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CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
COURSE FILE OF SIGNALS & SYSTEMS
CONTENTS:
1. Syllabus.
2. Course Plan.
3. Course Objective.
4. Time Table.
5. Teaching Notes.
6. JNTU Question Papers.
7. Descriptive Question Bank.
8. Objective Question Bank.
9. Mid Exam Papers.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
1. SYLLABUS COPY :
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY
HYDERABAD
II Year B.Tech. ECE. I-Sem L T/P/D C
4 1/-/- 4
(53021) SIGNALS & SYSTEMS
Unit I : Signal Analysis
Analogy between Vectors and Signals, Orthogonal Signal Space, Signal approximation using
Orthogonal functions, Mean Square Error, Closed or complete set of Orthogonal functions,
Orthogonality in Complex functions, Exponential and Sinusoidal signals, Concepts of
Impulse function, Unit Step function, Signum function.
Unit II : Fourier Series Representation of Periodic Signals
Representation of Fourier series, Continuous time periodic signals, Properties of Fourier
Series, Dirichlet’s conditions, Trigonometric Fourier Series and Exponential Fourier Series,
Complex Fourier spectrum.
Unit III : Fourier Transforms :
Deriving Fourier Transform from Fourier Series, Fourier Transform of arbitrary signal,
Fourier Transform of standard signals, Fourier Transform of Periodic Signals, Properties of
Fourier Transform, Fourier Transforms involving Impulse function and Signum function,
Introduction to Hilbert Transform.
Unit IV : Signal Transmission Through Linear Systems
Linear System, Impulse response, Response of a Linear System, Linear Time Invariant (LTI)
System, Linear Time Variant (LTV) System, Transfer function of a LTI system, Filter
characteristics of Linear Systems, Distortion less transmission through a system, Signal
bandwidth, System bandwidth, Ideal LPF, HPF and BPF characteristics, Causality and Paley-
Wiener criterion for physical realization, Relationship between Bandwidth and Rise time.
Unit V : Convolution and Correlation of Signals
Concept of convolution in Time domain and Frequency domain, Graphical representation of
Convolution, Convolution property of Fourier Transforms, Cross correlation and Auto
Correlation of functions, Properties of Correlation function, Energy density spectrum,
Parseval’s Theorem, Power density spectrum, Relation between Auto Correlation function
and Energy/Power spectral density function, Relation between Convolution and Correlation,
Detection of periodic signals in the presence of Noise by Correlation, Extraction of signal
from noise by filtering.
Unit VI : Sampling
Sampling theorem – Graphical and analytical proof for Band Limited Signals, Impulse
Sampling, Natural and Flat top Sampling, Reconstruction of signal from its samples, Effect
of under sampling – Aliasing, Introduction to Band Pass sampling.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Unit VII : Laplace Transforms
Review of Laplace Transforms (L.T.), Partial fraction expansion, Inverse Laplace Transform,
Concept of Region of Convergence (ROC) for Laplace Transforms, Constraints on ROC for
various classes of signals, Properties of L.T., Relation between L.T. and F.T. of a signal,
Laplace Transform of certain signals using waveform synthesis.
Unit VIII : Z-Transforms
Fundamental difference between Continuous and Discrete time signals, Discrete time signal
representation using Complex exponential and Sinusoidal components, Periodicity of
Discrete time signal using complex exponential signal, Concept of Z-Transform of a Discrete
Sequence, Distinction between Laplace, Fourier and Z Transforms, Region of Convergence
in Z-Transform, Constraints on ROC for various classes of signals, Inverse Z-transform,
Properties of Z-transforms.
Text Books :
1. Signals, Systems & Communications – B.P. Lathi, 2009, BSP
2. Signals and Systems – A. Rama Krishna Rao – 2008, TMH.
3. Signals and Systems – A.V. Oppenheim, A.S. Willsky and S.H. Nawab, 2ed., PHI.
References :
1. Signals & Systems – Simon Haykin and Van Veen, Wiley, 2 ed.
2. Introduction to Signal and System Analysis – K. Gopalan 2009, CENGAGE
Learnings.
3. Fundamentals of Signals and Systems – Michel J.Robert, 2008, MGH International
Edition.
4. Signals, Systems and Transforms – C.L. Philips, J.M.Parr and Eve A.Riskin, 3 ed.,
2004, PE.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
2. COURSE PLAN:
SUBJECT: SIGNALS AND SYSTEMS
BRANCH: B.Tech II Year ECEI Semester 2012 -2013
SECTION: C
NAME OF THE FACULTY: B.CHAKRADHAR
Unit / Topic No. of
periods
required
Cumulative
periods
Planned
Date
Cover
ed
Date
Unit-1: Signal Analysis
Introduction 1 1 2/7/2012
Classification of signals & systems 1 2 3/7/2012
Operations on Signals 2 4 4/7/2012
Analogy between vectors and signals 1 5 7/7/2012
Orthogonal vector space & Orthogonal signal
space
1 6 7/7/2012
Mean square Error & Orthogonality in
complex functions
1 7 9/7/2012
Numerical Problems 2 9 10/7/2012
Unit II: Fourier series Representation of
Periodic Signals
Introduction to Representation of Fourier
Series
1 10 117/2012
Trigonometric Fourier series 2 12 11/7/2012
16/7/2012
Exponential Fourier series 2 14 17/7/2012
18/7/2012
Relationship between Trigonometric
&Exponential Fourier series.
1 15 18/7/2012
Dirichlet’s Conditions 1 16 21/7/2012
Complex Fourier Spectrum 1 17 21/7/2012
Numerical Problems 2 19 23/7/2012
24/7/2012
Unit-III: Fourier Transforms
Introduction 1 20 25/7/2012
Deriving Fourier Transform from Fourier
series
1 21 25/7/2012
Fourier Transform of standard signals 1 22 28/7/2012
Fourier Transform of Periodic Signals 1 23 30/7/2012
Properties of Fourier Transform 3 26 31/7/2012
1/8/2012
Introduction to Hilbert Transform 1 37 4/8/2012
Numerical Problems 2 29 4/8/2012
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Unit-IV: Signal Transmission Through
Linear Systems
Introduction to Linear systems 1 30 6/8/2012
Impulse Response, Response of Linear
Systems
2 32 7/8/2012
8/8/2012
Linear Time Invariant & Linear Time Variant
systems
1 33 8/8/2012
Filter Characteristics of Linear system 2 35 13/8/2012
14/8/2012
Distortion less Transmission through
systems, Signal Bandwidth, System
Bandwidth
1 36 18/8/2012
Ideal LPF,HPF,BPF characteristics 1 37 18/8/2012
Relationship between Bandwidth and Rise
time.
1 38 20/8/2012
Numerical Problems 2 40 21/8/2012
Unit-V: Convolution and Correlation of
Signals
Concept of Convolution in Time domain &
Frequency Domain
1 41 22/8/2012
Graphical representation of Convolution 2 43 22/8/2012
25/8/2012
Convolution Property of Fourier Transforms 1 44 25/8/2012
Cross Correlation and Auto Correlation
Functions
2 46 1/9/2012
Properties of Correlation function 2 48 3/9/2012
4/9/2012
Energy Density Spectrum, Parsevals
Theorem
1 49 5/9/2012
Relation Between Convolution and
Correlation
1 50 5/9/2012
Detection of periodic signals in the presence
of Noise by Correlation
1 51 10/9/2012
Extraction of signal from Noise by filtering 1 52 11/9/2012
Numerical Problems 2 54 12/9/2012
Unit-VI: Sampling
Introduction to Sampling 1 55 15/9/2012
Sampling Theorem 1 56 15/9/2012
Impulse sampling, Natural and Flat top
Sampling
1 57 17/9/2012
Reconstruction of signal from its samples 1 58 18/9/2012
Aliasing Effect, Introduction to Band Pass
Sampling
1 59 22/9/2012
Numerical Problems 2 61 22/9/2012
24/9/2012
Unit-VII: Laplace Transforms
Introduction 1 62 25/9/2012
Partial Fraction Expansion 1 63 26/9/2012
Concept of ROC for Laplace Transforms 1 64 26/9/2012
Constraints on ROC for various classes of 1 65 29/9/2012
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
signals
Properties of Laplace Transforms 2 67 29/9/2012
1/10/2012
Inverse Laplace Transforms 2 69 3/10/2012
Relation Between Laplace Transform and
Fourier Transform
1 70 6/10/2012
Laplace Transform using Waveform
Synthesis
1
71
8/10/2012
Numerical Problems 1 72 9/10/2012
Unit-VIII: Z-Transforms
Introduction & Differences between
Continuous and Discrete time signals
1 73 10/10/2012
Discrete time signal representation using
Complex Exponential and Sinusoidal
components
1 74 10/10/2012
Concept of Z-Transform of a Discrete time
Sequence
2 76 17/10/2012
Properties of Z-Transforms 2 78 20/10/2012
ROC of Z-Transform 1 79 22/10/2012
Inverse Z-Transform 1 80 23/10/2012
Constraints on ROC for Various classes of
signals
1 81 27/10/2012
Relation between Laplace, Fourier and Z-
Transforms.
