1

Lecture 1B (01/07)Signal Modulation

• Introduction • Signal Analysis• Advantages of Signal Modulation • Amplitude Modulation• Amplitude Modulation of Digital Signals • Reducing Power and Bandwidth of AM Signals • Frequency Modulation • Power and Bandwidth of FM Signals • Conclusion

2

Introduction

• The tasks of communication is to encode information as a signal level, transmit this signal, then decode the signal at the receiver.

• An analog signal varies continuously with time, and has an infinite number of possible signal levels.

• A discrete signal changes only once during a certain time interval. The signal value during this time interval (sampling period) is one sample. Each sample can occupy a infinite number of possible levels.

• A digital signal is discrete, but each sample has a finite number of possible signal levels. The limited number of levels means that each sample transmits a single information. It also means that each sample can be represented as digital data, a string of ones and zeroes.

• A digital signal is preferred in computer communications because computers already store and process information digitally.

3

 

S(t)

t

 

S(t)

tSampling Period

  

S(t)

tSampling Period

00000001001000110100010101100111100010011010101111001101

• In an analog signal, noise are added to the signal during transmission. The received signal will never be identical to the original signal. • A digital signal will still be subject to noise, but the difference between signal levels (an “0101” and an “0110”) will be sufficiently large so that the receiver can always determine the original signal level in each sampling period. The regenerated, so that the received digital data is an exact replica of the original digital data. • Analog to digital conversion reduces the amount of information of the signal by approximating the analog signal with a digital signal. The sampling period and number of levels of the digital signal should be selected in order to capture as much information of the original signal as possible

4

  

S(t)

t

Sampling Interval

• A periodic signal satisfies the condition:

• “Period” of the signal.

• “Aperiodic.” signals

• An even function S (t) = S (-t)

• “Symmetric”.

• The phase of the signal.

tTtStS );()(

Signal Properties

5

S(t)

A

T

t

A · cos( 2 · f · t )

A · cos( 2 · f · t - )

f

A

1/T

A

f

A

1/T

=0

1/T f

A

=/4

tftftftS mmmm

10cos

5

16cos

3

12cos

4V5)(

A

T

t

T

20

Square Wave

6

Digital signals in the time and the frequency domain.

• The time domain is the signal level expressed as a function of time. • The frequency domain is comprised of amplitude and a phase for an

infinite number of sinusoidal functions. These correspond to the superposition of an infinite number of sinusoidal waveforms in the time domain.

• To convert from the time domain to the frequency domain, we take the Fourier transform of the time domain representation. 

• The frequency domain representation of a signal does not change with time, but the Fourier transform for the signal over an infinitely long time period would be impossible. Instead, we assume that the signal has some finite duration, over a sampling interval:

1. We assume that outside of the sampling interval, the signal repeats itself, so that the signal is periodic.

2. An alternative assumption is that the time-domain signal has a zero value outside of the sampling interval, so that the signal is aperiodic.

7

Fourier Series Expansion of Periodic Signals

a. There are a finite number of discontinuities in the period T.

b. It has a finite average value for the period T.

c. It has a finite number of posit. and negat. Max. in the period T.

1

00

0 )2(sin)2(cos)(n

nn

n tfnbtfnatS

f0 : Fundamental

frequency, n · f0 : Frequency of each

terman , bn : Fourier series

coefficients :

tdtST

aT

0

000 )(

1

tdtnftST

aT

n 0

00

0

)2(cos)(2

dttnftST

bT

n 0

0 00

)2(sin)(2

1

00 2cos)(n

nn tnfcatS

8

A

T

t

T

20

Square Wave

tftftftS mmmm

10cos

5

16cos

3

12cos

4V5)(

c(f)

ffrequency

c1

c2

c3

c4

c5 c

6c

7c

8c

9c

10c

11c

12

f0 2f

0 3f0 4f

0 5f0 6f

0 7f0 8f

0 9f0 10f

0 11f0 12f

0

tdtST

VT

RMS 0

2)(1

9

c(f) V

f frequency

5th harmonic missing

10th harmonic missing

f0 2f0 3f0 4f0 5f0 6f0 7f0 8f0 9f0 10f0 11f0 12f0

c(f) rms

f frequency

5th harmonic missing 10th harmonic missing

f0 2f0 3f0 4f0 5f0 6f0 7f0 8f0 9f0 10f0 11f0 12f0

tdtST

VT

RMS 0

2)(1

10

n = 1n = 3

n = 1 n = 5

n = 4 1

S ( t )

t

11

According to Fourier analysis, any composite signal can be represented as

a combination of simple sine waves with different frequencies, phases, and

amplitudes.

