−2

y(x)

12

x

Figure C.24: Convolution product y(x) = rect(x) ∗ rect(x) ∗ rect(x)

(b) There are two ways to calculate the spectrum F { triang(t/T )}.• By Fourier transformation of triang(t/T ) explicitly.

F { triang(t/T )} =

∫ ∞

−∞triang(t/T ) e− j 2πft d t (C.244)

This method won’t be shown here.

• We know that

triang(t/t) = rect(t/T ) ∗ rect(t/T ) (C.245)

applies, as well as (see Tab. A.3)

rect(t) si(πf) (t, f– dimensionless) (C.246)


C.2 Solutions 125

F { triang(t/T )}T

f− 1T

1T

Figure C.25: Spectrum of the triangular function F { triang(t/T )}

By exploiting the convolution and scaling properties we get (see Fig. C.25)

F { triang(t/T )} = T si(πfT ) T si(πfT ) (C.247)

= T 2 si2(πfT ) (C.248)

13. The spectrum is given as

H(f) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

A f2+ff2−f1

for − f2 ≤ f < −f1A for − f1 ≤ f < f1

A f2−ff2−f1

for f1 ≤ f < f2

0 otherwise

(C.249)

There are (at least) 2 solutions.

• With the help of the inverse Fourier transformation, the impulse response h(t) can becomputed directly.

h(t) =

∫ ∞

−∞H(f) ej 2πft d f (C.250)

da H(f) = H(−f)

= 2

∫ ∞

0

H(f) cos 2πft d f (C.251)

= 2A

∫ f1

0

cos 2πft d f + 2A

∫ f2

f1

f2 − f

f2 − f1cos 2πft d f (C.252)

With the help of the relations∫x cos ax d x =

cos ax

a2+x sin ax

a(C.253)

cos x− cos y = −2 sinx+ y

2sin

x− y

2(C.254)

gives

h(t) =A

πtsin 2πft

∣∣∣∣f10

+A

πt

f2f2 − f1

sin 2πft

∣∣∣∣f2f1

− A

πt

(cos 2πft

2π(f2 − f1)t+f sin 2πft

f2 − f1

)∣∣∣∣f2f1

(C.255)


126 C EXCERCISES

=A

πt

cos 2πf1t− cos 2πf2t

2π(f2 − f1)t(C.256)

= A(f1 + f2) si(π(f1 + f2)t) si(π(f2 − f1)t) (C.257)

• Convolution of two rect functions generates the triangle function. The two square pulsesare developed differently in a trapezoidal process. Thus H(f) can be also representedas (fa > fb)

H(f) =A

2fbrect

(f

2fa

)∗ rect

(f

2fb

)(C.258)

The corner frequency can be given as

f1 = fa − fb or fa =f1 + f2

2(C.259)

f2 = fa + fb or fb =f2 − f1

2(C.260)

Since convolution in frequency domain corresponds to a multiplication in time domain,the impulse response h(t) can be indicated as (see Tab. A.1)

h(t) =A

f2 − f1F−1

{rect

(f

f1 + f2

)}F−1

{rect

(f

f2 − f1

)}(C.261)

With help of the correspondence of the rect function (see Tab. A.3) and of the scalingproperty (see Tab. A.1) finally results in

h(t) = A(f1 + f2) si(π(f1 + f2)t) si(π(f2 − f1)t) (C.262)

The first solution method always leads (slowly but surely) to the goal, however is at the costof cpu time. The second version is clearly faster and more elegant, requires however a certainamount of experience. Usually only a few correspondences to the Fourier transformation arenecessary in order to solve more complicated problems. Fig. C.26 shows the process of h(t)with f2 = 2f1.

14. (a) The output signal y(t) results from the convolution of the signals x(t) and h(t).

y(t) = (x ∗ h)(t) =∫ ∞

−∞x(τ)h(t− τ)︸︷︷︸

m(τ,t)

d τ (C.263)

It is now to be demonstrated that the convolution of time-limited signals results inagain a time-limited signal. The signal y(t) can be zero for any time-limited signal x(t)and h(t), when m(τ, t) = 0 is valid for any τ , i.e.

∀x(·),h(·),τ m(τ, t) = x(τ)h(t− τ) = 0 (C.264)

Therefore x(·) = 0 or h(·) = 0. It applies

x(τ) = 0 when 0 > τ > T1 (C.265)

h(t− τ) = 0 when 0 > t− τ > T2 (C.266)


C.2 Solutions 127

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−0.5

0

0.5

1

1.5

2

2.5

3

h(t)/

(A f 1)

t/f1

Figure C.26: Impulse response h(t) of the low-pass with finite area

⇒ t− T2 < τ < t (C.267)

Thereby two coherent intervals develop in each case

x(τ) = 0 for τ ∈ (−∞, 0) ∨ (T1,∞) (C.268)

h(t− τ) = 0 for τ ∈ (−∞, t− T2) ∨ (t,∞) (C.269)

and m(τ, t) = 0 Hence by combining (intersection) these intervals

m(τ, t) = 0 for τ ∈ (−∞,max(0, t− T2)) ∨ (min(T1, t),∞) (C.270)

The convolution integral y(t) will be zero when

m(τ, t) = 0 for τ ∈ (−∞,∞) (C.271)

gilt, d.h.max(0, t− T2) > min(T1, t) (C.272)

This is fulfilled for

t > T1 + T2 (C.273)

t < 0 (C.274)

This yields finallyy(t) = 0 for 0 > t > T3 = T1 + T2 (C.275)

Fig. C.27 should clarify these situations.

(b) The impulse response h(t) is for −2T ≤ t ≤ −T and t = 6T non zero. Thus the signaly(t) results as follows at time t = 0:

y(0) =

∫ ∞

−∞x(τ) · h(−τ) d τ (C.276)


128 C EXCERCISES

x(t)

t0 T1

h(t)

t0 T2a) Input signal x(t) and impulse response h(t)

x(τ), h(t− τ)

τt− T2 t 0 T1b) 1. Fall: t < 0 ⇒ y(t) = 0

x(τ), h(t− τ)

τt− T2 t0 T1c) 2. Fall: 0 ≤ t ≤ T3 = T1 + T2

x(τ), h(t− τ)

τt− T2 t0 T1d) 3. Fall: t > T3 = T1 + T2 ⇒ y(t) = 0

Figure C.27: Convolution of time-limited signals

= x(−6T ) +

∫ 2T

T

x(τ) d τ (C.277)

Hence it follows that the signal x(t) at the time t = −6T and in the interval [T, 2T ]must be known, in order to compute the output signal y(t) at time t = 0.

15. (a) The evaluation of convolution shows

y(t) =

∫ ∞

−∞x(τ) h(t− τ) d τ (C.278)

=A

Ts

∫ t

t−TS

x(τ) d τ =A

Ts

∫ Ts

0

x(t− τ) d τ (C.279)

⇒ Short-time integrator, moving average, low-pass filter

(b) Since both signals are causal, i.e. x1(t) = 0 and h(t) = 0 for t < 0, it simplifies theintegration limit of the convolution (applies generally)

y(t) = (x1 ∗ h)(t) =∫ t

0

x1(τ) h(t− τ) d τ (C.280)


C.2 Solutions 129

Fig. C.28 illustrates the convolution graphically. The calculation yields (see result inFig. C.29)

y1(t) =A

Ts

∫ max(min(t,Ti),0)

min(max(0,t−Ts),Ti)

d τ (C.281)

h(t− τ)

x1(τ)

ATs

1

τt− TS 0 t Ti

∫ t

0

Figure C.28: Graphical illustration of the convolution operation

y1(τ)

A Ti

Ts

t0 Ti TS Ti + TSa) 1. Fall: Ti < TS

y1(τ)A

t0 TS Ti + TS = 2TSb) 2. Fall: Ti = TS

y1(τ)A

t0 TS Ti Ti + TSc) 3. Fall: Ti > TS

Figure C.29: Result from convolution of two square pulses

Ti ≥ TS must be valid for the input impulse x1(t), in order to get the full amplitude A.The output signal y1(t) has a duration of Ti + TS.

(c) See Fig. C.30

(d) From the last subtasks it becomes evident that the input impulse must reach at leastthe duration of the impulse response.

TG =1

2fg(C.282)


130 C EXCERCISES

y2(t)

t0

A

−A

TS 2TS 3TS 4TS 5TS

Figure C.30: Output signal y2(t)

Ti ≥ TG =1

2fg= 50 ns (C.283)

16. The function XT (t) =+∞∑

n=−∞δ(t− nT ) can be extented as periodic signal with the period T

into Fourier series:

XT (t) =∞∑

n=−∞cn e

j 2π ntT (C.284)

mit

cn =1

T

∫ T2

−T2

XT (t) e− j 2π nt

T d t (C.285)

=1

T

∫ ∞

−∞δ(t) e− j 2π nt

T d t (Masking property) (C.286)

=1

Te− j 2π n·0

T (C.287)

=1

T(C.288)

and hence

XT (t) =1

T

∞∑n=−∞

ej 2πnTt (C.289)

F {XT (t)} =1

T

∞∑n=−∞

δ(f − n

T

)(C.290)

=1

TX 1

T(f) (C.291)

XT (t)1

TX 1

T(f) (C.292)

17. The voice signal contains speech information in frequency range 0.3 < f < 3.4 kHz (thuscan be regarded as BP signal). However, part of speech, interfering signals and/or noise canbe present above the fundamental frequency fg = 3.4 kHz. Therefore an anti-aliasing filteris needed before the sampler.


C.2 Solutions 131

(a) minimum sampling rate (Nyquist rate)

fg = 3.4 kHz ⇒ fs =1

T= 2fg = 6.8 kHz (C.293)

Since LP filters with finite slope are used as anti- aliasing filter, the sampling rate is tobe selected accordingly (for telephone signals fs = 8 kHz is selected (PCM technology)).

(b) • f1 ≥ fg, otherwise distortions develop by pass-band droop (linear distortion of thepass band)

• fs − fg ≥ f2 ⇒ fs ≥ fg + f2, otherwise distortions occurred due to insufficientimage rejection (see also Fig. C.31)

|H(f)|, |Ss(f)|

f0 fg fs − fg f2 fs

Figure C.31: Signal reconstruction with LP-Filter H(f) of finite steepness

18. Sampling (time discretization) results images, i.e. the original spectrum is continued peri-odically. The value discretization (quantization) is left unconsidered.

(a) The spectrum |SV (f)| of the sampled signals (bird voice) is shown in Fig. C.32.

|SV (f)|

f/kHz-96 -48 48 96Figure C.32: Amplitude spectrum of bird voice

(b) That connected analog filter of D/A converter is a reconstruction filter and it suppressesthe images (spectral copies) of the former time-discrete signal. Since the original signalis to be reconstructed here, a low-pass filter with a maximum fundamental frequencyfg = fs/2 = 24 kHz is needed.

(c) The spectrum |SH(f)| of the sampled signal (dog whistle) is shown in Fig. C.33.

|SH(f)|

f/kHz-96 -48 48 96Figure C.33: Amplitude spectrum of dog whistle


132 C EXCERCISES

(d) Since no anti-aliasing filter is attached with the A/D converter, aliasing errors mayoccur. Exactly what happens here. An image (image frequency) of the dog whistlefh = 43 kHz lies in the audible range fH = fs − fh = 5 kHz. This tone hides the birdvoice due to its loudness. Fig. C.34 shows this superimposition.

|SA(f)|

f/kHz-48 -24 24 48Figure C.34: Amplitude spectrum of the superimposed signals (bird voice and dog whistle)

(e) An anti-aliasing filter is needed before the A/D conversion. This is a low-pass filterwith a maximum fundamental frequency fg = fs/2 = 24 kHz.

(f) Since you cannot hear tones above 16 kHz, the whistle may work within the range16 kHz ≤ f < 24 kHz. Since moreover, the range 24 kHz ≤ f < 32 kHz drop behindby aliasing into latter range, the permitted range of the dog whistle is 16 kHz ≤ f <32 kHz.

(g) In principle it is possible to take up the bat voices and can be accurately reproduced.The LP filter as anti-aliasing or Reconstruction filters becomes now bandpass filterswith the cut-off frequencies fg,u = 144 kHz, fg,o = 168 kHz (the loudspeaker mustbe able to work within this range). This procedure is called under-sampling and thefollowing conditions must be kept: 1) the sampling rate must at least be double of thesignal bandwidth B = fupper − flower,

2) no integral multiple of the half sampling rate may lie in the desired frequency range.

Here both conditions are fulfilled.

19. (a) The real bandpass signal s(t) can be represented through its quadrature componentsuI(t) = Re{u(t)} (in phase component) and uQ(t) = Im{u(t)} (quadrature phase com-ponent) (see Fig. C.35).

s(t) = Re{u(t) ej 2πfct} (C.294)

= Re{(uI(t) + juQ(t)) (cos 2πfct+ j sin 2πfct)} (C.295)

= uI(t) cos 2πfct− uQ(t) sin 2πfct (C.296)

For the computation of the originating spectrum proven in eqn. (C.99) in context with

Re{x} =1

2x+

1

2x∗ (C.297)

is used. Consequently in time domain applies

s(t) = Re{u(t) ej 2πfct} (C.298)

=1

2u(t) ej 2πfct +

1

2u∗(t) e− j 2πfct (C.299)


C.2 Solutions 133

×

×+

Re{u(t)} = uI(t)

Im{u(t)} = uQ(t)

cos 2πfct

− sin 2πfct

s(t)

Figure C.35: Up-mixing of a complex low-pass signal u(t) into a real band-pass filter signals(t)

The Fourier transformation of s(t) results with x∗(t) = X∗(−f) (see Tab. A.1)

S(f) =1

2U(f − fc) +

1

2U∗(−(f + fc)) (C.300)

Fig. C.36 shows exemplary the spectrum U(f). In practical applications |fTP,min| =|fTP,max| is valid, however, more general case will be considered here. The spectra S(f)

U(f) = URe(f) + jUIm(f)

ffTP,min fTP,max

U∗(f) = URe(f)− jUIm(f)

ffTP,min fTP,max

U∗(−f) = URe(−f)− jUIm(−f)

ffTP,min fTP,max

Figure C.36: Real part (solid line) and imaginary part (dotted line) of U(f), U∗(f), U∗(−f)

is yield by right or left shift of the spectra U(f) or U∗(−f). That real part of S(f)is an even function, as well as the imaginary part of S(f) is an odd function (seeFig. C.37). The corresponding time function s(t) is therefore (as expected) pure real.fTP,min < f < fTP,max should be valid for the spectrum U(f) with fTP,min < 0 and

S(f) = SRe(f) + jSIm(f)

f−fc fc

1

2U∗(−(f + fc))

1

2U(f − fc)

−fc − fTP,min fc + fTP,min

Figure C.37: Real (solid line) and imaginary part (dotted line) of S(f)

fTP,max > 0. Then applies for the spectra U(f − fc): fc + fTP,min < f < fc + fTP,max orfor U∗(−(f + fc)): −fc − fTP,max < f < −fc − fTP,min. Hence these two spectrums arenot overlapping, must consequently applies

−(fc + fTP,min) < fc + fTP,min ⇒ fc > |fTP,min| (C.301)


134 C EXCERCISES

(b) A phase synchronous down-conversion takes place in the receiver (the pertinent low-pass filtering will be considered later)

r(t) = rI(t) + j rQ(t) = s(t) e− j 2πfct (C.302)

From the (Frequency) shifting theorem (see Tab. A.1)we get

r(t) R(f) = S(f + fc) (C.303)

The spectra RI(f) = F {rI(t)} and RQ(f) = F {rQ(t)} are yield by utilizing eqn. (C.99,C.101) similar to

rI(t) = Re{r(t)} =1

2

(r(t) + r∗(t)

)(C.304)

RI(f) =1

2

(S(f + fc) + S∗(−(f − fc))

)(C.305)

and

rQ(t) = Im{r(t)} =1

2 j

(r(t)− r∗(t)

)(C.306)

RQ(f) =1

2 j

(S(f + fc)− S∗(−(f − fc))

)(C.307)

(c) With

S(f) =1

2

(U(f − fc) + U∗(−(f + fc))

)(C.308)

yields

R(f) =1

2

[U(f) + U∗(−(f + 2fc))

](C.309)

RI(f) =1

2

[1

2

(U(f) + U∗(−f))︸︷︷︸UI(f)=F{uI(t)}

+1

2

(U(f − 2fc) + U∗(−(f + 2fc))

)](C.310)

RQ(f) =1

2

[1

2 j

(U(f)− U∗(−f))︸︷︷︸UQ(f)=F{uQ(t)}

− 1

2 j

(U(f − 2fc)− U∗(−(f + 2fc))

)](C.311)

By superimposition of the two spectrums U(f) and U∗(−f), is it not possible to re-construct the entire complex signal from rI(t) or rQ(t) (see Fig. C.38). For the recon-struction of u(t) one needs r(t). This can be done via a simple low-pass filter with afundamental frequency of fg = fc. Filtering of rI(t) or rQ(t) leads to the reconstructionof uI(t) or uQ(t). In communications the principle described in this task is called I/Qdemodulation (see Fig. C.39).

20. (a) The receipt signal can be disassembled into its quadrature components rI(t) and rQ(t).

r(t) = sBP (t) e− j(2πfct+ϕ0)

∣∣∣|f |<fc

(C.312)


C.2 Solutions 135

R(f) = RRe(f) + jRIm(f)

f−2fc 2fc

RI(f)

f−2fc 2fc

RQ(f)

f−2fc 2fc

Figure C.38: Real (solid line) and imaginary part (dotted line) of the spectra R(f), RI(f),RQ(f)

×

×

TP

TP

fg = fcs(t)

cos 2πfct

− sin 2πfct

rI(t)

rQ(t)

12uI(t) =

12Re{u(t)}

12uQ(t) =

12Im{u(t)}

Figure C.39: phase synchronous down-mixture – I/Q-Demodulator

= (sI(t) cos 2πfct− sQ(t) sin 2πfct)(cos(2πfct+ ϕ0)− j sin(2πfct+ ϕ0))∣∣∣|f |<fc

(C.313)

rI(t) = Re{r(t)} (C.314)

= sI(t) cos(2πfct) cos(2πfct+ ϕ0)− sQ(t) sin(2πfct) cos(2πfct+ ϕ0)∣∣∣|f |<fc

(C.315)

rQ(t) = Im{r(t)} (C.316)

= −sI(t) cos(2πfct) sin(2πfct+ ϕ0) + sQ(t) sin(2πfct) sin(2πfct+ ϕ0)∣∣∣|f |<fc

(C.317)


136 C EXCERCISES

Hence yields

rI(t) =1

2sI(t) cosϕ0 +

1

2sQ(t) sinϕ0 (C.318)

and

rQ(t) = −1

2sI(t) sinϕ0 +

1

2sQ(t) cosϕ0 (C.319)

21. (a) In two-sideband amplitude modulation (DSB) the following condition must be kept(A >0)

s(t) + A ≥ 0 ∀t (C.320)

i.e. the information of the signal s(t) is exclusively contained in the envelope of themodulated signal m(t). This is equivalent to the following condition

μ =max(|s(t)|)

A≤ 1 (C.321)

Hence

max(|a(cos 2πf1t+ cos 4πf1t)|)A

≤ 1 (C.322)

2a

A≤ 1 (C.323)

a

A≤ 1

2(C.324)

(b) The modulated signal results as

m(t) = (a1 cos 2πf1t+ a2 cos 2πf2t+ A) cos 2πfct (C.325)

= a(cos 2πf1t+ cos 4πf1t+ 4) cos 20πf1t (C.326)

Fig. C.40 shows the spectrum M(f) m(t), which can be easily indicated withcorrespondences from Tab. A.2. It gives

M(f) =a

4

(δ(f + 8f1) + δ(f + 9f1) + δ(f + 11f1) + δ(f + 12f1) + δ(f − 8f1)

+ δ(f − 9f1) + δ(f − 11f1) + δ(f − 12f1))+2a

(δ(f + 10f1) + δ(f − 10f1)

)(C.327)

M(f)2a

a4

f0−10f1 10f1

Figure C.40: Spectrum M(f) of the transmitted signal m(t)


C.2 Solutions 137

(c) As it was presupposed here that no over modulation takes place, i.e. μ ≤ 1 can be used

|m(t)| = | (s(t) + A)︸︷︷︸>0

cos 2πfct| (C.328)

= (s(t) + A) | cos 2πfct| (C.329)

=2

π(s(t) + A)

(1− 2

∞∑n=1

(−1)n

(2n− 1)(2n+ 1)cos 2π2nfct

)(C.330)

=2

π

(A+ s(t) + 2A

∞∑n=1

(−1)n+1 cos 2π2nfct

4n2 − 1+ 2s(t)

∞∑n=1

(−1)n+1 cos 2π2nfct

4n2 − 1

)(C.331)

By Fourier transformation yields

MB(f) = F {|m(t)|} (C.332)

=2

π

(Aδ(f) + S(f) + A

∞∑n=1

(−1)n+1 δ(f + 2nfc) + δ(f − 2nfc)

4n2 − 1(C.333)

+S(f) ∗∞∑n=1

(−1)n+1 δ(f + 2nfc) + δ(f − 2nfc)

4n2 − 1

)(C.334)

=2

π

(Aδ(f) + S(f) + A

∞∑n=1

(−1)n+1 δ(f + 2nfc) + δ(f − 2nfc)

4n2 − 1(C.335)

+∞∑n=1

(−1)n+1S(f + 2nfc) + S(f − 2nfc)

4n2 − 1

)(C.336)

The amplitude spectrum of the rectified signal is can be visualized from Fig. C.41.

|MB(f)|8aπ

f0−20f1 20f1

∼ 1n2

· · ·

Figure C.41: Amplitude spectrum |MB(f)| of the rectified signal |m(t)|

22. (a) We get the following signal by mixing

g(t) = m(t) cos 2πfM t (C.337)


138 C EXCERCISES

= (A1 + s1(t)) cos 2πfct cos 2πfM t+ (A2 + s2(t)) cos 2πfc,St cos 2πfM t (C.338)

Using addition theorem

cosα cos β =1

2(cos(α− β) + cos(α + β)) (C.339)

originates

g(t) =A1 + s1(t)

2(cos(2π(fc − fM)t) + cos(2π(fc + fM)t))

+A2 + s2(t)

2(cos(2π(fc,S − fM)t) + cos(2π(fc,S + fM)t)) (C.340)

With the stipulations y1(t) = A1 + s1(t) and y2(t) = A2 + s2(t), as well as Y1(f) =F {y1(t)} and Y2(f) = F {y2(t)} gives the spectrum G(f) at the output of the mixeras

G(f) =1

4(Y1(f + (fc − fM)) + Y1(f − (fc − fM)) + Y1(f + (fc + fM)) + Y1(f − (fc + fM))

+ Y2(f + (fc,S − fM)) + Y2(f − (fc,S − fM)) + Y2(f + (fc,S + fM)) + Y2(f − (fc,S + fM)))(C.341)

Fig. C.42 shows exemplarily the spectrum G(f) at the output of the mixer, where onlyused-signal channel 1 is shown for simplification. The influence of the interference-signalchannel 2 will be considered later with image frequency fc,S.

|G(f)|

f−(fc − fM ) fc fM fc + fM(fc − fM )−fc−fM−(fc + fM )

IF-BPIF-BP

Figure C.42: Receipt spectrum |G(f)| at the mixer output with fM > fc, fZF = fM − fc whileconsidering the simplified form y2(t) = A2 + s2(t) = 0

Thus one receives two different possible intermediate frequencies fZF , of which one liesbelow and one above the carrier frequency fc.

fZF,1 = |fM + fc| (C.342)

fZF,2 = |fM − fc| (C.343)

Naturally one selects fZF < fc. For a selected intermediate frequency fZF also twodifferent heterodyne frequencies are possible

fM,1 = |fc + fZF | (C.344)

fM,2 = |fc − fZF | (C.345)

As will be shown later that the heterodyne frequency fM is selected mostly above thecarrier frequency fM > fc.


C.2 Solutions 139

(b) The antenna is usually interpreted as highly broadband and receives therefore (in-tended) the entire MW range. At the mixer input therefore not only the utilizablechannel (here indicated as s1(t)), but also all adjacent channels would rest. If the socalled image frequency fc,S lies with same power for such a interference channel (heremarked as s2(t)), both channels at the output of the mixer would be superimposed.

In order to clarify this effect, the spectrum at the output of the mixer (eqn. C.341)will be considered. A superposition of the spectra of Y1(·) and Y2(·) occurs if due to

|G(f)|

ffZF fc fM fc,S−fZF−fc−fM−fc,S

Figure C.43: Receipt spectrum |G(f)| at the Mixer output with fM > fc, fZF = fM − fc < fcand interference channel with image frequency fc,S = 2fM − fc, both channelsare superimposed at fZF and are not separable any more

the symmetrical spectrum one of the two possible intermediate frequency of signal 1corresponds with one of the two possible intermediate frequency of signal 2.

fZF,Signal 1 = fZF,Signal 2 (C.346)

From eqn. (C.342, C.343)

fZF = |fM ± fc| = |fM ± fc,S| (C.347)

Here a distinction of cases with fM , fZF , fc, fc,S > 0 and fc �= fc,S must be accomplished.It becomes evident that the frequency pair fM + fZF and |fM − fZF | are always formedfrom the carrier and the corresponding image frequency.

fc,S = 2fM − fc for fM > fZF (C.348)

fc,S = 2fM + fc for fM < fZF (C.349)

From Fig . C.43 it follows that the unwanted superposition and distortions comesfrom intermediate frequencies. This disturbance can be avoided, if a bandpass filter isattached to the mixer around fc.

(c) Hence the image frequency fc,S lies always outside of the utilizable region, so that thefrequency pair must belong together

fM + fZF and |fM − fZF | (C.350)

with the difference

|fc − fc,S| = 2fZF for fM > fZF (C.351)

|fc − fc,S| = 2fM for fM < fZF (C.352)

be larger than the utilizable region fc,max − fc,min. Since a down-mixture always takenplace in the receiver, the interesting case is fM > fZF and therewith

fZF >fc,max − fc,min

2(C.353)


140 C EXCERCISES

MW: 0.5 MHz ≤ fc ≤ 1.5 MHz ⇒ fZF > 0.5 MHzUKW: 88 MHz ≤ fc ≤ 108 MHz ⇒ fZF > 10 MHz

(d) MW: fZF = 0.5 MHz (in practice mostly fZF = 0.455 MHz is selected – Standardfilter)

fM,min = fc,min − fZF = 0 (C.354)

fM,max = fc,max − fZF = 1 MHz (C.355)

or

fM,min = fc,min + fZF = 1 MHz (C.356)

fM,max = fc,max + fZF = 2 MHz (C.357)

Here only the adjustable region fM ∈ [1 MHz, 2 MHz] is important.

USW: fZF = 10 MHz (in practice mostly fZF = 10.7 MHz is selected – Standard filter)

78 MHz ≤ fM ≤ 98 MHz (C.358)

98 MHz ≤ fM ≤ 118 MHz (C.359)

Here both adjustable regions are realizable. As the ration of variation fM,max/fM,min

should be kept as small as possible, usually

fM > fc (C.360)

is selected.

(e) The bandpass filter (around fc) prior to the mixer must suppress the image frequencysignals sufficiently, meanwhile bandpass filter (around fZF ) following the mixer accom-plishes the actual channel splitting. So

fg = 5 kHz ⇒ B = 2fg = 10 kHz (C.361)

The ratio of bandwidth and center frequency of the bandpass filter determines itsnecessary quality. The smaller the ratio the better is the quality. Therefore as small anintermediate frequency as possible is aimed at in order to keep the quality of necessaryfilter small.

23. From h(t) H(f) we get

h(nT ) =

∫ ∞

−∞H(f) ej 2πnTf d f (C.362)

=∞∑

k=−∞

∫ 12T

− 12T

H

(f +

k

T

)ej 2πnT(f+

kT ) d f (C.363)

=

∫ 12T

− 12T

∞∑k=−∞

H

(f +

k

T

)︸︷︷︸

Hp(f)

ej 2πnTf ej 2πnTkT︸︷︷︸

≡1

d f (C.364)


C.2 Solutions 141

=

∫ 12T

− 12T

Hp(f) ej 2πnTf d f (C.365)

As Hp(f) is a periodic function of 1/T , it can be expanded in a Fourier series

Hp(f) =∞∑

m=−∞cm ej 2πmTf (C.366)

cm = T

∫ 12T

− 12T

Hp(f) e− j 2πmTf d f (C.367)

From eqn. (C.365) follows

cm =

{Th(0) for m = 0

0 otherwise(C.368)

And hence (see Fig. C.44)

Hp(f) =∞∑

n=−∞H

(f − n

T

)= Th(0) = const. (C.369)

Such an impulse h(t) after eqn. (C.69) s designated also as Nyquist-1-pulse. A filter havingimpulse response which satisfy this condition is called Interpolator filter.

H(f − n

T

)Th(0)

12T

0− 12T

f

Hp(f)

Figure C.44: 1. Nyquist condition in frequency domain

24. (a) see Fig. C.45

s(t)A0

−A0

−2T −T T 2T 3T 4T 5T 6T 7T 8T 9T t

×

× ×

×

×

× ×

×{0} {1} {1} {0} {1} {0} {0} {1}

Figure C.45: Transmitted signal s(t)


142 C EXCERCISES

d{k} → b{k} → b(t)

DAC

hT (t)s(t)

Transmitting filterChannel

r(t)hR(t)

rE(t)

Receiving filter

kT

ADC

rE{k}Detector d{k}

Figure C.46: Block diagram of Data transmission system

(b) see Fig. C.46

(c) As r(t) = s(t) is valid, it is possible also to write

rE(t) = sE(t) = r(t) ∗ hR(t) = s(t) ∗ hR(t) (C.370)

=

( 7∑k=0

b{k}T hT (t− kT )

)∗ hR(t) (C.371)

=7∑

k=0

b{k}T h(t− kT ) (C.372)

with

h(t) = hT (t) ∗ hR(t) (C.373)

=A0

Ttriang

(t

T

)(C.374)

gives

rE(t) = sE(t) = A0

7∑k=0

b{k} triang(t/T − k) (C.375)

See sketch in Fig. C.47

rE(t) = sE(t)A0

−A0

−2T −T T 2T 3T 4T 5T 6T 7T 8T 9T t

×

× ×

×

×

× ×

×{0} {1} {1} {0} {1} {0} {0} {1}

Figure C.47: Filtered receipt signal rE(t) = sE(t) = s(t) ∗ hR(t)

(d)

PsE =1

N

N−1∑k=0

s2E{k} (C.376)

= A20 (C.377)


C.2 Solutions 143

d{k} → b{k} → b(t)

DAC

hT (t)s(t)

Transmitting filterChannel

+

n(t)

r(t)hR(t)

rE(t)

Receiving filter

kT

ADC

rE{k}Detector d{k}

Figure C.48: Block diagram of Data transmission system

(e) see Fig. C.48

(f)

SnE ,nE(f) = |HR(f)|2 Sn,n(f) (C.378)

= |F {hR(t)} |2N0/2 (C.379)

=

∣∣∣∣F{1

Trect

(t

T

)}∣∣∣∣2 N0

2(C.380)

=N0

2si2(πfT ) (C.381)

(g)

PnE=

∫ ∞

−∞SnE ,nE

(f) d f (C.382)

=N0

2

∫ ∞

−∞si2(πfT ) d f (C.383)

by substituting a = πT and with the help of eqn. (C.83)

PnE=N0

2

π

πT=N0

2T(C.384)

(h)

SNR = 10 lgPuseful-signal

Pnoise

dB = 10 lgPsE

PnE

dB (C.385)

= 10 lg2TA2

0

N0

dB (C.386)

25. (a) see Fig. C.49

(b) see Fig. C.49

(c) In BPSK the imaginary axis is the separation function between the two decision aresd{k} = d0 and d{k} = d1, only the sign of real part of the receipt value rE(kTs) =Re {rE(kTs)} is needed for the decision. The transmitter phase diagram is rotated by thenon-ideal phase synchronization in reference to that of the receiver. Thus the resultingamplitude of the information signal is reduced with unchanged noise amplitude at thereceiver input (see Fig. C.50).

s′0 = Re{s0 e

jϕ}= s0 cos

π

4(C.387)


144 C EXCERCISES

Im {rE}

Re {rE}••

d{k} = d0d{k} = d1

s0s1

Im {rE}

Re {rE}•• •

•

•••••••• •

•••

•••

••••••

••••

•

d{k} = d0d{k} = d1

s0s1

frE |d0(rE|d0)frE |d1(rE|d1)1

s1 s0 rE

P (d{k} = d0|d{k} = d1)

frE |d0(rE|d0)frE |d1(rE|d1)1√2πσ

s1 s0 rE

Figure C.49: Phase diagram at the detector input (above) and amplitude density functionfrE (rE) with rE = Re

{rE

}(below) for ideal noise free transmission (left) and

for additive noise (right)

=1√2s0 (C.388)

s′1 =1√2s1 (C.389)

This corresponds to a decrease of the signal-to-noise ratio by factor 2

Im {rE}

Re {rE}

•••

••

•••• ••

••

••

••

••••

••

•••

• •••

••

••

••••

•

•

•••

•••• •

• •••

s0s′0s1 s′1

ϕ

Figure C.50: Phase diagram with a phase offset ϕ = π4


C.2 Solutions 145

SNR′ =s′0

2

σ2=

s202σ2

(C.390)

=SNR

2(C.391)

or about 3 dB reduction in logarithmic scale (see Fig. C.51).

SNR′ = SNR− 10 lg 2 dB ≈ SNR− 3 dB (C.392)

−15 −10 −5 0 5 10

BER

100

10−1

10−2

10−3

10−4

10 lg(SNR) dB

3 dB

Figure C.51: BER with a phase offset of ϕ = π4 (Ideal curve for ϕ = 0 is shown as dotted.)

(d) From (see eqn. (8.48))

BER =1

2erfc

(s0√2σ

)(C.393)

anderfc(x) = 1− erf(x) (C.394)

follows

BER =1

2− 1

2erf

(s0√2σ

)(C.395)

s0 =√2σ erf−1

(2

(1

2−BER

))(C.396)

=√2 0.5V erf−1(2(0.5− 0.1)) (C.397)

= 0.6408V = 640.8mV (C.398)

An MATLAB example 20.