1 82 27/10/2012
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
3. COURSE OBJECTIVE
COURSE TITLE: SIGNALS AND SYSTEMS
ODD SEMESTER: 4 Credits
2012-13 CATALOG DATA:
Signal Analysis
Fourier series Representation of periodic signals
Fourier Transforms
Signal Transmission through Linear Systems
Convolution and Correlation of Signals
Sampling
Laplace Transforms
Z-Transforms
TEXT BOOKS:
1. Signals, Systems & Communications- B.P.Lathi,2009,BSP.
2. Signals & Systems- A.V. Oppenheim, A.S. Willsky and S.H.Nawab, PHI.
3. Signals & Systems-Simon Haykin and Van Veen, Wiley 2ed.
REFERENCE BOOKS:
1. Signals & Systems- Ramesh Babu, Anandnatarajan- SciTech.
2. Introduction to Signals and System Analysis – K.Gopalan 2009, CENGAGE
Learning.
3. Signals and Systems – A. Rama Krishna Rao – 2008, TMH
4. Signals, Systems and Transforms – C.L.Philips, J.M. Parr and Eve A.Riskin, 3
ed.,2004,PE
COORDINATOR/INSTRUCTOR: Mr. B.CHAKRADHAR, Asst. Professor, ECE Dept.
COURSE OBJECTIVE: Signals & Systems is one of the core subjects of ECE Branch and
is an essential subject of the curriculum. It covers the complete set of signals in Electronics
Engineering. It also covers the linear & non-linear systems and their behaviour.
Main learning objectives of the course are,
1. To make the students aware of the problems in analysis and manipulation of various
signals and their processing through linear systems.
2. To make the students understand the techniques of using specific analytical methods
and transforms for different signals and their applications.
3. To make the students conversant with extraction of signals in the presence of noise.
4. To give the students a thorough exposure to various types of transforms like F.T., L.T.
& Z.T’s.
5. To provide the students with an understanding of sampling techniques and make the
student appreciate and prepare him to learn digital signal processing.
6. To equip the student with convolution & correlation techniques used in signal
processing.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
EVALUATION METHODS: Internal exam: Descriptive-10 marks
Objective-10 marks
Assignment-5 marks
External exam: 75 marks
PROGRAMME EDUCATION OUTCOMES:
After the successful completion of the course on Signals & Systems, the student
should be able to demonstrate,
1. An ability to use the knowledge of Signals & Systems to analyse the signals, he
encounters in his future endeavours.
2. An ability to select proper transform methods to fulfill the task of processing the
signals.
3. An ability to differentiate between various linear and non-linear systems and design
an optimum system for a specific purpose.
4. An ability to apply correlation & convolution techniques to extract the signal in the
presence of noise.
5. The ability to optimize the system performance with respective the particular signal.
6. An ability to convert analogue signal to digital signal & vice versa through sampling
& reconstruction processes.
7. An ability to confidently participate in competitive examinations.
(B.CHAKRADHAR)
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
5. Teaching Notes:
Unit-I : Signal Analysis.
Unit-II : Fourier series Representation of periodic signals.
Unit-III : Fourier Transforms.
Unit-IV: Signal Transmission through Linear Systems.
Unit-V : Convolution and Correlation of Signals.
Unit-VI: Sampling.
Unit-VII: Laplace Transforms.
Unit-VIII: Z-Transforms.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
UNIT-I
SIGNAL ANALYSIS
CONTENTS:
1.1. Introduction.
1.2. Classification of signals.
1.3. Standard signals.
1.4. Operations on signals.
1.5. Analogy between vectors and signals.
1.6. Orthogonal signal space.
1.7. Evaluation of Mean Square Error.
1.8. Repreentation of a signal by complete set of orthogonal functions.
1.9. Orthogonality in complex functions.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
UNIT – 1
SIGNAL ANALYSIS
1.1. Introduction:
- Signals are quantities which describe a variety of phenomena.
- Signal contains information.
- Information in a signal is contained in a pattern of variations of some form.
Ex :
A signal can be defined as a fn. of one or more variables which conveys information on the
nature of the physical phenomenon.
A simple RC cct with VS as input An automobile moving by the
application
and VC as the output (Voltages). of a force f and a retarding frictional
force
V. Proportional to the Velocity.
A system is an entity that manipulates one or more (variables) signals to accomplish a fn.
thereby yielding new signals.
- Signals are represented mathematically as fn. of one or more variables.
Ex :
A speech signal is acoustic pressure as a fn. of time i.e. a fn. of (mathematically) acoustic
pressure with time.
Signals can be classified into continuous time and discrete time signals.
In our study of signals and systems we deal with signals whose independent variable in time.
i) A continuous time signal in one which is defined or which has values at every
instant of time.
ii)
Week – Weekly stock market index. Discrete time signal.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Graphical representation of continuous time and discrete time signals.
a) Continuous time signal
b) Discrete time signal
1.2 Signals can be classified into various types.
i) Continuous time and discrete time signals.
ii) Deterministic and random signals
iii) Periodic and aperiodic signals
iv) Even and odd signals
v) Energy and power signals
A continuous time signal is defined for every value of time “t” and is represented by
x(t), f(t), g(t) etc.
Some of the continuous time signals are
(i) Sinusoidal signals
(ii) Exponential signals
(iii) Unit step fn.
(iv) Unit ramp fn.
(v) Unit parabolic fn.
(vi) Unit impulse fn.
(vii) Rectangular pulse fn.
(viii) Triangular pulse fn.
(ix) Signum fn.
(x) sinc fn.
(II) a) Deterministic signal is one whose magnitude and phase can be determined
accurately at any instant of time by a mathematical eqn.
Ex : x(t) = A sin (t + )
b) Random signal is a signal whose occurrence & value are uncertain.
(III) Periodic signal is characterized by the condition.
x(t) = x(t + T) for all values of t.
The smallest value of T for which the above equation is satisfied is called the
“fundamental period”.
Similarly for discrete time signals.
x[n] = x [n + N] for all n and N is the fundamental period-Exponential and sinusoidal
signals are examples of continuous complex time periodic signals.
Consider a sinusoidal signal x(t) = A sin (ot + )
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
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Where A, o & , are amplitude, frequency and phase of the signal respectively and o = 2
fo.
x(t) = x (t + T) for a periodic signal.
For x(t) = A sin (ot + ) = A sin [o (t + T) + ]
= A sin [ot + oT + ] = A sin (ot + + 2) because T = of
1
= A sin [ot + ]
Consider complex exponential :
x(t) = tJ oe
for a periodic signal x(t) = x(t + T)
x(t) = tJtJtJTJTtJ ooooo eeeee
===
+.1.
)( 12sin2cos2 =+== Jee JTJ o
Find the fundamental period of :
1. x(t) = JeJ5t → eJ5t = tJ oe
→ o = 5 ; 5
2=
T So
=
2
51
T or T = =
4.0
5
2 Sec.
2. x(t) = sin 50t → sin ot → o = 50 → T
2= 50 → T =
50
2 =
25
1 Sec.
3. x(t) = 20 cos (t + /6) = 20 cos (ot + /6) → o = ; T
2 = →
2= T → 2
sec.
Where a sum of signals is given to find its periodicity
a) all the ratios of the period of the first signal in the sum to the period of the other
signals in the sum should be rational.
b) Then the period of the sum signal is T0 = T01 n1 where T01 is the period of the first
signal and n1 is the LCM of the denominator of the remaining ratios.
(1)
i.e. 02
01
01 T
T
T
PeriodIst = rational number ;
03
01
T
T = rational number.
Then 03
01
02
01 &T
T
T
T are converted into integral numbers and LCM of T02 & T03 is
evaluated and it is n1.
Then the period of the sum signal is T01 n1
For a discrete signal
x [n] = A cos [on + ] ; o = N
2 or N =
o
2
and oN should be an integral multiple of 2 then only the discrete time signal is
periodic.