12

Introduction

A network and transport layers basic services comprise the end-to-end transport of the bit streams over a set of routers (switches).

They are produced using six basic mechanisms: a. Multiplexing b. Switching c. Error control d. flow control e. congestion control f. and resource allocation.

13

Multiplexing

Combines data streams of many such users into one large bandwidth stream for long duration. Users can share communication medium.

a. Switch b. Multiplexer/Demultiplexer

a. Fully connected network b. network with shared links.

14

Signal Modulation is: • The process of encoding a baseband (data), source

signal Sm (t) onto a carrier signal.

• The carrier waveform is varied in a manner related to the baseband signal.

• The carrier can be a sinusoidal signal or a pulse signal. The result of modulating the carrier signal is called the modulated signal.

15

Signal Modulation Sc ( t ) = A · cos ( 2 f t+ )

A. Amplitude Modulation (AM), or Amplitude Shift Keying (ASK): The amplitude A of the carrier signal changes in proportion to the baseband signal.

B. Phase Modulation (PM), or Phase Shift Keying (PSK): The phase of the carrier signal changes in proportion to the baseband signal.

C. Frequency Modulation (FM), or Frequency Shift Keying (FSK): The frequency f of the carrier changes in proportion to the baseband signal.

16

Amplitude, Phase, and Frequency Modulation of

a digital baseband signal

AM

1 0 0 0 1 0 1 1 0 0

PM

1 0 0 0 1 0 1 1 0 0

FM

1 0 0 0 1 0 1 1 0 0

17

Advantages of Signal Modulation

• Radio transmission of the signal.

m103

s

11000

s

m103

;s

11000 5

8

f

cf

Sc ( t ) = A · cos ( 2 fc ) range (fc  + f)  and (fc  - f) .

m30

s

11010

s

m103

;s

11010

6

8

6

f

cf

5GHz 6 cm

FDMWDM

18

FDM Original Signals

M

Sc1 (t)

Sm1 (t)

M

Sc2 (t)

Sm2 (t)

M

Sc3 (t)

Sm3 (t)

DM

Sc1 (t)

DM

Sc2 (t)

DM

Sc3 (t)

Received Signals

Sm1 (t)

Sm2 (t)

Sm3 (t)

Transmission Line

Guard band

f

fc1-fm1 fc1 fc1+fm1 fc2-fm2 fc2 fc2+fm2 .......................

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Amplitude Modulation

• AM is simply the multiplication of the baseband signal with the carrier signal.

)2(cos)( tfAtS ccc

)2(cos)(1

)( tftSAk

tS cmc

)2(cos)()2(cosoffsetDC

)2(cos)(offsetDC)(11

1

tftSAtfA

tftSAtS

cmckcmck

cmmck

•Modulation coefficient k describeshow efficiently the modulator device multiplies the two signals.

Sm ( t ) X

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DSBTC AM• DSBTC AM. DC = Ac .

• Modulation index i of a DSBTC AM signal

• The simplest case of DSBTC AM, where the baseband signal is a sinusoidal signal with:

DC = Ac

21

)2(cos)2(cos1)( 1 tftfA

AAAtS cm

c

mcck

c

m

A

Ai

)2(cos)2(cos1)(2

tftfik

AtS cm

c

)(cos)(cos2

1)(cos)(cos yxyxyx

componentfrequencycarrierNoncomponentfrequencyCarrier

tfftffik

Atf

k

A

tftfik

Atf

k

A

tftfik

AtS

mcmcc

cc

cmc

cc

cmc

))(2cos())(2cos(2

1)2cos(

)2cos()2cos()2cos(

)2(cos)2(cos1)(

22

22

2

c

m

A

tSi

)(max (without DC offset)

22

Power spectrum of DSBTC AM signal with

sinusoidal baseband signal, i = 100%, k = 1.