% Communications: Digital Modulation -- BPSK

% Solution with MATLAB lsg_bpsk.m

%

% Default standard deviation

20For further studies see Proakis, J. G. and Masoud S.: Contemporary Communications Systems Using MAT-

LAB, Boston: PWS 1998


146 C EXCERCISES

% in V

sigma = 0.5 ;

% Default bit error rate

BER = 0.1 ;

% inverse error function

s_0 = sqrt(2)*sigma*erfinv(2*(0.5-BER)); % in V

%

% graphical presentation

%

% Default range

r_E = -2:0.05:2;

% Amplitude density function

f = 1/(sqrt(2*pi)*sigma)*exp(-(r_E-s_0).^2/(2*sigma^2));

% Draw

plot(r_E,f);


147

D Exams

D.1 Exam SS 2006

Problem 1: Signal Theory and LTI Systems

Note: Question a) ,b) and c) can be solved independently from each other.

(a) (7 Points) A LTI system is characterized by the following equation

h(t) =

N2∑n=N1

(n+ 1) δ(t− nT ) , N1 < N2

(a1) What is h(t) and its Fourier transform H(f) usually refered to?

(a2) Calculate H(f) !

(a3) Sketch h(t) for N1 = −3 and N2 = 4 !

(a4) The system is excited with an arbitrary signal s(t). Calculate the output signal y(t)depending on s(t) !

(a5) Now N1 = 0 and N2 = 1. Sketch the output signal y(t) fors(t) = 1

2Trect( t

2T) !

(Note: It is helpful sketching s(t) first.)

(b) (3 points) The figure below shows the block diagram of a radio transmission chain.

Sender vTL RL

RPTP1SNR 2SNR

1 0.1NP pW= 2 90NP pW=

Demodulation

AL

The channel attenuation LA is assumed to

LA = 100dB + 20 log10(d/[km])dB

where d stands for the distance between transmitter and receiver.

(b1) Calculate the receive power level LR in dBm with d = 10km and a transmit powerPT = 1W?

(b2) How large can be the distance d chosen maximally, when the transmit power PT isdegraded to 10mW and a receive power level LR of −90dBm is needed?

(c) (5 points) The following considerations on the SNR are related to the same transmissionchain as used in part b).


148 D EXAMS

(c1) Let the SNR in front of the amplifier be SNR1 = 20 dB. Calculate the needed amplifi-cation v (in dB), to ensure an SNR before the demodulation of at least SNR2 ≥ 10 dB.

(c2) The SNR before the amplifier(SNR1) is now 20 dB. Calculate the theoretically max-imum value of SNR2 after the amplifier! Give reasons for your answer! (Note: Theamplifier can be adjusted between −100 dB and +100 dB.)

Notes:

log(a · b) = log(a) + log(b) ; log(a/b) = log(a)− log(b); log(ab) = b log(a)

x 0.1 0.25 1 2 5 10 100 1000

log10(x) −1 −0.6 0 0.3 0.7 1 2 3

Question 2: Analog-digital conversion

(a) (5 Punkte) Consider the following spectrum S(f) of a real-valued, analog signal s(t):

Lowpass

This signal is now sampled with sampling frequency fs = 6 kHz and then reconstructed witha low pass filter.

(a1) Draw the spectrum Sa(f) of the signal after sampling in the range −15 kHz ≤ f ≤15 kHz! Note: Make sure you label the axes of the diagram correctly!

Now assume that the signal s(t) is disturbed through a real-valued, analog 11 kHz cosinesignal.

(a2) Why is it now impossible to reconstruct the original signal s(t) from sa(t)?

(a3) Draw the spectrum of the signal s(t) after the low pass filter in the range −3 kHz ≤f ≤ 3 kHz.

(a4) Due to technical reasons, the sampling frequency can be chosen within the range4 kHz ≤ fs ≤ 10 kHz. How should fs be chosen, so that s(t) yields a perfect re-construction of s(t)?

(a5) How should the above sampling circuit be extended, to allow a perfect reconstructionof s(t) out of sa(t) while keeping the sampling frequency fixed to fs = 6 kHz?

Now the time-discrete signal x[k] is supposed to be quantized with 3 bit. For this, a linear mid-risequantizer is used. The input range of the quantizer is defined as −4�s ≤ xin ≤ 4�s, where �s isthe step size of the quantizer.

(b) (5 Punkte)


D.1 Exam SS 2006 149

(b1) Draw the output signal xout as a function of the input signal xin of the quantizer! Makesure the axes of the diagram are labelled correctly.

(b2) The SNR shall now be increased by 6 dB. The following parameters can be adjusted:

- Number of quantization bits b.

- Sampling frequency fs.

- Cut-off frequency of the reconstruction low pass fr.

Decide for each stated parameter, whether it is suitable for the desired SNR improve-ment, and to which value the parameter would have to be adjusted! Please justify youranswer.

Question 3: Analog modulation

(a) (5 points) A signal corresponding to

x(t) = cos(2πf1t) +1

2cos(2πf2t)

with f2 = 2f1, is transmitted via double-side band amplitude modulation (DSB-AM)

s(t) = (A+ x(t)) cos(2πfct)

where A = 4 and fc = 10f1.

(a1) Calculate the modulation index μ of the above modulation system.

(a2) Draw the spectrum S(f) of s(t).

An envelope demodulator (see diagram below) is the simplest way to demodulate a double-side band amplitude modulated signal

s(t) x(t)

Rectifier Lowpass Highpass

s(t) x(t)

Rectifier Lowpass Highpass

(a3) Within which range should the modulation index μ be chosen, to enable a correctenvelope demodulation?

(a4) Explain the function of the high pass filter in the envelope demodulator!


150 D EXAMS

2cos(2πfgt) 2cos(2πfct)

|f |< fgLowpass

2sin(2πfgt) 2sin(2πfct)

|f |< fgLowpass

fc>> fg

x1(t)

x(t) s(t)

x2(t)2cos(2πfgt) 2cos(2πfct)

|f |< fgLowpass

2sin(2πfgt) 2sin(2πfct)

|f |< fgLowpass

fc>> fg

x1(t)

x(t) s(t)

x2(t)

-fg fg

X(f)

f

1

S(f)2

-fc fc f-fc+fg fc -fg-fg fg

X(f)

f

1

S(f)2

-fc fc f-fc+fg fc -fg

(b) (6 points) The following figure shows a transmit system for single-side band amplitudemodulation (SSB-AM) with a suppressed carrier. Advantages of single-side band modulationcompared to double-side band modulation are that only half of the transmission bandwidthis required, and that the suppression of the carrier leads to a more efficient system. Below,the frequency domain representations of the input signal, X(f), and the output signal, S(f),are given.

(b1) Draw a block diagram of a suitable receiver structure (including quantities such asfrequencies etc.), which can obtain x(t) out of s(t) again.

(b2) Draw the spectra X1(f) andX2(f)

jof the signals x1(t) and x2(t).

Note: cos(2πfct)◦–•12[δ(f − fc) + δ(f + fc)] sin(2πfct)◦–•−j

2[δ(f − fc)− δ(f + fc)]

Question 4: Digital modulation

(a) (6 points) Consider the time-domain representation of the following two digitally modulatedsignals x1(t) and x2(t). In both cases, all possible symbols are shown.


D.1 Exam SS 2006 151

0 10 20 30 40 50 60 70 80

−5

0

5

t in ms

x 1(t)

0 10 20 30 40 50 60 70 80−2

−1

0

1

2

t in ms

x 2(t)

(a1) Which digital modulation schemes were used?

(a2) Draw the constellation diagram (phase diagram) of the modulation scheme 4-ASK.

(a3) Which data rate (in bit/s) can be achieved if 4-ASK with a symbol length of 0.5ms isemployed?

(a4) Which modulation scheme could be used to achieve twice the data rate of 4-ASK whilemaintaining the same symbol length?

(a5) We assume that the transmitted signal is disturbed through additive white Gaussiannoise, so that some symbols will be detected wrongly. Find a suitable mapping of databits to the constellation points in the 4-ASK constellation diagram that minimizes thebit error rate.

(b) (8 points) We now observe a transmitter that uses the modulation scheme 4-QAM. Groupsof two data bits b(2k) ∈ {0, 1}, b(2k + 1) ∈ {0, 1} are mapped onto time-discrete symbolssI(k), sQ(k) and then transformed into time-continuous information signals mI(t), mQ(t).

1

Transmitter

Channel

n(t)

-sin(2πfct)

mQ(t) xQ(t)

UpconversionMapper

b(2k)

sQ(k)

Puls shaper

y(t)

cos(2πfct)mI(t) xI(t)0 sI(k)

x(t)22−

10

22−b(2k+1)

Ts

Ts

1

sT

1

sT

(b1) Draw a brief sketch of a suitable receiver for this modulation scheme and name thechosen components! Assume here that there is no phase shift between the oscillators ofthe transmitter and receiver.

QAM-systems are often described by their equivalent, complex-valued baseband signalss(k) = sI(k) + j · sQ(k). We assume here that we have the signal

r(k) = s(k) + n(k)

at the input of the detector, where n(k) is complex Gaussian noise with one-dimensionalvariance σ2.


152 D EXAMS

(b2) Draw the constellation diagram (phase diagram) of the input signal r(k) of the detectorfor σ2 = 0 with the corresponding bit values b(2k), b(2k + 1). Add suitable decisionborders such that the bit error rate is minimized, assuming that the data bits areequally probable, i.e. ∀l : P (b(l) = 0) = P (b(l) = 1) = 1

2.

The bit error rate (BER) of such a 4-QAM transmission system can be calculated throughthe following equation, where Es = E{|s(k)|2} is the average signal power of the transmittedsymbols s(k), and the complementary error function erfc(x) is given in the following figure.

BER =1

2erfc

(√Es

4σ2

)

0 0.5 1 1.5 210−3

10−2

10−1

100Complementary error function

x

erfc

(x)

(b3) Calculate the average signal power Es of the symbols s(k).

(b4) Calculate the bit error rate (BER) of the 4-QAM-system for σ2 = 1.

(b5) The symbol error rate (SER) of a 4-QAM-system corresponds to the probability thatone of the two bits (or both) has been detected wrongly. Derive an equation that allowsto calculate the symbol error rate for a given bit error rate, and then calculate thesymbol error rate for BER = 0.1.


D.2 Solutions: Exam SS 2006 153

D.2 Solutions: Exam SS 2006

Question 1: Signal theory and LTI Systems

(a) (a1) (1 point)h(t) Impulsantwort (impulse response)H(f) Ubertragungsfunktion (transfer function)

(a2) (1 point)

F{δ(t− nT )} = e−j2πfnT

H(f) = F{h(t)} =

N2∑n=N1

(n+ 1)F{δ(t− nT )} =

N2∑n=N1

(n+ 1)e−j2πfnT

(a3) See figure belowqualitatively correct: (1 point)quantitatively correct: (1 point)

-1-2

4321

-3

h(t)

t/T

5

4

3

2

-1

-3

-2

(a4) (1 point)

y(t) =

∫ +∞

−∞s(t− τ)

N2∑n=N1

(n+ 1)δ(τ − nT )dτ

=

N2∑n=N1

(n+ 1)

∫ +∞

−∞s(t− τ)δ(τ − nT )dτ

=

N2∑n=N1

(n+ 1)s(t− nT )

Use shifting property of the Dirac impulse

(a5) y(t) is obtained through the superposition of two scaled and time-shifted rectangularimpulses of width 2T and height 1

T


154 D EXAMS

-T T 2T t

y(t)

32T

1T

12T

qualitatively correct: (1 point)quantitatively correct: (1 point)

(b) (b1) (1 point)

LR = LT − LA

LR = 10log10(PT

mW)dBm− 100dB − 20log10(

d

km)dB

LR = 30dBm− 100dB − 20dB = −90dBm

(b2)

LR = −90dBm = 10log10(PT

mW)dBm− 100dB − 20log10(

d

km)dB

−90dBm = 10log10(10)dBm− 100dB − 20log10(d

km)dB

−90dBm = 10dBm− 100dB − 20log10(d

km)dB

20log10(d

km)dB = 0dB

d = 100km = 1km

approach correct (1 point)result correct (1 point)if an alternative approach is chosen, points are distributed accordingly

(c) (c1) (3 points)

SNR1 = 20dB = 10log10(PR

PN1

)dB

PR = 100PN1

SNR2 = 10dB = 10log10(vlinearPR

vlinearPN1 + PN2

)dB

10 =vlinearPR

vlinearPN1 + PN2

vlinear =10PN2

PR − 10PN1

=10PN2

100PN1 − 10PN1



vlinear =10 ∗ 90 ∗ 10−12W

90 ∗ 10−13W= 100

vdB = 10log10(vlinear)dB = 20dB

1 point for the correct calculation of the noise before demodulation: PN = vlinearPN1 +PN2

1 point for the correct calculation of PR = 100PN1 = 10pW1 point for the result vdB = 20dBif another approach is chosen or the noise term is not mentioned explicitly, but theresult is correct, both points are obtained

(c2) (2 point)

SNR2 = 10log10(vlinearPR

vlinearPN1 + PN2

)dB = 10log10(PR

PN1 +PN2

vlinear

)dB

SNR2 is maximized, if the denominator is minimized; the denominator is minimized,if the amplification is maximized

vdB = +100dB

vlinear = 1010 → PN2

vlinear≈ 0

SNR2,max = 10log10(PR

PN1

) = SNR1 = 20dB

1 point for result and 1 point for calculation or explanation

Question 2: Analog-digital conversion

(a) (a1) Diagram qualitatively and quantitatively correct (1 point)

(a2) Due to the sampling we observe the effect of aliasing. (1 point)

(a3) Diagramm qualitatively and quantitatively correct (1 point)


156 D EXAMS

(a4) The sampling frequency must be chosen in the range 7 kHz ≤ fs ≤ 8 kHz.Any choice within this range valid. (1 point)

(a5) Anti-aliasing low pass before sampling with the cut-off frequency 2 kHz < |fg| ≤ fs2(1

point)

(b) (b1) See figure (1 point qualitative, 1 point quantitative)

outx7

2sΔ

inx

5

2sΔ

3

2sΔ

2

sΔ

7

2sΔ−

5

2sΔ−

3

2sΔ−

s−Δ3 s− Δ 2 s− Δ4 s− Δ sΔ 3 sΔ2 sΔ 4 sΔ

(b2) Following parameters are suitable for increasing the SNR by 6 dB:

- Increase number of bits b by 1See calculation of the SINR as a function of the number or quantization bits b, asderived in lecture. (1 point)

- Increase sampling frequency fs by a factor of 4The quantization noise power remains constant and can be assumed as distributedover the frequency range −fs

2< f ≤ fs

2. Thus, increasing fs by a factor of 4

reduces the quantization noise density by a factor of 4, corresponding to a 6dBSNR improvement. (1 point)

- The cut-off frequency of the reconstruction low pass is not a suitable parameter,as the in-band noise power cannot be reduced by filtering, without also affectingthe desired signal. (1 point)

Question 3: Analog modulation

(a) (a1) Maximum value of |x(t)|:

x(t) = cos(2πf1t) +1

2

[2 cos2(2πf1t)− 1

]=

[cos(2πf1t) +

1

2

]2

− 3

4

where −1 ≤ cos(2πf1t) ≤ 1 → max[|x(t)|] = 32



Modulation index μ: (1 point)

μ =max[|x(t)|]

A

=3

8

(a2) Spectrum according to following figure.qualitatively correct (1 point)quantitatively correct (1 point)

10f1

S(f)

f

2

11/21/4

11f19f1 12f18f1-10f1 -9f1-11f1 -8f1-12f1 10f1

S(f)

f

2

11/21/4

11f19f1 12f18f1-10f1 -9f1-11f1 -8f1-12f1

(a3) (0 <)μ < 1 (1 point)

(a4) The high pass filter removes the DC component from the carrier signal. (1 point)

(b) (b1) Receiver according to following diagram.qualitatively correct: Down converter and low pass filter (each 1 point)quantitatively correct (1 point)

cos(2π(fc-fg)t)

|f |< fgLowpass x(t)s(t)

cos(2π(fc-fg)t)

|f |< fgLowpass x(t)s(t)

(b2) Spectra according to following diagram.qualitatively correct (each 1 point)quantitatively correct (1 point)


158 D EXAMS

X1(f)

f

1

fc -fg fc fc +fg-fc -fg -fc -fc +fg

X1(f)

f

1

fc -fg fc fc +fg-fc -fg -fc -fc +fg

X2(f) / j

f

1

-fg-2fg

fg 2fg

-1

X2(f) / j

f

1

-fg-2fg

fg 2fg

-1

Problem 4: Digital Modulation

(a) (a1) x1(t): 8-ASK (1 point), x2(t): 2-FSK (1 point) (2 points)

(a2) 4-ASK has 4 symbol amplitudes either on the I branch or on the Q branch possible,e.g.

I

Q

(1 Punkt)

(a3) 4-ASK allows 2bit/symbol, i.e. R = 2 · 2000 = 4000bit/s (1 point)

(a4) correct are 16-QAM, 16-ASK, 16-PSK, 16-FSK etc. (1 answer is enough) (1 point)

(a5) neighbouring symbols only divers in 1 bit, e.g.

00 01 11 10

I

Q

(1 point)

(b) (b1) sketch should be similar to the figure below:

Kanal

n(t)

Empfänger

-2sin(2πfct) kTs

( )0

sT

dt•∫

AbwärtsmischerIntegrierer

Entscheider

Abtaster

b(l+1)^

y(t)

2cos(2πfct) kTs

( )0

sT

dt•∫b(l)^

x(t)

Komponenten vorhanden (1 point)order and names correct (1 point)



(b2) constellation diagram:

Entscheidungsgrenzen

2

2−

22−

1101

00 10

I

Q

constellation correct: (1 point)decision borders correct: (1 point)

(b3) Es = (√2)2 + (

√2)2 = 4 (1 point)

(b4) BER = 12erfc(1) ≈ 0.07 (A range of 0.06 ... 0.08 is accepted) (1 point)

(b5) Ps = 1− (1− Pb)2 (1 point)

Ps = 1− (0.9)2 = 0.19 (A range of 0.18 ... 0.2 is accepted) (1 point)


160 D EXAMS

D.3 Exam WS 2006/2007


Note: Question a), b) and c) can be solved independently from each other. Please note the hintsat the end of this problem on the next page (back side).

(a) (3 Points) A System is characterized by the following relation between the input signal x(t)and output signal y(t):

y(t) = [x(t) + x(t+ 2)]2 + 3

(a1) Is the system linear?

(a2) Is the system time invariant?

(a3) The system is not causal. How could the above relationship be modified in order toensure causality?

(b) (4 Points) The figure below shows the block diagram of a radio transmission link between amobile and a base station.

Mobile vTL RL

RPTP

131 10NP W−=

Demodulation

AL

The channel attenuation LA (in dB) is assumed to be given by

LA = 100dB + 10α log10(d/km)dB

,

where d stands for the distance between transmitter and receiver and α denotes the pathlosscoefficient, which depends strongly on the characteristics of the environment.

(b1) The mobile typically has a transmit power of 1W. Calculate the received power in Wat a base station located at a distance of 3 km! (α = 4)

(b2) Calculate the signal-to-noise ratio (SNR) at the input of the amplifier!

(b3) To ensure reliable transmission, the SNR at the output of the amplifier must be SNR ≥0 dB. Calculate the maximum allowable distance between the mobile and the basestation!

Note: The pathloss coefficient is now α = 3!

(c) (8 Points) A filter is characterized by the following impulse response:

h(t) = a1δ(t) + a2δ(t− T )


D.3 Exam WS 2006/2007 161

(c1) Draw h(t) for a1 = 1 and a2 = 0.5 !

Note: Make sure you label the axes of the diagram correctly!

(c2) Calculate |H(f)| and draw it in the range − 1T< f < 1

Tfor a1 = a2 = 1 !


(c3) Now, only the frequencies between − 12T

< f < 12T

are of interest. What is the char-acteristic of the filter in this frequency range (high pass, low pass, band pass, bandstop)?

(c4) The filter is excited with one half-wave of the sine function:

x(t) =

⎧⎨⎩ sin(

π

2

t

T) if 0 < t < 2T

0 else

Draw x(t) !


(c5) Sketch the output signal for a1 = 1 und a2 = 0.5 !

Hints:cos2x = 1

2(1 + cos(2x)) ;

log(a · b) = log(a) + log(b) ; log(a/b) = log(a)− log(b);log(ab) = b log(a)

x 0.1 0.25 1 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.3 0.5 0.7 1 2 3

Problem 2: Analog-Digital Conversion

(a) (5 Points) The real-valued, analog signal s(t) with the spectrum S(f) is sampled and after-wards reconstructed.

S(f)

ffg- fg

(a1) Sketch the arrangement needed for the distortion-free sampling and reconstruction andname all components!

(a2) What is the minimal sampling frequency fs,min which ensures a distortion-free recon-struction of the signal s(t)?

(a3) Which effect arises when the sampling frequency is chosen to be fs < fs,min, and hencea distortion-free reconstruction of s(t) is not possible any more?

(a4) Sketch the spectrum of the signal s(t), sampled with a sampling frequency fs < fs,min!Mark the effect described in question (a3) in the diagram!


162 D EXAMS

outx

inx

3

2sΔ

2

sΔ

3

2sΔ−

s−Δ2 s− Δ sΔ 2 sΔ

a) Characteristics of the 4 stepmid-rise quantizer

b) probability density functionof the amplitude of xin(k)

(b) (5 Punkte) The discrete-time signal xin[k] is now quantized by a linear, 4 step mid-risequantizer with the following characteristics.

(b1) What is the signal-to-noise ratio (SNR) at the output of the quantizer, when the am-plitude of the input signal xin[k] is uniformly distributed with xmax = 2Δs (see figureabove)!

Note: Equation is sufficient!

(b2) Give one advantage and one disadvantage of the linear mid-rise quantizer!

Now xin[k] is still uniformly distributed between [−xmax, xmax] (see figure above), but nowxmax � Δs.

(b3) Calculate the signal power Pout = E{x2out[k]} of the signal xout[k] at the output of thequantizer, where E{.} denotes the statistical expectation!

Note: Draw first the probability density function of the amplitude of the output signalxout[k]!

Problem 3: Analog Modulation

(a) (6 Points) A transmitter based on amplitude modulation is transmitting the band pass signal

sBP (t) = (1 + μx(t)) · cos(2πfct) ,

where x(t) is the actual, real-valued information signal to be transmitted (with the property∀t, |μx(t)| < 1) and fc is the carrier frequency. The spectrum of x(t) is shown in the followingplot:

f

|X(f)|

fg-fg

1 fg < fc


D.3 Exam WS 2006/2007 163

(a1) Derive an equation for the spectrum SBP (f) of the generated band pass signal, andplot the absolute value of this spectrum! Make sure the axes are labelled correctly!

(a2) Now draw a block diagram of an envelope demodulator, which is able to obtain thesignal x(t) from the signal sBP (t)! Please label all components and briefly state theirfunction!

(a3) In our example, a DC component is added to the information signal before upconversion.Why is this necessary or helpful, if

• an envelope demodulator

• a synchronous demodulator

is used, respectively?

(b) (5 Points) We assume that the signal sBP (t) from question (a) is transmitted without any at-tenuation, and the receiver uses the following synchronous demodulator. As you can see, thereceived signal is downconverted with g(t) = cos(2πfct+φ), i.e. the oscillators of transmitterand receiver have a carrier phase offset of φ.

Ideale Band pass(surpression of

adjacent channels) g(t)=cos(2� fct+�)

Ideale low pass|f| � fg

Ideale high pass|f| > 0

sBP(t)x(t)

(b1) Derive the signal x(t) at the output of the synchronous demodulator! Here, the relationcos(a)cos(b) = 1/2(cos(a− b) + cos(a+ b)) might be useful.

(b2) A measure for the quality of transmission is the signal-to-noise-ratio (SNR) after trans-mission and demodulation. We assume that the signal x(t) has an average power ofEs = E{|x(t)|2}, we have a perfect transmission without signal attenuation, and mea-sure an additive white noise with power En at the output of the demodulator. Derivethe SNR at the output of the demodulator as a function of φ!

(b3) We now assume that the synchronous demodulator downconverts the received signalwith g(t) = cos(2π(fc+Δf)t), i.e. transmitter and receiver use slightly different carrierfrequencies. What is the consequence for the signal x(t)? (Short comments sufficient;please name at least two consequences)


(a) (5 Points) A digital transmission system has a symbol duration TS = 1ms, where the datarate is 3000 bits/s.

(a1) How many bits are represented by one symbol?

(a2) How many symbols contains the symbol alphabet? Give a modulation scheme whichfullfills this requirement!


164 D EXAMS

(a3) Now the data rate shall be doubled while keeping the symbol duration constant. Howmany bits must be represented by one symbol now? Give an appropriate modulationscheme!

(b) (7 Points)

(b1) Sketch the constellation diagram of the modulation scheme 8-PSK!

(b2) Write in the diagram of question (b1) a possible mapping of the bits to the constellationpoints (symbols)!

Now we consider the modulation scheme 4-PSK. To enable the use of a given receiver, thereal axis should be the decision threshold for the first bit and the imaginary axis for thesecond bit. The symbols have a magnitude of

√2.

(b3) Draw a constellation diagram which fullfills these requirements! Write in the mappingof the bit pairs to the symbols.

If the transmission is corrupted by additive Gaussian noise with noise power 2σ2, the biterror rate Pb (BER) in such a QPSK transmission system can be calculated by the followingequation:

Pb =1

2erfc

(1√2σ2

).

The complementary Gaussian error function erfc(x) is depicted in the figure below.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)


D.3 Exam WS 2006/2007 165

(b3) Calculate Pb for σ2 = 1/8 !

(b4) Calculate σ2 for Pb = 1/4! Give also the derivations!

(c) (2 Points) The following three constellation diagrams of binary digital modulations are given:

1-1 1

1

0 2I

Q

I

Q

I

Q1) 2) 3)

(c1) Which contellation is best suitable for the data transmission in presence of additivegaussian noise, when the bit error rate shall be kept as low as possible?

Decide for exactly one answer, and give reasons for your decision! Apply additionalcriteria if necessary.


166 D EXAMS

D.4 Solutions: Exam WS 2006/2007

Problem 1: Signal Theory und LTI Systeme

(a) (a1) (1 Point)nonlinear

(a2) (1 Point)time invariant

(a3) (1 Point)function could be shifted by 2 in positive temporal direction:y(t) = [x(t− 2) + x(t)]2 + 3

(b) (b1) (1 Point)

LT = 10log10PT

mWdBm = 30dBm

LR = LT − LA = 30dBm− 100dB − 40log103dB

LR = 30dBm− 100dB − 20dB = −90dBm

PR = 10−9mW = 10−12W

(b2) (1 Point)

LN = 10log10PN

mWdBm = −100dBm

SNR = LR − LN = −90dBm− (−100dBm)

SNR = 10dB

(b3) (2 Point)The SNR is not influenced by the amplifier: SNRin = SNRout

SNR = LR − LN = 0dB

LR = LN = −100dBm

LR = −100dBm = LT − LA = 30dBm− 100dB − 30log10d

kmdB

30dBm = 30log10d

kmdB

d = 10km

1 Point if SNRin = SNRout

1 Point for correct distance d

(c) (c1) (1 Point)


D.4 Solutions: Exam WS 2006/2007 167

1 2

h(t)

t/T

1

0.5

-1

qualitative and quantitative correct: 1 Point

(c2) (3 Point)first H(f) and |H(f)| must be calculated:

H(f) = 1 + e−j2πfT

|H(f)| =√(1 + cos(2πfT ))2 + sin2(2πfT )

|H(f)| =√1 + 2cos(2πfT ) + cos2(2πfT ) + sin2(2πfT )

|H(f)| =√2 + 2cos(2πfT )

|H(f)| =

√4(1

2(1 + cos(2πfT )))

|H(f)| = 2|cos(πfT )|

f

|H(f)|

2

1

12T

1T

- 12T

- 1T

1 Point for |H(f)|1 Point: Sketch qualitatively correct1 Point: Sketch quantitatively correct

(c3) (1 Point)Low pass

(c4) (1 Point)


168 D EXAMS

t/T

x(t)

1

1/2

1 2-1

(c5) (2 Point)The output signal is the superposition of 2 sine half-waves, where the second half-waveis shifted by T and scaled by 1/2.

t/T

y(t)

1

1/2

1 2-1 3 4

2 half-waves correct: 1 Pointsuperposition correct: 1 Point


(a) (a1) Arrangement (with or w/o anti-aliasing filter) correct (1 Point)Names of components correct (1 Point)

LP

Sampling Reconstructionlow pass

LP

Anti-Aliasing

Filter

(a2) fs,min > 2 · fg (1 Point)

(a3) Aliasing (1 Point)

(a4) Diagram qualitatively correct and spectral overlapping due to aliasing correctly marked(1 Point)



Sa(f)

ffg- fg- fs- 2fs fs 2fs

Aliasing Aliasing

... ...

(b) (b1) Number of quantisation bits b = 2SNR ≈ 6 dB b = 12 dB (1 Point)

(b2) Advantage: symmetric characteristic (1 Point)Disadvantage: No quantization step for xout[k] = 0 (1 Point)→ Noise at output of quantizer even if xin[k] = 0

(b3) Amplitude distribution of xout[k] see figure (1 Point)

Pout =12

(Δs2

)2+ 1

2

(− Δs2

)2= Δs2

4(1 Point)


(a) (a1) SBP (f) =12δ(f − fc) +

12δ(f + fc) +

μ2X(f − fc) +

μ2X(f + fc)

f

|SBP(f)|

-fc-fg

0.50.5μ

-fc+fg fc-fg fc+fgfc-fcEquation correct (1 Point)Sketch and descriptions correct (1 Point)

(a2) Magnitude demodulator should be sketched as follows:

Band pass(Surpression of

adjacent channels)

Rectifier(Reconstruction of

envelope)

Low pass(Cancellation

of carrier)

High pass(Cancellation of

DC offset)

constellation correct (bandpass optional): (1 Point)Description and explanation of components correct: (1 Point)

(a3) The DC offset servers in the magnitude demodulator to avoid phase leaps in the bandpass signal due to zero-crossings of the base band signal. (1 Point)


170 D EXAMS

The DC offset serves in synchron demodulators for detection of the carrier signal.(1 Point)

(b) (b1) The signal at the output of the demodulator can be calculated by:

x(t) = (1 + μx(t)) · cos(2πfct) · cos(2πfct+ φ)|0<|f |≤fg

= 12(1 + μx(t)) · (cos(4πfct+ φ) + cosφ)

∣∣0<|f |≤fg

= 12(1 + μx(t)) · cosφ∣∣

0<|f |= μ·cosφ

2x(t)

Derivation correct: (1 Point)Result correct: (1 Point)

(b2) SNR = E{|x|2}En

= μ2·cos2φ·Es

4·En(1 Point)

(b3) Each correct answer one point, maximum 2 points total: (max. 2 Points)

• The spectrum of the information signal x(t) is shifted in comparison to x(t).

• The DC offset in the base band signal can not be surpressed.

• High frequency components of the information signal are surpressed partly by fil-tering with |f | ≤ fg.


(a) (a1) 2 Bits/Symbol (1 Point)

(a2) 22 = 4 Symbols (1 Point)Example: 4-PSK (1 Point)

(a3) 4 Bits/Symbol (1 Point)Example: 16-FSK (1 Point)

(b) (b1) Sketch can be like this:

I

Q



Qualitatively correct (1 Point)

(b2) Every mapping of all possible binary triples is correct. (1 Point)

(b3) Drawing can be like this:

I

Q

1

1-1

-1

10 11

00 01

Qualitatively correct (1 Point)Quantitatively correct (1 Point)Bit mapping correct (1 Point)

(b4) Pb ≈ 2,3 · 10−3, are considered as correct 2 · 10−3 ≤ Pb ≤ 3 · 10−3 (1 Point)

(b5) For Pb = 1/4:

1

2= erfc

(1√2σ2

)1

2≈ 1√

2σ2

4 ≈ 2σ2

σ2 ≈ 2

Derivation and solution: (1 Point)

(c) Two answers are possible:

The constellation 1) is best suited, because it has the symbol distance 2 and a lower meansignal power than constellation 3).

The constellation 3) is best suited, because it has the symbol distance 2 and it can be usedwith simpler transmitter and receiver circuits 3).

Solution: (1 Point)Explanation: (1 Point)


172 D EXAMS

D.5 Exam SS 2007


Note: Questions a), b) and c) can be solved independently from each other. Please note the hintsat the end of this problem on the next page (back side).

(a) (2 Points) A system is characterized by the following relation between input signal x(t) andoutput signal y(t):

y(t) = |x(t+ 1)|.

Which of the following properties has the system:

• linear,

• causal,

• time invariant?

(b) (6 Points) The figure below shows the impulse response h(t) of a LTI-system.

h(t)

t

1

1-1 2-2

(b1) Is the system causal?

(b2) Express h(t) using the triangular function triang(t)!

(b3) Determine the transfer function H(f) of this LTI-system!

(b4) The system is excited by the signal x(t) = δ(t) + δ(t − 3). Draw the output signal ofthe LTI system! Make sure you label the axes of the diagram correctly!

Notes: triang(t) =

{1− |x| fur −1 < x < 1

0 fur sonstF{triang(t)} = si2(πf)

(c) (7 Points) The signal arriving at a radio receiver with the receiving level LR = −60dBmhas to be amplified before demodulation. Considering the 2 available amplifiers, which ofthem are connected in an amplifier chain? For each amplification a certain amount of noiseis added to the signal by the amplifier itself. The resulting block diagram can be found inthe figure below.


D.5 Exam SS 2007 173

Demodulation

10LNAv dB=

LNAv

10, 9 *10N LNAP W−=

30PAv dB=

PAv

7, 10N PAP W−=

60RL dBm= − PL

Power AmplifierLow Noise Amplifier

PP

The 2 amplifiers differentiate showed in the picture in their gain and in the noise powerthey cause. One of them is a low noise amplifier (LNA) with a low noise power of PN,LNA =9 · 10−10W and a low gain of vLNA = 10dB. The other one is a power amplifier (PA) with astrong noise power of PN,PA = 10−7W and high gain of vPA = 30dB.

(c1) How large is the signal-to-noise ratio (SNR) before the power amplifier?