A cos [on + oN + ] = cos [on + + oN]
oN = m . 2 or N = mo
.2
;
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
=
o
m
2
Ex : x[n] = 5 sin 2n → x[n] = A sin [on + ]
here A = 5 ; o = 2 & = 0.
So N = =
=
2
22
o →
2
12 i.e. N is not an integral multiple of 2 hence not
periodic.
(a) (b)
Find whether x(t) = 2 cos (10t + 1) – sin (4t – 1) is periodic
for (a) 01 = 10 or 01
2
T
= 10 or
01
1
T =
2
10 or T01 =
10
2 =
5
sec.
for (b) 02 = 4 or 02
2
T
= 4 or
20
1
T =
2
4 or T02 =
4
2 =
2
sec.
5
2
2502
01 =
=T
T is a rational number hence periodic.
5 T01 = 2 T02 =
T = =
=
2
25
5 secs is the period of the sum signal.
Find whether periodic :
1. cos 2 n → o = 2 → N
2 = 2 → N = 1→N is an integer
2. eJ6n → o = 6 → N
2 = 6
= N = 3
1
6
2=
N = 3
1 (m) = 1 → N becomes an integer for a value of 3 for m. Hence the fn.
is periodic.
Signals which do not satisfy the conditions :
a) x(t) = x (t + T) for continuous time signals. And
b) x[n] = x[n + N] for discrete time signals
are not periodic in the case of (b) N should be positive integer.
Even and Odd Signals :
(i) A continuous time signal is said to be even (or symmetric) if it satisfies the condition.
x(-t) = x(t) for all values of t.
ex : x(t) = A cos t
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
(ii) Anti symmetric : A continuous time signal is said to be odd (anti symmetric) if it
satisfies the condition
x(-t) = -x(t) for all values of t
Ex : x(t) = A sin t
Similarly for discrete time signal.
x[-n] = x[n] for even signal
for all values of n.
x[-n] = -x [n] for odd signal
A signal x(t) can be expressed as a sum of even and odd signals.
x(t) = xe(t) + x0(t) --------- (1)
x(-t) = xe(t) – x0(t) --------- (2)
Adding x(t) + x(-t) = 2xe(t) --------- (3)
xe(t) = 2
)()( txtx −+
and x(t) – x(-t) = 2 x0(t)
x0(t) = 2
)()( txtx −−
Similarly for a discrete time signal the even and odd parts can be found as
xe [n] = ][][2
1nxnx −+
x0 [n] = ][][2
1nxnx −−
Find even and odd parts of the following :
x(t) = sin t + 2 sin t + 2 sin2t cos t
xe(t) = tttttttttxtx
cossin2sin2sincossin2sin2sin2
1
2
)()( 22 +−−++=−+
= 2 sin2t cos t.
x0(t) = tttttttttxtx
cossin2sin2sincossin2sin2sin2
1
2
)()( 22 −++++=−−
= sin t + 2 sin t
Find the time period of discrete time sum signals :
(a) (b)
nJnJ
ee 4
3
3
2
+
3
2 = 01 → N01 =
3
2
2
m = 3(m) when m = 1 N01 = 3
4
3 = 02 → N02 =
3
8
4
3
2=
m (m) → N02 = 8 when m = 3
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
8
3
2
1 =N
N → 8N1 = 3N2
N = 8 3 = 3 8
N = 24
Energy and power signals :
- Consider a voltage v(t) across a resister R producing i(t).
The instantaneous power p(t) = v(t) . i(t)
= v(t) . R
tv
R
tv )()( 2
=
p(t) = i2(t) R
- For a resister of 1 the instantaneous power p(t) = the square of the signal.
- On integration of the instantaneous power over a period | t | T. We can express the
total
energy and average power of a signal as
Total energy = E = −
→
T
T
dttiT
Lt)(2 Joules for R = 1
Average power = P = −
→
T
T
dttiTT
Lt)(
2
1 2 watts
Thus for a signal x(t)
The average power is defined as
P = −
→
T
T
dttxTT
Lt2|)(|
2
1 watts
And the total energy E = −
→
T
T
dttxT
Lt2|)(| Joules
For a discrete time signal the total energy is defined as
E = −=
n
nx 2|][|
The average power is defined as
P = −=
+→
N
Nn
nxNN
Lt2|][|
12
1
Definition :
(i) A signal x(t) is called energy signal if the energy E satisfies the condition 0 < E <
and P=0.
(ii) A signal x(t) is called a power signal if the average power P satisfies the condition
0 < P < and E =
Ex : Determine the power and RMS values of the signal :
x(t) = A cos (ot + )
P = −
→
T
T
dttxTT
Lt2|][|
2
1
= −
+→
T
T
o dttATT
Lt)(cos
2
1 22 Cos 2 = 2 cos2 - 1
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Cos2 = 2
2cos1 +
= −
++→
T
T
o dttA
TT
Lt)]22cos(1[
22
1 2
= 2
222
10
22
1 222 AT
A
TT
Ltt
A
TT
LtT
T
=→
=
+
→−
P = 2
2A
RMS Value = 22
2 AA=
Energy = E = =→
=→
−
TA
T
Ltdttx
T
Lt T
T
2.2
|)(|2
2
i.e. energy of a power signal is .
Classify the signals into power, energy or none.
(i) x(t) = e-3t u(t)
E =
=
=
→=
−→=
→=
→
−
−−−
TT
Tt
t
T
t
e
e
T
Lte
T
Ltdte
T
Ltdte
T
Lt
0
0
6
0
66
0
23
6
1
6
0
6|||| |
= 6
1 Joules hence energy signal.
(ii) x(t) = )4/2( +tJe
P = WattTTT
Ltdt
TT
Ltdte
TT
Lt T
T
T
T
tJ 122
1
2
1
2
1 2)4/2( =
→=
→=
→ −−
+
Energy of the signal =→
=→
−
TT
Ltdt
T
Lt T
T
2
(iii) x1 [n] =
n
3
1 u[n]
E =
2
0 3
1
→
n n
N
Lt
E =
→
n n
N
Lt
0 9
1
= Joulesn
n
8
9
9
11
1
9
1
0
=
−
=
=
P = −=
+→
N
Nn
nxNN
Lt2|)(|
12
1
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=
+→=
+→
N N nn
NN
Lt
NN
Lt
0 0
2
9
1
12
1
3
1
12
1
= 08
9
12
1=
+→ NN
Lt
1.3. Standard Signals:
1.3.1.Sinusoidal signal:
x(t) x(nT)
x(t) = A sin t discrete time signal
x(t) = A sin (t + )
1.3.2.Exponential Signals :
x(t) = A eat When both ‘A’ & ‘a’ are real they are called real exponential signals.
for a = 0 for a > 0 for a < 0
d.c. signal exponentially growing signal exponentially decaying
signal
Complex exponential signal :
x(t) = est where S is a complex variable.
S = + J
So x(t) = et . eJt
Using Euler’s identity x(t) = et [cos t + J sin t]
Case (i) if both & = 0. Then the signal is a pure d-c.
S = 0
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Case (ii) = 0 ; S = then x = et.
This is a decaying exponential for < 0 and growing exponential for > 0.
Case (iii) If = 0 then S = J. x(t) = eJt
This becomes a sinusoidal signal where real part is cos t and imaginary is sin t.
Case (iv)
Case (v)
(1) x(t) = cos t (2) x(t) = A e-t
Sinusoidal fn. Exponential fn.
1.3.3.Unit step signal:
u(t) = 1 for t 0
= 0 for t < 0
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unit step fn. delayed unit step
1.3.4.Unit ramp fn:
r(t) = t for t 0
= 0 for t < 0
The unit ramp fn. can be obtained by applying a step fn. to an integration.
i.e. r(t) = u(t) dt = t in the interval 0 t or t 0 as a corollary u(t) = dt
trd )(.
1.3.5.Unit parabolic fn:
p(t) = 2
2t for t 0
= 0 for t < 0
p(t) = 2
2t u(t)
Unit parabolic fn. can be obtained by integrating ramps.
p(t) = r(t) dt = t dt = 2
2t for t 0
or r(t) = dt
tpd )(
1.3.6.Unit impulse fn:
=
1)( dtt and (t) = 0 for t 0
Properties of unit impulse fns. :
1.
−
= )0()()( xdtttx ------ (1)
Consider the product of x(t) & (t) .
Let x(t) be continuous at t = 0.
The value of x(t) at t = 0 x(0).
But the impulse exists only at t = 0. Hence x(t) (t) = x(0) (t).