 Vrms

ffrequencyfc + fmfc - fm

2

fc

Ac2

2· 2

Ac2

Ac2

2· 2

23

Sm (t)

t

t

Sc (t)

t

S(t)

Ac

24

)2(cos)2(cos)( tftfAAtS cmcc

))(2cos())(2cos(

2

1)2(cos)2(cos)(

2

2

tfftffA

tftfAtS

mcmcc

cmc

Vrms

ffrequencyfc + fmfc - fm fc

Ac2

2· 2

Ac2

2· 2

Sm ( t ) = Ac · cos ( 2 fm t)

DSBSC AM

25

Sm (t)

t

t

Sc (t)

t

S(t)

26

Amplitude Modulation of Digital Signals

Baseband Signal

Baseband Signal with DC shift

tftftftS mmmm

10cos

5

16cos

3

12cos

4V5)(

tftS cc 2cosV5)(

Carrier Signal

tftftftS mmmm

10cos

5

16cos

3

12cos

4V5V5)(

27

Transmitted Signal

sidebandupper

tfftfftffk

V

sidebandlower

tfftfftffk

V

frequencycarrier

tfk

VtS

mcmcmc

mcmcmc

c

...52cos5

132cos

3

12cos

92.15

...52cos5

132cos

3

12cos

92.15

)2cos(25

)(

2

2

2

28

DSBTC AM signal with square wave baseband signal, i = 10

0%, k = 10.

 

Vrms

ffrequencyfc - fm

2

fc

2.50V

fc - 3fm

2

1.59V

2

1.59V

fc + 3fmfc + fm

2

0.53V

2

0.53V

2

0.32V

2

0.32V

fc - 5fm fc + 5fm

29

  Sm (t)

t

t

Sc (t)

t

S(t)

5V

5V

10V

50V 1/2

1/2

625504

10

22

10

2

1

21

21

22

A

PPP

PowersignalModulated

Full

30

DSBSC AM

tftftftS mmmm

10cos

5

16cos

3

12cos

4V5)(

sidebandupper

tfftfftffk

V

sidebandlower

tfftfftffk

VtS

mcmcmc

mcmcmc

...52cos5

132cos

3

12cos

92.15

...52cos5

132cos

3

12cos

92.15)(

2

2

Vrms

ffrequencyfc - fm fcfc - 3fm

2

1.59V

2

1.59V

fc + 3fmfc + fm

2

0.53V

2

0.53V

2

0.32V

2

0.32V

fc - 5fm fc + 5fm

31

t

Sm (t)

t

Sc (t)

t

S(t) 180 degree phase shifts

32

Binary Phase Shift Keying and Quadrature Phase Shift Keying

V5)(if180

V5)(if0)(

)(2cosV25

)(

tS

tSt

ttfk

tS

m

m

c

Since the phase of the transmitted signal can take one of two values (phase jumping), this type of modulation is called binary phase-shift keying (BPSK). It is a constant-amplitude method of modulation.

33

If we add together two BPSK signals that are offset by a 90 degree phase shift, there are four possible phase jumping during every sampling period. Instead of one possible shift of 180º, there are three possible transitions: +90º, 180º, and ‑90º (+270º). The signal can be represented by the following formula:

V5)(V,5)(if315

V5)(V,5)(if225

V5)(V,5)(if135

V5)(V,5)(if45

)(

)(2cosV25

)(

21

21

21

21

tStS

tStS

tStS

tStS

t

ttfk

tS

mm

mm

mm

mm

c

34

t

S(t) phase shifts

t

S1 (t)

t

S2 (t)

0 1 0 1 0

0 1 1 0 0

00 11 01 10 00

180º 90º 180º -90º

Time domain representation of QPSK signal, created: by combining BPSK signals S1 and S2 .

35

Reducing Power and Bandwidth of AM Signals

• In DSBTC AM, the carrier frequency component carries no information.

• In all forms of amplitude modulation, the two sidebands contain identical information.