(c2) The received signal must have a power of at least PP = 0.1mW before demodulation.Is this with the given amplifiers possible? If not, to which value has the gain of thepower amplifier vPA (in dB) to be changed?

(c3) How large is the SNR before demodulation for a power amplifier gain of vPA = 30dB?

(c4) The SNR before demodulation has to become maximum. For this reason the order ofthe amplifiers in the amplifier chain can be exchanged. Which oder is the best one?Give Reasons! For this, look at the SNR before demdodulation in case that the poweramplifier is located in front of the low noise amplifier in the amplifier chain!

Notes:log(a · b) = log(a) + log(b) ; log(a/b) = log(a)− log(b);log(ab) = b log(a)

x 0.1 0.25 1 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.3 0.48 0.7 1 2 3


(a) (2 Points) Digital signals can be attained from analog signals by means of analog-digital-conversion. Specify the difference between digital signals and analog signals!

(b) (4 Points) Consider the analog signal s(t) with the frequency spectrum S(f) shown in figureD.1. During an analog-digital conversion s(t) is perfectly sampled with sampling frequencyfA = 25 kHz. The signal resulting from this sampling is denoted by sA(t). Assume that thereis no anti-aliasingfilter employed.

(b3) Draw the frequency spectrum SA(f) of the signal sA(t) in the range of−60 kHz ≤ f ≤ 60 kHz! Pay attention to correct axis labeling!

(b4) Can the analog signal s(t) be recovered from the signal sA(t) without any error? Givereasons and refer to the given sampling frequency fA!


174 D EXAMS

��

�

�

� ��

Figure D.1: Frequency spectrum S(f)

��

�

�

� ��

Figure D.2: Frequency spectrum N(f)

(b5) Now, there is an additive noise signal n(t), that interferes the analog signal s(t) beforethe sampling. The frequency spectrum N(f) of the additive noise signal n(t) is shownin Figure D.2. What is the minimum sampling frequency fA,min that can be allowed, ifwe just want to recover the signal s(t) without any error using a low-pass filter afterthe sampling?

(c) (4 Points) Consider the quantizer of an analog-digital converter with the quantization char-acteristic shown in figure D.3. x(t) denotes the input signal and xq(t) denotes the outputsignal of the quantizer. The quantization step size is Δx.

��

��

!

"#

$ "#

%!

"#

&'

()

"* #

%!

"+ #",#$ ", #

Figure D.3: Quantization characteristic


D.5 Exam SS 2007 175

(c1) Is it a mid-rise or mid-tread quantizer? Give reasons!

(c2) Draw the absolute value of the quantization error signal |e(t)| as a function of the inputsignal x(t) in the range of −2Δx ≤ x(t) ≤ 2Δx! Pay attention to correct axis labeling!

(c3) Assume the input signal of the quantizer to be x(t) = 3 cos(2πft). Furthermore, as-sume Δx = 1.5. Determine the signal-to-noise-ratio SNR resulting at the output of thequantizer in this case!


(a) (7 points) A transmitter which is based on amplitude modulation generates the band passsignal

sBP (t) = (1 + μx(t)) · cos(2πfct)where x(t) is the actual, real-valued information signal to be transmitted, and fc is thecarrier frequency. The frequency spectrum X(f) of the information signal x(t) is given asfollows:

f

X(f)

fg-fg

1 fg << fc

(a1) Compute the frequency spectrum SBP (f) of the band pass signal and plot this spectrum!Please make sure that the axes are correctly assigned!

(a2) We now assume that the information signal is x(t) = sin(2πfit). How large can thefactor μ be maximally chosen, if an envelope demodulator is supposed to be used atthe receiver side? Please state briefly, why an appropriate choice of μ is important inthis case.

(a3) A disadvantage of this kind of amplitude modulation is that only a portion of thetransmitted power is actually connected to the information signal x(t). Please compute- assuming that x(t) = sin(2πfit) and μ = 0.5 - the ratio of the power of the scaledinformation signal μx(t) over the power of the entire band pass signal sBP (t)!

(b) (4 points) We now assume that the signal sBP (t) from problem (a) is transmitted error-free,and the following demodulator is used at the receiver side:

Idealer Bandpaß(Nachbarkanal-unterdrückung)

Idealer Tiefpaß|f| � fg

Idealer Hochpaß|f| > 0

Idealer Tiefpaß|f| � fg

Idealer Hochpaß|f| > 0

Phasen-schätzung

ˆ ( )Ix t

ˆ ( )Qx t

ϕϕ ϕ≈

( )cos 2 cf tπ ϕ+

( )sin 2 cf tπ ϕ− +

( )BPs t


176 D EXAMS

(b1) Compute the signals xI(t) and xQ(t) at the output of the demodulator as a function ofϕ (Note: cos(a)cos(b) = 1/2(cos(a− b) + cos(a+ b)) and cos(a)sin(b) = 1/2(−sin(a−b) + sin(a+ b))).

(b2) We now assume that ϕ can be estimated at the receiver side through ϕ ≈ ϕ. Pleaseextend the demodulator architecture in such a way that the original information signalx(t) can be reconstructed from xI(t), xQ(t) and ϕ as accurately as possible (Note:∀ϕ : cos2ϕ + sin2ϕ = 1). Please ensure that all used components and signals areadequately labelled.


(a) (6 Points)

The following two diagrams show two digital modulated band-pass signal sBP,1(t), sBP,2(t)in time domain. In both cases, all possible symbols are shown.

0 1 2 3 4 5 6 7−4

−2

0

2

4

t/T

s BP

,1(t)

Figure D.4: sBP,1(t)

0 1 2 3 4 5 6 7−1

−0.5

0

0.5

1

t/T

s BP

,2(t)

Figure D.5: sBP,2(t)

(a1) Which digital modulation schemes were used? Give the names and their order.

(a2) Draw the I/Q diagram (constellation diagram) of the modulation scheme of sBP,1(t)!

(a3) Sketch the impulse response of the pulse shape filter, used for the modulation of sBP,1(t)!

(a4) Which data rate R1 (in bit/s) can be achieved with the modulation scheme of sBP,1(t)when a symbol length T = 2ms is employed?

(a5) For doubling the data rate (R3 = 2R1), without changing the modulation scheme, thesymbol length T must be adapted. How T must be changed to achieve the doubleddata rate?

(b) (6 Points)

In figure D.6 and D.7 the block diagram of a QPSK-Modulators is given with the correspond-ing symbol mapping. The impulse former may have the impulse response φ(t) = rect

(tT− 1

2

).

The data stream d[k] = [‘01‘, ‘10‘, ‘11‘, ‘11‘] is transmitted.


D.5 Exam SS 2007 177

Sym

bol-

map

per

Figure D.6: QPSK-Modulator

d[k] a[k] = I[k] + j Q[k]

‘00‘ 1

‘01‘ j

‘11‘ −1

‘10‘ −jFigure D.7: Symbolmapper

(b1) Draw the signals I(t) and Q(t) in the time range 0 ≤ t < 4T !Note: Pay attention to complete axes labels!

(b2) Draw the resulting bandpass signal sBP (t) in the time range 0 ≤ t < 4T for the givendata stream, where the carrier frequency is given by fc · T = 2!Note: Pay attention to complete axes labels!

(b3) Draw the I/Q-diagram (constellation diagram) of the used modulation scheme and alsothe decision thresholds!

(c) (2 Points)

The bit error rate (BER) of such a QPSK transmission can be computed through the fol-lowing Equation:

BER =1

2erfc

(√ Es

4σ2

)where Es = E

[|S[k]|2] denotes the average signal power of transmitted symbols S[k] andσ2 denotes the variance of the channel noise. The complementary error function erfc(x) isgiven in the following figure.

(c1) Compute the bit error rate (BER) with an average symbol energy Es = 1 and a noisevariance σ2 = 1

16!

(c2) How will change the bit error rate (BER) of the QPSK modulation scheme if there isa is a unknown phase offset between the transmit and receive oscillator? Give reasons!


178 D EXAMS

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)





(a) The system is:non-linear (computation of magnitude is a nonlinear operation)non-causal (e.g. for computation of y(0), knowledge of x(1) is necessary)time invariant (time shift of x(t) by τ results in a time shift of y(t) by τ , too)

2 properties correct: (1 point)all 3 properties correct: (2 points)

(b) (b1) (1 point)non causal (in the negative time h(t) has values which are not zero)

(b2) (1 point)Superposition of 3 time shifted triangles or one large triangle minus one small triangleh(t) = triang(t) + triang(t− 1) + triang(t+ 1) = 2triang(t/2)− triang(t)

(b3) Use of linearity and time shifting of fourier transformation:

H(f) = si2(πf) + si2(πf)e−j2πf + si2(πf)ej2πf = si2(πf)(1 + 2cos(2πf))

or

H(f) = 4si2(2πf)− si2(πf)

The 2 solutions are equivalent as it can be seen below:

4si2(2πf)− si2(πf) = 4sin2(2πf)

4π2f 2− sin2(πf)

π2f 2=

(2sin(πf)cos(πf))2 − sin2(πf)

π2f 2

=sin2(πf)

(πf)2(4cos2(πf)− 1) =

sin2(πf)

(πf)2(4 · 1

2(1 + cos(2πf))− 1)

=sin2(πf)

(πf)2(2 + 2cos(2πf)− 1) = si2(πf)(1 + 2cos(2πf))

H(f) is the fourier transfomred of h(t) (1 point)result correct (1 point)

(b4) 2 shifted trapezoids superimpose to resulting signal

y(t)

t

1

1-1 2-2 3 4 5

Partial trapezoids quantitative and qualitative correct: (1 point)Complete signal quantitative and qualitative correct: (1 point)


180 D EXAMS

(c) (c1) (1 point)

LN,LNA = 10lg(9 ∗ 10−10W

10−3W)dBm = (10lg(9) + 10lg(10−7))dBm

LN,LNA = (9.6− 70)dBm = −60.4dBm

SNR = LP + vLNA − LN,LNA = −60dBm+ 10dB − (−60.4dBm)

SNR = 10.4dB

(c2) (2 points)

LP = 10lg(10−4

10−3)dBm = LR + vLNA + vPA

vPA = LP − LR − vLNA = −10dBm− (−60dBm)− 10dB = 40dB

The given amplifiers are not enough. The power amplifier has to be enhanced to a gainof 40dB.

(c3) (2 points)Power level of received signal in front of demodulation:Lp = LR + vLNA + vPA = −60 dBm + 10dB + 30dB = −20dBmPower level of first noise signal nLNA(t) in front of demodulation:

Ln,1 = 10 lg(

PN,LNA

1mW

)+ vPA = (4.8 + 4.8− 70)dBm+ 30dB = −30.4dBm Signal power

of first noise signal in front of demodulation:Pn,1 = 10Ln,1/10mW = 9 · 10−7WTotal signal power of the noise signals nLNA(t) and nPA(t) before demodulation:Pn,ges = Pn,1 + Pn,PA = 10−6WSNR before demodulation:SNR2 = Lp − 10 · lg

(Pn,ges

1mW

)= 10dB

(c4) (2 points)

SNR2 = 10lg(vLNA ∗ vPA ∗ PR

vLNA ∗ PN,PA + PLNA

)dB

SNR2 = 10lg(10−5W

10−6W + 9 ∗ 10−10W)dB

Compared to SNR1 from c3) here the term 9 · 10−10 is added to the denominator.Thus, the SNR drops under the 10dB from c3). Therefore, it is better to use low noiseamplifier first and then power amplifier in the amplifier chain.


(a) Analog signals: continous in time and valueDigital signals: discrete in time and value (1 Point)

(b) (b1) Spectrum SA(f) as shown in figure below6Diagram qualitatively correct - curve (1 Point)Diagram quantitatively correct - axis labeling (1 Point)



��

�

�

� ��

. . .. . .

(b2) Yes, s(t) can be recovered from sA(t) without any error.Reasons: The sampling freqeuncy fA = 25 kHz is greater than twice the upper criticalfrequency fG,s = 10 kHz of s(t), i.e. fA > 2 fG,s. Consequently, there is no aliasing. (1Point)

(b3) The frequency spectra S(f) and N(f) do not overlap. After sampling the resultingfrequency spectra SA(f) and NA(f) must not overlap, too. Aliasing within NA(f) isirrelevant (see figure).

��

�

�

� �� ! ! � �! �� ! �� ! "� "! #� #! !� !! $� �! "� "! #� #! !� !! $�

. . .. . .

If fG,s and fG,n denote the upper critical frequencies of s(t) and n(t), respectively, than:

fA ≥ fG,s + fG,n = 10 kHz + 20 kHz

Hence, the minimum sampling frequency that can be allowed is fA,min = 30 kHz. (1Point)

(c) (c1) It is a mid-rise quantizer.Reasons: xq(t) ∈

{±Δx2· (1, 3, . . .)}, i.e. there is no quantization step xq(t) = 0 in the

quantization characteristic. (1 Punkt)

(c2) |e(t)| = |xq(t)− x(t)| as a function of x(t) as shown in figure belowDiagram qualitatively correct - curve (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

%&'()%

*'()+ ,-

./

,-

,-,0-+ ,0 -

(c3) With 2Δx = 3 and x(t) = 3 cos(2πft) the full range of 4 quantization steps is used,which corresponds to a resolution of b = 2Bits. For a sine signal the signal-to-noise-ratioat the output of the quantizer is given by:

SNR = (1.76 + 6.02 b/Bits) dB

Subsequently: SNR = 13.8 dB. (1 Point)


182 D EXAMS


(a) (a1) SBP (f) =12δ(f − fc) +

12δ(f + fc) +

μ2X(f − fc) +

μ2X(f + fc)

f

|SBP(f)|

-fc-fg

0.50.5μ

-fc+fg fc-fg fc+fgfc-fcEquation correct: (1 point)Plot qualitive correct: (1 point)Plot quantitive correct: (1 point)

(a2) ∀t : 1 + μx(t) > 0 ⇒ μ < max|x(t)| ⇒ μ < 1 (1 point)One of the following aspects is mentioned: (1 point)

• Information signal may only be contained in the envelope of sBP (t)

• Phase jumps in sBP (t) must be avoided

(a3) Power of the information signal in the band pass is si,BP (t) = μx(t) cos(2πfct):Pi,BP = μ2 Px

12= 0.52 · 1

2· 12= 1

16

Total power of the band pass signal sBP (t):Ps,BP = μ2 Px

12+ 1

2= 9

16

Power ratio:Pi,BP

Ps,BP= 1

9

Correct approach: (1 point)Correct result: (1 point)

(b) (b1) The signals at the output of the demodulator can be calculated as

xI(t) = (1 + μx(t)) · cos(2πfct) · cos(2πfct+ ϕ)|0<|f |≤fg

= 12(1 + μx(t)) · (cos(4πfct+ φ) + cosϕ)

∣∣0<|f |≤fg

= 12(1 + μx(t)) · cosϕ∣∣

0<|f | =μ·cosϕ

2x(t)

xQ(t) = (1 + μx(t)) · cos(2πfct) · (-sin(2πfct+ ϕ))|0<|f |≤fg

= 12(1 + μx(t)) · (−sin(4πfct+ ϕ)− sin(−ϕ))∣∣

0<|f |≤fg

= 12(1 + μx(t)) · sinϕ∣∣

0<|f | =μ·sinϕ

2x(t)


(b2) The following solution is possible:



ˆ ( )Ix t

ˆ ( )Qx t

ˆ( )x t

( )2 sin ϕμ

( )2 cos ϕμ

Correct principle: (1 point)Correct annotation: (1 point)


(a) (a1) left: 8-ASK (1 point)rigth: 4-FSK (1 point)

(a2) I/Q-Diagram

Q[k]

I[k]× × × × × × × ×

qualitive und quantitive correct (1 point)

(a3) Impulse response of the impulse former φ(t) = rect(

tT− 1

2

)(1 point)

(a4) 8-ASK: Order M = 8 → ld(M) = 3 Bits/SymbolR1 = ld(M)

[Bits

Symbol

] · 1T

[Symbol

s

]= 1500Bits/s (1 point)

(a5) The symbol length must be divided in half. See problem a4)! (1 point)

(b) (b1) Diagrams of I(t) and Q(t):

1 2 3 4 1 2 3 4

1

-1

1

-1


184 D EXAMS

every diagram qualitive and quantitive correct (1 point)

(b2) Diagram qualitive correct (1 point)quantitve correct (1 point)

0 1 2 3−1

−0.5

0

0.5

1

t/T

s BP(t)

(b3) I/Q-DiagramQ[k]

I[k]×

×

×

×

Decisionthresholds

qualitive and quantitive correct (1 point)decision thresholds correct (1 point)

(c) (c1) BER = 12erfc(2)

erfc(2) = 4 · 10−3...6 · 10−3

BER = 2 · 10−3...3 · 10−3 (1 point)

(c2) The bit error rate increases.Reason:The phase offset between the transmit and receive oscillator leads to a rotation ofthe I/Q-diagram in the receiver with unknown angle. By this the constellation pointsapproximate to the decision thresholds or even cross them. Additionally due to thechannel noise the sampled values in the receiver vary around the constellation points,leading to so called noise clouds. Therefore the decreasing distance of the constellationpoints to the decision thresholds results in more wrong decisions of the decider andthus the bit error rate increases. (1 point)


D.7 Exam WS 2007/2008 185

D.7 Exam WS 2007/2008


Note: Questions a), b) and c) can be solved independently from each other. Please note the hintsat the end of this problem on the back side.

(a) (5 Points) The first digital cellular network in Germany was built in 1992. The maximaltransmit power of the mobiles is PTx = 2W. The path loss LA is defined by:

LA = 130 dB + 30 log10

( d

km

)dB .

(a1) Compute the maximal transmit power level in dBm of the mobiles!

(a2) The mobile transmits to the base station with a constraint of having a minimal re-ceived power level of LRx = −106 dbm at the antenna of the base station. What is themaximum allowed distance dmax between the mobile and the base station?

The received signal form the mobile at the base station is composed by the attenuatedtransmitted signal and the thermal noise, which is added by the receiver front-end. Thenoise power level is Ln = −107 dBm (See figure below). Furthermore, in order to havecorrect demodulation of the received signal, the signal to noise ratio at the front-end of thedemodulator must have a minimum SNR of 10 dB.

Modulator Demodulator

Base stationMobile

NoiseLA

PTx PRx

(a3) What is the maximum distance which can be allowed between the mobile and the basestation?

(b) (4 Points) A LTI system with the impulse response h(t) and the transfer function H(f) isgiven as shown in the figure below:

LTIx(t) y(t)

(b1) Give the relation between the input signal x(t) and the output signal y(t) in timedomain and between the related signal spectrum X(f) and Y (f) in frequency domain?

(b2) An amplifier has the relation between the input signal x(t) and the output signal y(t):y(t) = x(t)− 0.2x2(t)− 0.01x3(t). Decide whether the amplifier is a LTI system! Givereasons!

(b3) A LTI-System have the impulse response h(t) as shown in the figure below:


186 D EXAMS

−3 −2 −1 0 1 2 3−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t [s]h(

t)

Decide and give reasons whether the LTI system can be realized in practise or how h(t)must be modified, in order to implement the LTI system in practise!

(c) (6 Punkte) Two low pass filters, TP1 and TP2, are given with the transfer functions H1(f)and H2(f), respectively:

H1(f) = rect( f

B1

), H2(f) = rect

( f

B2

)where B1 = 10 kHz and B2 = 20 kHz.

(c1) Draw the transfer function H1(f) of the low pass filter TP1 within the range −30 kHz <f < 30 kHz! Make sure you label the axes of the diagram correctly!

(c2) Give the Fourier integral for the calculation of the impulse response h1(t) of the lowpass filter TP1!

(c3) Calculate the real- and imaginary part of h1(t) by solving the Fourier integral!

Now, a band pass filter should be constructed, having the lower cut frequency fg,u = 5kHzand the upper cut frequency fg,o = 10 kHz. For the construction of this band pass filterthe two described low pass filters, TP1 and TP2, and adding and subtracting elements areavailable.

(c4) Sketch a suitable construction of the band pass filter, using the described filters andelements.

Notes:log(a · b) = log(a) + log(b) ; log(a/b) = log(a)− log(b);log(ab) = b log(a)

x 0.1 0.25 1 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.3 0.48 0.7 1 2 3


(a) (2 points) Analog signals can be translated to digital signals by means of analog-digitalconversion. This enables digital signal processing. Give two advantages of digital signalprocessing over analog signal processing!


D.7 Exam WS 2007/2008 187

(b) (4 points) Analog-digital converters with very high sampling rate typically have only lowquantization resolution. The analog-digital converter depicted in the following picture hasthe sampling rate fA = 1/TA and a quantization resolution of b = 1bit.

Analog-Digital Converter

Sampling Quantization( )s ta ( )s td( )x t

The sampling is carried out perfectly at time instances t = nTA with n ∈ Z and canbe expressed by a Dirac pulse train. The quantization characteristic of the analog-digitalconverter is given by

q(x(t)) =

{1 for x(t) ≥ 0

0 for x(t) < 0,

where x(t) denotes the continuous-valued signal before quantization.

(b1) Give the mathematical expression including the sampling rate fA and the quantizationcharacteristic q(·) that describes the relation between the input signal sa(t) and theoutput signal sd(t) of the analog-digital converter!

(b2) Assume that the input signal of the analog-digital converter is sa(t) = cos(2π fA2t)! Draw

the related output signal sd(t) in the range of 0 ≤ t ≤ 8TA! Pay attention to the correctaxis labeling!

(c) (4 points) The analog low-pass signal sTP(t) has a cut-off frequency of fG = 10 kHz andvalues in the range of −10 ≤ sTP(t) ≤ 10. To digitize sTP(t), an analog-digital converter isrequired.

(c1) What is the minimum sampling rate fA,min that the analog-digital converter has toprovide to enable analog-digital conversion of sTP(t) without aliasing?

(c2) How can be avoided, that arbitrary interfering signals with frequencies f > 10 kHz leadto aliasing errors, when sTP(t) is analog-digital converted using the minimum samplingrate fA,min?

(c3) Determine the quantization step size Δs, when the analog-digital converter employs alinear mid-rise quantizer with a resolution of b = 3bit and a dynamic input range thatis equal to the range of values of sTP(t)!

(c4) What is the signal-to-noise ratio SNR that results from the analog-digital conversion ofsTP(t) due to the quantization resolution of b = 3bit? Assume sTP(t) to have a uniformamplitude density over its whole range of values!


188 D EXAMS

Problem 3: Analoge Modulation

(a) (7 Points) A transmitter based on amplitude modulation transmits the band pass signal

sBP (t) = (1 + μx(t)) · sin(2πfct),

where x(t) is the real-valued information signal to be transmitted, and fc denotes the carrierfrequency. Please note that the information signal is here upconverted with asinus - as opposed to the approach stated in the lecture!. The spectrum X(f) of theinformation signal x(t) is shown here - separated into its real and imaginary component:

ffg-fg

1fg << fc

0,5

Re{X(f)}

ffg-fg

10,5

Im{X(f)}

(a1) Please derive the spectrum SBP (f) of the band pass signal as a function of arbitraryX(f).

(a2) Draw the spectrum of SBP (f) for given X(f) and μ = 1! Please make sure that the axesare appropriately annotated and real and imaginary signal components are separated!

(a3) Assume the information signal is now x(t) = 12cos(2πfit). How large can μ be maximally

chosen, if an envelope demodulator is supposed to be used at the receiver side? Brieflyargue why an appropriate choice of μ is important for this kind of receiver.

(a4) Please state an advantage and a disadvantage of this modulation scheme.

(b) (4 Points) We now assume that the band pass signal sBP (t) = xI(t) · cos(2πfct) − xQ(t) ·sin(2πfct) is received at the antenna of the following demodulator:

Ideal band pass(Suppression of

neighboring channels)

cos(2�fct+�)Ideal low pass

|f| � fgsBP(t)

-sin(2�fct+�)Ideal low pass

|f| � fg

ˆ ( )Ix t

ˆ ( )Qx t

(b1) Derive the signals xI(t) and xQ(t) at the output of the demodulator as a function of ϕand arbitrary real-valued information signals xI(t) and xQ(t). Note:

cos(a)cos(b) = 1/2(cos(a+ b) + cos(a− b))

sin(a)sin(b) = 1/2(cos(a− b)− cos(a+ b))

cos(a)sin(b) = 1/2(sin(a+ b)− sin(a− b))

(b2) If the receiver knows the phase shift ϕ, the following extension to the demodulatorcan be used to reconstruct the information signal xI(t). Please draw a sketch of acorresponding extension of the demodulator that can reconstruct xQ(t). Make sureyour sketch is appropriately annotated.


D.7 Exam WS 2007/2008 189

ˆ ( )Ix t

ˆ ( )Qx t

( )Ix t

( )2sin ϕ−

( )2cos ϕ


(a) (3 Points)

(a1) Why is digital modulation necessary?

(a2) Sketch the block diagram of a digital transmission line beginning at the send bit seriesup to the received, decoded bit series!

(b) (6 Punkte)

(b1) Sketch one constellation-(I/Q)-diagram for each of the following modulation schemes:8-Phase Shift Keying (PSK) and On-Off Keying (OOK)!

(b2) Draw the decision thresholds into the I/Q-diagrams of problem b1)!

(b3) The symbol duration of both modulation schemes (OOK and 8-PSK) shall be 2ms.How large is the achievable data rate in bit/s at each modulation scheme?

(c) (5 Punkte) A bit series {bk}, k ∈ N, bk ∈ {0, 1} shall be transmitted across a channel withadditive white Gaussian noise (AWGN-channel). The bandpass signal s(t), which is sendafter digital modulation, is superimposed by the noise n(t). As modulation scheme binaryphase shift keying (BPSK) is used. The symbols are given by sk = 1 − 2bk. A rectangularimpulse shaping in applied.

(c1) The bit series bk = {0110} , k = 0, 1, 2, 3 shall be transmitted. Draw the correspondingsend bandpass signal s(t) in the interval from 0 to 4T! The bandpass signal must havethe following parameters: signal amplitude A, symbol duration T, carrier frequencyfc =

2T. Pay attention to completely labeled axis!

(c2) At the receiver the send bandpass signal is superimposed by noise. Calculate the biterror rate of the received signal! At the receiver there is an SNR of 6dB and all sendsymbols have the same probability. (Also be aware of the hints at the end of thisproblem.)

(c3) A bit error rate of 1.5 · 10−3 must be achieved. Calculate the necessary SNR (in dB!)at the receiver entrance!

(c4) The channel is no simple AWGN-channel anymore. Additional to the noise every symbolexperiences independently from each other a random phase shift. How large is the biterror rate now with and without noise?

Hint: If the transmission is corrupted by additive Gaussian noise, the bit error rate Pb

(BER) in such a BPSK transmission system can be calculated by the following equation:

BER =1

2erfc

(√SNRlinear

2

)


190 D EXAMS

The complementary Gaussian error function erfc(x) is given in the following draw.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)





(a) Power level calculation

(a1) Power level of the transmitted signal

LTx = 10 log10

( PTx

1mW

)dBm

= 10 log10

(2 · 1000mW

1mW

)dBm

= (3 + 30) dBm = 33 dBm

Result correct (1 Point)

(a2) Maximal distanceMaximal path loss: LA = LTx − LRx = 33, dBm− (−106 dBm) = 139 dBand thus:

LA = 139 dB = 130 dB + 30 log10

( d

km

)dB

9

30dB = log10

( d

km

)dB

100.3 =d

kmdmax = 2km

approach correct (1 Point)result correct (1 Point)

(a3) Required minimal receive power level:

LRx,min = SNR + Ln = 10 dB + (−107 dBm) = −97dBm

Maximal path loss: LA,max = LTx − LRx,min = 33 dBm− (−97 dBm) = 130 dBand thus:

130 dB = 130 dB + 30 log10

( d

km

)dB

0 = log10

( d

km

)d = 1km

approach correct (1 Point)result correct (1 Point)

(b) (b1) LTI System:Time domain: y(t) =

(h ∗ x)(t)

Frequency domain: F{y(t)

}= Y (f) = F{

h(t)} · F{

x(t)}= H(f) ·X(f)

Both equations correct (1 Point)


192 D EXAMS

(b2) The amplifier is not a LTI system, because his characteristic contains quadratic andcubic terms. (1 Point)

(b3) The LTI system can not be implemented in pratice, because h(t) �= 0 for t < 0 and thusis not causal. By shifting the impulse response by t0 ≥ 3 s to the right: h(t) = h(t− t0)the causal impulse response h(t) can be derived.Correct decision (yes/no) (1 Point)Reson and modification (Shifting of h(t) by t0 ≥ 3 s) correct (1 Point)

(c) (c1) See figure below

B1/2-B1/2f

H1(f)1

figure correct (1 Point)

(c2) Inverse Fourier integral:

h1(t) =∞∫

−∞H1(f) e

j2πft d f

Equation correct (1 Point)

(c3)

h1(t) =

∞∫−∞

rect( fB1

)ej2πft d f =

−B12∫

B12

cos(2πft

)+ j sin

(2πft

)d f

Re{h1(t)} =[sin(2πft)

2πt

]B1/2

−B1/2

= 2sin

(2πB1t

2

)2πt

= B1 sinc(B1t) = B1 si(πB1t)

Re{h1(t)} =[− cos(2πft)

2πt

]B1/2

−B1/2= 0

Approach correct (2 Points)Result correct (1 Point)

(c4) Bandpass filter

H1(f)

H2(f)+

-

Constellation correct (1 Point)




(a) Advantages of digital signal processing are (among others): (2 points)- independent of technology and temperature- easier to implement compared to analog signal processing- simple to parameterize and reconfigure- some signal modifications can not be done by analog signal processing- repeatable results- simple signal storage

(b) (b1) Mathematical expression as given in equation below:Sampling using Dirac pulse train correct (1 point)Quantization using quantization characteristic q(·) correct (1 point)

sd(t) = q

(sa(t)

∞∑n=−∞

δ(t/TA − n)

)= q

(sa(t)

∞∑n=−∞

δ(t fA − n)

)

(b2) Signal sd(t) as shown in figure below:Diagram qualitatively correct - curve (1 point)Diagram quantitatively correct - axis labeling (1 point)

sd(t )

1

t /TA210 3 4 5 6 7 8

. . .0

(c) (c1) fA,min > 2 fG = 20 kHz (1 point)

(c2) To avoid aliasing errors caused by interfering signals with frequencies f > 10 kHz, ananti-aliasing filter, i.e. a low-pass filter with cut-off frequency fG = 10 kHz has to beapplied before the analog-digital converter. (1 point)

(c3) The dynamic input range of the linear mid-rise quantizer is symmetric w.r.t 0. Thewidth of the dynamic input range is equal to the width of the range of values of sTP(t),which yields max(sTP(t))−min(sTP(t)) = 20. The total number of quantization stepsis given by q = 2b/bit = 23. Consequently, the quantization steps size results in:

Δs =max(sTP(t))−min(sTP(t))

q=

20

23= 2.5 (1 point)

(c4) The dynamic input range of the quantizer of the analog-digital converter is equal to therange of values of sTP(t). For an input signal with uniform amplitude density over thewhole dynamic input range, the signal-to-noise ratio SNR at the output of the quantizeris given by:

SNR = 6.02 dB · b/bit


194 D EXAMS

Consequently, the signal-to-noise ratio SNR that results from the analog-digital-conversion of sTP(t) due to the quantization resolution of b = 3bit yields:

SNR = 18.06 dB ≈ 18 dB (1 point)


(a) (a1) SBP (f) =12δ(f − fc) +

12δ(f + fc) +

μ2X(f − fc) +

μ2X(f + fc)

f

SBP(f)

-fc-fg

0.25

-fc+fg fc-fg fc+fgfc-fcRe{X(f)}

Im{X(f)} j/2

-j/2

Equation correct: (1 point)Plot qualitive correct: (1 point)Plot quantitive correct: (1 point)

(a2) ∀t : 1 + μx(t) > 0 ⇒ μ < max|x(t)| ⇒ μ < 1 (1 point)One of the following aspects is mentioned: (1 point)

• Information signal may only be contained in the envelope of sBP (t)

• Phase jumps in sBP (t) must be avoided

(a3) Power of the information signal in the band pass is si,BP (t) = μx(t) cos(2πfct):Pi,BP = μ2 Px

12= 0.52 · 1

2· 12= 1

16

Total power of the band pass signal sBP (t):Ps,BP = μ2 Px

12+ 1

2= 9

16

Power ratio:Pi,BP

Ps,BP= 1

9


(b) (b1) The signals at the output of the demodulator can be calculated as

xI(t) = (1 + μx(t)) · cos(2πfct) · cos(2πfct+ ϕ)|0<|f |≤fg

= 12(1 + μx(t)) · (cos(4πfct+ φ) + cosϕ)

∣∣0<|f |≤fg

= 12(1 + μx(t)) · cosϕ∣∣

0<|f | =μ·cosϕ

2x(t)

xQ(t) = (1 + μx(t)) · cos(2πfct) · (-sin(2πfct+ ϕ))|0<|f |≤fg

= 12(1 + μx(t)) · (−sin(4πfct+ ϕ)− sin(−ϕ))∣∣

0<|f |≤fg

= 12(1 + μx(t)) · sinϕ∣∣

0<|f | =μ·sinϕ

2x(t)


(b2) The following solution is possible:



1( )x t

2ˆ ( )x t

ˆ ( )Ix t

( )2sin ϕ−

( )2cos ϕ

1( )x t

2ˆ ( )x t

ˆ ( )Qx t

( )2cos ϕ

( )2sin ϕ

Correct principle: (1 point)Correct annotation: (1 point)


(a) (a1) Digital modulation is necessary in order to transmit digital signals across an analogchannel. (1 Punkt)

(a2)

symbolmapper

impulseshaping

filter

up conversion

channel

down conversion

Rreceiving filter/

Matched filter/

Integrator

samplingdecision maker

symboldemap-

per

{ }sb k

{ }Rb k

at least 3 components correct: (1 point)3 more components correct: (1 point)

(b) (b1)

I

QOOK

I

Q8-PSK

A-A

A

-A

A

per diagram: (1 point)

(b2)


196 D EXAMS

I

QOOK

A

decision thresholds

I

Q8-PSK

A-A

A

-A

per diagram: (1 point)

(b3) Data rate D = bit/symbol * symbol/second8-PSK: 2ms result in 500 symbols per second; 1 symbol transmits 3 bits

D = 3bit/symbol ∗ 500symbols/sD = 1500bit/s

result correct (1 point)

OOK: 2ms result in 500 symbols per second; 1 symbol transmits 1 bit

D = 1bit/symbol ∗ 500symbols/sD = 500bit/s

result correct (1 point)

(c) (c1)

0 T 2T 3T 4T-A

-A/2

0

A/2

A

BPSK modulated bandpass signal

diagram qualitatively correct: 1 pointdiagram quantitatively correct: 1 point

(c2) The logarithmic SNR must be transformed into linear SNR.