So the integral (1) can be written as
−
−
= dttxdttx )()0()()0( .
= x(0) since
−
=1)( dtt provided x(t) is continuous at t =
0.
2. x(t) (t-t0) = x(t0) (t – t0)
Let the signal x(t) be continuous at t = t0. and the value of x(t) at t = t0 is x(t0).
(t-t0) is an impulse at t = t0.
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Hence x(t) (t – t0) = x(t0) (t – t0)
3.
−
−
−
−=−=− dttttxdttttxdttttx )()()()()()( 00000
= x(t0)
Put (t – t0) = dt = d
So x(t0)
−
= )()( 0txd i.e.
−
=− )()()( 00 txdttttx .
4. (at) = )(
1
a (t)
Consider the integral.
−
dtattx )()( for a > 0
Let at = then a dt = d or dt = a
d.
=
−
−
=
=
)0(
1)(
1)( x
ad
ax
aa
d
ax
Let us consider :
a be -2 then we write 2t = ; t = 2
and dt =
2
d
)()(2
)(2
)2()(
−=−
=−
−
−
d
xdtttx
(i) = 2
1 x(0) =
||
1
a x(0) for a < 0
−
== 0)0(1
)0(1
)()( aforxa
xa
dtattx
Consider )0(||
1x
a we know that x(0) =
−
dtttx )()(
a
1 x(0) =
−
−
= dtt
atxdtttx
a)(
||
1)()()(
||
1
(at) = )(||
1t
a
Evaluate :
(i)
−
− − dtte t )10(2
(v)
−
−− dttt )3()3( 2
(ii)
−
− dttt )3(2 (vi) dttttt
−
−+ sin)1(cos)(
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(iii)
5
0
2sin)( dttt (vii)
−
− dtet Jwt)(
(iv)
−
−+ dtet t)3( (viii) =
+0
)3()( dtttx
1.3.7. Rectangular pulse :
(t) = 1 for |t| ½
= 0 else where
1.3.8. Triangular pulse :
a (t) =
−
at
atfora
t
)(0
)(||
1
1.3.9. Signum function. :
Sgn (t) =
−
=
01
00
01
tfor
tfor
tfor
This also can be expressed as
Sgn (t) = -1 + 2 u(t)
1.3.10. Sinc function :
Sinc t = t
tcsin - < t <
1.4. Basic operations on signals :
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(1) Time shifting (2) Time reversal (3) Time scaling
(4) Amplitude scaling (5) Signal multiplier (6) Signal addition
1.4.1. Time Shifting :
Let the signal be x(t)
Time shifting of x(t) may delay or advance the signal in time.
i.e. y(t) = x(t – T) is T is +ve, it is a delay by T units
if T is –ve it is an advance by T-units.
Original Signal Advance by T-units Delay by T-units
1.4.2. Time reversal :
Original signal Time reversed signal
Reversed and delayed by 2 units Reversed and advanced by 2 units
1.4.3. Amplitude scaling : Amplitude scaled signal is obtained by scaling the
amplitude of signal at each and every point.
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Ex : y(t) = 3 x(t)
1.4.4.Time scaling: This can be accomplished by replacing t by “at” or “t/a”. Where a
is a +ve integer.
(1) Case 1 – ‘t’ replaced by at
Y(t) = x(t) Y(t) = x(2t) Y(t) = x(t/2)
Original signal Compressed Expanded
1.4.5. Signal Addition x3 (t) = x1 (t) + x2 (t)
Sketch the following signals :
x(t) find x(t – 2) x(-t)
x (2t+3) x(-t + 1)
x (3/2 t)
(1) u(t) – u(t – 2) (4) r(t) – 2 r(t-1) + r (t – 2)
(2) (t – 1/2 ) (5) r(t) u [2 – t]
(3) 12
1−+
− t
t (6) r [-0.5t + 2]
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Odd and even signals discrete times :
Odd Signal x[n] = -x [-n] Even Signal x[n] = x[-n]
Even and odd composition original signal
Even composition xe [n] = 2
][][ nxnx −+
Odd composition
2
][][][0
nxnxnx
−−=
1.5. Analogy between vectors and signals :
To have a better understanding of any topic we resort to an example. In this case
signals can be understood well comparing them with vectors which we are conversant with.
- A vector is specified by its magnitude and direction. Consider two vectors V1 and V2.
Let the component of vector V1 along V2 be given by C12V2.
- Geometrically the component of V1 along V2 is
obtained by drawing a vertical from the end of V1
on to the vector V2.
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The vector V1 = C12V2 + Ve
- Actually there are many ways of expressing V1 in terms of V2, tending to .
- Thus among the three representations of V1 along V2 Ve,
1eV & 2eV are the error
vectors. From the geometry Ve is the smallest error in representing V1 along V2.
- If the component of the vector V1 along V2 is C12V2 then the magnitude C12 is an
indication of similarity between V1 and V2.
i) If C12 is zero then V1 is ⊥r to V2 and hence V1 has no component along V2 and such
vectors are called orthogonal vectors.
ii) Thus orthogonal vectors are independent vectors and C12 = 0,
dot product of two vectors in A.B = AB Cos.
A . B = B . A
The component A along B is ACos = B
BA..
The component B along A is BCos = A
BA.
Similarly the component of V1 along V2 is 2
21.
V
VV = C12 V2.
Therefore C12 = 22
21
.
.
VV
VV.
Signals :
The concept of vector comparison and orthogonality can be extended to signals also.
Let us consider two signals f1(t) and f2(t). Suppose we want to approximate f1(t) in
terms of f2(t) over a certain interval t1<t<t2.
Then f1 C12 f2(t) over (t1 < t < t2)
How do we choose C12 to achieve the best approximation.
We should choose C12 such that the error (difference) between the actual fn. and the
approximated fn is minimum (ideally zero) over the interval t1<t<t2.
Thus the error fn : fe(t) = f1(t) – C12 f2(t) should be minimum over the interval
t1<t<t2.
One possible criterion to make the error minimum is to minimize the average value of the
error fe(t) over the interval.
That is minimize −−
2
1
.)]()([)(
12121
12
t
t
dttfCtftt
But this does not give the correct results.
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For example : We approximate sint with a fn f(t) = 0. Over the interval “0 to 2”. The
average error is zero indicating that sint can be approximated to zero order the interval 0 to
2 without any error. This is an absurd result.
This situation can be corrected if we choose to minimize the average of the mean
square error instead of the average of the error.
Let us denote =− )(
)(
12
2
tt
tfe. and fe(t) = f1(t) – C12 f2(t) over the interval t1<t<t2.
i.e. −−
=2
1
.)]()([)(
1 2
2121
12
t
t
dttfCtftt
To minimize the error we carry out 12dC
d and equate it to zero.
i.e.
=−−
2
1
0)]()([)(
1 2
2121
1212
t
t
dttfCtfttdC
d
changing the order of integration and differentiation.
We have ( ) ( ) .0)()()(2)(1 2
1
2
1
2
1
2
2
2
12
12
2112
12
2
1
1212
=
+−
− t
t
t
t
t
t
dttfCdC
ddttftfC
dC
ddttf
dC
d
tt
= 0)(2)()(202
1
2
1
2
21221 =
+−
t
t
t
t
dttfCdttftf .
Or =2
1
2
1
)()()(2
21221
t
t
t
t
tfCtftf or
=
2
1
2
1
)(
)().(
2
2
21
12 t
t
t
t
dttf
dttftf
C .
We can see a similarity between the vectors and the signals. By analogy with vectors
f1(t) has a component of wave form f2(t) and the component has a magnitude C12. If C12
vanishes then the signal f1(t) contains no component of signal f2(t) and f1(t) and f2(t) are
orthogonal. Over the interval (t1, t2). It implies that f1(t) and f2(t) are orthogonal if 2
1
t
t
f1(t)
f2(t) dt = 0 in the interval
(t1 < t < t2).
Thus sin not and sin mot are orthogonal over any interval (t0, 0
0
2
+t ) for integral values
of n and m.
Consider the integral I =
+
00
0
2
sinsin
t
t
oo dttmtn .
I = dttmntmn
t
t
oo
+
+−−0
0
0
2
)cos()cos(2
1
.
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I =
0
0
0
2
)(sin)(
1)(sin
)(
1
2
1
t
ttmn
mntmn
mnoo
o
+
+
+−−
−.
Since n and m are integers (n-m) and (n+m) are also integers. In such a case the
integral I is zero. Hence the two fns. are orthogonal over the interval t1<t<t2. Similarly sin n
ot and
cos m ot ; cos n ot and cos m ot are all orthogonal.