• Eliminating one of the two sidebands. • There are four possibilities for transmitting an amplitude

modulated signal • A. Dual sideband, transmitted carrier (DSBTC AM)• B. Vestigial sideband• C. Dual sideband, suppressed carrier (DSBSC AM)• D. Single sideband• Selecting the type of transmission can depend on the following

factors:• Power limitations• Bandwidth limitations• Simplicity of Demodulation Equipment

36

• The power required for transmitting a sinusoidal baseband signal with DSBTC AM is:

Pt : Transmitted power, Ac :Carrier Signal Amplitude

• i : Modulation index

• i=100%, the Pt =1.5 times the power squared of the carrier frequency. Each sideband contributes 0.25 times the power of the carrier frequency.

• Vestigial sideband uses only one of the two sidebands of the signal.

21

222

12

22

1

22

222

iP

AAAP

c

cmct

37

Four possible ways to transmit an AM signal.

fc

Vrms

(a) DSBTC AM

ffc

Vrms

(c) DSBSC AM

f

fc

Gain

ffc

Vrms

f

Bandpass Filter

fc

Vrms

(b) Vestigial sideband AM

ffc

Vrms

(d) Single sideband AM

f

Bandpass Filter

38

Example Example We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations.

SolutionSolution An AM signal requires twice the bandwidth of the original signal:

BW = 2 x 4 KHz = 8 KHz

Concept 1:

The total bandwidth required for AM can be determined from the bandwidth

of the audio signal: BWt = 2 x BWm.

39

Frequency Modulation • Additive noise• Frequency modulation is a special case of phase

modulation, where the phase of the transmitted signal accumulates according to the amplitude of the baseband signal:

t

mcc dSktfAtS )(2cos)(

k : Constructional coefficient of the modulator

40

• To find the maximum possible frequency deviation of an FM signal, we can assume the baseband signal is a constant voltage of +Am , then the integral will reduce to a linear function of t :

tAk

fAtAktfAtS mccmcc

2

2cos2cos)(

2max

mc

Akf

Sm ( t ) = Am · sin (2 fm t )

)2(cos2

)2(sin)()(0

tff

AkdfAkdSkt m

m

mt

mm

t

m

m

mf f

Akk

2

)2(cos2cos)( tfktfAtS mfcc

41

• Equation can be expanded using Bessel's trigonometric identities into a series of phase-shifted sinusoidal terms.

• The amplitude of each term is determined by the Bessel function of the first kind Jn ( kf ), where kf  is the modulation index.

n

mcfnmfcn

fnftkJtfktf2

2cos)()2(cos2cos

n

mcfncn

fnftkJAtS2

2cos)()(

)()1()( xJxJ nn

n

22cos

22cos)(

2cos)()(

1

0

ntfnf

ntfnfkJA

tfkJAtS

mc

n

mcfnc

cfc

42

kf

1.0

0.6

0.4

0.8

0.2

0

-0.2

-0.4

J0 ( kf )

J1 ( kf ) J2 ( kf ) J3 ( kf ) J4 ( kf ) J5 ( kf )

10 6 8 4 2

First five Bessel coefficients for varying values

of kf .

43

Power and Bandwidth of FM Signals

• Calculating the power of an FM signal is much simpler in the time domain.

1

220

22

)(2)(2

1

2 fnfcc

FM kjkjAA

P

12 fT kBB

In this equation B is effective bandwidth of baseband signal, but BT is absolute bandwidth of FM signal

)2(cos2cos)( tfktfAtS mfcc

44

Ac  · cos ( 2 fc  + kf  cos ( 2 fm  ) ) at various values of kf . Vrms

f

kf = 0.5

Vrms

f

kf = 1

Vrms

f

kf = 2

Vrms

ffc +6fm

kf = 10

fc +9fm fc -9fm fc -6fm

fc fc + 3fm fc - 3fm

fc fc + 3fm fc - 3fm

fc fc + 3fm fc - 3fm

fc fc +3fm fc -3fm

45

Example 1Example 1 We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations.

SolutionSolution An FM signal requires 10 times the bandwidth of the original signal:

BW = 10 x 4 MHz = 40 MHz

Concept 2:

The total bandwidth required for FM can be determined from the bandwidth

of the audio signal: BWt = 10 x BWm.