SNRdb = 6dB = 3dB + 3dB = 10lg(SNR1) + 10lg(SNR1)



SNRlin = SNR1 · SNR1 = 2 · 2 = 4

BER =1

2erfc

(√SNRlin

2

)

=1

2erfc

(√2)

=1

2erfc (1.4)

In the diagram an x = 1.4 results in erfc = 5 · 10−2

BER =1

2· 5 · 10−2 = 2.5 · 10−2

correct is: 2 · 10−2 ≤ Pb ≤ 3 · 10−2 (1 point)

(c3)

BER = 1.5 · 10−3 =1

2erfc

(√SNRlin

2

)

3 · 10−3 = erfc

(√SNRlin

2

)

In the diagram a erfc(x) = 3 · 10−3 results in x = 2.1√SNRlin

2= 2.1

SNRlin

2= 4.41

SNRlin = 8.82 ≈ 9

SNRdb = 10lg(9)dB ≈ 10dB

(1 point)

(c4) Because of the random phase shift detection at BPSK becomes impossible. In theI/Q-diagram it can be seen independently from noise half of the symbols slip onto theother side of decision threshold. Noise does not play any role anymore. Therefore, BER(with and without noise) is:BER = 0.5 = 50% (1 point)


198 D EXAMS

D.9 Exam SS 2008


Hint: Parts (a), (b) and (c) can be solved independently.

(a) (5 points)

Given the following signal:

x(t) =

N2∑n=N1

(−1)n · n · δ(t− nT ) , N1 < N2

(a1) Draw x(t) for N1 = −3 and N2 = 3! Pay attention to correct axis labeling!

(a2) Consider now the more general case when N1 = −N2, where N2 > 0. Answer thefollowing questions!

i. Can the signal x(t) be a valid impulse response of a causal system?

ii. Is the signal x(t) odd?

iii. Is the spectrum X(f) of the signal x(t) real valued?

(b) (5 points)

Two LTI systems with impulse response h1(t) = rect(t − 1/2) and h2(t) = rect(t) areserially connected, resulting in an overall system with impulse response h(t), as shown inthe following figure.

h1(t) h2(t)

h(t)

x(t) y(t)

(b1) Draw the impulse response h(t) of the overall system! Pay attention to correct axislabeling!

(b2) Determine the transfer function H(f) = F{h(t)}!(b3) Draw the output signal y(t), when the input signal x(t) has the following shape:


D.9 Exam SS 2008 199

x(t)

t

1

2

3

1

2

−1

−2

−3

(c) (5 points)

The picture bellow shows the block diagram of a radio transmission link.

LT LR SNR1 SNR2

vPT

LA

PR

PN1 PN2

Transmitter Demodulator

The power attenuation LA over a distance d can be approximated by the following equation:

LA = 95 dB + 25 log10

( d

km

)dB

The power of the transmitted signal is PT = 1W. The noise powers at the receiver arePN1 = 0.1 pW and PN2 = 90 pW. The amplifier has a gain of v = 20 dB.

(c1) What is the power level LR in dBm for a distance of 4 km?

(c2) Determine the signal-to-noise ratio SNR1 at the input of the amplifier as a function ofd!

(c3) What is the maximum distance between transmitter and receiver if the signal-to-noiseratio SNR2 at the input of the demodulator should be at least 10 dB?

Hints:

x 0.1 0.25 1 1.5 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.18 0.3 0.48 0.7 1 2 3


200 D EXAMS

Problem 2: Analog-to-Digital Conversion

(a) (7 points)

A mono audio signal s(t) shall be digitized, stored and reconstructed later on with the equip-ment given below. Unfortunately, the signal s(t) is superimposed by an interference signali(t) at the microphone. The associated signal spectra S(f) = F{s(t)} and I(f) = F{i(t)}are depicted in the figure below. The sampling frequency is fs = 40 kHz. At first, quantiza-tion errors shall be neglected. The reconstruction low pass filter TPR shall be considered asideal, having a cutoff frequency of fR = 10kHz.

Sampling Quantization TPRMemory DA

s(t) + i(t) sd(t) sq(t)

fsfs

a) Block diagram of the audio recorder/player

fR

1.01.2

0 10−10 38 40−38−40 f/kHz

I(f) I(f)S(f)

b) Spectra S(f) and I(f)

(a1) Derive an equation that relates the ideally sampled signal sd(t) to the analog signalss(t) and i(t). Derive also the equation of the spectrum Sd(f) = F{sd(t)} in function ofS(f) and I(f)!

(a2) Draw the spectrum Sd(f) in the range of −60 kHz ≤ f ≤ 60 kHz! Pay attention tocorrect axis labeling!

(a3) Now, the sampling frequency is changed to fs = 28 kHz. Draw the spectrum Sd(f) inthe range of −42 kHz ≤ f ≤ 42 kHz!Pay attention to correct axis labeling!

(a4) Due to this reduced sampling frequency aliasing is avoided. Determine the minimumsampling frequency fs,min to avoid aliasing.

(b) (3 Points)

Assume the amplitudes of the sampled audio signal sd(t) of problem (a) to be uniformlydistributed in the range of −Amax ≤ sd(t) ≤ Amax. Now, sd(t) shall be quantized usinga linear mid-tread quantizer with a resolution of 3 bits and a symmetrical dynamic range[−Amax, Amax]. Note that one quantization level remains unused.

(b1) Sketch the quantization characteristic of the mid-tread quantizer described above!

(b2) For the quantization of sd(t) a linear mid-rise quantizer with a resolution of 3 bitsand a symmetrical dynamic input range [−Amax, Amax] can also be used. Which of thedescribed quantizers causes less quantization noise power? Give reasons!


D.9 Exam SS 2008 201


(a) (3 points)

Analog modulation is understood as imposing an analog source signal x(t) on a carrier signalsc(t) = A cos(2πfc t+ φ).

(a1) Explain why analog modulation is required to transmit signals in communications sys-tems!

(a2) Which analog modulation schemes exist? Denominate the modulation schemes andspecify the parameters of sc(t) that are modulated.

(b) (4 points)

The following figure shows the plot of an analog modulated signal sm(t), where the analogsource signal is given by x(t) = Ax cos(2πfx t) and the carrier signal is given by sc(t) =cos(2πfc t).

0 1 2 3 4 5 6 7 8−5−4−3−2−1

012345

t/s

s m(t

)

(b1) What is the value of the frequency fc of the carrier signal sc(t)?

(b2) What is the value of the frequency fx of the analog source signal x(t)?

(b3) Compute the modulation index μ and decide, whether an envelope detector can be usedto recover the source signal x(t) from the modulated signal sm(t)!

(c) (4 points)

A single sideband modulated received signal sBP (t) shall be demodulated with a heterodynereceiver. The magnitude spectrum |SBP (f)| of sBP (t) is shown in the figure below.


202 D EXAMS

f / MHz

|SBP (f )|

-600 -590 590 600

2

1

The next figure shows the structure of the heterodyne receiver. The frequency of the firstmixer is given by fLO = 450MHz and the frequency of the second mixer is given by fIF =150MHz.

IF bandpass filter Lowpass filtersBP (t) y(t )

cos(2πfLO t) cos(2πfIF t)

100MHz<|f |< 200MHz |f |< 15MHz

z (t)

(c1) Draw the magnitude spectrum |Z(f)| of the IF signal z(t) at the input of the secondmixer in the range of −250MHz ≤ f ≤ 250MHz! Pay attention to correct axis labeling!

(c2) Draw the magnitude spectrum |Y (f)| of the demodulated received signal y(t) in therange of −50MHz ≤ f ≤ 50MHz! Pay attention to correct axis labeling!


A customer requested an order to you in order to develop a wireless digital communicationssystem. For this purpose, you can choose between two different modulation schemes: QPSK and4-ASK. Due to source and channel coding schemes, all symbols can be assumed to be transmittedwith the same probability. The symbol amplitudes of QPSK and 4-ASK are given in the followingconstellation (IQ) diagrams, where A is a parameter:

I

QQPSK

I

Q4-ASK

A

j A

-A

-j A

A 2 A 3 A

(a) (6 points)


D.9 Exam SS 2008 203

(a1) With QPSK or 4-ASK, more than 1 bit can be transmitted per symbol. By that, thebits can be assigned to the symbols in such a way that the bit error rate is minimizedat a constant symbol error rate. Add mapping schemes of bits to symbols to the givenconstellation diagrams of QPSK and 4-ASK which minimize the bit error rates!

(a2) Add the decision thresholds of the detector to the constellation diagrams of QPSK and4-ASK!

(a3) Compute the average symbol energy ES of both modulation schemes assuming thatall symbols are transmitted with equal probability! (Hint: ES = E{|dk|2}, dk is theamplitude of each symbol.)

(b) (4 points)

In order to realize the wireless digital communications line you will need transmitters andreceivers. The block diagram of a QPSK transmitter is shown below:

QP

SK-M

apper

Pulse Shaping Filter

sin(2 )Cf tπ−

cos(2 )Cf tπ

lb ( )s tST

ST

( )Im t

( )Qm t

bl denotes the bit stream to be transmitted, TS is the symbol duration, s(t) is the transmittedbandpass signal and mI(t) and mQ(t) denote the inphase and quadrature phase componentsof the transmitted signal s(t) in the baseband. The pulse shaping filter has an rectangularimpulse response: rect( t

TS− 1

2).

(b1) Design and sketch a suitable QPSK receiver for the given QPSK transmitter and nameits components!

(b2) What is the basic difference between a 4-ASK transmitter and the given QPSK trans-mitter?

(b3) Baseband signals with zero mean value are preferred in communications engineering.Do the baseband signals of the given QPSK and 4-ASK have a zero mean value withthe given pulse shaping filter?

(c) (2 points)

Consider the 4-ASK. There are conditional probability density functions of the amplitudesof the sampled receive signal rk at the input of the detector at the receiver. These probabilitydensity functions are denoted by p(rk|dk). dk denotes the kth 4-ASK symbol which is definedby the constellation diagram and can take the following values: dk = {0, A, 2A, 3A}.(c1) Sketch (qualitatively) the conditional probability density functions p(rk|dk) of all 4

symbols of the 4-ASK in one diagram for a channel without noise.


204 D EXAMS

(c2) In your communications systeme there is an AWGN (Additive White Gaussian Noise)channel present. Sketch (qualitatively) the resulting conditional probability densitydistributions p(rk|dk) of all 4 symbols of the 4-ASK in one diagram for an AWGNchannel.

(d) (2 points)

Consider the QPSK. The bit error rate (BER) of a QPSK transmission over an AWGNchannel can be computed via the following formula:

BER =1

2erfc

(√ES

4σ2

)

ES denotes the average symbol energy. σ2 is the noise variance of the AWGN channel givenby σ2 = 1. The complementary Gaussian error function erfc(x) is given by the diagrambelow.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)

(d1) What is the value of the amplitude A of the symbols in your QPSK transmission systemthat is required to achieve a bit error rate not greater than 1,5% if the transmission isdone over an AWGN channel?





(a) (a1)

1 3

−3 −1 2

−2

1

−1

−2

−3

2

3

x(t)

t/T

Positions and absolute values of the pulses correct: (1 point)All signs correct: (1 point)

(a2) i. No, the system would not be causal, since x(t) = 0 for t < 0 is not satisfied.(1 point)

ii. Yes, the signal is odd, since x(−t) = −x(t). (1 point)

iii. No, X(f) is imaginary valued (and odd), since x(t) is real valued and odd. (1 point)

(b) (b1)

h(t)

t−1/2 1/2 3/2

1

Shape correct: (1 point)Axis labels correct: (1 point)

(b2) Fromh(t) = rect(t− 1/2) ∗ rect(t)

follows

H(f) = F{rect(t− 1/2)}F{rect(t)} = e−jπf si(πf) si(πf) = e−jπf si2(πf)

(1 point)


206 D EXAMS

(b3)

y(t)

t0.5 1.5 2.5 3.5 4.5

1

2

−1

−2

−3

Individual triangles correct: (1 point)Superposition correct: (1 point)

(c) (c1)

LR = LT − LA = 10 log10

(1000mW

1mW

)dBm− 95 dB− 25 log10 (2 · 2) dB

= 30 dBm− 95 dB− 2.5(3 + 3) dB = −80 dBm

Result correct: (1 point)

(c2) General power level at the receiver:

LR = −65 dBm− 25 log10

(d

km

)dB

Noise power level at the amplifier input:

LN1 = 10 log10

(10−10 mW

mW

)dBm = −100 dBm

This gives:

SNR1 = LR − LN1 = 35 dB− 25 log10

(d

km

)dB

Calculation method correct: (1 point)Result correct: (1 point)



(c3) Overall power level at the demodulator input:

Ltotal = LR + v = LR + 20 dB

Overall noise power level at the demodulator input:

LNtotal = 10 log10

(PN1 · vlinear + PN2

mW

)dBm = 10 log10

(0.1 pW · 100 + 90 pW

mW

)dBm

= 10 log10(10−7

)dBm = −70 dBm

This gives:

SNR2 = 10 dB = Ltotal − LNtotal = −65 dBm− 25 log10

(d

km

)dB + 20 dB− (−70) dBm

10 log10

(d

km

)dB =

15 dB

2.5= 6 dB

d = 4km

Calculation method correct: (1 point)Result correct: (1 point)


(a) (a1) Equation for s(t), i(t) and sd(t) in time domain:

sd(t) =1fs

∞∑k=−∞

(s(t) + i(t)

)δ(t− k/fs

)(1 point)

Equation for S(f), I(f) and Sd(f) in frequency domain:

Sd(f) =(S(f) + I(f)

) ∗ ∞∑n=−∞

δ(f − nfs

)(1 point)

(a2) Graphic qualitatively correct: (1 point)Graphic quantitatively correct: (1 point)

10-10 0-2 2 503040

38 42-30-50-40

-42 -38

1.0

1.2

f/kHz

Sd(f)

(a3) Graphic qualitatively and quantitatively correct: (1 point)


208 D EXAMS

10-10 0 12 28-28 1816-12

4038-16-18

-40 -38

1.0

1.2

f/kHz

Sd(f)

(a4) In order to avoid aliasing the following conditions must be fulfilled:

Sampling theorem: fs ≥ 2fg|fu +mfs| > fR, m = ±1,±2, ...|fo + nfs| > fR, n = ±1,±2, ...where fu = 38 kHz and fo = 40 kHz denote the lower and upper corner frequency of i(t).

m = −1 leads to fs ≤ fu − fR and n = −2 leads to fs ≥ fo+fR2

.

Thus, the minimum sampling frequency to avoid aliasing is fs,min = 25 kHz.Approach and solution correct: (2 points)

(b) (b1) Graphic qualitatively correct (1 point)

Amax−Amax

Δs

Δs

sd(t)

sq(t)

(b2) Decision: (1 point)The mid-rise quantizer causes less quantization noise power.

Explanation: (1 point)The quantization step size of the mid-rise quantizer is lower than that of the mid-treadquantizer.



Detailed Explanation:Due to its symmetric dynamic input range the mid-tread quantizer comprises only

q = 2b − 1 = 7

quantization levels and the quantization step size is Δs = 2Amax

q= 2

7Amax. The mid-rise

quantizer comprises

q = 2b = 8

quantization levels and the quantization step size is Δs = 2Amax

q= 2

8Amax.

For input signals the amplitudes of which are uniformly distributed within the dynamicinput range of the quantizer, the quantization noise power Pe is given by (see script):

Pe =1

12

(Δs

)2Thus, the linear mid-rise quantizer causes less quantization noise power than the mid-tread quantizer.


(a) (a1) Main reasons for analog modulation: (1 point)

1) Matching the source signal to the propagation channel.

2) Transmitting different analog source signals using the same propagation channel,e.g. by frequency multiplexing.

Points : At least one correct answer required!

(a2) Amplitude modulation (AM): A→ A(x(t)) (2 points)Phase modulation (PM): φ→ φ(x(t))Frequency modulation (FM): fc → f(x(t))

Points : 1 point for modulation schemes, 1 point for modulated parameters.

(b) (b1) fc = 4/1s = 4Hz (1 point)

(b2) fx = 1/2s = 0.5Hz (1 point)

(b3) With sm(t) = [A0 + x(t)] cos(2πfc t) and μ =max |x(t)|

A0

, where (1 point)

max |x(t)| = Ax = 2 and A0 = 3 follows: μ = 2/3 = 0.666.

Yes, an envelope detector can be used, since the source signal x(t) is completely (1 point)represented by the envelope of sm(t) (μ < 1).

(c) (c1) Magnitude spectrum |Z(f)| as shown in figure below. (2 points)


210 D EXAMS

f / MHz

|Z (f )|

1

140 150-140-150-250 250

0.5

Points : 1 point for correct shape, 1 point for correct axis labeling

(c2) Magnitude spectrum |Y (f)| as shown in figure below. (2 points)

f / MHz

|Y (f )|

-10 10

0.5

20 30-20-30-40-50 40 50

Points : 1 point for correct shape, 1 point for correct axis labeling


(a) (a1) The following mappings are possible. (2 points)Important is to change only one bit between neighboring symbols.

I

QQPSK

I

Q4-ASK

00

01

10

11 00 10 11 10

Per correct mapping: 1 point

(a2) The following decision thresholds are correct. (2 points)

I

QQPSK

I

Q4-ASK

Decision Thresholds

Per qualitatively correct diagram: 1 point

(a3) All symbols have the same probability: probability of one symbol is 14. (2 points)

sk is the amplitude of each of the 4 possible symbols.

ES = E{|sk|2} =1

4|s0|2 + 1

4|s1|2 + 1

4|s2|2 + 1

4|s3|2



ES =1

4(|s0|2 + |s1|2 + |s2|2 + |s3|2)

QPSK : ES =1

4(|A|2 + |jA|2 + | − jA|2 + | − A|2) = 1

4· 4A2 = A2

4− ASK : ES =1

4(|0|2 + |A|2 + |2A|2 + |3A|2) = A2

4(0 + 1 + 4 + 9) = 3.5A2

(b) (b1) Basic receiver block diagram: (2 points)

Receive Filter, Integrator,

Matched Filter

Dem

apper( ) ( )R Sh t h t= −

DetectorSampling

SkT

sin(2 )Cf tπ−

cos(2 )Cf tπ

( )b k( )r t( ) ( )R Sh t h t= −

(b2) The quadrature phase branch (or inphase branch) of the QPSK transmitter (1 point)(and also receiver) can be removed. Transmission and processing is only carried outin the quadrature phase (or inphase).

(b3) QPSK is free of a mean value. Due to the same probability of all symbols, (1 point)positive and negative parts are of same size and nullify in average.

The given 4-ASK is not free of a mean value, because all symbols are positive anddo not nullify each other in average.

(c) (c1) Conditional amplitude probability density function without any noise: (1 point)

0 0d = 1 Ad = 2 2Ad = 3 3Ad = Rr

0( | )Rp r d1( | )Rp r d 2( | )Rp r d 3( | )Rp r d1

(c2) Conditional amplitude probability density function with AWGN: (1 point)

0 0d = 1 Ad = 2 2Ad = 3 3Ad = Rr

0( | )Rp r d 1( | )Rp r d 2( | )Rp r d 3( | )Rp r d

(d) (d1) (2 points)

0.015 =1

2erfc

(√ES

4σ2

)


212 D EXAMS

0.03 = erfc

(√ES

4σ2

)

From the diagram it can be found that x = 1.5 for erfc(x) = 0.03 = 3 ∗ 10−2.

1.5 =

√ES

4σ2

2.25 =ES

4σ2

9 = ES

ES = E{|sk|2} = E{|A|2}

Due to the fact that all QPSK symbols have the same absolute amplitude A, the averagesymbol amplitude is the same as the amplitude of any arbitrary QPSK symbol.

A =√ES = 3

Correct x from the complementary Gaussian error function: 1 pointCorrect symbol amplitude: 1 point


D.11 Exam WS 2008/2009 213

D.11 Exam WS 2008/2009


Hint: Parts (a), (b) and (c) can be solved independently.

(a) (4 points)

In communications, LTI sytems are often considered, for which there exist specific relation-ships between input signals x(t) and output signals y(t). The mappings f : x(t) → y(t) ofLTI systems therefore have to fulfill certain properties.

x(t) y(t)H

(a1) Which properties do linear systems have to fulfill with respect to their mappings?Give also a counter example of a mapping f : x(t) → y(t) that is impossible for a linearsystem!

(a2) Which properties do source free systems have to fulfill with respect to their mappings?Give also a counter example of a mapping f : x(t) → y(t) that is impossible for a sourcefree system!

(b) (6 points)

Consider the signal

x(t) = t · σ(t) , with σ(t) =

⎧⎪⎪⎨⎪⎪⎩1 if t > 0

1/2 if t = 0

0 otherwise

and two positive constants t0 and T .

(b1) Draw the signal x(t+t0T

)within the range −2 t0 < t < 2 t0! Pay attention to correct

axis labeling!


214 D EXAMS

(b2) Show by means of a drawing how the signal

y(t) = triang(t/T )

can be expressed as a superposition of several shifted and scaled instances of the signalx(t)! Pay attention to correct axis labeling! Give also a formula for this relationship!

Hint:

triang(t) =

{1− |t| if − 1 < t < 1

0 otherwise.

(b3) Determine the Fourier transform Y (f) = F{triang(t/T )}!

(c) (5 points)

The following figure shows a device for adding the power of two incoming signals se,1(t) undse,2(t). The power Pa of the output signal sa(t) is equal to the sum of the powers Pe,1 andPe,2 of the input signals.

s1(t)

s2(t)

se,1(t)

se,2(t)

sa(t)

n1(t)

n2(t)

Power Adder

Pa = Pe,1 +Pe,2

(c1) Consider first the case that two signals s1(t) and s2(t) with powers P1 = 200mW andP2 = 800mW are combined noisefree, i.e. n1(t) = n2(t) = 0. Determine the power levelsL1, L2 and La of the signals s1(t), s2(t) and sa(t) in dBm!

(c2) Now let the signals s1(t) and s2(t) from (c1) be disturbed by additive noise n1(t)and n2(t) with powers Pn1 = 100mW and Pn2 = 300mW, respectively (see figure).Determine the signal-to-noise ratio SNRa at the output of the device in dB!

Hints:

x 0.1 0.25 1 1.5 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.18 0.3 0.48 0.7 1 2 3


D.11 Exam WS 2008/2009 215

Promblem 2: Analog-to-Digital Conversion

(a) (2 points)

Analog signals can be converted to digital signals by means of analog-to-digital conversion.What are the advantages of representing analog signals in the digital domain. Give at leasttwo advantages!

Consider the configuration of an analog-to-digital converter depicted in figure (A). At the inputof the analog-to-digital converter, the desired signal s(t) is disturbed by an interfering signal i(t).The respective frequency spectra S(f) und I(f) are shown in figure (B).

Sampling Quantizerza(t) zq(t)s(t)+i(t)

Analog-to-Digital-Converter

(A)

f / MHz

S(f )

-10

2

1

10 20 305 2515 35-5-15-20-25-30-35

I(f ) I(f )

(B)

(b) (4 points)

(b1) Draw the frequency spectrum Za(f) of the signal za(t) resulting after sampling withinthe range −35HMz ≤ f ≤ 35MHz assuming a sampling frequency of fa = 25MHz!Pay attention to correct axis labeling!

(b2) For technical reasons, the sampling frequency can not be higher than 50MHz. Explaintwo technically different solutions that nevertheless allow for recovering the desiredsignal s(t) from the sampled signal za(t) without errors! Give reasons for both solutions!Hint: An additional component can be used.

(c) (4 points)

(c1) Draw the quantization characteristic of the linear mid-rise quantizer, the input rangeof which is [−4, 4] and which has a quantization resolution of b = 3bits. Pay attentionto correct axis labeling!

(c2) The signal za(t) at the input of the quantizer takes only values within the range −2 ≤za(t) ≤ 2 with uninform amplitude distriubution. Determine the signal-to-noise ratioSNR at the output of the quantizer with the quantization characteristic given in (c1)!


216 D EXAMS


(a) (4 points)

(a1) Which are the symmetry properties of the real and imaginary part of the frequencyspectrum of a real-valued time domain signal?

Figure (A) shows the magnitude spectra A1(f) = |S1(f)| and A2(f) = |S2(f)| of the twotime domain signals s1(t) and s2(t) with frequency spectra S1(f) = F{s1(t)} and S2(f) =F{s2(t)}.

ff

A1(f) = |S1(f)| A2(f) = |S2(f)|

(A)

(a2) Based on A1(f), decide whether s1(t) is a real-valued time domain signal! Give reasons!

(a3) Based on A2(f), decide whether s2(t) can definitely be identified as real-valued timedomain signal!

(b) (7 points)

The amplitude modulated baseband signal s3(t) of a radio station is received and demodu-lated by the receiver depicted in Figure (B).

s3(t) m(t)

2 cos(2πfLOt) 2 cos(2πfZF t) −A

|f |>180 kHz|f |<220 kHz |f | < 40 kHz

Demodulator

(B)

The received signal is given by s3(t) = (A+ μx3(t)) cos(2πfct), where the carrier frequencyis fc = 1MHz and the modulation index is μ = A

|max{x3(t)|} < 1. The frequency spectrum

S3(f) = F{s3(t)} is shown in Figure (C).

f/MHz f/kHz

0.98 1.0 1.02−1.02−1.0 −0.98

S3(f) I(f)

1122

4

580 600 620−620 −600 −580

(C) (D)


D.11 Exam WS 2008/2009 217

The receiver down-converts the received signal to the intermediate frequency fZF = 200 kHz.

(b1) Which two oscillator frequencies fLO of the first mixer stage are feasible to down-converts3(t) to fZF?

(b2) Draw the frequency spectrum M(f) = F{m(t)} of the signal after the first mixer stagefor the case of fLO < fc within the range −250 kHz ≤ f ≤ 250 kHz! Pay attention tothe correct axis labeling!

Due to a defect, the amateur radio transceiver of your neighbor transmits an interferingsignal i(t) to the antenna of your receiver. The corresponding frequency spectrum I(f) isshown in Figure (D).

(b3) Sketch the spectrum M(f), resulting from the superimposed reception of s3(t) and i(t)for the case of fLO < fc within the range −250 kHz ≤ f ≤ 250 kHz!

(b4) What kind of filter (high-pass, low-pass, band-pass, band-stop) with which cut-off fre-quencies has to be employed between the antenna and the first mixer in order to assurean interference-free demodulation of s3(t) with the interfering signal i(t) at the antenna?

(b5) The demodulation of m(t) can also be done by a different method than shown in Figure(B). Sketch a different type of demodulator that is suitable for the demodulation ofm(t)and name its components!


A wireless communications system can be used with two different modulation schemes: either with8-PSK or with 4-QAM. Dependening on the quality of the channel and the resulting bit error rateone of the modulation schemes is chosen in such a way that the data rate is maximized whilereliability of data transmission is ensured.

(a) (4 points)

(a1) Sketch an IQ-diagram (constellation-diagram) of the 8-PSK modulation!

(a2) Add a possible assignment of bits to the symbols in the IQ-diagram of the 8-PSK whichminimizes the bit error rate at constant symbol error rate! What is the name of thiskind of mapping?

(a3) Add the decision thresholds of the detector to the IQ-diagram of the 8-PSK assumingthat each of the symbols is sent with the same probability!

(b) (2 points)

It is possible to switch from the 4-QAM mode to the 8-PSK mode if the bit error rateof the 4-QAM transmission across an AWGN channel falls below 5 · 10−3. For a 4-QAMtransmission across an AWGN channel, the bit error rate (BER) can be computed via thefollowing formula:

BER =1

2erfc

(√ES

4N0

)SNR =

ES

N0

SNR[dB] = 10 log(SNR)


218 D EXAMS

ES denotes the average symbol energy. N0 is the noise power of the expected AWGN channel.The complementary Gaussian error function erfc(x) is given in the diagram below.

(b1) At the receiver input there is a signal-to-noise ratio of SNR = 12 dB. Is it possible toswitch to the 8-PSK mode in this case?Hint: Make sure not to mix the linear and the logarithmic SNR!

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)

(c) (8 points)

The figure below shows the block diagram of a receiver for 4-QAM transmission and theIQ-diagram the detector uses for bit detection including the bit to symbol assignments.

Integrator Detector, DeciderSampling

SkT

π θ− +sin(2 )Cf t

π θ+cos(2 )Cf tˆ(2 )d k

+

∫

( 1)

()S

S

k T

kT

d t

1( )Sr kT

Demapper

+ˆ(2 1)d k2( )Sr kT

( )s t

+

∫

( 1)

()S

S

k T

kT

d t

I

Q

C-C

-jC

jC`10´ `00´

`01´`11´

s(t) denotes the sent and received bandpass signal, fC is the carrier frequency, TS is thesymbol duration and θ denotes the phase offset between the transmitter and receiver oscil-lator. fC is related to TS by fC = 5

TS. In the IQ-diagram, C denotes a constant factor that

is not specifed in more detail.


D.11 Exam WS 2008/2009 219

(c1) Sketch a suitable 4-QAM transmitter that corresponds to the given 4-QAM receiverstarting at the bit series d(k) and name the components!

Now, assume the signal s(t) = −A cos(2πfCt)− A sin(2πfCt) to be transmitted.

(c2) Compute the signals r1(kTS) and r2(kTS) in the inphase and quadrature phase branchof the receiver after integration under the condition θ = 0◦ for the transmitted signals(t)! See also the hints given below.

(c3) Which 2 bits will be decoded in case of the transmitted signal s(t) according to theIQ-diagram of the detector?

(c4) Which 2 bits will be decoded if there is a phase offset of θ = −90◦ between the trans-mitter and receiver oscillator. Give reasons for your decision!