Ex : f(t) =
−
ntA
t
21
01
Approximate this fn. by a waveform sint over (0, 2) such that the mean square error is
minimum.
Solution : The fn. will be approximated over the interval (0, 2) as f(t) C12 sint.
We shall find approximate value of C12 which will minimize the mean square error in the
approximation.
Thus C12 =
2
0
2
2
0
sin2
1
sin)(2
1
dtt
dtttf
=
2
0
2sin dttDr
Nr =
−+0
2
sinsin dttdtt = 4.
−=
2
0
)2cos1(2
1dtt
−=
2
02
2sin
2
1 tt
== 22
1Dr
=
+
−
2cos
0cos tt = - [-1-1] + [1-(-1)] = 4
Hence C12 =
4
So f(t)
4 sin t.
1.6. Orthogonal Signal Space :
A fn can be expressed as a sum of its components along a set of mutually orthogonal
fns. which form a complete set.
Let us consider a set of fns. g1(t), g2(t)……..gn(t) which are orthogonal to each other
over the interval t1 to t2. i.e.
2
1
)()(
t
t
kj dttgtg = 0 for J K.
And let =2
1
)(2
t
t
jj Kdttg
We know that
cos 2t = 1 – 2 sin2t
2 sin2t = 1 – cos 2t
sin2 t = 2
2cos1 t−
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Let an arbitrary fn ; f(t) be approximate over an interval (t1, t2) by a linear combination of
these n mutually orthogonal fns.
i.e. f(t) c1g1(t) + c2g2(t) + ……… ckgk(t) + …….. cngn(t) = =
n
r
rr tgC1
)( .
For the best approximation in the interval t1 < t < t2 we must find proper values for the
constants C1, C2, …….. Cn such that , the mean square of fe (t) is minimized.
fe (t) = f(t) - =
n
r
rr tgC1
)(
dttgCtftt
t
t
n
r
rr
2
112
2
1
)()()(
1
−
−=
=
It is evident that “” is a fn. of C1, C2…….. Cn and to minimize “” we must have
0...............21
=
=
=
=
nr CCCC
.
(t2 – t1) is a constant So
dtdttgCtfC
t
t
n
r
rr
j
2
1
2
1
)()(
−
=
= 0 --------- (1)
When we expand the integrand we note that all the terms arising due to dot product of
the orthogonal fns. are zero (1) i.e. all the terms of the form gj(t) . gx(t) dt = 0.
(ii) |||ly all terms not containing Cj are zero because jdC
dfor terms not containing Cj is zero.
i.e.
=
=
=
2
1
2
1
2
1
0)()()()(222
t
t
t
t
t
t
rr
j
rr
jj
dttgtfCC
dttgCC
dttfC
This leaves only two non zero terms. From eqn. (1)
2
1
t
tjC [-2 Cj f(t) gj (t) + Cj
2 gj2 (t)] dt = 0
Changing the order of differentiation and integration.
dttgCtgtfCdC
djjjj
t
t j
)()()(222
2
1
+− .
= =+−
2
1
2
1
0)(2)()(22
t
t
t
t
jjj dttgCtgtf
i.e. Cj =
=
2
1
2
1
2
1 )()(1
)(
)()(
2
t
t
j
j
t
t
j
t
t
j
dttgtfK
dttg
dttgtf
Summary :
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Given a set of n fns. g1(t), g2(t), ………… gn(t) mutually orthogonal over the interval t1<t<t2
it is possible to approximate an arbitrary fn, f(t) over the interval by a linear combination of
these fns.
f(t) C1 g1(t) + C2 g2(t) + ………….. Cn gn(t) = =
n
r
rr tgC1
)(
and for best approximation we should choose C1, C2…….Cn as per
Cj =
=
2
1
2
1
2
1 )()(1
)(
)()(
2
t
t
j
j
t
t
j
t
t
j
dttgtfK
dttg
dttgtf
where =2
1
)(2
t
t
jj dttgK
The set of signals which forms a complete set of orthogonal signals is called orthogonal
signal space.
1.7. Evaluation of Mean Square Error :
Let us find the optimum value of ‘’ when optimum values of coefficients C1,
C2…….Cn are to be evaluated, as explained earlier
dttgCtftt
t
t
n
r
rr
2
112
2
1
)()()(
1
−
−=
=
=
−+
− = =
2
1
2
1
2
11 1
222
12
)()(2)()()(
1t
t
n
r
t
t
t
t
r
n
r
rrr dttgtfCdttgCdttftt
------ (1)
In the above we know,
==
2
1
2
1
2
1 )()(1
)(
)()(
2
t
t
j
j
t
t
j
t
t
j
j dttgtfK
dttg
dttgtf
C
i.e. Cj Kj = =2
1
2
1
)(;)()(2
t
t
t
t
jjj dttgKdttgtf ------------ (2)
Putting (2) in (1) we have
−+
−=
= =
2
11 1
222
12
2)()(
1t
t
n
r
n
r
rrrr KCKCdttftt
=
−
− =
2
11
22
12
)()(
1t
t
n
r
rr KCdttftt
=
++−
− 2
1
2
2
2
21
2
1
2
12
......()()(
1t
t
nn KCKCKCdttftt
------ (4)
The equation helps us in evaluating the mean square error.
1.8. Representation of a signal by complete set of orthogonal functions :
From the equation (4) as the no. of orthogonal fns. by which f(t) is approximated is
made larger and larger the error will become smaller and smaller.
But “” is a +ve quantity by definition.
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Hence in the limit as the no. of terms is made infinity the sum =
n
r
rr KC1
2may
converge to the integral 2
1
)(2
t
t
dttf then “” vanishes.
Thus =
=
1
222
1
)(r
rr
t
t
KCdttf ------- (5)
Under these conditions f(t) is represented by the infinite series.
f(t) = C1g1(t) + C2 g2(t) + …….. Crgr(t) + ………… --------- (6)
The infinite series of (5) will converge to f(t) such that the mean square error is “0”.
This series is said to “Converge in the mean” in such case f(t) is exact.
A set of fns. g1(t), g2(t) ……….. gr(t) mutually orthogonal over the interval “t1 to t2”
is said to be complete or closed set if there exists no fn. x(t) for which =2
1
0)()(
t
t
k dttgtx for
K = 1, 2, 3 ………
If a fn. x(t) can be found such that the above integral is zero then obviously x(t) is
orthogonal to each member of the set “gr(t)” and consequently x(t) is itself a member of the
set. So the set can not be complete without x(t) being its member.
Summary :
For a set gr(t) ; r = 1, 2, 3…….. mutually orthogonal over the interval “t1<t<t2”
=
=
2
11
0)()(
t
t
nmnmif
nmifdttgtg
If the set is complete then any f(n), f(t) can be expressed as
f(t) = C1g1(t) + C2g2(t) + …….. Crgr(t) + …….
Where Cr = r
t
t
r
K
dttgtf2
1
)()(
when Kr = 2
1
)(2
t
t
r dttg
Let us see as to how the approximation improves when large no. of mutually orthogonal fns.
are used.
Ex : Consider the fn. f(t) =
−
21
01
tA
t
We know that sin not and sin mot are mutually orthogonal. Over the interval
(to, to +o
2 ) for all integral values of n&m.
Hence it follows that sin t, sin 2t, sin 3t etc. are mutually orthogonal over the interval
(0, 2).
Hence the given rectangular pulse can be approximated by finite series of sinusoidal
fns.
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i.e. f(t) C1 sin t + C2 sin 2t + C3 sin 3t + ……… Cn sin nt.
The constants Cr can be evaluated by Cr =
2
0
2
2
0
sin
sin)(
dtrt
dtrttf
Cr = r
4 when r is odd and = 0 when r is even.
For a set of complex fns. [gr(t)], r = 1, 2, ……. which are mutually orthogonal over
the interval (t1, t2).
=
=
2
1
0)()(
*
t
t m
nmnmforK
nmfordttgtg
If the set of fns. is complete then any fn. f(t) can be expressed as
f(t) = c1g1 (t) + c2g2 (t) + ……… cr gr(t) + ………..
Where Cr = 2
1
)()(1 *
t
t
r
r
attgtfK
.
If the set of fns. is real then gr*(t) = gr(t) and the results are as discussed earlier.
Integration by parts :
xn cos m x dx = 22
1
2
)1(cos
sin−
− −−+ n
nn
Im
nnmxx
m
n
m
mxx
i.e. vdu = uv - udv.