Hints:sin(x) · cos(y) = 1

2(sin(x− y) + sin(x+ y))

sin(x) · sin(y) = 12(cos(x− y)− cos(x+ y))

cos(x) · cos(y) = 12(cos(x− y) + cos(x+ y))

cos(x) = sin(π2± x)

sin(x) = cos(π2− x)


220 D EXAMS



(a) (a1) If x1(t) → y1(t) and x2(t) → y2(t) the following relations must be satisfied (linearity):x1(t) + x2(t) → y1(t) + y2(t) and a · x1(t) → a · y1(t) , ∀a ∈ R (1 point)

Counter example: x(t) → y(t) = |x(t)|. (1 point)

(a2) If x(t) = 0 for all t < t0 (source freeness): y(t) = 0 must hold for all t < t0. (1 point)

Counter example: x(t) → y(t) = 2x(t) (1 point)

(b) (b1)

x(t+t0T

)

t0/T

t−t0

Shape and labeling correct: (2 Points)

(b2)

y(t)

−T T t

1

−2

x1(t)

x2(t)

x3(t)

Shape and labeling correct: (2 Points)

y(t) = x1(t) + x2(t) + x3(t) = x

(t+ T

T

)− 2x

(t

T

)+ x

(t− T

T

)(1 Point)



(b3) (1 point)

Y (f) = T si2(πfT )

(c) (c1) (3 points)

L1 = 10 log10

(2 · 100mW

1mW

)dBm = 20 dBm + 3dB = 23 dBm

P2 = 4P1 ⇒ L2 = L1 + 6dB = 29 dBm

Pa = P1 + P2 = 5P1 = 1/2 · 10P1 ⇒ La = L1 − 3 dB + 10 dB = 30 dBm

In general we have:La = 10 log10

(10L1/10 + 10L2/10

)(c2) (2 points)

SNRa = 10 log10

(P1 + P2

Pn1 + Pn2

)dB

= 30 dBm− 10 log10

(4 · 100mW

1mW

)dBm = 30 dBm− (

20 dBm + 6dB) = 4 dB


(a) Advantages: Signal processing becomes independent of integration technology andtemperature; storage of digital signals is easier; digital signal processing allows foradditional / some special signal manipulations (2 points)

(b) (b1) Frequency spectrum Za(f) according to diagram belowDiagram qualitatively correct: (1 point)Diagram quantitatively correct (axis labeling): (1 point)

f / MHz-10

2

1

10 20 305 2515 35-5-15-20-25-30-35

(b2) Applying an anti-aliasing / low-pass filter at the input of the analog-to-digitalconverter with cut-off frequency 10MHz < fG < 25MHz, such that the interferingsignal is removed before the analog-to-digital conversion (1 point)

Dedicated sampling frequency: 45MHz < fa < 50MHz, such that the frequencyimages of the interfering signal do not superimpose with the frequency spectrum ofthe desired signal. (1 point)

(c) (c1) Quantization characteristic accordig to diagram belowDiagram qualitatively correct: (1 point)Diagram quantitatively correct (axis labeling): (1 point)


222 D EXAMS

1.5

2 31 4 za(t)

zq(t)

2.5

0.5

3.5

-2.5

-1.5

-3.5

-0.5-3 -2-4 -1

(c2) Only half of the input range of the quantizer is used by the input signal za(t). Hence,only 4 out of 2b/bit = 8 quantization levels are used. This corresponds to an effectivequantization resolution of b′ = 2bits. (1 point)

In case of an input signal with uniform amplitude distribution the signal-to-noiseratio at the quantizer output is given by

SNR = 6.02 dB · b′/BitThus, the signal-to-noise ratio resulting from the quantization of za(t) derives asSNR ≈ 12 dB. (1 point)

Solution 3: Analog Modulation

(a) (a1) The real part of the frequency spectrum of a real-valued time domain signal is axialsymmetric to f = 0 (even function) and the imaginary part is point symmetric tof = 0 (odd function). (1 point)

(a2) The signal s1(t) is complex-valued. Due to the symmetry conditions described in (a1)for frequency spectra of real-valued time domain signals, the magnitude spectrum ofany real-valued time domian signal is axial symmetric to f = 0.Decision correct: (1 point)Reasoning correct: (1 point)

(a3) It can not clearly be concluded, whether s2(t) is real- or complex-valued. As explainedin (a2), the magnitude spectrum of a real-valued signal is axial symmetric. Hence,s2(t) could be real-valued. However, the real and imaginary part of the frequency spec-trum S2(f) = F{s2(t)} could also be as depicted in the following figure, which wouldcorrespond to a complex-valued time domain signal.

ff

Re{S2(f)} Im{S2(f)}



Decision correct: (1 point)

(b) (b1) Oscillator frequency of the first mixer stage:1. possibility: fLO = fc + fZF = 1.2MHz2. possibility: fLO = fc − fZF = 0.8MHzCalculation and values correct: (1 point)

(b2) Frequency spectrum M(f) according to figure below

f/kHz

180 200 220−220 −200 −180

M(f)

1

2

4

Diagram qualitatively correct: (1 point)Axis labeling correct: (1 point)

(b3) Frequency spectrum M(f) according to figure below

f/kHz

180 200 220−220 −200 −180

M(f)

1

2

4

Diagram qualitatively correct: (1 point)

(b4) Either a band-stop filter with lower cut-off frequency fg,u = 1.38MHz and uppercut-off frequency fg,o = 1.42MHz or a band-pass or low-pass filter with uppercut-off frequency fg,o = 1.38MHz has to be inserted between the antenna and thefirst mixer. (1 point)

(b5) An envelope demodulator can be applied:

Band pass(Surpression of

adjacent channels)

Rectifier(Reconstruction of

envelope)

Low pass(Cancellation

of carrier)

High pass(Cancellation of

DC offset)

Configuration correct: (1 point)Labeling and description of the components correct: (1 point)


224 D EXAMS


(a) (a1) IQ-diagram as shown belowIt is important that there are 8 constellation points arranged on a circle. (1 point)

j

1−1

− j

I

Q

(a2) Assignment as shown belowIt is important that adjacent symbols do only differ by one bit. (1 point)This kind of mapping is called GRAY mapping. (1 point)

I

Q

'000'

'001''011'

'010'

'110'

'111' '100''101'

(a3) Decision thresholds as shown belowIt is important that all thresholds lead through the origin of ordinates. (1 point)

Decision Thresholds

I

Q

(b) (b1) Remember that 3 dB corresponds to 2 in the linear domain and 12dB = 4 · 3dB.SNR[db] = 12dB = 4 · 3dB = 4 · 10 · 0.3dB

0.3 = log10(2) (see also table of logarithms in problem 1)

SNR = 104·10·0.3

10 = (100.3)4 = 24 = 16

BER =1

2erfc

(√ES

4N0

)

BER =1

2erfc

(√SNR

4

)



BER =1

2erfc

(√16

4

)

BER =1

2erfc (2)

In the diagram x = 2 corresponds to erfc(2) = 5 · 10−3. Hence, we get:

BER =1

2· 5 · 10−3 = 2.5 · 10−3

This is smaller than 5 · 10−3. So, the communciations system can switch to the 8-PSKmode.Linear SNR correct: (1 point)Result correct: (1 point)

(c) (c1) General transmitter architecture as shown below (2 points)

4-QA

M-

Mapper

Puls shaping filter

sin(2 )Cf tπ−

cos(2 )Cf tπ

( )d k ( )s tST

ST

1( )m t

2( )m t

Up conversion

(c2)

r1(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) · (cos(2πfCt))dt

r1(kTS) = −A∫ (k+1)TS

kTS

cos(2πfCt) cos(2πfCt)dt− A

∫ (k+1)TS

kTS

sin(2πfCt) cos(2πfCt)dt

r1(kTS) = −A∫ (k+1)TS

kTS

1

2(cos(0) + cos(4πfCt))dt− A

∫ (k+1)TS

kTS

1

2(sin(0) + sin(4πfCt))dt

r1(kTS) = −A∫ (k+1)TS

kTS

1

2(1 + cos(4πfCt))dt− A

∫ (k+1)TS

kTS

1

2(sin(4πfCt))dt

According to fC = 5TS, sinus and cosinus are integrated over over full periods. Therefore,

these integrals become zero.

r1(kTS) =

∫ (k+1)TS

kTS

−A2dt = −A

2((k + 1)TS − kTS) = −ATS

2

Similar to the computation of r1(kTS):

r2(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) · (−sin(2πfCt))dt


226 D EXAMS

r2(kTS) = A

∫ (k+1)TS

kTS

cos(2πfCt) sin(2πfCt)dt+ A

∫ (k+1)TS

kTS

sin(2πfCt) sin(2πfCt)dt

r2(kTS) = A

∫ (k+1)TS

kTS

1

2(sin(0) + sin(4πfCt))dt+ A

∫ (k+1)TS

kTS

1

2(cos(0)− cos(4πfCt))dt

r2(kTS) = A

∫ (k+1)TS

kTS

1

2sin(4πfCt)dt+ A

∫ (k+1)TS

kTS

1

2(1− cos(4πfCt))dt

r2(kTS) =

∫ (k+1)TS

kTS

A

2dt =

A

2((k + 1)TS − kTS) =

ATS2

Approach via the two integrals correct: (1 point)Integrals over full sinus and cosinus periods become zero: (1 point)Results for r1(kTS) and r2(kTS) correct: (1 point)

(c3) The results from the previous solution can be reused. r1(kTS) is the inphasecomponent and negative. r2(kTS) is the quadrature phase component and positive.This corresponds to the symbol in the upper left quadrant of the IQ-diagram whichwill be decoded as ’10’. (1 point)

(c4) Simple way: A phase offset of θ = −90◦ = −π2results in a virtual rotation of the

IQ-diagram at the detector by 90◦ in mathematically negative direction (clockwise).So, over the received symbol, which lies in the upper left quadrant of the transmitterIQ-diagram, there is the rotated demapper IQ-diagram (see also diagrams below). Inthe rotated demapper IQ-diagram the received signal can be found in the lower leftquadrant and is recognized as −ATS

2− jATS

2, which is decoded as ’11’.

j

1−1

− j

I

Q

j

1

−1

− j

I

Q

Virtual turn of Receiver-IQ-diagram

by -90°

Transmitter side Receiver side

Complex solution (computation):

r1(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) (cos(2πfCt− π

2))dt

r2(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) (−sin(2πfCt− π

2))dt

−sin(x− π

2) = sin(−(x− π

2)) = sin(

π

2− x) = cos(x)

cos(x− π

2) = cos(−(x− π

2)) = cos(

π

2− x) = sin(x)



r1(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) sin(2πfCt)dt

r2(kTS) =

∫ (k+1)TS

kTS

(−A cos(2πfCt)− A sin(2πfCt)) cos(2πfCt)dt

After multiplying out, applying addition theorems and integration (partially over fullperiods of sinus and cosinus like in (c2)):

r1(kTS) = −ATS2

r2(kTS) = −ATS2

The received symbol is thus in the lower left quadrant of the I/Q diagram and decodedas ’11’.

Correct bit combination: (1 point)Reason for decision (explanation, diagram or computation): (1 point)


228 D EXAMS

D.13 Exam SS 2009

Problem 1: Signal Levels and LTI Systems

Note: Questions (a), (b) and (c) can be solved independently of one another.

(a) (7 points)

In LTI systems, which are often used in communications technology, the output signal y(t)is distinctly related to the input signal x(t). We assume the following LTI system

h(t)x(t) y(t)

with the impulse response

h(t) =

{e−at , t ≥ 0

0 , otherwise.

(a1) Give the general relation between x(t), y(t) and h(t).

(a2) Draw a sketch of h(t)!

(a3) Calculate the transfer function H(f) for the given impulse response h(t).

We now assume the input signal to be:

x(t) = a1 sin(2πf1t)− a2 cos(2πf2t)

(a4) Draw a sketch of the magnitude spectrum |X(f)| for a1 < a2 and f1 < f2!

(a5) Calculate and draw the magnitude spectrum |Y (f)| of the output signal for the followingparameter values: a = 4, a1 = 10, a2 = 11, f1 = 3/(2π) and f2 =

√105/(2π)! Pay

attention to correct axis labeling!

(b) (5 points)

In wireless communications systems, data transmission between a transmitter and a receiveris often disturbed by signals of other transmitters that send at the same time using thesame carrier frequency. In this case, the data signal is impaired by both, the thermal noiseat the receiver and the signals of other transmitters (termed interference). We consider thefollowing situation where transmitter 1 transmits the desired data to the receiver, while thesignals of transmitter 2 cause interference to the data transmission.

Receiver

Transmitter 1

Demodulator

Lg = 20dB

PN

d1

d2Transmitter 2

LS1

LS2

L’E LE


D.13 Exam SS 2009 229

The attenuation of the radio channel is given as

LA = 20dB + 10 log10((d/m)4

)dB,

where d is the distance between transmitter and receiver.

(b1) Calculate the power level LS1 of transmitter 1 assuming that its transmit power isPS1 = 250mW!

(b2) Calculate the power level LE1 of the received signal from transmitter 1 at the input ofthe demodulator given that the power level of transmitter 1 is LS1 = 30dBm and thedistance between transmitter 1 and receiver is d1 = 20m!

(b3) How long is the distance d2 between transmitter 2 and receiver, if both transmitters usethe same transmit power, the ratio of signal power and interference power (SIR, signal-to-interference-ratio) at the receiver is known to be 12 dB and the distance betweentransmitter 1 and the receiver is d1 = 57m?

Note:

x 0.1 0.25 1 1.5 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.2 0.3 0.5 0.7 1 2 3

(c) (3 points)

The same scenario as in problem (b) is assumed.

(c1) What is the degradation in dB of the signal-to-disturbance ratio (ratio of the signalpower to the power of the disturbance), if the interference of transmitter 2 is considered,as opposed to the case where the thermal noise at the receiver is the only considereddisturbance?

Calculate the degradation for the assumption that the power level of the receivedsignal of transmitter 1 is L′

E1 = −66dBm, the ratio of signal power to interferencepower (signal-to-interference-ratio, SIR) at the receiver is given by 9dB and the powerof the thermal noise is PN = 0.01nW (and thus the noise power level LN = −80dBm)!

Note: The degradation equals the difference of the SINR (signal-to-interference-and-noise-ratio) and the SNR (signal-to-noise-ratio) in dB.


(a) (1 point)

Numerous communications systems use analog-to-digital conversion to transform analogsignals to digital signals. Name one of the main reasons!

(b) (4 points)

Assume the signal s(t) at the input of an analog-to-digital converter. The real-valued spec-trum S(f) of the signal s(t) is shown in the following figure.


230 D EXAMS

� � kHz

��

�

� ��

(b1) What is the minimum sampling frequency fs for the signal s(t) to be recoverable withouterrors?

(b2) Name the effect which prevents recovering the signal s(t) without errors when thesampling frequency fs is chosen too low!

(b3) Now, the sampling frequency is fs = 4kHz. Draw the spectrum Ss(f) of the sampledsignal ss(t) in the interval f ∈ [−8kHz, 8kHz]! Pay attention to correct axis labeling!

(c) (5 points)

The sampled signal ss(t) is quantized. For this, a linear mid-rise quantizer is used with aninput range defined as Ulower ≤ ss(t) ≤ Uupper with Ulower = −4V und Uupper = 4V. The stepsize of the quantizer is Δs.

(c1) Compute Δs, in case the resolution of the quantizer is 3 bits! Draw the quantizationcharacteristic in the intervall [−4Δs, 4Δs]! Pay attention to correct axis labeling!

(c2) We now assume the quantizer to have a resolution of 5 Bits. Compute the maximumabsolute quantization error emax.

(c3) What are the two options to increase the SNR (signal-to-noise ratio) of the quantizationby +12dB?


D.13 Exam SS 2009 231


(a) (3 points)

Shaping the waveform of a carrier signal by an analog source signal is known as analogmodulation.

(a1) Why is analog modulation used for transmitting analog signals in communications?

(a2) Compare amplitude modulation to frequency modulation in terms of the required am-plifiers and the occupied bandwidth when the same analog source signal is transmitted.

(b) (4 points)

Consider a signal s(t) that is transmitted by means of amplitude modulation. The modulatedband-pass signal is given as sBP (t) = (A + s(t)) cos(2πfc t). The carrier frequency is fc =100 kHz.

(b1) Draw sBP (t) in the range 0μs ≤ t ≤ 80μs for the case of A = 2 and s(t) = cos(2πf1 t)with f1 = 25 kHz! Pay attention to correct axis labeling!

(b2) Consider the signal s(t) to be given as s(t) = cos(2πf1 t) +14cos(2πf2 t), where f1 =

25 kHz and f2 = 12.5 kHz. Which value of A is required to allow for demodulation ofs(t) with an envelope detector at the receiver and to spend as little transmit power aspossible for the carrier signal?

(c) (4 points)

A modulated single-sideband signal sESB(t) shall be demodulated with a simple direct con-version receiver. The figure below shows the magnitude spectrum |SESB(f)| of sESB(t).

f / MHz

|SESB (f )|

-120 -100 110 120

1.0

0.5

100-110

The next figure shows the block diagram of the direct conversion receiver. The mixer fre-quency is fLO = 120MHz.

sESB (t)

cos(2πfLO t)

sBB (t) Low-pass filter

|f |<fg

y(t)


232 D EXAMS

(c1) Give an equation that relates the frequency spectrum SBB(f) of the signal sBB(t) afterthe mixer to SESB(f)! Draw the magnitude spectrum |SBB(f)| in the range −40MHz ≤f ≤ 40MHz! Pay attention to correct axis labeling!

(c2) Consider a signal i(t) that interferes with sESB(t) at the receiver input. The figure belowshows the magnitude spectrum |I(f)| of i(t). What is the maximum cut-off frequencyfg of the low-pass filter at the receiver that can be allowed to remove i(t) from thereceived signal?

f / MHz

|I (f )|

-95 -90 9590

1.0

0.5


Consider two binary modulation schemes (A) and (B). Their constellation diagrams (IQ-diagrams),including the mapping of the input bits, are illustrated in the following figure:

1

I

Q

I

Q(A) (B)

A A

A

0

0

0 1

(a) (4 points)

(a1) Name two further common binary modulation schemes!

(a2) Now the bit sequence {bk} = {0, 1, 1, 0} is transmitted. The symbol duration is Ts and arectangular function with hTs

(t) = rect(t/Ts− 1/2) is used for pulse shaping. Draw theinphase and quadrature phase components I(t) and Q(t) of the modulated basebandsignals s(t) for the cases (A) and (B)! Pay attention to correct axis labeling!

(b) (4 points)

(b1) One of the two modulation schemes (A) and (B) is going to be used for data transmissionat a carrier frequency fc. The real valued bandpass signal at the receiver input is rBP (t).Sketch a general receiver architecture that can be used for both modulation schemes(A) and (B)! Name its components!

(b2) Draw the decision boundaries into the constellation diagrams of (A) and (B) in such away, that the bit error probability is minimized when the transmitted symbols occurwith equal probability.


D.13 Exam SS 2009 233

(c) (6 points)

For achieving a higher data rate, a higher order modulation scheme (C) is considered, havingthe following constellation diagram:

−2A

Q(C)

I

−2A

−A

A

2A

2AA−A

(c1) There are different ways of mapping the input bits to the symbols. Gray mappingdenotes a mapping rule that minimizes the bit error probability at a constant symbolerror probability.Which property characterizes Gray mapping in general? Insert such a bit mapping intothe constellation diagram (C) given above!

(c2) Now the average energies ES,A and ES,B are considered for the modulation schemes (A)and (B), respectively. In both cases the same amplitude A is used and the transmittedsymbols are assumed to occur with equal probability. Determine the ratio of the averageenergies ES,A /ES,B!

(c3) The symbol error probability is usually dominated by the pairwise error probabilitybetween symbols of lowest distance dmin. For the AWGN channel the pairwise errorprobability of two symbols with distance d in the constellation diagram is given by

PS =1

2erfc

(√d2

8σ2

),

where σ2 denotes the variance of the noise and the complementary error function erfc(x)is given in the figure below.

How large does the amplitude A need to be for case (B) to achieve a symbol errorprobability of 10−3 for σ = 1/2?


234 D EXAMS

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)





(a) (a1) y(t) = x(t) ∗ h(t) = ∫ ∞−∞ x(t− τ)h(τ)dτ =

∫ ∞−∞ x(τ)h(t− τ)dτ (1 point)

(a2) Sketch - see figure below.Sketch qualitatively correct: (1 point)

0

1

t

h(t)

(a3) Correct transfer function: (1 point)

H(f) =

∫ ∞

−∞x(t)e−j2πftdt =

∫ ∞

0

e−ate−j2πftdt

=

∫ ∞

0

e−t(a+j2πf)dt = − 1

a+ j2πf

[e−t(a+j2πf)

]∞0

=1

a+ j2πf

(a4) Sketch - see figure below.Sketch qualitatively correct: (1 point)

a1/2

f1

a2/2

f2-f2 -f1

|X(f)|

f0

(a5) Magnitude spectrum correct: (1 point)

|H(f)| = 1√a2 + (2πf)2


236 D EXAMS

|H(f1)| =|H(−f1)| = 1√42 + 32

=1

5

|H(f2)| =|H(−f2)| = 1√42 + 105

=1√121

=1

11

|Y (f)| =a12|H(f1)|

(δ(f − f1) + δ(f + f1)

)+a22|H(f2)|

(δ(f − f2) + δ(f + f2)

)=(δ(f − f1) + δ(f − f1)

)+

1

2

(δ(f − f2) + δ(f + f2)

)Plot - see figure below.Qualitatively correct: (1 point)Quantitatively correct (axis labeling): (1 point)

1/2

f1

1

f2-f2 -f1

|Y(f)|

f0

(b) (b1) Result and unit correct: (1 point)

LS1 = 10 log10(250)dBm = 10 log10(1000 · 0.25)dBm = (30− 6)dBm = 24dBm

(b2) Channel attenuation correct: (1 point)

LA1 = 20dB + 10 log10(204)dB = 20dB + 40 log10(2 · 10)dB

= (20 + 52)dB = 72dB

Result and unit correct: (1 point)

LE1 = LS1 − LA1 + Lg = 30dBm− 72dB + 20dB = −22dBm

(b3) Equations correct: (1 point)Result and unit correct: (1 point)

SIR = 12dB = LS − LA1 + Lg︸︷︷︸signal power

−(LS − LA2 + Lg︸︷︷︸interference power

)

= LA2 − LA1

= 20dB + 40 log10(d2)− 20dB− 40 log10(57)

⇒ 3dB = 10 log10

(d257

)⇒ d2 = 2 · 57m = 114m



(c) (c1) Solution approach correct: (1 point)

SINR

SNR=P ′

E1/(PN + P ′E2)

P ′E1/(PN)

=PN

PN + P ′E2

=PN

PNI

difference in dB:

SINR[dB] − SNR[dB] = LN − LNI

Combined disturbance power correct: (1 point)

L′E2 = L′

S1 − SIR = −75dBm = −80dBm + 5dB → P ′E2 = 3 · 10−8mW

PNI = PN + P ′E2 = 4 · 10−8mW

Degradation in decibels correct: (1 point)

LNI = 10 log10(PNI)dBm = (−80 + 3 + 3)dBm = −74dBm

LN − LNI = −80dBm− (−74dBm) = −6dB

Answer : The ratio of signal power to disturbance power degrades by 6dB, due to theinterference.

Problem 2: Analog-Digital-Conversion

(a) Advantages of digital signals: (1 point)Small sensitivity to distortion and disruption; simple storage of signals;simple signal processing and manipulation; independence of technology and temperature

(b) (b1) Nyquist criterion: fs ≥ 2fg = 6kHz (1 point)

(b2) Aliasing (1 Point)

(b3) Spectrum - see figure below

Diagram qualitatively correct (S(f) recuring): (1 point)Diagram quantitatively correct (labeling of the axes): (1 point)

f / kHz

2

1 3-1-3

Ss(f)

75-5-7

1

(c) (c1) Given the size of the input range as Δu = Uupper − Ulower, the number of steps q andthe quantization resolution b in bits, the results are


238 D EXAMS

q = 2b ⇒ q = 23 = 8

Δs =Δu

q⇒ Δs =

Uupper − Ulower

8= 1V (1 point)

Quantization characteristic - see figure below (1 point)Diagram qualitatively correct:Diagram quantitatively correct (labeling of the axes):

sa(t)

sq(t)

� �

��

��

� � � � ��

� �

� �

�

� ��

Uupper

Ulower

(c2) Given the maximum absolute quantization error as emax = Δs/2 the results are

q = 2b ⇒ q = 25 = 32

Δs =Δu

q⇒ Δs =

Uupper − Ulower

32= 250mV

emax =250mV

2= 125mV (1 point)

(c3) Increase the number of bits used for quantization by 2; (2 points)increase the sampling frequency fs of the input signal s(t) by a factor of 16


(a) (a1) Main reasons for analog modulation: (1 point)

1) Matching the analog source signal to the propagation medium, i.e. the transmissionchannel.

2) Transmitting different analog source signals using the same propagation channel,e.g. by frequency multiplexing.

Points: At least one correct answer required.

(a2) AM amplifier for AM needs to be linear, (1 point)FM amplifier can be non-linear

AM bandwidth is equal to the bandwidth of the analog source signal (2 fg), (1 point)FM bandwidth is larger than AM bandwidth (2 fg + 2ΔF , Carson rule)



(b) (b1) Signal sBP (t) - see figure below.Qualitatively correct: (1 point)Quantitatively correct (axis labeling): (1 point)

0 20 40 60 80−3

−2

−1

0

1

2

3

t / μs

s BP(t)

(b2) To allow for an envelope detector at the receiver the modulation index needs to be

μ =A

max |s(t)| ≥ 1

To spend as little as possible energy for the carrier signal, A needs to be a small (1point)as possible, i.e.:

μ =A

max |s(t)|!= 1

From this requirement and with max |s(t)| = 54, for e.g. t = 0, follows: (1 point)

A = max |s(t)| = 5

4.

(c) (c1) Equation for frequency spectrum SBB(f): (1 point)

SBB(f) =1

2SESB(f − fLO) +

1

2SESB(f + fLO)

Magnitude spectrum |SBB(f)| - see figure below.Qualitatively correct: (1 point)Quantitatively correct (axis labeling): (1 point)

f / MHz

|SBB (f )|

-10 20

1.0

0.5

10-20


240 D EXAMS

(c2) Down-conversion with fLO = 120MHz leads to the following frequency spectrum at theinput of the low-pass filter:

f / MHz

|SBB (f )|

-30 -25 3025

1.0

0.5

-10 2010-20

To remove the down-converted interfering signal with a frequency band lower-boundedby fu = 25MHz, the cut-off frequency fg of the low-pass filter must not be greaterthan fu, i.e.: (1 point)

fg ≤ 25MHz


(a) (a1) BPSK, 2-FSK (1 point)

(a2) s(t) = I(t) + j Q(t)

(A) qualitatively and quantitatively correct: (1 point)(B) qualitatively correct: (1 point)(B) quantitatively correct: (1 point)

(B)

Q(t)

t

Q(t)

t

(A)

I(t)

t

I(t)

tT 3T

A A

T 3T

T 3T 4T

A

(b) (b1) Basic receiver structure: (2 points)



rBP (t)

rI(t)

rQ(t)

cos(2πfct)

− sin(2πfct)

h∗T (−t)

h∗T (−t)

kT

kT

rI(k)

rQ(k)

bk

Receiver filterIntegrator

Matched filter

Sampling Detector


242 D EXAMS

(b2) (A) correct: (1 point)(B) correct: (1 point)

1

I

Q

I

Q(A) (B)

A A

A

0

0

0 1

(c) (c1) Gray-mapping is characterized by the property that neighboring symbols onlx differ(1 Point)in a single bit. This way it is ensured that the most likely symbol errors cause onlya single bit error.

One possible mapping: (1 Point)

100

Q

I

−2A

−A

A

2A

2AA−A−2A

000

001

011

010

110

111

101

(c2) Approach correct: (1 point)Result correct: (1 point)

ES,A = 12(0 + A2) = A2

2

ES,B = 12(A2 + A2) = A2 ⇒ ES,A

ES,B= 1

2

(c3) Approach correct: (1 point)



Result correct: (1 point)

2PS = 2 · 10−3 = erfc

(√d2

8σ2

)

From the curve one obtains with d2 = 2A2:√d2

8σ2=

√2A2

2≈ 2.2 ⇒ A ≈ 2.2


244 D EXAMS

D.15 Exam WS 2009/2010

Problem 1: LTI-Systems und Signal Theory

Note: Question a), b) and c) can be solved independently from each other. Please also note thehints at the end of this problem.

(a) (8 points)

Consider the following LTI-system

h(t)x(t) y(t)

with the impulse response

h(t) = rect

(t

T− 1

2

).

(a1) Give the general relation between x(t), y(t) and h(t) in frequency domain.

(a2) Is the given system causal? Give a reason for your answer!

(a3) Calculate the transfer function H(f) and draw its magnitude spectrum in the range− 3

T≤ f ≤ 3

T. Pay attention to correct axis labeling!

(a4) Let x(t) be

x(t) =

{ATt+ A −T ≤ t ≤ 0

0 otherwise.

A

x(t)

t-T

Determine the output signal y(t) and give its equation.


D.15 Exam WS 2009/2010 245

(b) (3 points)

The following figure depicts the radio communication link between a transmitter and areceiver.

Receiver

Transmitter Demodulator

Lg = 20dB

PN

d

LS

LE LD

The transmitted signal is subject to attenuation on the radio channel. This attenuation isdirectly related to the distance d (in meters) between transmitter and receiver. The rela-tionship is given as follows

LA = 20dB + 10 log10((d/m)4

)dB.

At the receiver, the received signal is subject to additive noise (with power Pn), is amplifiedand reaches the demodulator which tries to reconstruct the transmitted information fromthe received signal.

(b1) Calculate the transmit power PS (in Watts), if the transmitter uses LS = 46dBm fortransmission.

(b2) Calculate the received power level LD at the input of the demodulator if the transmitpower level is LS = 43dBm, Pn = 0 and the receiver is located at a distance of d = 250mfrom the transmitter.

(c) (4 points)

We consider the same radio communication link as considered in the problem above.

(c1) Many practical systems use antennas that have a certain directional characteristic (re-ferred to as antenna pattern). This means that transmit signals are amplified or atten-uated depending on the transmit direction. We now assume the following positioningof transmitter and receiver.

Transmitter

Receiver

α

Furthermore, we assume the transmitter uses an antenna with a directional amplifica-tion that depends on the angle α (angle between the main direction of the transmitantenna and the receiver) in the following way:

Lant = 24 cos(α) dB.


246 D EXAMS

Calculate the signal-to-noise-ratio (SNR) at the input of the demodulator for thissetup, given the following parameters: the transmitter uses a transmission power ofLS = 20dBm, the angle between the main antenna direction of the transmitter and thereceiver is α = 60◦, the radio channel has an attenuation of LA = 122dB and the noiselevel at the receiver is given with Ln = 10 log10(Pn) = −104dBm.

Hint: cos 0 = 1, cos π6= 1

2

√3, cos π

4= 1

2

√2, cos π

3= 1

2, cos π

2= 0

(c2) The noise power at the receiver is directly related to the bandwidth of the system.It is often assumed that the noise density is -174dBm/Hz. The density units in thelogarithmic domain have the following relation to the units in the linear domain:10 log10(P in mW/Hz) = L in dBm/Hz. Given the above mentioned noise density, whatis the maximum system bandwidth B, if the received power is LE = −98dBm and theminimum required signal-to-noise-ratio (SNR) is 3dB?

Hint:

x 0.1 0.25 1 1.5 2 3 5 10 100 1000

log10(x) −1 −0.6 0 0.2 0.3 0.5 0.7 1 2 3

Aufgabe 2: Analog-digital conversion

(a) (1 Point)

In analog-digital (AD) conversion, an analog signal s(t) is sampled with 48kHz and quantizedwith a resolution of b = 3 Bit. Which data rate is achieved at the output of the ADconversion?

(b) (4 Points)

The following figure shows the real-valued spectrum S(f) of the desired signal s(t). Thesignal is sampled and stored. Afterwards, it is reconstructed again.

f / kHz

S(f )

1

1 3-1-3

0,5

(b1) Which minimal frequency fs,min has to be used for sampling in order to reconstruct thesignal s(t) without errors?

(b2) The signal s(t) is fed into the sampler with the frequency fs ≤ fs,min. What is the nameof the effect which prevents reconstructing s(t) without errors in this case?

(b3) We assume the sampling frequency fs = 4kHz. Draw the resulting spectrum Ss(f) ofthe sampled signal s(t) in the interval f ∈ [−7kHz, 7kHz]! Pay attention to correct axislabeling!


D.15 Exam WS 2009/2010 247

(c) (5 Points)

The following figure presents the quantization characteristics of an analog-digital converter.

��

� ��s(t )

sq(t)

��

��

��

��

��

��

��

�

(c1) The signal s(t) is quantized and sq(t) denotes the output signal of the quantizer. Whichtype of quantizer is shown in the figure above? Give reasons for your answer!

(c2) Δs is the distance between two adjacent quantization steps. Draw the absolute valueof the quantization error |e(t)| as a function of the input signal s(t) in the interval−4Δs ≤ s(t) ≤ 4Δs assuming the given quantization characteristics!

(c3) We assume a constant amplitude density for s(t). The signal is quantized with a reso-lution of 3 bits. What is the general equation of the signal-to-quantization-noise ratioSNR occurring with analog-digital conversion? Calculate the corresponding SNR!


248 D EXAMS


(a) (4 points)

An analog signal s(t) having a bandwidth of 48 kHz (cut-off frequency fG = 24 kHz) shallbe transmitted over a radio link. You can choose between amplitude modulation (AM) andfrequency modulation (FM). In case of frequency modulation, the peak frequency deviationis ΔF = 5kHz.

(a1) The radio channel allows a bandwidth of 50 kHz. Which modulation scheme can beused (AM, FM or both)? Give reasons!

(a2) Now, consider the case when the bandwidth of the channel does not matter. Which ofthe two modulation schemes (AM or FM) is better suited to allow for a high-quality sig-nal transmission that is robust to distance variations between transmitter and receiver?Give reasons!

(b) (5 points)

The analog signal s(t) = 0.5+sin(2πf1t) shall be transmitted by means of amplitude modu-lation, where f1 = 10MHz. The modulated bandpass signal is sBP(t) = (A+s(t)) cos(2πfc t).

(b1) Give an equation for the modulation index μ!

(b2) Consider a receiver with an envelope detector that has certain tolerances and thereforerequires a modulation index μ ≤ 0.8. Which value of A is then required?

(b3) Calculate the frequency spectrum SBP(f) of the modulated bandpass signal sBP(t) interms of f1, fc und A! Note: sin(a) · cos(b) = 0.5 (sin(a− b) + sin(a+ b)).

(b4) What is the minimum frequency fc of the carrier signal cos(2πfc t) that is required totransmit an arbitrary signal s(t) with a bandwidth of 10MHz without distortions?

(c) (2 points)

A modulated single-sideband signal sESB(t) shall be demodulated with a heterodyne receiver.The figure below shows the magnitude spectrum |SESB(f)| of sESB(t).

f / MHz

|SESB (f )|

-1200 -1100 1150 1200

2.0

1.0

1100-1150

The next figure shows the block diagram of the heterodyne receiver. The frequency of thetwo mixers is fLO = fZF = 650MHz.

sESB (t)

cos(2πfLO t)

Bandpass filter

350MHz < |f |< 650MHz

y(t)sZF (t)

cos(2πfZF t)

Lowpass filter

|f |< 250MHz


D.15 Exam WS 2009/2010 249

(c1) Draw the magnitude spectrum |Y (f)| of the demodulated signal y(t) in the range−400MHz ≤ f ≤ 400MHz! Pay attention to correct axis labeling!


(a) (6 Points) Given is a segment of the time-domain representation of a digitally modulated sig-nal sBP (t) in the bandpass domain, as shown in the following figure. All possible transmittedsymbols occur within this segment.

0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

3

t/Ts

s BP(t)

(a1) Specify the modulation scheme that is used, including the modulation order!

(a2) Draw the corresponding constellation diagram (I/Q-diagram)! Pay attention to correctaxis labeling! Draw a bit assignment into the diagram that satisfies the Gray mappingrule! What is the bit sequence that corresponds to the depicted signal segment?

(a3) In general the signal sBP (t) can be written as

sBP (t) = Ts

∞∑k=−∞

dkhT (t− kTs) cos(2πfct) .

Specify the impulse response hT (t) of the used transmit filter and the carrier frequencyfc!

(b) (4 Points) Consider now the three modulation schemes (A), (B), and (C) of order four,having the following constellation diagrams (IQ-diagrams):

I

Q(B)

I

Q(C)

I

Q(A)

A1

A1−A1

−A1

A2

A2

A3

A3

−A3

−A3


250 D EXAMS

(b1) Given is the following receiver structure:

rBP (t)

rI(t)

rQ(t)

cos(2πfct)

− sin(2πfct)

h∗T (−t)

h∗T (−t)

kT

kT

rI{k}

rQ{k}

dI{k}

dQ{k}

b{n}SymbolDemapper

The detectors in the inphase- and quadrature branches make a decision based on therule

dI/Q{k} =

{1 if rI/Q{k} > 0

−1 otherwise..

Specify for which of the three modulation schemes (A), (B), and (C) the minimumpossible bit error rate can be achieved with this receiver structure if all transmit symbolsoccur with equal probability (multiple selections are possible)! Give reasons for youranswer!

(b2) Assume now that an optimal receiver with ideal decision boundaries is used. For thiscase the modulation schemes (A) and (B) shall be compared, given that all transmitsymbols again occur with equal probability.How large does the amplitude A2 have to be chosen if A1 = 1 and the same bit errorprobability is desired for (A) and (B)? Which of the modulation schemes requires inthis case a larger average signal energy ES? Give reasons for your answer!

(c) (4 Points) Consider now a BPSK transmission, were the constellation appears rotated bythe angle ϕ at the receiver due to a phase shift.

I

Q

ϕd

For an optimal receiver the bit error rate is given by

BER =1

2erfc

(√d2

8σ2

),

where σ2 denotes the noise variance and the complementary error function erfc(x) is given inthe figure below. But we only have access to a conventional BPSK receiver, which is designedfor the case ϕ = 0 and unable to perform a phase correction.


D.15 Exam WS 2009/2010 251

(c1) Draw the decision boundary of the receiver designed for ϕ = 0 into the depicted con-stellation diagram!