‘’ in these approximations :
We know that =
−−
− 2
1
.....)()(
12
2
21
2
1
2
12
t
t
KCKCdttftt
=
2
0
4sin)(
rdtrttf
=
−+0
2
sinsin dtrtdtrt
In the present case t2 – t1 = 2 and f(t) =
−
21
01
t
t =
r
4
So 2
0
2 )( dttf = 2 → 1)1()( 22
0
2
=−=
+
tfdtdt = + [2 - ] = 2.
Also Cr =
evenisrif
oddisrifr
0
4
=
−=
2
0
2
0
2
2
2cos1sin dt
rtdtrt
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Kr = =
dtrtSin
2
0
2 Hence Cr =
evenrfor
oddrforr
dtrt
dtrttf
0
4
sin
sin)(
2
0
2
2
0
=
=
So for one term approximation 1 = 19.04
22
12
=
−
For two terms approximations 2 = 1.03
442
2
122
=
−
−
For three terms it is 0.0675 & for four terms it is 0.051 & so on.
1.9. Orthogonality in complex fns :
So far we have considered only the real fns. of the real variables. If f1(t) and f2(t) are complex
fns. of real variable then it can be shown that f1(t) can be approximated by C12 f2(t) over the
interval (t1, t2) i.e. f1(t) = C12 f2(t).
The optimum value of C12 to minimize the mean-square error is given by C12 =
2
1
2
1
)()(
)()(
*
22
*
21
t
t
t
t
dttftf
dttftf
Where f2*(t) is the complex conjugate of f2 (t)
It is evident that the two complex fns. above are orthogonal to each other if over the
interval (t1, t2).
==2
1
2
1
0)(.)()(.)( 2
*
1
*
21
t
t
t
t
dttftfdttftf
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UNIT – II
FOURIER SERIES REPRESENTATION OF
PERIODIC SIGNALS
CONTENTS:
2.1. Trigonometric Fourier series.
2.2. Symmetry conditions in evaluation of Fourier Coefficients.
2.3. Cosine form representation of Trigonometric Fourier series.
2.4. Exponential Fourier series.
2.5. Dirichlet’s conditions.
2.6. Properties of Fourier series.
2.7. Fourier Spectrum.
2.8. Solved Numerical Problems.
2.9. Objective questions.
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INTRODUCTION:
We have studied earlier that any given fn, f(t) can be expressed as a sum of linear
combination of orthogonal fn. and also we have derived the mean square error and proved
that it approaches zero as the series becomes larger and larger.
The same principle can be applied to a series called Fourier series.
2.1. TRIGNOMETRIC FOURIER SERIES:
(i) A linear combination of “harmonically related” sine and cosine terms can express a given
fn. which is periodic.
i.e. a given periodic fn. can be expressed as a linear combination of harmonically
related sine and cosine fns. such a series is called trigonometric Fourier series.
i.e. f(t) = a0 + a1 cos ot + a2 cos 2ot + ------------ + an cos not.
+ b1 sin ot + b2 sin 2ot + -------- + bn sin not
Where ao = T
dttfT
0
)(1
; an = T
o dttntfT
0
cos)(2
bn = T
o dttntfT
0
sin)(2
A signal is periodic when x(t) = x(t + T) for all “t”.
Where T = o
2 ; for f(t) = A sin ot.
f(t) = a0 + =
+k
n
onon tnbtna1
)sincos( ----------- (1)
Where a0, a1, a2 …. an and b1, b2 ….. bn are Fourier constant and o is the fundamental
frequency.
If the signal has to be periodic.
x(t + T) = a0 + =
+++k
n
onon TtnbTtna1
)(sin)(cos
= a0 + =
++
+
k
n
onon TT
ntwnbTT
ntwna1
.2
sin).2
(cos
= a0 + =
=+k
n
onon tftnbtna1
)()(sin)(cos
Hence a summation of sine and cosine fns. of frequencies o, o, 2o ……. Ko is a periodic
signal of period T.
Eq. (1) implies that by changing the values of an’s and bn’s we can construct any periodic
signal with period T.
If k → the eq. (1) becomes Fourier series representation of a periodic signal f(t).
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If Any periodic signal can be represented as a sum of infinite no. of sine and cosine fns.
which are harmonically related and are themselves periodic with angular frequencies of o, o,
….ko. This series of sine and cosine terms is known as trigonometric Fourier series and can
be written as f(t) = ao +
=
+1
)sincos(n
onon tnbtna --------- (2)
Where “ao, a1 ……… an & b1 ………. bn are called Fourier coefficients”.
ao is called the d.c. component or the average value of the periodic signal f(t).
a1 cos 1t + b1 sin 1t - 1st harmonic
a2 cos 2t + b2 sin 2t - 2nd harmonic
2.1.1 Evaluation of Fourier constants :
To evaluate a0 integrate both sides of eq. (2) starting at any arbitrary time to and over one
period to + T.
+ +
=
+
=
+
++=
Tt
t
Tt
t n
Tt
t n
onono
Tt
t
o
o
o
o
o
o
o
o
dttnbdttnadtadttf1 1
sincos)(
+
++=
Tt
t
o
o
o
ooTadttf )(
Hence ao = +
=
Tt
t
T
O
o
o
dttfT
dttfT
)(1
)(1
To evaluate an and bn :
We make use of the result
==
=+
02/
0coscos
andnmforT
nmfordttmtn
Tt
t
oo
o
o
------- (3)
And +
=
Tt
t
oo
o
o
nmallfordttmtn &0cossin -------
(4)
==
=+
0
2/
0sinsin
nmforT
nmfordttmtn
Tt
t
oo
o
o
------- (5)
To find an :
Multiply eq. 2 by cos mot and integrate over one period.
(a) (b) (c)
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+
=
++
=
+
++=
Tt
t n
Tt
t
oono
Tt
t n
Tt
t
onooo
o
o
o
o
o
o
o
o
dttntnbdttmtnadttmadttmtf
11
cos.sincoscoscoscos)( ----
(6)
(a) is zero because a sinusoid integrated over a period.
(2) using identities of (3), (4) & (6) as per (4) – (c) of (6) is zero and as per (3) – (b) = an .
T/2.
So +Tt
t
o
o
tf )( cos m ot dt = an T/2 Or
an = + +
=
Tt
t
Tt
t
oo
o
o
o
o
dttntfT
dttmtfT
cos)(2
cos)(2
To find bn :
Multiply (2) by sin mot
i.e. (a) (b) (c)
+
=
+
=
++
++=
Tt
t n
Tt
t n
Tt
t
oonoon
Tt
t
ooo
o
o
o
o
o
o
o
o
dttmtmbdttmtnadttmadttmtf1 1
sinsinsincossinsin)(
Here (a) → 0 & (b) → 0 & (c) = bn . T/2
Hence bn = +Tt
t
o
o
o
dttntfT
sin)(2
Find the trigonometric F.S. of the periodic fns. x(t)
Ans. ao = 0 ; an = 02
sin4
=
nb
n
n ao = 0 ; an = 0 ; bn =
−n
ncos
2
Here T = 4 Here T = 2
o = 2/4
22=
=
T o = =
=
2
22
T
Hence x(t) = =
+
+
1 2sin
2cos
n
nno tn
btn
aa x(t) = =
++
1
sincosn
nno tnbtnaa
f(t) = 1 -1 < t < 1
= -1 1 < t < 3
ao = − −
+
=
−==
3
1
1
1
3
1
04
1)(
4
1)(
1dtdtdttfdttf
T
Tt
t
o
o
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an =
−
−
=
−
+ 1
1
3
12
cos2
cos4
2cos)(
2dtt
ndtt
ndttntf
T
Tt
t
o
o
o
=
+
=
−
−
2/
2sin2
2sin2
2
1
2/
2sin
2
1
2/
2sin
2
1
3
1
1
1
n
nn
n
tn
n
tn
= 2
sin4
n
n
bn =
−
=
−
+ 1
1
3
12
sin2
sin2
1sin)(
2dtt
ndtt
ndttntf
T
Tt
t
o
o
o
= 02/
2cos
2
1
2/
2cos
2
1
3
1
1
1
=
−
−
−
−
n
tn
n
tn
2.2. Symmetry conditions in Fourier coefficients :
We know that a given signal (wave form) can be resolved into its odd and even parts.
i.e. f(t) = fe(t) + fo(t)
fe(t) = 1/2 [f(t) + f(-t)]
fo(t) = 1/2 [f(t) – f(-t)
These relations can be used to find the Fourier coefficients. To have a convenient
form, we choose –T/2 to T/2 as the interval of integration.
an = −
2/
2/
cos)(2
T
T
o dttntfT
and bn = −
2/
2/
sin)(2
T
T
o dttntfT
Substituting f(t) = fe(t) + fo(t) in the above eq.
an =
+
− −
2/
2/
2/
2/
cos)(cos)(2
T
T
T
T
oooe dttntfdttntfT
------ (1)
|||ly bn =
+
− −
2/
2/
2/
2/
sin)(sin)(2
T
T
T
T
oooe dttntftntfT
------ (2)
We know that odd fn. odd fn. = even fn.
even fn. even fn. = even fn.
even fn. odd fn. = odd fn.
and for even fn. −
=o
o
ot
t
t
o
ee dttfdttf )(2)(
and for odd fn. −
=o
o
t
t
o dttf 0)(
If f(t) is an even fn. then fo(t) = 0 substituting this.