(c2) Give an expression for the resulting bit error rate for −π/2 < ϕ < π/2!

(c3) How large does d have to be chosen such that a bit error rate of 0.5% is achieved forϕ = 45◦ and σ = 0.5?

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)


252 D EXAMS


Problem 1: LTI-Systems and Signal Theory

(a) (a1) Y (f) = X(f) ·H(f) (1 Point)

(a2) The system is causal, because h(t) = 0 for t < 0. (1 Point)

(a3) H(f) = F {h(t)} = F {rect

(tT− 1

2

)}From the correspondance table: F {rect(t)} = si(πf)

Applying the scaling property: F {rect

(tT

)}= T si(Tπf)

With the time shifting property: H(f) = F{rect

(t−T

2

T

)}= T si(Tπf)e−jπfT

(1 Point)

Drawing as follows:

-3/T -2/T -1/T 0 1/T 2/T 3/T0

T

f

|H(f)

|

Drawing qualitatively correct: (1 Point)Correct axis labeling: (1 Point)

(a4) Aside from the trivial case when both signals do not overlap, there are two cases todistinguish to calculate the convolution. They are depicted in the sketch below

-T t t-T 0

Case 1 Case 2



Calculation for Case 1:

y(t) =

∫ t

−T

A

Tτ + A dτ =

A

2Tt2 + At− A

2T

Calculation for Case 2:

y(t) =

∫ 0

t−T

A

Tτ + A dτ = − A

2Tt2 +

A

2T

Thus, it follows:

y(t) =

⎧⎪⎪⎨⎪⎪⎩

A2Tt2 + At− A

2T −T ≤ t < 0

− A2Tt2 + A

2T 0 ≤ t < T

0 otherwise

Cases correctly distinguished: (1 Point)Correct result for Case 1: (1 Point)Correct result for Case 2: (1 Point)

(b) (b1) PS = 40W

Correct result and unit: (1 Point)

(b2) LA = 20dB + 10 log10(2504)dB = 20dB + 4 · 24dB = 116dB

Channel attenuation correct: (1 Point)

LE = LS − LA + Lg = 43dBm− 116dB + 20dB = −53dBm

Correct result and unit: (1 Point)

(c) (c1) Lant = 24 · cos π3dB = 12dB

Directional antenna amplification toward user correct: (1 Point)

SNR = LS + Lant − LA − Ln = 20dBm + 12dB− 122dB− (−104dBm) = 14dBCorrect result and unit: (1 Point)

(c2) Ln = LE − SNR = −98dBm− 3dB = −101dBmCorrect noise level: (1 Point)

It follows: B = −101dBm− (−174dBm/Hz) = 73dB Hz = 20MHzCorrect result and unit: (1 Point)

Aufgabe 2: Analog-digital Conversion

(a) Data rate with analog-digital conversion: 144 Kbps (1 Point)

(b) (b1) Nyquist criterium: fs,min > 2fg = 6kHz (1 Point)


254 D EXAMS

(b2) Aliasing (1 Point)

(b3) Drawing as follows

Diagram qualitatively correct - S(f) periodically continued (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

f / kHz

S(f )

1

1 3-1-3

2

-7 -5 75

1,5

(c) (c1) Mid-rise quantizer sq(t) ∈ {±Δs2· (1, 3, 5, ...q − 1)

}because the quantization charac-

teristic does not include the quantization step sq(t) = 0 (1 Point)

(c2) | e(t) |=| sq(t)− s(t) | as function of s(t) corresponds to following figure

|e(t)|

��

s(t )

�

� � � � � ��

Diagram qualitatively correct - curve (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

(c3) SNR = 6.02b dB (1 Point)SNR = 6.02dB/bit× 3bit = 18.06 dB (1 Point)


(a) (a1) AM (1 point)Reasons: (1 point)The bandwidth of the AM signals is only 48 kHz, i.e. less than 50 kHz. Itdoes not exceed the bandwidth of the radio channel. The bandwidth of the FMsignal derives from carson rule as B + 2ΔF = 58 kHz. Hence, FM would exceed thebandwidth of the radio channel.

(a2) FM (1 point)Reasons: (1 point)AM modulates the amplitude of the carrier signal. The pathloss of the radiochannel causes an attenuation that depends on the distance between transmitter andthe receiver and hence effects the quality of an AM signal. FM modulates frequencyof the carrier signal, which is not affected by the pathloss.



(b) (b1) μ =max |s(t)|

A(1 point)

(b2) For μ ≤ 0.8 follows with max |s(t)| = 1.5: A ≥ 15

8= 1.875. (1 point)

(b3) SBP(f) = (0.5 · A+ 0.25) (δ(f + fc) + δ(f − fc)) (2 points)+ 0.25 j (δ(f + f1 − fc)− δ(f − f1 + fc) + δ(f + f1 + fc)− δ(f − f1 − fc))

(b4) fc ≥ 5 MHz (1 point)

(c) (c1) Magnitude spectrum |Y (f)| - see figure below.Qualitatively correct: (1 point)Quantitatively correct (axis labeling): (1 point)

f / MHz

|Y (f )|

100 200-150 -100

0.50

0.25

-200 150


(a) (a1) 4-ASK (1 Point)

(a2)

3 I

Q11 01 00 10

−3 −1 1

The figure shows an example assignment that follows the Gray mapping rule. In totalthere exist the following correct assignments:

dk = −3 dk = −1 dk = 1 dk = 3

00 01 10 11

00 10 01 11

01 00 10 11

10 00 01 11

11 01 00 10

11 10 00 01

01 11 10 00

10 11 01 00

From the depicted signal segment follows: d0 = 1, d1 = 3, d2 = −1 and d3 = −3. Forthe example assignment the bit sequence is equal to 0, 0, 1, 0, 0, 1, 1, 1

Constellation diagram correct: (1 Point)


256 D EXAMS

Bit assignment correct (Gray mapping): (1 Point)Bit sequence correct: (1 Point)

(a3) hT (t) = 1/Tsrect(t/Ts − 1/2) (1 Point)

fc = 3/Ts (1 Point)

(b) (b1) The receiver is only optimal for (C). In case (A), I and Q cannot be decided indepen-dently. In case (B), a threshold equal to A2/2 is required in the detectors.

Correct choice: (1 Point)Explanation correct: (1 Point)

(b2) For equal bit error rate the possible distances between constellation points must beequal. Thus, we have

√2A1 = A2 and hence A2 =

√2. (1 Point)

Average energy: (A) ES =4A2

1

4= 1 (B) ES =

0+A22+A2

2+2A22

4= A2

2 = 2

⇒ for equal bit error rate a larger (the double) average energy is required in case (B).(1 Point)



(c) (c1) Decision boundary: (1 Point)

I

Q

ϕd

(c2) The receiver uses the inphase part of the signal, which reduces the distance to d cosϕ

⇒ BER = 12erfc

(√(d cosϕ)2

8σ2

). (2 Points)

(c3) We have: (d cosϕ)2 = d2/2

⇒ 0.01 = erfc(√

d2/(2 · 2))= erfc (d/2)

From the curve we have: d/2 ≈ 1.8 ⇒ d ≈ 3.6 (1 Point)


258 D EXAMS

D.17 Exam SS 2010



(a) (3 points)

A discrete time system is defined by

y[n] = 0.5x[n] - 0.25nx[n− 1] ,

where n is the discrete time index.Answer the following questions with reason!


(a2) Is the system time-invariant?

(a3) Is the system causal?

(b) (6 points)

An LTI system, as described below in the figure

h(t)x(t) y(t)

has an output of y(t) = rect( t2) when an impulse, x(t) = δ(t) is applied at the input.

(b1) Determine the frequency response H(f) of the system! Also sketch the waveform ofH(f)!

(b2) Find an analytical expression and sketch the output signal y(t), when the input signalis x(t)=rect(t)!

(b3) Determine the frequency spectrum of the output signal y(t), when the input signal isx(t)=rect(t)!


D.17 Exam SS 2010 259

(c) (6 points)

The figure shows the block diagram of a radio transmission link between a mobile and abase station.


LA = 100dB + 10 log10((d/km)4

)dB


(c1) The transmit power of the mobile is 1W. Calculate the received power, PR in Watt atthe base station located at a distance of d = 1000m!

(c2) Calculate the Signal to Noise Ratio (SNR) in dB at point A, the input of the amplifier!

(c3) The minimum SNR required at pointB, the input of the demodulator is 10dB. Calculatethe maximum allowable distance between the mobile and the base station!

Note:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3

Problem 2: Analog-digital conversion

Analog-digital (AD) converters transform analog signals to digital signals. Following figure showsthe connection.

Analog signal Sampling Quantisation Digital

signals(t) ss(t) sq(t)

Analog-digital converter

(a) (2 Points)


260 D EXAMS

(a1) Which sampling frequency fS is neccessary to reconstruct the original signal s(t) withfg = 2kHz from the sampling values without errors. Motivate the answer!

(a2) We consider the AD conversion of the signal s(t). Quantisation is done with1024 quantisation steps. Is it possible to reconstruct the original signal s(t) fromthe resulting signal sq(t) without errors? Motivate the answer!

(b) (4 Points)

S(f) indicates the spectrum of the analog signal s(t). S(f) is defined by the following functionwith fg = 1kHz.

S(f) = rect

(f

2fg

).

s(t) is sampled with the sampling frequency fS = 3kHz.

(b1) Sketch the spectrum of the sampled signal ss(t) in the interval [−4kHz, 4kHz]!

(b2) Now, we consider the single carrier signal s(t) = A cos(2πft) with A = 1 and f = 2kHz.

Draw the spectrum of the sampled signal ss(t) in the range [−4kHz, 4kHz]!

(c) (4 Points)

We consider s(t) = sin(2π fS4t) as input signal of the AD conversion and the

sampling frequency fS = 1TS. Here, two alternative quantisers are available. ss(t) denotes

the sampled signal before the quantisation.

(c1) The quantisation of ss(t) is applied by a mid-rise quantiser at a resolution of 4 bits, astep width of Δs = 50mV and a symmetric recording level [−Umax, Umax]. Calculate thewidth of the recording level at the input of the quantiser and the maximum absolutevalue of the quantization error emax!

(c2) The second quantiser has following quantisation characteristic q(·).

q(ss(t)) =

⎧⎪⎪⎨⎪⎪⎩1 for ss(t) > 0

0 for ss(t) = 0

−1 for ss(t) < 0 .

Draw the output signal sq(t) = q(ss(t)) for 0 ≤ t ≤ 6TS! Make sure you label the axesof the diagram correctly!


D.17 Exam SS 2010 261

Task 3: Analog Modulation

(a) (3 Points)

(a1) Explain the term analog modulation!

(a2) Give two reasons why it is advantageous to transmit modulated carriers compared tobase band transmission!

(b) (8 Points)

An analog signal sQ(t) = cos(2πf1t) + cos(8πf1t) shall be transmitted through a carriersc(t) = A cos(2πfct) by amplitude modulation. The amplitude modulated signal is givenas sAM(t) =

(A+ sQ(t)

) · cos(2πfct) .(b1) Determine the modulation index μ of the amplitude modulated signal for A = 2.5 and

A = 0.4! Reason whether or not the signal can be demodulated correctly by an envelopedemodulator!

(b2) Compute the square of the amplitude modulated signal s(t) = s2AM(t) for A = 1! Com-pute the spectrum of s(t). Sketch the spectrum for −8f1 ≤ f ≤ 8f1! Assume fc � f1.

Hint: cos2(x) = 12

(cos(2x) + 1

)In the following, the analog signal sQ(t) shall modulate the phase of the carrier sc(t) insteadof its amplitude. The phase modulated carrier shall be denoted by sPM(t) .

(b3) Give an equation for the phase modulated carrier sPM(t)!

(b4) Determine an equation for the frequency of the phase modulated carrier sPM(t)!


A model pilot wants to develop a remote control system for his newest plane. He decides to use adigital data transmission method. The remote control transmits data packets to the model.

(a) (4 points)

A channel with a bandwidth B = 10kHz is available for the transmission. One packet has asize of 25 bytes.

(a1) State the general relationship between symbol duration T and bandwidth B for linearmodulation!

The bandwidth of a transmit pulse shall now be given as B =2

T.

(a2) How many packets can be transmitted per second when on-off-keying (OOK) is used?


262 D EXAMS

(a3) Tests have shown that at least 80 packets per second are necessary for accurate control.However, the channel bandwidth cannot be increased. Is there a modulation schemethat fulfills these requirements? If yes, give the name of the method and sketch itsconstellation diagram!

(b) (3 points)

After implementing a better compression method, it’s now sufficient to use a two-valuedmodulation scheme. The following figure shows a possible transmitter for a BPSK-signal.

(b1) Draw a brief sketch of a suitable receiver for this modulation scheme from the antennato the estimated bitstream!

(b2) Is your receiver able to compensate a random phase rotation that is caused by thechannel? State a reason and explain the required enhancements if not!

(c) (7 points)

The data is now modulated using BPSK with the symbols s0 and s1. Both symbols aresent with equal probability. The signal is distorted by additive white Gaussian noise withvariance σ2. The decision boundary of the receiver is D. It is then possible to calculate thebit error rate using the following equation:

BER =1

4erfc

(s0 −D√

2σ2

)+

1

4erfc

(−s1 −D√

2σ2

).

The complementary error function erfc(·) is shown on the next page. Furthermore, s0 =−s1 = 4mV, σ2 = 2(mV)2 and D = 0mV are given.

(c1) Give one advantage of BPSK over OOK!

(c2) Calculate the bit error rate of the receiver!

(c3) State a reason whether or not the decision boundary D = 0mV minimizes the bit errorrate!

Due to tolerances of the electrical components the values of all received symbols are shiftedby a constant value c given in [mV] with 0mV ≤ c < s0. Assume that the receiver algorithmsare not aware of this shift and therefore they don’t compensate it.Hint: c < s0 implies that the symbols are not shifted over the desicion boundary.

(c4) Extend the equation for the bit error rate by the influence of c!


D.17 Exam SS 2010 263

(c5) The error correction capabilities of the system only work well if the bit error rate doesnot exceed BER = 4 · 10−2. What is the maximum value of c such that this limit is notexceeded?Hint: The smaller of the two erfc(·) terms can be omitted for c > 1mV.

0 0.5 1 2 310−5

10−4

10−3

10−2

10−1

100

x

y=er

fc(x

)Complementry Error Function


264 D EXAMS



(a) (a1) For two input sequences x1[n] and x2[n], the outputs are

y1[n] = 0.5x1[n] - 0.25nx1[n− 1]

y2[n] = 0.5x2[n] - 0.25nx2[n− 1] .

A linear combination of the two input sequences results in the output

y3[n] =T [a1x1[n] + a2x2[n]]

y3[n] =0.5a1x1[n] - 0.25a1nx1[n− 1] + 0.5a2x2[n] - 0.25a2nx2[n− 1] .

Again, the linear combination of the two outputs results in

a1y1[n] + a2y2[n] = 0.5a1x1[n] - 0.25a1nx1[n− 1] + 0.5a2x2[n] - 0.25a2nx2[n− 1] .

From the above two equations, we can say that the system is linear. (1 Point)

(a2) The response of the system to x[n− k] is

y[n, k] = 0.5x[n] - 0.25nx[n− k − 1] .

Now, if we delay y[n] by k units in time, we obtain

y[n− k] = 0.5x[n− k] - 0.25(n− k)x[n− k − 1] .

Hence from the above we see that the system is NOT time invariant. (1 Point)

(a3) The system is causal, as the present output depends only on the present input andpast inputs and not on any future inputs. (1 Point)

(b) (b1) The impulse response is

h(t) = rect( t2) .

The Fourier transform yields

H(f) =∫ ∞−∞ rect( t

2)e−j2πftdt =

∫ 1

−11 · e−j2πftdt

H(f) = sin(2πf)πf

= 2 si(2πf)

(1 Point)



(1 Point)

(b2) The output signal is given by

y(t) = h(t) ∗ x(t) = ∫ ∞−∞ rect( τ

2)rect(t− τ)dτ

y(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 t ≤ −32∫ t+1/2

−1dτ = 3

2+ t −3

2≤ t ≤ −1

2∫ 1

0dτ = 1 −1

2≤ t ≤ 1

2∫ 1

t−1/2dτ = 3

2− t 1

2≤ t ≤ 3

3

0 32≤ t

.

(2 Points)

(1 Point)

(b3) The spectrum of the output is given as


266 D EXAMS

Y (f) = H(f) ∗X(f) = sin(2πf)πf

· sin(πf)πf

= sin(πf) sin(2πf)π2f2 .

(1 Point)

(c) (c1) The transmit power is 1 W, which corresponds to 30 dBm

LA = 100 + 10 log10(1)4 = 100 + 40(0) = 100 dB.

The received power level at the receiver is

LR = LT − LA = 30 dBm−100 dB= −70 dBm.

The received power in Watt is

PR = 10−10W .

(2 Points)

(c2) The SNR at the input of the amplifier is given as

SNR = PR

PN1= 10−10W

10−14W= 104 = 40 dB.

(1 Point)

(c3) The SNR at the output of the demodulator is 10 dB.Therefore the signal power is

10 · PB = 10(gPN1 + PN2) = 10−11 W.

The signal power before the amplifier is

10−11

10= 10−12 W = 10−9 mW.

The signal power level before the amplifier is −90dBm.If the minimum distance is x, then the signal power at the receiver before the amplifieris

LR = 30− (100 + 40 log10(x)) = −70− 40 log10(x) .

Therefore we have

−70− 40 log10(x) = −90log10(x) =

2040

= 0.5x = 3 .

(2 Points) for correct derivation

The maximum allowable distance between the mobile and the base station is 3 km.

(1 Point) for the correct value



Problem 2: Analog-digital conversion

(a) (a1) The original signal s(t) can be reconstructed from the sampling values without errorsif the sampling follows fS > 2fg (sampling theorem). (1 Point)

(a2) Quantisation is always lossy. Therefore it is not possible to reconstruct s(t) from sq(t)without errors. (1 Point)

(b) (b1) Spectrum | Ss(f) | is the periodic continuation of the original spectrum S(f).

Diagram qualitatively correct - (| Ss(f) |) periodically continued (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

f / kHz

1

1 3-1-3

|Ss(f)|

-2-4 2 4

(b2) Spectrum | Ss(f) | according to the figure below.

Diagram qualitatively correct - (aliasing) (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

f / kHz1 3-1-3

|Ss(f)|

-2-4 2 4

1

(c) (c1) Assuming the maximum absolute quantisation error emax, the width of the recordinglevel 2U , the step width Δs and a resolution in bit b, calculation results in

q = 2b = 16q = 2U

Δs

2U = q ·Δs ⇒ 2U = 16 · 50mV = 800mV (1 Point)emax = Δs

2⇒ emax = 25mV . (1 Point)

(c2) Signal sq(t) according to the figure below.

Diagram qualitatively correct - curve (1 Point)Diagram quantitatively correct - axis labeling (1 Point)

t / TS1 3

sq(t )

2 4

1

5 60

-1


268 D EXAMS


(a) (a1) Analog modulation is understood as imposing an analog source signal on a carriersignal. (1 Point)

(a2) Reasons for transmitting modulated carriers are (2 Points)

• Adjustment of the source signal into the transmitting medium

• Possibility of transmission of multiple signals over the same medium (frequencymultiplexing).

(b) (b1) The modulation index is defined as μ =max |sQ|

A. For A = 2.5 and A = 0.4 we have

(1 Point) for two correct values

μ1 = 2 · 25=

4

5

μ2 = 2 · 52= 5 .

Since the information is contained not only in the amplitude but also in the phase ofthe carrier if μ > 1, an envelope detector cannot be used in the second case whereμ = 5. (1 Point)

(b2) The squared signal s(t) = s2AM(t) for A = 1 is given as

s(t) = a(t) · [1 + cos(4πfct)]

where

a(t) =

[1+cos(2πf1t)+cos(8πf1t)+cos(2πf1t) cos(8πf1t)+

1

4cos(4πf1t)+

1

4cos(16πf1t)

].

(1 Point)

For the spectrum S(f) we obtain

S(f) = A(f) + A(f − 2fc) + A(f + 2fc)

where

A(f) = δ(f) +1

2

[δ(f − f1) + δ(f + f1) + δ(f − 4f1) + δ(f + 4f1)

]+

+1

4

[δ(f − 3f1) + δ(f + 3f1) + δ(f − 5f1) + δ(f + 5f1)

]+

+1

8

[δ(f − 2f1) + δ(f + 2f1) + δ(f − 8f1) + δ(f + 8f1)

].



18

12

∣∣S(f)∣∣

ff10 2f1 3f1 4f1 5f1 8f1−5f1−8f1 −4f1−3f1−2f1−f1

1

14

Figure D.8: Amplitude spectrum of the signal s2AM(t) in the range −8f1 ≤ f ≤ 8f1

(2 Points)

Since fc � f1 we do not need to consider the terms A(f − 2fc) and A(f + 2fc) in thesketch. The spectrum of the signal in the range −8f1 ≤ f ≤ 8f1 is depicted in FigureD.8.

(1 Point) if drawing corresponds to the spectrum equation obtained above (even ifequation is not correct), correct axes, correct range.

(b3) In case of a phase modulation the modulated signal is given as (1 Point)

sPM(t) = A cos(2πfct+ kPM · sQ(t) + φ0

)= A cos

(2πfct+ kPM · ( cos(2πf1t) + cos(8πf1t)

)+ φ0

).

(b4) With the phase of the modulated signal φ(t) = 2πfct+kPM

(cos(2πf1t)+cos(8πf1t)

)+φ0

the frequency writes as (1 Point)

f(t) =1

2π

dφ(t)

dt= fc + kPM(f1 sin(2πf1t) + 4f1 sin(8πf1t)).


(a) (a1) B ∼ 1

T(1 Point)


270 D EXAMS

(a2) With M = 1 bit per symbol, the bit rate R is given as: (1 Point)

R =M

T=M · B

2= 5kbit · s−1

With 25 · 8 = 200 bits per packet it is possible to transfer 5000/200 = 25 packets persecond!

(a3) The data rate must be increased by factor 80/25 = 3.2. Quadrupling the rate yields theclosest result. Possible modulation schemes are therefore: 16-PSK, 16-QAM or schemeswith even more bits per symbol. (1 Point)

Diagram correct: (1 Point)

(b) (b1) Valid receivers: I/Q-receiver, receiver with only I-branchBlock diagram correct (1 Point)Labels correct (1 Point)

(b2) In case of an I/Q-receiver:The phase rotation can be estimated and mitigated using a complex multiplication inthe digital basebandFor receiver with only an I-branch:No, for growing phase rotation, the resulting signal power will decrease. If the rotationreaches 90◦ the signal power becomes zero and no detection is possible. The problemcan be solved by adding a Q-branch which transforms the receiver to an I/Q-receiver.Explanation correct (1 Point)

(c) (c1) Possible reasons: (1 Point)

• Smaller BER with the same transmit power

• Smaller transmit power while retaining the BER



(c2) BER =1

4erfc

(4mV− 0mV

4(mV)2

)+

1

4erfc

(−4mV− 0mV

4(mV)2

)=

1

2erfc(2) (1 Point)

erfc(2) = 4, 7 · 10−3 (Correct range: [4 . . . 5] · 10−3)BER = 2, 3 · 10−3 (Correct range: [2 . . . 2, 5] · 10−3) (1 Point)

(c3) Yes, the distance to s0 and s1 is maximized and equal. Therefore, the BER is minimized!(1 Point)

(c4) BER =1

4erfc

(s0 + c−D√

2σ2

)+

1

4erfc

(−s1 + c−D√

2σ2

)(1 Point)

(c5) Using the approximation we get: BER ≈ 1

4erfc

(−s1 + c√

2σ2

)=

1

4erfc

(4mV− c

2mV

)(1 Point)erfc(x) = 4 · 4 · 10−2 = 1, 6 · 10−1 at roughly x = 14mV− c

2mV= 1� c = 2mV

The maximum value for c is 2mV! (1 Point)


272 D EXAMS

D.19 Exam WS 2010/2011



(a) (5 points)

Given the following signal

x(t) =5∑

n=−5

(−1)n · (n− 1) · δ(t− n)


(a1) Draw x(t)! Pay attention to correct axis labeling!

(a2) Can the signal x(t) be a valid impulse response of a causal system?

(a3) Is the signal x(t) odd or even?

(a4) Is the spectrum X(f) of x(t) real valued?

(b) (5 points)

Two LTI systems with impulse responses h1(t) =rect(t+ 12) and h2(t) = 2rect(t) are serially

connected, as shown in the figure below, with overall impulse response h(t).

h(t) = h1(t) ∗ h2(t)

(b1) Draw the impulse response of the overall system! Pay attention to correct axis labeling!

(b2) Determine the transfer function H(f) of the system!

(b3) Draw the output signal y(t) when the input signal x(t) is

x(t) = −δ(t) + δ(t− 1)− 0.5δ(t− 2) + 1.5δ(t− 3)

Pay attention to correct axis labeling!

(c) (5 points)



D.19 Exam WS 2010/2011 273


LS = 20 log10

(100d

m

)dB,

where d is the distance between the transmitter and receiver in meter.

(c1) Calculate the maximum distance, d that ensures a minimum signal to noise ratio (SNR)of 5dB at point B, the input of the demodulator!

(c2) Calculate the minimum power level LT (in dBm) necessary, in order to reach a SNR of20dB at point B, at a distance of d = 50m!

Note:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3


In general, digital signals are stored after the analog-digital (A/D) conversion in a digital memory,e.g. the main memory. In order to convert digital signals back to analog signals, digital-analog(D/A) converters are needed. This relationship is illustrated in the figure below.

(a) (2 points)

(a1) We consider x′(t) the recovered, analog signal after the sampling and quantization.Why is an errorless reconstruction of the analog signal x(t) not possible even assumingideally working components?


274 D EXAMS

(a2) Name two advantages of digital signals over analog signals!

(b) (4 points)

(b1) x1(t) denotes a bandwidth-limited, analog signal, as shown in the figure below.

��

��

��

In the sampling, x1(t) is multiplied by the Dirac-comb function XT (t).

XT (t) =∞∑

n=−∞δ(t− nT ).

The resulting sampled data sequence xs1(t) results from xs1(t) = x1(t)XT (t).Draw xs1(t) in the region −4T ≤ t ≤ 4T ! Pay attention to correct axis labeling!


D.19 Exam WS 2010/2011 275

In sampling x2(t), the sampling frequency fs2 = 5kHz is used. Draw the amplitudespectrum of the sampled signal xs2(t) in the interval f ∈ [0kHz, fs2]! Name and highlightthe effect caused by the low sampling frequency! Pay attention to correct axis labeling!

(c) (4 points) Now, we consider a quantizer with distorted (biased) quantization characteristic,shown in the figure below. Δs denotes the step width of the quantizer.

��

�q��

��

��

��

��

��

��

��biased

(c1) How large is the dynamic range 2U of the quantizer with Δs = 50mV? Determine thewidth of the dynamic range at the input of the quantizer and the resolution b of thequantizer in bit!

(c2) The error signal e(t) = x(t) − xq(t) is the deviation of the quantized signal xq(t)from the original signal x(t). Draw e(t) using the quantizer shown above in the region−4Δs ≤ t ≤ 4Δs! Pay attention to correct axis labeling!


276 D EXAMS


A signal sQ(t) = AQ sinc (t) is transmitted by a carrier c(t) = Ac cos(2πfct) via amplitude modu-lation. The amplitude modulated signal is given as

s1(t) =(Ac + sQ(t)

)cos(2πfc1t).

(a) (2 points)

(a1) Calculate the modulation index μ of s1(t)!

(a2) Specify the range of the amplitude AQ, for which the signal s1(t) exhibits no phasejumps!

(b) (4 points)

The receiver consists of the components depicted in Figure D.9, where a band pass filterand mixer are used for channel selection. In addition to s1(t), the signal s2(t) = (Ac +sQ(t)) cos(2πfc2t) from a neighboring channel is received at the antenna and the sum s(t) =s1(t) + s2(t) is fed into the mixer. In the following we assume AQ = Ac = 1, fc1 +

12Hz <

fM < fc2 − 12Hz.

(b1) Explain briefly the purpose of the low pass filter and the high pass filter in the demod-ulator in Figure D.9!

(b2) Sketch the amplitude spectrum |S(f)| of the signal s(t) at the mixer input for f > 0!

(b3) Sketch the amplitude spectrum |S ′(f)| of the signal s′(t) at the mixer output for f > 0!

(c) (5 points)

Consider the band pass filter in Figure D.9 as being ideal with a bandwidth of 1Hz and denoteits center frequency by fbp. For a certain range of the mixer frequency fM it is possible thatthe signals s1(t) and s2(t) spectrally overlap after the bandpass filter.

(c1) State the conditions on the mixer frequency fM such that s1(t) and s2(t) do NOTspectrally overlap after the band pass filter! Distinguish the cases fbp = fM − fc1 andfbp = fM + fc1. Also take into account the bandwidth of the signals s1(t) and s2(t)!

(c2) State another simple signal processing technique at the receiver to avoid spectral overlapof s1(t) and s2(t) after mixing!


(a) (6 points)

The following figure shows the plots of three different digitally modulated transmit signals.All possible transmit symbols are shown in each plot


D.19 Exam WS 2010/2011 277

Figure D.9: Receiver block diagram

0 1 2 3 4−1

0

1

t/Ts

y/A

0 1 2 3 4−1

0

1

t/Ts

y/A

0 1 2 3 4−1

0

1

t/Ts

y/A

(a1) Which digital modulation schemes are shown? Give the names and the valency!

(a2) A radio system shall transmit a video stream with a data rate of 768kbit/s. The symbolduration of the system is fixed at Ts = 5μs. Is there a modulation scheme that achievesthe required data rate with the given symbol duration? Give the name of the schemeand the resulting maximum data rate of the system!

(a3) Sketch the constellation diagrams of QPSK and 8-PSK!

The following figure shows a transmitter that is to be used for transmission of digital data.Two incoming information bits are assigned to one baseband symbol and all possible symbolsoccur with equal probability. A filter with rectangular impulse response is used for pulseshaping such that the kth symbol is transmitted in the time interval [k · Ts, (k + 1) · Ts]. Tsdenotes the symbol duration.

(b) (4 points)

(b1) Draw a brief sketch of a suitable receiver for this modulation scheme and give the namesof all components!


278 D EXAMS

(b2) The data sequence {01, 10, 00, 11} shall be transmitted. fc = 1kHz and Ts = 2ms aregiven. Draw the resulting signal x(t) in the interval 0 ≤ t ≤ 4Ts! Pay attention tocorrect axis labeling!


D.19 Exam WS 2010/2011 279

(c) (4 points)

The signal r(k) at the input of the detector can be written as follows:

r(k) = s(k) + n(k)

n(k) is white Gaussian noise with variance σ2. The resulting bit error rate is then given bythe following formula:

BER =3

8erfc

(√A2

8σ2

)

Furthermore, the average symbol energy is Es = E {|s(k)|2} and the signal-to-noise ratio isSNR = Es/σ

2.

(c1) The system allows for dynamic adaptation of the parameter A to the current transmis-sion conditions. An attached error protection scheme (not shown) is able to compensatea maximum bit error rate of 0.375 · 10−2. On the other hand, choosing a very large Amight be a waste of transmit power. Let the noise variance be σ2 = 0.5. Calculate thesmallest possible A for which the required bit error rate is reached!

(c2) When using the system in a different scenario you calculate that A = 1 is required forreliable operation. For this value, the average symbol energy (according to the formulagiven above) is Es = 3.5. However, regulatory conditions state an upper limit of Es ≤ 2.The system allows you to change the mapping between bits and symbols in the mapper.However, you are neither allowed to change the data throughput nor the error toleranceof the system. Give a new mapping that fulfills all these requirements! Give the averagesymbol energy for the new mapping for A = 1!


280 D EXAMS

0 0.5 1 2 310−5

10−4

10−3

10−2

10−1

100

x

y=er

fc(x

)

Complementry Error Function





(a) (a1) The plot of the signal is as follows

Shape correct (1 point)Axis labels correct (1 point)

(a2) No, the system would not be a causal system as x(t) = 0 for t < 0 is not satisfied.(1 point)

(a3) The signal is neither odd nor even, as x(−t) �= −x(t) and also x(−t) �= x(t). (1 point)

(a4) No, X(f) is complex valued as x(t) is neither odd nor even. (1 point)

(b) (b1) The overall impulse response of the system is


282 D EXAMS


(b2) From

h(t) = rect(t+ 1/2) ∗ 2 rect(t)follows

H(f) = 2ejπf · si(πf) · si(πf)(1 point)

(b3) The output signal is


(c) (c1) The noise power level at the input of the demodulator is

LN,dem = LN + g2 = −50dBm + 60dB = 10dBm

The signal power level at the input of the demodulator is

LT,dem = LT − LS + g1 + g2 = 15dBm− 20 log10 (100d/m)dB + 20dB + 60dBLT,dem = 95dBm− 20 log10 (100d/m) dB

Correct equations (1 point)

The SNR at the input of the demodulator is

SNR = LT,dem − LN,dem ≥ 5dB95dBm− 20 log10 (100d/m)dB− 10dBm ≥ 5dB

log10 (100d/m) ≤ 4d ≤ 100m



Therefore maximum distance d=100m.Correct equations till the penultimate step (1 point)Correct value (1 point)

(c2) Solving for LT

SNR ≤ LT − LS + g1 + g2 − LN − g2LT ≥ SNR + LS − g1 + LN

LT ≥ 20dB + 20 log10 (5000) dB− 20dB− 50dBmLT ≥ 24dBm

Therefore the minimum transmitter power level LT = 24dBm.Correct equations (1 point)Correct value (1 point)


(a) (a1) Quantization noise (&saturation). (1 point)

(a2) Digital signals are easily stored, technology-and temperature-independent. (1 point)

(b) (b1) The sampled data sequence xs1(t) preserves the shape of the corresponding analogsignal in principle.