In eq. (2) we have
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bn = −
=
2/
2/
0sin)(2
T
T
oe dttntfT
even odd = odd fn. odd fn. = 0
over a period.
Hence bn = 0
And an = 2/
cos)(4
T
o
o dttntfT
And ao = 2/
)(2
T
o
dttfT
i.e. the Fourier series expansion of an even periodic fn. contains only the d.c. terms and
the cosine terms.
If f(t) is an odd fn.
f(t) = fe(t) + fo(t) = fo(t) because fe(t) = 0.
So an = −−
+
2/
2/
2/
2/
cos)(2
cos)(2
T
T
oe
T
T
oo dttntfT
dttntfT
odd even fn. Hence = 0 0 because fe(t) = 0
Hence for an odd fn. an = 0
ao = 0)(1
2/
2/
=−
dttfT
T
T
o because an odd fn.
and bn = dttntfT
o
T
T
sin)(2
2/
2/
−
odd odd = even
Hence bn = dttntfT
o
T
o
sin)(4
2/
Half wave symmetry :
A periodic signal which satisfies the condition
f(t) = - f (t T/2) is said to have half wave symmetry.
Fourier series expansion of such a signal contains odd harmonics only.
2.3. Cosine representation of Trigonometric Fourier series :
an cos not + bn sin not can be written as
An cos (n ot + n)
Thus f(t) = Ao +
=
+1
)(cosn
non tnA
Ao = ao
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An = 22
nn ba +
n = - tan-1
n
n
a
b
The number of An is called the amplitude coefficients of cosine trigonometric F.S. and n’s
are called phase coefficients.
Conditions for periodic signals and their symmetry :
S.No. Type of Condition Example
Symmetry
=
2/
)(2
T
o
o dttfT
a
1 Even f(t) = f(-t) =
2/
cos)(4
T
o
on dttntfT
a
bn = 0
ao an bn
2 Odd f(t) = f(-t) 0 0
2/
sin)(4
T
o
o dttntfT
bn
3 Half wave f(t) = -f (t T/2) 0 2/
cos)(4
T
o
o dttntfT
2/
sin)(4
T
o
o dttntfT
for odd n - only
2.4. Exponential F.S. :
Cosine form of F.S.
f(t) = Ao
=
+1
)(cosn
non tnA Where Ao = ao
An = n
nnnn
a
baba 122
tan; −−=+
By using Euler’s identity :
An cos (not + n) = An
++−+
2
(( nono tnJtnJee
Using this
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f(t) = tnJJ
n
ntnJJ
n
no
onon eeA
eeA
A )(
11 22
−−
=
=
++
Put n = - k in the second summation and the limits become -1 to -.
Thus f(t) = tkJ
k
JktnJ
n
Jno
okon eeA
eeA
A
−
−=
=
+
+
11 22
Here An = Ak & -n = k for n>0 and k<0.
Define Co = Ao ; Cn = nJn eA
2 for n > 0.
So f(t) = tnJ
n n
JntnJJno
onon eeA
eeA
A
=
−
−=
+
+
1 1 22
f(t) =
−=
n
tnJ
noeC
This equation is known as exponential F.S.
We observe that the expression is for +ve and –ve frequencies. Mathematically –ve
frequency has meaning, but in practical term it has no meaning.
-ve frequency helps in arriving at the real sinusoid for the signal.
Now f(t) =
−=n
tnJ
noeC
Where o =
T
2
Multiply both sides by tkJ oe
− and integrate over a period.
dteCdteeCdtetf
Tt
t n
tknJ
n
tkJ
Tt
t
Tt
t n
tnJ
n
tkJo
o
oo
o
o
o
o
oo +
−=
−−
+ +
−=
−==
][.)(
We know that +
−
Tt
t
tknJo
o
o dte][
= O for k n
= T for k = n
Thus +
−=
Tt
t
k
tkJo
o
o TCdtetf
)( Or
Ck = +
−
Tt
t
tkJo
o
o dtetfT
)(
1
Cn = +
−
Tt
t
tnJo
o
o dtetfT
)(
1
Cn = −
T
o
tnJdtetf
To)(
1
The F.S. pair of exponential F.S. are f(t) =
−=
n
tJn
noeC
Cn = −
T
o
tnJdtetf
To)(
1
Co = ao
Cn = nn Jba −2
1
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an = [Cn + C-n]
bn = J[Cn – C-n]
C-n = nn Jba +2
1
2.5. DIRICHLET’S CONDITIONS:
The F.S. can exist for a fn. provided it satisfies the Dirichlet’s conditions.
A periodic signal can be represented by F.S. if the fn. satisfies Dirichlet’s conditions. They
are
1. Over any period f(t) must be absolutely integrable i.e.
T
o
dttf )( . If this condition is satisfied.
Then ak’s of F.S. coefficients take finite values i.e.
|ak| =−
T
o
T
o
tkJdttf
Tdtetf
To |)(|
1|)(|
1
So if
T
o
dttf |)(| then |ak| <
Ex : A periodic signal that violates this condition is f(t) = t
1 ; 0 < t 1
2. In any finite interval of time f(t) has finite no. of maxima and minima. That is in any
single period there are no more than a finite no. of maxima & minima.
A signal which violates the condition is
f(t) = Sin t
2 0 < t 1
3. In any finite interval of time there are only finite no. of discontinuities.
The F.S. converges to the average of the RHS & LHS at the Pt of discontinuity.
Prob : Compute the exponential F.S. of the given periodic signal.
Period is T = 4
Ans :
−−−
−
22
1)1(
2
11
1 ne
Jn
Jn
2.6. Properties of F.S.
2.6.1. Linearity property :
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Let us represent F.S. coefficients by ak.
f(t) is a periodic signal and ak are the F.S. coefficients.
Let f1 (t) and f2 (t) denote two periodic signals, with period T and which have F.S.
coefficients denoted by ak and bk respectively.
Then f1(t) ⎯⎯→ ..SF ak & f2(t) ⎯⎯→ ..SF bn.
Then as f1(t) and f2(t) have the same period T ; then any linear combination of the
signals will also have the same period.
Also “F.S. coefficients Ck” of the linear combination of f1(t) and f2(t)
f3 (t) = Af1 (t) + B f2(t) are given by the same linear combination of the F.S.
coefficients ak & bk of f1(t) & f2(t).
i.e. f3(t) = A f1(t) + B f2 (t) ⎯⎯→ ..SF Aak + Bbk.
2.6.2. Time shifting property :
When a time shifting is applied to a periodic signal f(t) the period T of the signal is
preserved.
i.e. f(t – to) ⎯⎯→ ..SF ootkJe
− ak
If we denote F.S. coefficients of f(t – to) by bk.
Then bk = ootkJe
− ak.
Proof :
bk = −
−
T
o
tJk
o dtettfT
o)(1
Put (t – to) = then dt = d
bk = +−
T
o
tJkdtef
Too )(
)(1
bk = −−
T
o
JktJkdeTf
Te ooo
)(1
bk = k
tJ
T
o
JktJaedtef
Te ooooo −−−
=)(1
2.6.3. Time Reversal :
The period T of a periodic signal f(t) remains unchanged when the signal under goes
time reversal.
i.e. when f(t) is time reversed, it becomes f(-t)
Then the F.S. eq. becomes f(-t) =
−=
−
k
tT
Jk
k ea
2
Substitute k = -m
f(-t) =
−=
−−
−
m
tT
mJ
m ea
2)(
it is the F.S. expansion of f(-t)
Hence bk = a-k where bk is F.S. coefficients of f(-t)
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That is if f(t) ⎯⎯→ ..SF ak
then f(-t) a-k
i.e. Time reversal applied to a continuous time periodic signal results in a time
reversal of the corresponding sequence of the F.S. coefficients. This shows that if f(+t) is
even its F.S. coefficients are even and if f(t) is odd its F.S. coefficients are also odd.