Qualitatively correct: xs1(t) (1 point)Quantitatively correct: (axis label) (1 point)

��

��

� ��

(b2) Amplitude spectrum | Xs2(f) | according to figure.Qualitatively correct: (aliasing visible) (1 point)Quantitatively correct: (axis label) (1 point)

� kHz

��

�

Aliasing

� ��


284 D EXAMS

(c) (c1) With the width of the dynamic range 2U , the number of quantization steps q, the stepwidth Δs and the resolution b in bit follows

q = 7,b = �ld q� = 3, (1 point)2U = q ·Δs ⇒ 2U = 7 · 50mV = 350mV. (1 point)

(c2) Error signal e(t) according to figure (shifted with Δs/2).Qualitatively correct: (Curve) (1 point)Quantitatively correct: (Axis label) (1 point)

��

��

��


(a) (a1) The modulation index is defined as μ =max(sQ(t))

Ac. For sQ(t) = AQ sinc (t) we get

μ =AQ

Ac. (1 point)

(a2) It is required that μ ≤ 1, which implies AQ ≤ Ac. (1 point)

(b) (b1) Low pass filter: Reconstruction of the envelope. (1 point)

High pass filter: Removal of DC component. (1 point)

(b2) See Figure D.10 (1 point)

(b3) See Figure D.11 (1 point)

(c) (c1) To avoid spectral overlap for fbp = fM − fc1 and signal bandwidth 1 Hz it is requiredthat

fc2 − fM +1

2Hz < fM − fc1 − 1

2Hz or fc2 − fM − 1

2Hz > fM − fc1 +

1

2Hz

fM >1

2(fc2 + fc1 + 1Hz) or fM <

1

2(fc2 + fc1 − 1Hz)



(1 point) (1 point)

To avoid spectral overlap in case of fbp = fM + fc1 and signal bandwidth 1 Hz we need

fc2 − fM +1

2Hz < fM + fc1 − 1

2Hz or fc2 − fM − 1

2Hz > fM + fc1 +

1

2Hz

fM >1

2(fc2 − fc1 + 1Hz) or fM <

1

2(fc2 − fc1 − 1Hz)

(1 point) (1 point)

Note, that fM can only fulfill one of the conditions for each case, i.e., it is either greateror smaller than the invalid frequency range.

(c2) A low pass filter or band pass filter with cutoff frequency less than fc2 − 12Hz before

the mixer would remove s2(t) and prevent spectral overlap. (1 point)

Figure D.10: Amplitude spectrum |S(f)|


(a) (a1) QPSK, OOK, 3-FSK (2 points)Two correct: 1 point, All correct: 2 points

(a2) fs = 200kHz, i.e. at least 4 bits per symbol (e.g. 16-QAM, 16-PSK) (1 point)Reachable rate: 800kbit/s (1 point)

(a3) Arbitrary phase rotations also correct (2 points)


286 D EXAMS

Figure D.11: Amplitude spectrum |S′(f)|

−1 0 1−1

0

1

I

Q

QPSK

−1 0 1−1

0

1

I

Q8−PSK

(b) (b1) Drawing as follows, alternatively I/Q-receiver, also correct without demapper

Correct drawing (1 point)Correct labeling (1 point)

(b2) Drawing as follows



0 1 2 3 4−3

−2

−1

0

1

2

3

t/Ts

y/A

x(t)

Correct plot (1 point)Correct labeling (1 point)

(c) (c1)

0.375 · 10−2 =3

8· 10−2 =

3

8erfc

(√A2

8σ2

)

10−2 = erfc

(√A2

8σ2

)√

A2

8σ2= 1.8

Correctly read from plot (correct: 1.7 to 1.9) (1 point)

A = 1.8 · 2√2σ

A = 1.8 · 2 = 3.6

Correct result (correct: 3.4 to 3.8) (1 Punkt)

(c2) In order to preserve both data throughput and error tolerance, neither the amount northe distance between the symbols may change. Since the transmitter does not havea Q-branch, the constellation cannot be changed to QPSK. However, if all symbolsare moved along the I-axis by −1.5A, all properties are preserved. This results in thefollowing mapping:

00� −1.5A

01� −0.5A

11� 0.5A

10� 1.5A

Correct mapping (also correct without A) (1 point)The average symbol energy for A = 1 is:

Es =(−1.5)2 + (−0.5)2 + 0.52 + 1.52

4= 1.125


288 D EXAMS

Correct Es given (1 point)


D.21 Exam SS 2011 289

D.21 Exam SS 2011


Note: Questions (a), (b) and (c) can be solved independently of each other.

(a) (4 points)

A discrete time system with input signal x[n] and output signal y[n] is defined as

y[n] = 0.5x2[n] + 0.3x[n+ 1],





(a4) What is y[n] if x[n] = δ[n] + δ[n− 1]?Note: The digital unit pulse δ[n] is given as:

δ[n] =

{1 for n = 0

0 otherwise.

(b) (6 points)

An LTI system, as shown in the figure below, has an impulse response h(t) = rect( tT− 1

2).

An input signal x(t) = rect(2tT− 1

2) is applied to the system.

x(t) y(t)h(t)

(b1) Determine the transfer function H(f)! Draw |H(f)| with proper axis labeling!

(b2) Draw y(t) with proper axis labeling!

(b3) Determine the Fourier Transform Y (f) of the output signal y(t).


290 D EXAMS

(c) (5 points)


PTx = 5 W PRx

Mobile Base Station

LA

Radio Channel

PN = 8 pW

The transmit power is PTx = 5 W. The noise power at the receiver is given as PN = 8 pW.The attenuation of the radio channel is given as

LA = 70 dB + 10 log10

(d

km

)3

dB,

where d is the distance between the transmitter and the receiver in km.

(c1) Calculate the maximum distance d, if the minimum received transmit signal power levelat the receiver is −78 dBm.

(c2) Calculate the Signal to Noise Ratio (SNR) at point A, if the receiver is at a distanceof d = 5 km from the transmitter.

Note:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3


The illustrated, analog-digital (A/D) converter is used to transform the time-continuous and value-continuous signal s(t) into a time-discrete and value-discrete signal zq(k). The analog input signalis initially sampled with the sampling frequency f ′

s and then converted into a digital signal viaquantization. Finally, the sample rate is reduced to the sampling frequency fs in order to adjustthe signal to subsequent processing units.


D.21 Exam SS 2011 291

(a) (2 points)

(a1) If the sampling frequency is too small, the sampled signal will be distorted. What isthe name of this effect?

(a2) Which sampling frequency is necessary to be able to reconstruct the sampled signaldistortion-free?

(b) (4 points)

(b1) S1(f) denotes the spectrum of the bandwidth-limited, analog signal s1(t) illustrated inthe following figure.

Now, s1(t) is sampled using the above A/D-converter and the increased sampling fre-quency f ′

s is set to f′s = 9 MHz. Draw the spectrum Z ′

s(f) of the sampled signal z′s(k)in the interval f ∈ [−18 MHz, 18 MHz]! Pay attention to correct axis labeling!


292 D EXAMS

(b2) Now, the input of the A/D-converter is fed by the single-carrier signal s2(t) = sin(2πft)with f = 500 kHz. Furthermore, s2(t) is sampled with f ′

s = 80 MHz at the Sam-ple & Hold element. The sampled signal z′s(k) is discretized with the five-stage, linearmid-tread quantizer {Δs · (0, 1, 2, . . . , q/2− 1);−Δs · (1, 2, 3, . . . , q/2)} with Δs = 0.5.Moreover, the reduction of the sampling rate is applied with the factor f ′

s/fs = f ′s/2fg =

10.

• How many sampling values are created in each period of s2(t)?

• Sketch the sampling values in the interval k/fs = t ∈ [0 μs, 2 μs)!

Note:x 0 π/16 π/8 3π/16 π/4 5π/16 3π/8 7π/16 π/2

≈ sin(x) 0 0.20 0.38 0.56 0.71 0.83 0.92 0.98 1.00

(c) (4 points)Now, discretization of the sampled signal z′s(k) is done with a linear, mid-rise quan-tizer {±Δs

2(1, 3, 5, . . . , q−1)} in the A/D converter. The resolution amounts to 12 bit and the

dynamic range of the quantizer is defined as Umin = −2.048 V ≤ z′s(k) ≤ 2.048 V = Umax.

(c1) The Signal-to-Noise Ratio (SNR) is used as a measure for the interference suppressionin the A/D conversion. What can be changed in the circuit of the A/D converter toincrease the SNR at the quantizer output by 6 dB?

(c2) Calculate the amount of quantization steps q and the width of one quantization step Δs!

(c3) Now, the quantizer is planned to have a maximum, absolute quantization error emax =0.1 V. Determine the necessary number of quantization steps q!


D.21 Exam SS 2011 293


(a) (5 points)

The source signal x(t) should be transmitted by the use of amplitude modulation (AM).The associated magnitude spectrum of the source signal |X(f)| is depicted in the followingfigure.

|X(f)|

fg−fg

1

f

(a1) Formulate the common functional relationship between the the modulated transmitsignal s(t) and the source signal x(t) in time domain!Note: Consider x(t) as given.

(a2) Which common condition has to be fulfilled by the modulation index μ to avoid zerocrossings of the envelope of the amplitude modulated signal s(t)?

(a3) Calculate the bandwidth BAM of the amplitude modulated signal s(t)!

(a4) Let the carrier frequency fc = 3fg. Draw the magnitude spectrum |S(f)| of the depictedbaseband signal |X(f)| after modulation! Pay attention to correct axis labeling!


294 D EXAMS

(b) (3 points)

Another possible modulation technique is frequency modulation (FM). One application forFM is analog radio transmission. In this case, the available bandwidth for one radio stationis limited to Bmax = 200 kHz. Additionally, the frequency shift is upper bounded by ΔF =75 kHz.

(b1) Calculate the amplification factor kFM to achieve the given frequency shift ΔF ! Assumea maximal magnitude of Am = 3 for the source signal.

(b2) Imagine, you want to transmit your own radio signal in the free frequency range depictedin the following figure.

|S(f)|

fmin fmax

f

• Calculate the carrier frequency fc in a way that maximizes the available transmitbandwidth and minimizes the interference to neighboring channels!

• Calculate the maximum possible cut-off frequency fg for the source signal! Keepin mind the maximum bandwidth Bmax of the frequency modulated signal.

(c) (3 points)

In practice, multiple radio stations are transmitted in parallel. If the radio listener selectshis favoured station, the analog signal is firstly selected by a bandpass filter and afterwardsamplified. With a proper resonant circuit, the frequency modulated signal is then translatedto an amplitude modulated signal. This enables the recovery of the source signal by an AMenvelope demodulator. The explained FM demodulator concept is shown in the followingfigure.


D.21 Exam SS 2011 295

(c1) Sketch the structure of an AM envelope demodulator by means of a block diagram!Entitle each block of your sketch!

(c2) Describe the tasks of the individual blocks in the AM envelope demodulator!Note: You can do this in bullet point form.


(a) (6 points)

The following figure shows the constellation diagrams of two digital modulation schemes.

(a1) Give the names of both modulation schemes!

(a2) A transmission system reaches a throughput of 1200bits

with a symbol duration ofTs = 2.5 ms. Is it possible that one of the given modulation schemes was used? If yes,which one? Give reasons for your answer!

(a3) Assign bits to the symbols of modulation scheme (2) such that the bit error rate isminimized! Sketch the optimal decision boundaries, too!

(a4) Another system which also has a symbol duration of Ts = 2.5 ms reaches 1600bits. Give

the name of a modulation scheme that might have been used in the system! Give reasonsfor your answer! Sketch the constellation diagram of the modulation scheme, too!

Now, there is a system which uses 2-frequency shift keying (2-FSK). The system transmits a bitsequence dk. The bits are assigned to the frequencies f0 and f1 (f1 > f0) as follows:

s(t) =

{s0(t) = cos(2πf0t) fur dk = 0 t ∈ (kTs, (k + 1)Ts)

s1(t) = cos(2πf1t) fur dk = 1 t ∈ (kTs, (k + 1)Ts)

The symbol duration is Ts = 1 ms.

(b) (5 points)

(b1) Let f0 = 1 kHz and f1 = 2 kHz. The bit sequence dk which is being transmitted isgiven as dk = {0, 1, 1, 0, 1, 0}. Draw s(t) in the interval t ∈ [0, 6Ts]! Pay attention tocorrect axis labeling!


296 D EXAMS

(b2) The frequencies f0 and f1 are orthogonal if the following condition holds:∫ Ts

0

cos(2πf0t) · cos(2πf1t)dt = 0

Let f0 = 0.5 kHz and f1 = 2.5 kHz. Are those frequencies orthogonal? Give reason foryour answer with a calculation!Note: cos(a) · cos(b) = 1

2[cos(a− b) + cos(a+ b)]

(b3) Assume that the frequency f0 of s0(t) is fixed to a specific value. Find a rule for f1 asa function of f0 such that there are no phase discontinuities when switching betweenthe transmit symbols! Note: The phase is periodic with 2π. There are no phasediscontinuities if the cosine terms in s0(t) and s1(t) have equal arguments at the timest = kTs.

(c) (3 points)

The following figure shows the correlation receiver which is used to receive the 2-FSK modu-lated signal. The 2-FSK transmitter that was used in task (b) is also used here. The symbolduration remains Ts = 1 ms, too.

∫ (k+1)Ts

kTsr(t) · cos(2πf0t+ φ0)dt

∫ (k+1)Ts

kTsr(t) · cos(2πf1t+ φ0)dt

r(t)dk ∈ {0, 1}

r0(k)

r1(k)

A synchronisation error causes a random phase rotation of φ0 ∈ [0, 2π) in the correlationsignals (see figure above). Let f0 = 0.5 kHz and f1 = 2.5 kHz. Furthermore, the channelbetween transmitter and receiver is noiseless, i.e. r(t) = s(t).

(c1) At time k = 0 the bit d0 = 0 was transmitted. Give r0(0) and r1(0) as a function of φ0!

(c2) Does the random phase rotation influence the detection? If yes, how?





(a) (a1) The system is not linear.For the input ax1[n] + bx2[n], the output is y1[n] = 0.5(ax1[n] + bx2[n])

2 +0.3(ax1[n] +bx2[n]).If two inputs ax1[n] and bx2[n] are applied separately, then the final combined outputsignal is given as y2[n] = 0.5(ax1[n])

2 + 0.3ax1[n] + 0.5(bx2[n])2 + 0.3bx2[n].

As y1[n] �= y2[n], the system is nonlinear. (1 point)

(a2) Unshifted input: y1[n] = 0.5x2[n] + 0.3x[n+ 1]Input shifted by k: y2[n] = 0.5x2[n+ k] + 0.3x[n+ k + 1]The system is time invariant, since y2[n] = y1[n+ k]. (1 point)

(a3) The system is not causal since when x[n] = 0 for n < 0, it does not hold that y[n] = 0for n < 0. (1 point)

(a4) y[n] = 0.3δ[n+ 1] + 0.8δ[n] + 0.5δ[n− 1]. (1 point)

(b) (b1) H(f) is given as

H(f) = e−jπfTTsi(πfT )

(1 point)

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

fT

|H(f)

|/T

Shape correct (1 point)Axis correct (1 point)


298 D EXAMS

(b2) The output signal is given as

3/211/20

T/2


(b3) The frequency response of the output signal is given as

Y (f) = H(f) ·X(f)

Y (f) = e−jπfTTsi(πfT ) · e−jπf T2T2si(πf T

2)

(1 point)

(c) (c1) Solving for minimum distance d

37 dBm− (70 + 30 · log10 (d)) dBm = −78 dBm30 · log10 (d) = 45log10 (d) = 1.5d = 30 km

Correct equations (1 point)Correct value (1 point)

(c2) Signal level at the receiver is

LRx = LTx − LA = 10 log10

(5·103 mW1 mW

)dBm− (70 + 30 log10(5)) dB = −54 dBm

where 10 log10 5 = 10 log10 10− 10 log10 2 = 7 (1 point)SNR is

LRx − LN = −54 dBm− 10 log10 (8 · 10−9) dBm = 27 dB

Equations correct (1 point)Values correct (1 point)




(a) (a1) Aliasing (1 point)

(a2) fs > 2fg (sampling theorem) (1 point)

(b) (b1) Spectrum Z ′s(f) of the sampled signal z′s(k) with aliasing and overlaps in the spectrum

respectively.

Qualitatively correct: (Aliasing shown) (1 point)Quantitatively correct: (Axis label) (1 point)

(b2) • 16 sampling values (1 point)

• quantization values and quantized signal zq(k) respectively according to the figurebelow.

Qualitatively correct: (1 point)

(c) (c1) Increase number of necessary bits to represent the signal alphabet by one or samplewith fourfold sampling frequency f ′

s respectively (1 point)

(c2) With the width of the dynamic range ΔU = Umax − Umin, the number of quantizationsteps q, the step width Δs and the resolution in bit b follows


300 D EXAMS

q = 212 = 4096.

Δs =ΔU

q= 1 mV.

(2 points)

(c3) With the maximum, absolute quantization error emax, the number of quantizationsteps q, the step width Δs and the width of the dynamic range ΔU = Umax − Umin

follows

Δs =ΔU

q.

e ≤ Δs

2≤ emax.

q ≥ ΔU

2emax

.

q ≥ 21.

(1 point)


(a) (a1) s(t) = (1 + μx(t)) · cos(2πfct) (1 point)

(a2) |μx(t)| ≤ 1 or μ ·max [x(t)]− 1 ≥ 0 (1 point)

(a3) BAM = 2fg (1 point)

(a4) see figure magnitude spectrum |S(f)|.

|S(f)|

ffg

μ2

12

−3 3

Axis labeling correct. (1 point)Shape (position, amplitude, bandwidth) correct. (1 point)

(b) (b1) kFM = ΔFAm

= 25 kHz (1 point)

(b2) i. fc =fmax+fmin

2or fc =

Bmax

2+ fmin (1 point)

ii. from B ≈ 2(fg +ΔF ) → fg,max = 25 kHz (1 point)



(c) (c1) Realization of the FM receiver

Rectifier, LP (low pass) and HP (high pass) named. (1 point)Block diagram in the correct order. (1 point)

(c2) (1 point, if all three are correct)

component task

rectifier negative parts of the signal were removed (half-wave rec-tification)

low pass filter removes parts with high frequencies; envelope of the de-sired signal is reconstructed

high pass filter removes the direct current (dc) component of the signal


(a) (a1) (1): QPSK, (2): 8-PSK (12point each)

(a2) fs =1Ts

= 400 symbolss

1200 bits/symbol400 symbols/s

= 3 bitsymbol

(1 point)Three bits per symbol correspond to eight constellation symbols. 8-PSK might havebeen used.

(a3) Bit assignment with gray mapping (1 point)Boundaries sketched correctly (1 point)

000

100

110

010011

111

101

001

I

Q

(a4) 1600 bits/symbol400 symbols/s

= 4 bitsymbol

A modulation scheme with 16 symbols is required, e.g. 16-QAM. (1 point)Corresponding constellation diagram: (1 point)


302 D EXAMS

I

Q

(b) (b1) Qualitative correct (1 point)Quantitative correct (1 point)

0 1 2 3 4 5 6

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t / Ts

s(t)

(b2)∫ Ts

0cos(2πf0t) · cos(2πf1t)dt ?

= 0

12

∫ Ts

0

cos(2π(f0 − f1)t)dt︸︷︷︸=0

+12

∫ Ts

0

cos(2π(f0 + f1)t)dt︸︷︷︸=0

?= 0

f0 − f1 = −2 kHz, f0 + f1 = 3 kHz and fs =1Ts

= 1 kHz (Approach: 1 point)Integration is done over whole periods in both summands and is therefore 0 which isthen also true for the sum. The orthogonality condition is satisfied! (Conclusion:1 point)

(b3) To avoid phase discontinuities, the following must hold:

2πkTsf0 = 2πkTsf1 − nk2π ∀kThe term nk2π is allowed because the phase is periodic. This gives:

f1 = f0 +nk

kTs

Setting nk = nk gives:

f1 = f0 +n

TsIn words: f1 must be larger than f0 by integer multiples of the symbol frequencyfs = T−1

s

(1 point)

(c) (c1) r0(0) and r1(0) are calculated as:

r0(0) =

∫ Ts

0

cos(2πf0t) · cos(2πf0t+ φ0)dt



=1

2

∫ Ts

0

cos(φ0)dt+1

2

∫ Ts

0

cos(2π2f0t+ φ0)dt︸︷︷︸=0, since 2f0=fs=1 kHz

=Ts2cos(φ0)

(1 point)

r1(0) =

∫ Ts

0

cos(2πf1t) · cos(2πf0t+ φ0)dt = 0

A phase offset does not influence orthogonality. (1 point)

(c2) The distance between r0(0) and r1(0) is reduced. In a noisy channel this results inreduced error tolerance and increased error rate. In the limiting case φ0 = π

2it holds

that r0(0) = 0 and a desicion is not possible even in the noise-free case. (1 point)


304 D EXAMS

D.23 Exam WS 2011/2012


Hint: The following questions (a), (b), and (c) can be answered independently.

(a) (4 Points)


y[n] = x[n− 1] + x[n− 2] · x[n− 3],





(a4) Calculate y[n] for x[n] = δ[n] + δ[n− 1]!Hint: Here δ[n] is the unit impulse, which is defined as

δ[n] =

{1 if n = 0

0 else.

(b) (5 Points)

An LTI system with impulse response h(t) has an input signal x(t) and an output signaly(t), where

h(t) =

{1T

if 0 ≤ t ≤ 2T

0 otherwise

and

x(t) =

{tT

if 0 ≤ t ≤ 2T

0 otherwise.

(b1) Is the system causal? Give reason for your answer!

(b2) Derive the output signal y(t) between 0 ≤ t ≤ 2T !

(b3) Draw the output signal y(t) between 0 ≤ t ≤ 2T ! Pay attention to correct axis labeling!

(c) (6 Points)

A radio station in Dresden transmits with power PT = 500 kW. A radio receiver receivesthis signal in Calcutta, at a distance d = 7500 km away. The channel attenuation is given as

LA = 30 log10

(d

km

)dB,



D.23 Exam WS 2011/2012 305

(c1) Calculate the transmit power level, LPT, in dBm!

(c2) Calculate the channel attenuation, LA, between Dresden and Calcutta!

(c3) Calculate the received power level, LPR, at the radio receiver in Calcutta, in dBm!

(c4) Calculate the distance, d, between Dresden and Tokyo, if the received power in Tokyois 1

2μW!

Hint:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3


An analog signal s(t) with cut-off frequency fg = 4 MHz is digitized. The following figure illustratesthe spectrum S(f) of the signal s(t).

(a) (3 Points)

(a1) Sampling is applied with a sampling frequency fs = 3/2 · fg. Which disadvantageouseffect occurs in doing so? How must the sampling be changed in order to prevent thiseffect?

(a2) The signal s(t) is sampled with a sampling frequency fs = 8 MHz. The sampled signalis defined as x(t). Draw the spectrum X(f) in the interval [−12 MHz; 12 MHz]! Payattention to correct axis labeling!

(b) (4 Points)In the following, we still consider a sampling of the signal s(t). In doing so, the samplingfrequency is set to fs = 6 MHz.

(b1) Draw the spectrum X(f) of the sampled signal x(t) in the interval [−10 MHz; 10 MHz]!Pay attention to correct axis labeling!

(b2) After the sampling, filtering is applied and an output signal y(t) results. In the process,y(t) is considered as result of the convolution of the sampled signal x(t) with thefunction h(t), with h(t) = si(π · t/Δt) and Δt = 1/2fs. Draw the amplitude spectrum|Y (f)| for [−10 MHz; 10 MHz]! Pay attention to correct axis labeling!


306 D EXAMS

(c) (3 Points)The time discrete signal x[k], k ∈ Z is given after the sampling. Subsequently, x[k] is quan-tized. A linear mid-rise quantizer is used for the quantization. The dynamic range of thequantizer is given by −4·Δs ≤ xin ≤ 4·Δs. Here Δs denotes the step width of the quantizer.

(c1) Draw the characteristic of the quantizer xout as a function of xin! Pay attention tocorrect axis labeling!

(c2) Determine the maximum, absolute quantization error for the dynamic rangeΔU = 1.6 V !


(a) (4 Points)An arbitrary source signal s(t) is modulated on the frequency of the carrier signal sc(t). Thisis called frequency modulation (FM). The technique is widely applied to radio, television, andcommunications systems. One requirement for parallel transmission of multiple modulatedsignals is the band-limitation of each signal. The necessary bandwidth of a modulated signalcan be derived from its cut-off frequency fg.

(a1) Give one advantage and one disadvantage of FM!

(a2) The source signal s(t) with its maximum amplitude of |smax| = 2 is transmitted byusing FM. The maximal frequency deviation is limited to ΔF = 4 · fg. Calculate theamplification factor kFM as a function of fg!

(a3) Give an estimate about the absolute bandwidth BFM as a function of fg by applyingthe Carson rule!

(a4) The source signal is now transmitted by using AM instead of FM. How much will thebandwidth BAM of the AM signal be smaller or greater than the FM signal?

(b) (3 Points)The following figure shows the source signal s(t) in time domain. The modulated signal isgiven by

sFM(t) = cos

(2πfct+ 2π

ΔF

fgs(t)t

). (D.1)

In addition, the values for the carrier frequency fc = 2T, and the frequency deviation

ΔF = 1T· fg ·max (|s(t)|) are given.

(b1) Give the occurring frequencies f1 and f2 of the modulated signal sFM(t) as a functionof T!

(b2) Draw the shape of the modulated time domain signal sFM(t) in the interval t = [0; 4T ]!Pay attention to correct axis labeling!


D.23 Exam WS 2011/2012 307

(c) (4 Points)A speech of a famous book author is transmitted to the loudspeaker system with the help ofwireless microphones. For this purpose two microphones and a central receiver station areinstalled. For parallel usage of both microphones, separate frequency ranges have to be used.The cut-off frequency of the acoustic signal is given by fg = 40 kHz. The resulting bandwidthof the AM bandpass signal for each microphone is given by BAM = 2 · fg = 80 kHz.

(c1) Calculate the minimal bandwidth Bmin of the central receiver station for parallel recep-tion of two microphones using AM!

(c2) The following figure shows the magnitude spectrum |SAM(f)| of the signal at the inputof the central receiver station for one active microphone. The depicted receiver struc-ture for microphone no.1 shows, that the bandpass signal is down-converted into thebaseband (fc = 800 MHz) first and low-pass filtered afterwards.Draw the magnitude spectrum |SBB(f)| at the output of the receiver for microphoneno.1 in the interval f ∈ [−0.12; 0.12] MHz! Pay attention to correct axis labeling!

(c3) Assume that microphone no.2 is using frequencies in the range from 800.08 MHz to800.16 MHz.Give the maximal possible cut-off frequency fg,max of the low-pass filter at the centralreceiver station for receiving microphone no.1 without cross-talk from microphone no.2!


(a) (6 Points)


308 D EXAMS

The following figure shows the constellation diagrams and waveform plots of three digitalmodulation schemes.

−1 0 1

−1

0

1

I

jQ

Constellation A

−1.5 −0.5 0 0.5 1.5

−1.5

−0.5

0

0.5

1.5

I

jQ

Constellation B

−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7

−1

0

1

I

jQ

Constellation C

0 2 4 6 8

−1

−0.5

0

0.5

1

kT

s(t)

Signal 1

0 2 4 6 8−1.5

−1

−0.5

0

0.5

1

1.5

kT

s(t)

Signal 3

0 2 4 6 8−8

−6

−4

−2

0

2

4

6

8

kT

s(t)

Signal 2

(a1) Assign to each constellation diagram a fitting waveform! Give the names and orders ofeach modulation scheme!

(a2) A picture in a videostreaming system has an average size of 10 kByte. To avoid stut-tering it is required to transfer at least 30 pictures per second. The symbol duration isfixed at T = 1 μs. What is the minimum number of bits per transmit symbol in orderto fulfill the requirements? Which of the given modulation schemes reaches or surpassesthe requirements regarding data throughput?

(a3) Which of the given modulation schemes can be received using an envelope detector.

(b) (5 Points)

The following figure shows the receiver of a digital transmission system.


D.23 Exam WS 2011/2012 309

hr(t)

hr(t)

kT

kT

1 -¿ 0-1 -¿ 1

1 -¿ 0-1 -¿ 1

b2k

b2k+1

cos(2πfct)

−sin(2πfct)

(b1) Sketch the constellation diagram of the system’s modulation scheme! For each constel-lation point, give the corresponding bits in the form b2kb2k+1!

(b2) Sketch the block diagram of an appropriate transmitter! Give the name of each com-ponent!

(b3) How should hr(t) be chosen in order to minimize the bit error rate?Hint: To answer this question, use your components from (b2).

(c) (3 Points)

The symbol error probability can be approximated by the pairwise error probability betweenthe symbols with minimal distance dmin. For the AWGN channel, the error probability Ps

for two symbols with distance d is given as:

Ps =1

2erfc

(√(d/2)2

2σ2

),

where σ2 is the variance of the noise.

(c1) Let σ2 = 0.25. For the constellations A and C (see Task 4a), give the resulting symbolerror probabilities Ps!

(c2) Give the bit error rate Pb for constellation C using the given approximation under theassumption of an optimal bit assignment to the constellation symbols!

Hints:√

12≈ 0.7


310 D EXAMS

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)





(a) (a1) The system is not linear.For the input ax1[n] + bx2[n], the output isy1[n] = (ax1[n− 1] + bx2[n− 1]) + (ax1[n− 2] + bx2[n− 2]) · (ax1[n− 3] + bx2[n− 3]).If two inputs ax1[n] and bx2[n] are applied separately, then the final combined outputsignal is given as y2[n] = ax1[n−1]+a2x1[n−2]·x1[n−3]+bx2[n−1]+b2x2[n−2]·x2[n−3].As y1[n] �= y2[n], the system is nonlinear. (1 point)

(a2) The system is time invariant.If the input time index is changed to k + n, then the new output is given asy1[k + n] = x[k + n] + x[k + n− 2] · x[k + n− 3].If the output time index is changed, then we havey2[k + n] = x[k + n] + x[k + n− 2] · x[k + n− 3].As both the responses are same, the system is time invariant. (1 point)

(a3) The system is causal as y[n] = 0 for n < 0 is satisfied. (1 point)

(a4) y[n] = δ[n− 1] + δ[n− 2] + δ[n− 3]. (1 point)

(b) (b1) The system is causal as h(t) = 0 for all t < 0. (1 point)

(b2)

y(t) = h(t) ∗ x(t)=

∫ ∞

−∞h(τ) · x(t− τ) dτ

For t < 0, y(t) = 0.

For 0 ≤ t < 2T , y(t) =∫ t

01T

t−τT

dτ = t2

2T 2 .

For 2T ≤ t < 4T , y(t) =∫ 2T

t−2T1T

t−τT

dτ = 2− (t−2T )2

2T 2 .

For t ≥ 4T , y(t) = 0.

For the interval of interest, the output signal is given as

y(t) =t2

2T 2for t ∈ [0; 2T ]

Approach correct (1 point)Answer correct (1 point)


312 D EXAMS

(b3) The output signal is given as

0

0.5

2TT

2.0

y(t)

t


(c) (c1) The transmit power level is given as

LPT= 10 log10

(500 · 1031 · 10−3

)= 10 log10

(109

2

)= 87 dBm

(1 point)

(c2) The channel attenuation is

LA = 30 log10(7500) = 30 log10(3·25·100) = 30 log10

(3

4· 104

)= 30·(0.5−0.6+4) = 117 dB

(1 point)

(c3) The received power level in Calcutta is

LPR= LPT

− LA = 87 dBm− 117 dB = −30 dBm

(1 point)

(c4) The received power level in Tokyo is

LPR= 10 log10

(0.5 · 10−6

1 · 10−3

)= 10 log10(5·10−4) dBm = 10 log10

(10−3

2

)dBm = −33 dBm

LPRcorrect (1 point)

The channel attenuation suffered to transmit to Tokyo is

LA = LPT− LPR

= 87 dBm− (−33 dBm) = 120 dB



LA correct (1 point)If d is the distance between Dresden and Tokyo, then

30 log10(d) = 120

d = 104 km

Answer correct (1 point)


(a) (a1) Aliasing, fs ≥ 2fg (sampling theorem) (1 Point)

(a2) Spectrum X(f) of the sampled signal x(t) without aliasing and without overlaps in thespectrum, respectively.

Qualitatively correct: (no aliasing) (1 Point)Quantitatively correct: (axis labels) (1 Point)

(b) (b1) Spectrum X(f) of the sampled signal x(t) with aliasing and overlaps in the spectrum,respectively.

Qualitatively correct: (aliasing) (1 Point)Quantitatively correct: (axis labels) (1 Point)

(b2) Amplitude spectrum |Y (f)| results from the multiplication of |X(f)| and |H(f)|, withH(f) = 1

2fs· rect(f/2fs) and fs = 1/Δt.

Qualitatively correct: (X(f) filtered) (1 Point)Quantitatively correct: (axis labels) (1 Point)


314 D EXAMS



(c) (c1) Quantization characteristic according to the figure below.

Qualitatively correct: (1 Point)Quantitatively correct: (axis labels) (1 Point)

(c2) (1 Point)

Δs =ΔU

q=

1.6 V

8.

eabs,max =Δs

2= 0.1 V.


(a) (4 points)

(a1) Advantages: (if 1 element each is right, than 1 point)

• constant envelope of modulated signal

• transmit amplifier without back-off

• insensitive against interference

Disadvantages:

• high amount of bandwidth compared to AM

(a2) kFM = ΔFsmax

= 4fg2

= 2fg (1 point)

(a3) BFM ≈ 2 · (ΔF + fg) = 2 · (4fg + fg) = 10fg (1 point)

(a4) smaller, because BAM ≈ 2 · fg → ΔB = BFM − BAM = 2 ·ΔF = 8 · fg (1 point)

(b) (3 points)

(b1) f1 =1Tand f2 =

3T, because (1 point)

sFM(t) = cos

(2π

2

Tt± 2π

1

Ts(t)t

)= cos

(2π

(2

T± 1

Ts(t)

)t

)for t ∈ [0; 4T ]


316 D EXAMS

(b2) Axes labeling correct. (1 point)Shape correct. (1 point)

(c) (4 points)

(c1) Bmin = 2 · (2 · fg) = 160kHz (1 point)

(c2) Axes labeling correct. (1 point)Spectrum shape correct. (1 point)

(c3) |fg,max| = 80kHz (1 point)


(a) (6 Points)

(a1) A - 1 - QPSK (12Points per assignment / name)

B - 3 - 4-ASK (Total: 3 Points)C - 2 - 8-ASK

(a2) Calculating the required data rate gives

R = 10 · 8 · 30 = 2400 kBit/s.



For a sampling frequency of fs =1T= 1 MHz

nbit =R

fs= 2.4

bits per transmitted symbol are required. (1 Point)8-ASK can be used since it has 3 bits per symbol. (1 Point)

(a3) Only modulation scheme C has all information in the amplitude and not in the phase.For this reason, only scheme C can be received using an envelope detector. (1 Point)

(b) (5 Points)

(b1) Diagram and bit assignment as follows:

−1 0 1

−1

0

1

I

jQ

00

11

10

01

Diagram qualitatively correct (1 Point)Correct bit assignment (1 Point)

(b2) Block diagram as follows or similar

ht(t)

ht(t)

0 -¿ 11 -¿ -1

0 -¿ 11 -¿ -1

b2k

b2k+1

cos(2πfct)

−sin(2πfct)

Symbolmapper Pulseformer Mixer

Block diagram correct (1 Point)Correct names (1 Point)

(b3) Using the matched filter principle yields hr(t) = h∗t (−t), where ht(t) is the impulseresponse of the transmit filter (pulse former). (1 Point)


318 D EXAMS

(c) (3 Points)

(c1) Reading from the diagrams yields dmin,A =√2 and dmin,C = 1. Therefore,

Ps,A =1

2erfc

(√2

2

)=

1

2erfc (1) ≈ 0.16

2≈ 0.08

and

Ps,C =1

2erfc

(√1

2

)≈ 1

2erfc (0.7) ≈ 0.32

2≈ 0.16.

For each correct result (1 Point)

(c2) Constellation C exhibits 3 bits per symbol. Since the symbol error rate is dominatedfrom pairwise errors of neighboring symbols. Due to Gray Coding, all pairs of neigh-boring symbols differ in exactly one bit. Therefore,

Pb =Ps,C

3≈ 0.053.

Correct result (1 Point)


D.25 Exam SS 2012 319

D.25 Exam SS 2012


Hint: The following questions (a), (b), and (c) can be answered independently.

(a) (6 points)

WIFI services are provided inside a building. The access point (AP) can transmit at a powerof PT = 0.1 W, the receive signal power level at the distance of d = 10 m is LPR

= −57 dBm.The channel attenuation can be expressed as a function of the distance

LA = 40 dB + 10 log10

(d

d0

)α

dB,

where d is the distance between transmitter and receiver and d0 = 1 m is a reference distance.

(a1) Calculate the path-loss exponent α!

Assume α = 4 in the following.

(a2) Find the receive power level at distance d1 = 50 m!

(a3) Assume the noise power level is −140 dBm, and minimum required SNR = 40 dB atthe receiver. Calculate the maximum distance dmax the AP can cover!

(a4) Assume the same SNR requirement as in (a3). If we want to double the maximumachievable distance, by how many dB does the transmit power have to be increased?

Hint:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3

(b) (3 points)


y[n] = x2[n] + nx[n− 2],

where n is the discrete time index. Answer the following questions with reason!

(b1) Is the system linear?

(b2) Is the system time invariant?

(b3) Is the system causal?

(c) (6 points)

A system with input signal x(t) and output signal y(t) is defined as

Tdy(t)

dt+ y(t) = x(t),

where T is a constant.


320 D EXAMS

(c1) Calculate the transfer function H(f) of the system by the Fourier Transform!

Hint: The Fourier Transform theorem for differentiation dnx(t)dtn

is (j2πf)nX(f).

(c2) Calculate the impulse response h(t) of the system accordingly!

Hint: A Fourier Transform pair could be used: x(t) =

{e−

tT , t ≥ 0

0, t < 0and

X(f) = T1+j2πfT

.

(c3) If the input signal x(t) is a rectangle pulse as

x(t) = rect

(t− T

2

T

)=

{1, 0 ≤ t ≤ T

0, else,

calculate the output signal y(t) and sketch it!


(a) (2 points)Analog signals can be converted into digital signals by means of analog-to-digital converters.For this, the analog signal s(t) with its cutoff frequency fg is sampled first. The result is adiscrete time signal.

(a1) Which relation between sampling frequency fs and cutoff frequency fg has to be fulfilledin order to avoid aliasing errors?

(a2) The sampling procedure can be realized by the circuit depicted in the following figure.What is such a circuit called?

(b) (4 points)In the following, we are considering an analog speech signal s(t) with the magnitudespectrum |S(f)|, which has to be digitally transmitted. The sampling frequency is set tofs = 7 kHz. In order to avoid aliasing, the given signal s(t) needs to be initially limited inbandwidth by an anti-aliasing filter with its transfer function H(f).


D.25 Exam SS 2012 321

(b1) First sketch the magnitude spectrum of |S(f)| after low pass filtering!

(b2) Draw the magnitude spectrum of the sampled signal |Sd(f)| between−10.5 kHz ≤ f ≤ 10.5 kHz! Pay attention to correct axis labeling!

(b3) For reconstructing the signal s(t), an ideal low pass filter should be used. Give thecutoff frequency fg,D/A of this filter!


322 D EXAMS

(c) (4 points)The following figure shows a mid-rise-quantizer. Δs corresponds to the distance between thequantization stages. The input signal is given by s(t), the quantized output signal by sq(t).

(c1) The mid-rise quantizer has no quantization stage at sq(t) = 0. Justify briefly, why thiscould be a drawback for a speech signal with a lot of pauses in speech!Hint: The speech signal consists of the useful signal and a basic noise floor.

(c2) Draw the quantization error e(t) = sq(t)− s(t) as a function of the input signal s(t) inthe dynamic range −7Δs

2≤ s(t) ≤ 7Δs

2! Pay attention to correct axis labeling!

(c3) An input signal s(t), which has an uniformly distributed amplitude in the interval[−2Δs; 2Δs], is applied at the given quantizer. Give the signal-to-noise ratio SNR atthe output of the quantizer!


D.25 Exam SS 2012 323

0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.5

t

s(t)

(a)

0 0.5 1 1.5−1

−0.5

0

0.5

1

t

s(t)

(b)


(a) (2 points)A basic principle in communications is called ”modulation”.

(a1) What does the term ”modulation” describe in communications?

(a2) Why is modulation necessary in communications?

(b) (4 points)The following two figures are showing two typical modulation schemes.

(b1) Give the name of the applied modulation scheme for (a) and (b) !

(b2) Which of the two depicted modulation schemes is more power efficient for radio trans-mission?

(b3) Why is the chosen modulation scheme more power efficient?

(b4) Which scheme requires less bandwidth for transmission when the same source signal isused?

(c) (5 points)The following picture shows the magnitude spectrum |R(f)| of the receive signal r(t) at thereceiver input.

−fc fc f0

A

|R(f)|

B


324 D EXAMS

The receive signal r(t) should be translated from the original carrier frequency fc to anintermediate frequency fZF by means of the depicted receiver. The given parameters are asfollows: carrier frequency fc = 100 MHz and signal bandwidth B = 20 MHz.

H(f )

cos (2πfmt)

r(t) s(t) z(t)

(c1) Give the necessary input frequency fm of the mixer so that the intermediate frequencyof the output signal is fZF = 50 MHz!

(c2) Draw the magnitude spectrum |S(f)| of the signal s(t) at the output of the mixer inthe interval [−300, 300] MHz! Pay attention to the correct axis labeling!

(c3) Design the succeeding filter H(f) in the way that only the signal with carrier frequencyfZF is visible at the output of the receiver. Name the type and the cutoff frequency fgof this filter!

(c4) Sketch the magnitude spectrum |Z(f)| of the signal z(t) at the output of the receiverin the interval [−100, 100] MHz!


(a) (8 points)The following picture shows a plot of a digitally modulated passband signal sBP (t) and thecorresponding transmitted bits.

0 1 2 3 4

−3

−2

−1

0

1

2

3

t / ms

s BP(t)

10 00 11 01

(a1) Give the name of the modulation scheme! Sketch the constellation diagram! Add thegiven bit assignment to the constellation diagram!

(a2) The bit assignment is not optimal with regard to minimal bit error rate. Give reasonwhy this is the case! Give the optimal bit assignment for this constellation!


D.25 Exam SS 2012 325

(a3) Find the data rate of this modulation scheme!

(a4) Generally, the signal sBP (t) can be expressed mathematically as

sBP (t) = Ts

∞∑k=−∞

dkhT (t− kTs) cos(2πfct)

where Ts denotes the symbol duration and dk the k-th transmitted symbol. Give thecarrier frequency fc and the impulse response hT (t) of the transmit filter!

(b) (4 points)A new transmission system using BPSK is to be developed. The bit value ’0’ shall be assignedto the amplitude ’1’ and the bit value ’1’ to the amplitude ’-1’.

(b1) Sketch the transmitter of the system from the incoming bitstream to the transmitantenna! Give the names of the components!

(b2) The bit error rate of a BPSK transmission is given by

BER =1

2erfc

(√SNR

2

).

Furthermore

SNR[dB] = 10 · log10(SNR) .Calculate the bit error rate for an SNR of 9dB!Hints: Use the logarithm table of problem 1. The erfc-function is given by the diagrambelow.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 310−5

10−4

10−3

10−2

10−1

100

x

erfc

(x)


326 D EXAMS

(c) (2 points)Again consider a BPSK transmission. There are conditional probability density functions(PDF) of the amplitudes of the received signal rk at the input of the detector at the re-ceiver. They are denoted by p(rk|dk) where dk denotes the k-th transmitted BPSK-symbol.Therefore dk = {−1, 1}.(c1) Sketch the conditional PDFs p (rk|dk) of both BPSK-symbols as functions of rk in one

diagram for a channel without noise!

(c2) The transmit signal is now superimposed by additive white Gaussian noise (AWGN).Sketch the resulting conditional PDFs p (rk|dk) of both BPSK-symbols as functions ofrk in one diagram!





(a) (a1) The channel attenuation at d = 10 meters is

LPT= 10 log10

(PT

1mW

)dBm = 20 dBm

(1 point)

LA = LPT− LPR

= 20 dBm− (−57 dBm) = 77 dB

LA = 40 dB + 10 log10

(10

1

)α

dB = 40 dB + α · 10 log10 10 dB

So the path loss exponent is

α =77− 40

10= 3.7

(1 point)

(a2) The receive signal power level at d = 50 m is

LPR= LPT

− LA = 20 dBm− (40 + 10 log10 504) dB = −88 dBm

(1 point)

(a3) The signal-to-noise-ratio is given as

SNR = LPR− LPN

= LPT− LA − LPN

(1 point)To satisfy the SNR requirement

20 dBm− (40 + 10 log10 d4) dB− (−140 dBm) ≥ 40 dB

log10dmax =20 + 140− 40− 40

40= 2

dmax = 100 m

(1 point)

(a4) To satisfy the SNR requirement at the distance d1 = 2dmax = 200 m

LPT1− (40 + 10 log10(d1)

4) dB− (−140 dBm) = 40 dB

LPT1= (40 + 10 log10(d1)

4) dB + (−140 dBm) + 40 dB = 32 dBm

So at least we need to increase the tranmit power by LPT1− LPT

= 12 dB.

(1 point)


328 D EXAMS

(b) (b1) The system is not linear.For the input ax1[n] + bx2[n], the output isy1[n] = (ax1[n] + bx2[n])

2 + n(ax1[n− 2] + bx2[n− 2]).If two inputs ax1[n] and bx2[n] are applied separately, then the final combined outputsignal is given as y2[n] = (a2x21[n]− anx1[n− 2]) + (b2x22[n]− bnx2[n− 2]).As y1[n] �= y2[n], the system is nonlinear. (1 point)

(b2) The system is not time invariant.If the input time index is delayed by k, then the new output is given asy1[n− k] = x2[n− k]− nx[n− k − 2].If the output time index is delayed by k, then we havey2[n− k] = x2[n− k]− (n− k)x[n− k − 2].As y1[n+ k] �= y2[n+ k], the system is not time invariant. (1 point)

(b3) The system is causal as the output y[n] depends only on the current and past input.(1 point)

(c) (c1) According to the Fourier transform, the system can be described as

(j2πTf)Y (f) + Y (f) = X(f)

(1 point)So the system function is then

H(f) =Y (f)

X(f)=

1

1 + j2πTf

(1 point)

(c2)

h(t) =

{1Te−

tT , t ≥ 0

0, t < 0

(1 point)

(c3) The output signal can be computed by the convolution of input signal and the impulseresponse

y(t) = x(t) ∗ h(t)=

∫ ∞

−∞x(τ) · h(t− τ) dτ

For t < 0, y(t) = 0.

For 0 ≤ t < T , y(t) =∫ t

01 · 1

Te

τ−tT dτ = 1− e−

tT .



For t ≥ T , y(t) =∫ T

01 · 1

Te

τ−tT dτ = (e− 1)e−

tT .

The output signal is given as

y(t) =

⎧⎪⎨⎪⎩

0, t < 0

1− e−tT , 0 ≤ t < T

(e− 1)e−tT , t ≥ T

Approach correct (1 point)Answer correct (1 point)

The output signal is given as

−3T −2T −1T 0 1T 2T 3T 4T 5T0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

t

y(t)

output signal y(t)

0.63

(1 point)


(a) (a1) Aliasing is avoided, when fs ≥ 2fg (Nyquist sampling theorem) (1 Point)

(a2) sample and hold circuit (1 Point)

(b) (b1) Spectrum |S(f)| after lowpass filtering (refer to figure)

Qualitatively correct (filtered and symmetrical at ±2 kHz): (1 Point)


330 D EXAMS

(b2) Spectrum |Sd(f)| results from shifting |S(f)| by 7 kHz. No aliasing occurs.

Qualitatively correct (no aliasing): (1 Point)Quantitatively correct (axis labeling): (1 Point)

(b3) D/A conversion by means of low pass with cutoff frequency fg,D/A = 3.5 kHz (1 Point)

(c) (c1) If (due to pauses in speech) a very low input signal |s(t)| << Δs/2 or no signal(s(t)) = 0) is applied, there is however a noise signal generated at the output of thequantizer. (Since the approximately maximum quantization error occurs always, thenoise power is maximized to Δs2/4.) (1 Point)

(c2) characteristic curve of the quantization error (refer to figure).

Qualitatively correct: (1 Point)Quantitatively correct (axis labeling): (1 Point)

(c3) The dynamic range ±2Δs leads to an effective quantization with b = 2 bit.

SNR ≈ (6 · b) dB≈ 12 dB

(1 Point)




(a) (2 points)

(a1) Modulation describes the adaption of the source signal to the physical transmissionmedium, transmission channel or receiver. The baseband signal (to be transmitted) isoften modulated to carrier signal with a high frequency. (1 point)

(a2) Modulation is essential in communications because (min. 1 element, total 1 point)

• multiple source signals are transmitted without interferening each other,

• transmission is adapted to the characteristic of the transmission channel.

(b) (4 points)

(b1) Following modulation schemes: (1/2 point each, total 1 point)(a) AM - Amplitude modulation and(b) FM/PM - Frequency or Phase modulation (more general: Angle modulation).

(b2) Angle modulation (Frequency or Phase modulation) is more power efficient. (1 point)

(b3) Reason: (1 point)By using amplitude modulation more than 50% of the transmission power is used forthe carrier signal. This separat carrier is not necessary when using angle modulation.Additionally, the angle modulated signal shows a constant envelope. This fact can beused to increase the efficieny of the amplfifier at the transmitter.

(b4) BAM = 2fg < BWM ≈ 2 · (ΔF + fg) (1 point)Amplitude modulation requires less bandwidth during transmission of the same sourcesignal compare to angle modulation.

(c) (5 points)

(c1) fm = 50 MHz (or fm = 150 MHz) (1 point)

(c2) The magnitude spectrum of the signal s(t) as follows (for fm = 50 MHz):

−100 100f

MHz0

A2

|S(f)|

−200 200−300 300

The magnitude spectrum of the signal s(t) as follows (for fm = 150 MHz):

−100 100f

MHz0

A2

|S(f)|

−200 200−300 300


332 D EXAMS

Axis labeling correct. (1 point)Spectrum correct. (1 point)

(c3) The following filter: (min. 1 filter named, total 1 point)low pass filter with cut-off frequency 60 MHz< fg < 140 MHz orbandpass filter with 0 MHz< fg,u < 40 MHz and 60 MHz< fg,o < 140 MHz.

(c4) The magnitude spectrum of the signal z(t) as follows:

−100 100f

MHz0

A2

|Z(f)|

Spectrum correct. (1 point)


(a) (8 points)

(a1) It is 4-ASK (1 point)Constellation diagram: (1 point each for diagram and bit assignment)

−4 −2 0 2 4−1

0

1

I

Q

00 11 10 01

(a2) For AWGN channels, the most common error is a detection of a neighbouring symbol.To minimize the bit error rate Gray mapping should be used since neighbouring symbolsonly differ in one bit.Reason correct (1 point)Diagram correct (1 point)

−4 −2 0 2 4−1

0

1

I

Q

00 01 11 10

(a3) fs = 1/Ts = 1 kHz. For 2 bits per symbol the resulting rate is R = 2 kBit/s. (1 point)

(a4) fc = 5 kHz (1 point)

hT (t) = rect(

tTs

− 12

)(1 point)



(b) (4 points)

(b1) Transmitter as below or otherwise correct

Diagram correct (1 point)Names correct (1 point)

(b2) SNR[dB] = 10 · log10(SNR)0.9 = log10(SNR)0.3 + 0.3 + 0.3 = log10(SNR)SNR = 8 (1 point)BER = 1

2erfc(2)

BER ≈ 125 · 10−3

BER ≈ 2.5 · 10−3 (1 point)

(c) (2 points)

(c1) Diagram correct (1 point)

−2 −1 0 1 20

0.5

1

1.5

rk

p(r k|d

k)

p(rk|−1) p(rk|1)

(c2) Diagram correct (1 point)


334 D EXAMS

−2 −1 0 1 20

0.5

1

1.5

rk

p(r k|d

k)

p(rk|−1) p(rk|1)


D.27 Exam WS 2012/2013 335

D.27 Exam WS 2012/2013


Hint: The following questions (a), (b), and (c) can be answered independently.Remark on the awarding of points of a drawing: A second point for correct axis labeling isgiven only if the drawing is qualitatively correct.

(a) (5 Points)

The transmit signal power level, LT is 20 dBm.

The mobile device is situated at the distance of d meters (m) from the transmitter. Thesignal attenuation is given by the following formula,

LA = 40 dB + 20 log10

(d

m

)dB.

Consider the received power level, LR, at the mobile station to be −60dBm.

(a1) Give the transmit power PT, in mW!

(a2) Give the channel attenuation, LA, between transmitter and receiver!

(a3) Give the distance, d, between transmitter and receiver!

(a4) A thermal noise PN = 10−10 W is added at the receiver. Calculate the signal to noiseratio SNR in dB!

Hint:

x 0.1 1 1.5 2 3 10 100 1000

log10(x) −1 0 0.2 0.3 0.5 1 2 3

(b) (5 Points)

Consider a signal h(t) as follows,

−1

2

T 3T2T

h(t)

t

(b1) Give the analytical expression of h(t) using the rectangle (rect (t/T )) functions only,where

rect (t/T ) =

{1 if −T/2 ≤ t ≤ T/2

0 otherwise.


336 D EXAMS

(b2) Calculate the frequency responseH(f) of h(t) using the properties of Fourier transform!

(b3) Consider a time signal z(t), the real part is known to be odd while the imaginary partis even with respect to time.

What can be concluded about the real and imaginary part of its Fourier transformZ(f)?(Please choose only one option from (i) to (iii))

i. Z(f) is purely real.

ii. Z(f) is purely imaginary.

iii. Z(f) consists of both real and imaginary parts.

What can be concluded about the symmetry of its Fourier transform Z(f)?(Please choose only one option from (iv) to (vi))

i. Z(f) is an even function with respect to f .

ii. Z(f) is an odd function with respect to f .

iii. Z(f) consists of an even and an odd part with respect to f .

(c) (5 Points)

Now consider h(t) given in part (b) as an impulse response of the LTI system with x(t)as the input to the system and the output y(t). The input x(t) is applied and is given asfollows,

x(t) =

{1 if 0 ≤ t ≤ T

0 otherwise.

h(t)x(t) y(t)

(c1) Is the system causal? Please give reason for your answer!

(c2) Give the relation between x(t), y(t) and h(t) in time domain!

(c3) Give the expression for y(t)!

(c4) Sketch y(t)!


(a) (7 Points)An analog signal s(t) has to be digitized by means of an analog-digital converter. At its inputthe wanted signal s(t) is disturbed by an interference signal i(t). The described scenario aswell as the spectra of the signals are depicted in the following figures.


D.27 Exam WS 2012/2013 337

(a1) Which relation between sampling frequency fs and cutoff frequency fg of a signal hasto be fulfilled in order to avoid aliasing errors?

(a2) Give the cutoff frequency fg of the input signal z(t)!

(a3) Draw the magnitude spectrum |Zs(f)| at the sampler output for the case thatfs = 40MHz! The considered range is −50MHz ≤ f ≤ 50MHz! Pay attention to correctaxis labeling!

(a4) Is it possible to reconstruct the wanted signal s(t) out of |Zs(f)| without any error?Justify your answer!

(a5) Give the minimum sampling frequency fs,min for which the wanted signal s(t) can bereconstructed just error-free!

(a6) The input signal z(t) is now sampled keeping the sampling condition in (a1). For re-constructing the wanted signal s(t) an ideal lowpass filter should be used. Additionaly,by adjusting its cutoff frequency we want to remove the interference signal i(t), so thatthe following applies: z(t) = s(t).Give a valid filter function R(f)!Hint: The impact of the quantizer can be neglected!


338 D EXAMS

(b) (3 Points)An analog mono audio signal has to be digitized. The sampling rate of the analog-digitalconverter is fixed to 48 kHz, the number of quantization stages can be adjusted arbitrarily.

(b1) For the given audio signal we assume a uniformly distributed amplitude over[−2V,+2V]. In addition, the mean power (refered to 1Ω) of the quantizationerror is not allowed to exceed Pe = 1/48V2.How many bits are needed to quantize the signal?What is the signal-to-noise ration (SNR) of the quantized signal?

(b2) In the following, the audio signal is quantized with 16 bit. For each 24 byte audio data,an additional byte is needed to record subchannel data.How many megabyte (MB) free disk space has to be available in order to store a monoaudio signal of 3 minutes length?Hint: 1 megabyte (MB) = 1000 · 1000 byte


(a) (7 points)The signal x(t) = sin (2πfst), where fs =

fc8, is amplitude modulated (AM) for transmission.

The modulated band-pass signal is given as,

sAM(t) = (1 + μx(t)) cos (2πfct)

, where μ is the modulation index and fc the carrier frequency.The following figure shows one possible realization of the transmitted signal sAM(t).

−2

−1.5

−1.0

−0.50

0.5

1.0

1.5

2.0

0 2 4 6 8 10 12 14 16 18 20t in μs

s AM(t)

(a1) Give the modulation index μ and the carrier frequency fc of the shown signal!

(a2) Draw the transmit signal sAM(t) in the interval t ∈ [0, 5]μs given that fc = 2 MHz andμ = 1/2! Pay attention to correct axis labeling!

(a3) Draw the transmit signal |SAM(f)| in the interval f ∈ [−3/2fc, 3/2fc] given that fc =2 MHz and μ = 1/2! Pay attention to correct axis labeling!


D.27 Exam WS 2012/2013 339

(a4) Give the signal power Ps and the power of the carrier Pc of the transmit signal sAM(t)given that fc = 2 MHz and μ = 1/2!Note: Make use of the following equation to calculate the power of the signal andcarrier:

Px =

∫ ∞

−∞|X(f)|2 df.

(b) (4 points)The spectrum R(f) at the input of the receiver is shown in the following figure. The receivedsignal is known to be amplitude modulated with a modulation index μ and a carrier frequencyfc. The modulation index is μ = 1.

R(f)

f

fc + fs,1−fc − fs,1fc + fs,2−fc − fs,2

fc − fs,1−fc + fs,1fc − fs,2−fc + fs,2

−fc fc

1

0.5

0

−0.5

The useful part x(t) of the received signal r(t) is filtered by an ideal synchronous demodulatoras shown in the following figure. After the synchronous down-conversion, the signal is filteredby an ideal low-pass (LP) filter with cut-off frequency fs,2 < fg,TP < 2fc − fs,2, first, and thanfiltered by an ideal high-pass (HP) filter with cut-off frequency 0 < fg,HP < fs,1.

Synchronous down-conversion mixer

r(t) × u(t) ≈ x(t)LP HP

ideal BP at fc cos 2πfct

carrier recovery

(b1) Draw the spectrum of the demodulated signal X(f) in the interval f ∈ [−2fs,2, 2fs,2]!Pay attention to correct axis labeling!

(b2) Give the useful part of the signal x(t) at the output of the demodulator!

(b3) The cut-off frequency of the low-pass filter in the demodulator is to low due to processtolerances. In this case, the cut-off frequency is fg = (fs,1 + fs,2) /2.Give the useful part of the signal x(t) at the output of the demodulator!


340 D EXAMS


(a) (5 Points) The following figure shows the constellation diagrams of three different modulationschemes.

(a1) Name each modulation scheme! Draw the decision thresholds into the figure!

(a2) A video streaming system uses 8-PSK as modulation scheme.How many symbols have to be at least transmitted per second, if the data rate forunimpeded transmission is 360 bit/s?

(a3) Name a modulation scheme that achieves the rate of 360 bit/s with a symbol rate of60 symbols/s!

(b) (5 Points)

The following figure shows the receiver of a digital 4-ASK system.

(b1) Sketch and label a corresponding transmitter!

(b2) Is Gray mapping used in the symbol mapper? Please give reason for your answer!

(b3) Name one reason, why in practice rectangular filters are not used as impulse shapers!

(b4) 4-PSK is also a four-state modulation scheme. Why can the depicted receiver still notbe used for a 4-PSK system?

(c) (4 Points)The eye diagram is a graphical depiction of signals that is used to estimate the quality of areceived signal. It is constructed by overlaying multiple periods of the signal. The followingfigure shows the eye diagrams of the signals of two different systems after passing through


D.27 Exam WS 2012/2013 341

the receive filter.

−0.5 0 0.50.5

0

0,5

1

t/Ts

Am

plitu

de

−0.5 0 0.50.5

0

0.5

1

Am

plitu

de

t/Ts

BA

(c1) In which system can you expect less bit errors? Please give reason for your answer!

The following figure shows another eye diagram of a purely real signal.

−0.5 0 0.50.5−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

t/Ts

Am

plitu

de

(c1) Draw the constellation diagram of the modulation scheme that was used! Pay attentionto correct axis labeling!


342 D EXAMS



(a) (5 Points)

(a1) The transmit power PT is given as,

LT = 10 log10

(PT

1mW

)dBm

20 dBm = 10 log10

(PT

1mW

)dBm

PT

1 mW= 102

= 100 mW

(1 point)

(a2) The received power level is given as,

LR = LT − LA

LA = LT − LR

= 80 dB

(1 point)

(a3) The path loss between transmitter and receiver is given as,

LA = 40dB + 20 log10

(d

m

)dB.

As LA = 80 dB,

LA dB = 40dB + 20 log10

(d

m

)dB

80dB = 40dB + 20 log10

(d

m

)dB

80− 40 = 20 log10

(d

m

)dB

2 = log10

(d

m

)dB

d = 100 m

(1 point)



(a4) The noise level PN = 10−10 W and the received signal level is LR = −60 dBm. Firstwe calculate the noise power level in dBm.

LN = 10 log10

(PN

1mW

)= 10 log10

(10−7

)= 10 · (−7) log10(10)

= −70 dBm

Noise level in dBm (1 point)

SNR = LR − LN

= −60 dBm− (−70 dBm)

= 10 dB

SNR calculation (1 point)

(b) (5 Points)

(b1) The signal h(t) can be written as follows,

h(t) = −rect

(t− T/2

T

)+ 2 · rect

(t− 2T

2T

)(1 point)

(b2) As we know,rect(t/T ) T si(πfT )

also we know from above part,

h(t) = −rect

(t− T/2

T

)+ 2 · rect

(t− 2T

2T

)

First we find the Fourier transform of −rect(

t−T/2T

)

−rect

(t− T/2

T

)− Te−j2πfT/2 · si(πfT )

Now similarly we find the Fourier transform of second part of h(t)

2 · rect(t− 2T

2T

)4Te−j4πfT · si(2πfT )


344 D EXAMS

And finally using the linearity property of the Frourier transform

H(f) = −T si(πfT )e−jπfT + 4T si(2πfT )e−j4πfT

(1 + 1 points)

(b3) (ii) and (iii) are correct. As the signal z(t) has odd real and even imaginary part.This translates into odd imaginary and even imaginary parts in Fourier transformrespectively.

(1 + 1 points)

(c) (5 Points)

(c1) Yes the system is causal as h(t) = 0 for t < 0 is satisfied. (1 point)

(c2) The output of the system is given by the convolution of input x(t) and impulse responseh(t). The relation between x(t), y(t) and h(t) is gives as,

y(t) = x(t) ∗ h(t)=

∫ ∞

−∞h(t− τ)x(τ) dτ

=

∫ ∞

−∞x(t− τ)h(τ) dτ

(1 point)

(c3) The output y(t) is given as,For t ≤ 0 ⇒ y(t) = 0.

For 0 ≤ t ≤ T ,y(t) =

∫ t

0(1) · (−1) dτ = −t.

For T ≤ t ≤ 2T ,y(t) =

∫ T

−T+t(1) · (−1) dτ +

∫ t

T(1) · (2) dτ = 3t− 4T .

For 2T ≤ t ≤ 3T ,y(t) =

∫ t

−T+t(1) · (2) dτ = 2T .

For 3T ≤ t ≤ 4T ,y(t) =

∫ 3T

−T+t(1) · (2) dτ = 8T − 2t.

For t ≥ 4T ⇒ y(t) = 0.Answer correct (2 points)



2T

3T 4T

y(t)

2TT

−T

T

t

Sketch correct (1 point)


(a) (a1) Aliasing is avoided, if fs ≥ 2fg (Nyquist sampling theorem). (1 Point)

(a2) The (upper) cutoff frequency of |Z(f)| is fg = 25MHz. (1 Point)

(a3) Spectrum |Zs(f)| result from periodic shift of |Z(f)| by 40MHz. No aliasing occurs.

Qualitatively correct: (no aliasing) (1 Point)Quantitativly correct: (axis labeling) (1 Point)

(a4) Yes, despite subsampling of z(t), s(t) can be reconstructed error-free since there is nospectral overlap of the wanted and interference signal (no aliasing). (1 Point)

(a5) fs,min = 35MHz (1 Point)

(a6) A rectangular filter: R(f) = rect(

ffB

)with 20MHz ≤ fB ≤ 40MHz. (1 Point)

(b) (b1)

Pe =1

48V2 =

(Δs)2

12

From this the distance between the quanization stages follows: Δs = 12V

The number of bits b results in:

b = log2

(Umax − Umin

Δs

)bit

= log2

(2V + 2V

0.5V

)bit = log2

(23

)bit

= 3 bit


346 D EXAMS

(1 Point)

Signal-to-noise ration (SNR) is given by:

SNR ≈ (6 · b) dB≈ 18 dB

(1 Point)

(b2)

data rate r = sampling rate · (audio data + subchannel data)

= 48 kHz · (2B + 1/12B)

= (96 + 4) kB/s

data D = 100 kB/s · 180 s= 18MB

(1 Point)


(a) (7 points)

(a1) (1 point each, total 2 points)

• Modulation index μ is determined by the envelope (1+μx(t)) with max (x(t)) = 1for t = 2μs: (1 + μmax (x(t))) = (1 + μ · 1) = 7

4→ μ = 3

4.

• Carrier frequency is determined by the length of a period T of one oscillation. HereT = 1μs → fc =

1T= 1 MHz.

(a2) Signal sAM(t):

−1.5

−1.0

−0.5

0

0.5

1.0

1.5

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

t in μs

s AM(t)

Qualitatively correct: (signal shape) (1 point)Quantitatively correct: (axis labeling) (1 point)



(a3) Spectrum |SAM(f)|:

|SAM(f)|

f

fc + fc/8−fc − fc/8 fc − fc/8−fc + fc/8

−fc fc

0.5

0.25

0

−0.25


(a4) (1/2 point each, total 1 point)The power of the signal and the carrier can be determined by using the transfer functionSAM(f):

• Ps = 2 · 2 · (μ4

)2= μ2

4= 1

16

• Pc = 2 · (12

)2= 1

2

(b) (4 points)

(b1) Spectrum X(f):

X(f)

f

−fs,1−fs,2

fs,1fs,2

1

0.5

0

−0.5


(b2) Calculation of x(t): (1 point)

X(f) =1

2(δ(f + fs,2) + δ(f − fs,2)) +

3

4(δ(f + fs,1) + δ(f − fs,1))

x(t) = cos (2πfs,2t) +3

2cos (2πfs,1t)

x(t) = cos (2πfs,2t) +3

2cos (2πfs,1t)


348 D EXAMS

(b3) Calculation of x(t): (1 point)

x(t) =3

2cos (2πfs,1t)


(a) (5 Points)

(a1) A - 16-QAMB - 4-PSKC - 4-ASK

(12point for each correct name/ correct decision threshold) (Total: 3 Points)

(a2) The symbolrate can be calculated as

R =360 bit/s

3 bit/symbol= 120 symbols/s

(1 Point)

(a3) At 60 Symbols/s, a scheme with

nbit =360 bit/s

60 symbole/s= 6 bit

per constellation symbol is required. For example, 64-QAM can be used. (1 Point)

(b) (5 Points)

(b1) Diagram like depicted or similar



Diagram correct (1 Point)Labeling correct (1 Point)

(b2) No, because the symbols for 01 and 10 differ by 2 bit. (1 Point)

(b3) • Infinite bandwidth

• Difficult to implement

One correct reason (1 Point)

(b4) The quadrature phase must be demodulated separately, a second path is required.(1 Point)

(c) (4 Points)

(c1) For the right signal less errors can be expected, since the vertical and horizontal eyeopening is wider.Correct choice (1 Point)Correct reason (1 Point)

(c2) Diagram like depicted or similar:

Qualitatively correct: (I/Q points) (1 Point)Quantitatively correct: (axis labeling) (1 Point)


350

Notes


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