2.6.4. Time Scaling :
It is an operation which changes the period of the signal i.e. If f(t) is periodic with
period T and fundamental frequency o = T
2 then f( t) where is a +ve real number is
periodic with period T/ and frequency wo.
i.e. f( t) =
−=k
tJk
koea
)(
2.6.5. Multiplication :
If f1(t) ⎯⎯→ ..SF ak f(t) =
−=
+
k
tJk
koea
f2 (t) ⎯⎯→ ..SF bk ak = −
T
o
tJkdtetf
To
)(1
Then f1 (t) f2 (t) ⎯⎯→ ..SF hk =
−=
−
l
lkl ba
2.6.6. Conjugation and conjugate symmetry :
Taking the complex conjugate of the periodic signal f(t) has the effect of “Complex
Conjugation and time reversal” on the corresponding F.S. coefficients i.e. if
f(t) = ⎯⎯→ ..SF ak
Then f*(t) ⎯⎯→ ..SF *
ka−
Proof :
f(t) = tJk
k
koea
−=
------ (1)
Taking complex conjugation of both sides
f*(t) = tJk
k
koea
−
−=
*
Put K = - k
f*(t) = tJk
k
koea
−=
−
* ------- (2)
Comparing (1) & (2)
f(t) ⎯⎯→ ..SF ak
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f*(t) ⎯⎯→ ..SF *
ka−
2.6.7. Parseval’s relation for continuous time periodic signals :
−=
=
T
o k
kadttfT
22 |||)(|1
is Parseval’s relation.
LHS is average power (energy / unit time) in one period of the signal f(t).
Also
−=
−=
=
T
o k k
T
o
k
tJk
k dtaT
dteaT
o 22 ||1
||1
=
−=k
ka 2||
|ak|2 is the average power in the kth harmonic component of f(t). Thus Parseval’s
relation states that the total average power in a periodic signal equals the sum of the average
powers contained in all the harmonics.
Conjugate property (symmetry for real signals)
Case 1 :
f(t) real → ak = *
ka− → a-k = *
ka
Re ak = Re (a-k)
Im (ak) = -Im (a-k)
|ak| = |a-k|
Case 2 :
Where f(t) is real and even
then ak is real and even. ak = a-k
Case 3 :
Where f(t) is real & odd a-k = - ak
Power spectrum of a period fn. :
Average power = −
2/
2/
2 )(1
T
T
dttfT
But f(t) =
−=n
tJn
noeF
= dteFtfT
T
T n
tJn
no
−
−=
2/
2/
.)(1
We know that F-n = −
2/
2/
)(1
T
T
tJndtetf
To
Interchange the operations on the RHS.
i.e. power =
−=
−
−= −
=n
nn
n
T
T
tJn
n TFFT
dtetfFT
o )(1
)(1
2/
2/
TF-n
=
−=
−=
− =n
n
n
nn FFF 2||
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i.e. Power = F02 + F1
2 + F22 + …… Fn
2 + F-12 + F-2
2 + -….. F-n2
which is Parseval’s relation.
2.7 Fourier Spectrum :
A Fourier series expansion of a periodic fn. is really equivalent to resolving the fn. in term of
its components of various frequencies.
- A periodic fn. with period T has frequency components of angular frequencies o, 0,
2o……3o, now where o = T
2.
- i.e. the periodic fn. f(t) possesses a spectrum of frequencies.
- If we specify a fn. we can find its spectrum and conversely if we know the spectrum
we can find the fn.
- Hence we have two ways of specifying a periodic fn.
(i) The time domain representation of the fn. where f(t) is expressed as a fn. of
time.
(ii) The frequency domain representation of the fn. where the spectrum (i.e. the
amplitude of various frequencies and their phases) is specified.
(iii) The spectrum exists only at = o, = 2o, 3o…. Thus the spectrum is not a
continuous curve. It exists only at some discrete values of .
(iv) It is a discrete spectrum and is known as line spectrum.
(v) The discrete frequency spectrum thus appears as a series of equally spaced
vertical lines with heights proportional to the amplitude of the corresponding
frequency components.
2.8. NUMERICAL PROBLEMS:
2.8.1 : Full wave rectified voltage.
Sol : f(t) = tJtJtJtJtJtJ eA
eA
eA
eA
eA
eAA −−−
−
+−
−
−
−
−
642642
35
2
15
2
3
2
35
2
15
2
3
22
Here the spectrum exists at = 0, = 2, = 4……. with magnitudes
−
−
−
35
2,
15
2,
3
2,
2 AAAA.
Thus when shown as a spectrum on the graph.
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2.8.2 : Convert the given series into exponential form.
f(t) = 1+sin ot+2cos ot+cos (2ot+/4) =
++++
−+
−
−−
221
22
4
tJtJJ
tJtJtJtJ oo
oo
oo eeeee
j
ee
=
tJ
JJ
tJtJtJ oooo eee
eJ
eJ
e 2
442
222
111
2
11
−
−+
+
−+
++
Sol : F0 = 1 ; F1 = 1 + J2
1 ; F2 = 4
2
1J
e ;
F-1 = J2
11− ; F-2 = 4
2
1−J
e and
Fn = 0 for |n| > 2.
Phase spectrum Magnitude Spectrum
Real Imaginary
2.8.3. Suppose we are given the following information about a signal x(t) :
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1. is a real signal .
2. is periodic with period T = 6 and has Fourier coefficients ak .
3. = 0 for k = 0 and k > 2 .
4. = .
5. .
6. is a positive real number .
Show that , and determine the values of the constants A , B , and
C .
Solution
Since is a real signal, . But from the given hypothesis, for k > 2.
This implies that
for k > 2.
Also, it is given that . Therefore the only non-zero Fourier coefficients are ,
, and .
It is also given that is a positive real number. Therefore . Thus we have,
=
=
=
Since and are both periodic with period 3, we have
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2.8.4. Determine the Fourier series representations for the following signals:
(1) Each illustrated in the figures (a) - (f)
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(2) periodic with period 2 and for .
(3) periodic with period 4 and
(1)
(a)
It is periodic with period 2. So, consider segment between .
Let
be the Fourier expansion, where:
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For ,
But, in
Substituting in x(t)
(b)
It is periodic with period 6. So, consider segment between .
The function in this interval is:
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Let
be the Fourier expansion, where:
For ,
Substituting in x(t)
(c)
It is periodic with period 3, so take the segment
The function in the interval is:
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Let
be the Fourier expansion, where:
For ,
Substituting in x(t)
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(d)
It is periodic with period 2. So take segment
Here,
Let
be the Fourier expansion, where:
For
Substituting in x(t)
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(e)
It is periodic with period 6. So, take the segment
Here will be
Let
be the Fourier expansion, where:
For
Substituting in x(t)
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(f)
It is periodic with period 3. So, take the segment
Here will be
Let
be the Fourier expansion, where:
For
Substituting in x(t)
Substituting in x(t)
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(2) Let
be the Fourier expansion, where:
Given .
For
Substituting in x(t)
(3) Let
be the Fourier expansion, where:
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Given .
For
Substituting in x(t)
2.8.5. A continuous-time periodic signal is real valued and has a fundamental period
T=8 . The NON ZERO Fourier series coefficients for are specified as
,
Express in the form
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Solution
Given that the non zero coefficients are:
; ; ;
=>
Comparing with
We get the non-zero coefficients as:
;
;
2.9. OBJECTIVE QUESTIONS:
1. Which of the following is an “even” function of t ?
(A) 2t
(B) tt 42 −
(C) tt 3)2sin( +
(D) 63 +t
2. A “periodic function” is given by a function which
(A) has a period 2=T
(B) satisfies )()( tfTtf =+
(C) satisfies )()( tfTtf −=+
(D) has a period =T
3. Given the following periodic function, )(tf .
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The coefficient 0a of the continuous Fourier series associated with the above given
function )(tf can be computed as
(A) 9
8
(B) 9
16
(C) 9
24
(D) 9
32
4. For the given periodic function
=
=
)(62for4
20for2)(
Tt
tttf . The coefficient 1b of
the continuous Fourier series associated with the given function )(tf can be
computed as
(A) 6800.75−
(B) 5680.7−
(C) 8968.6−
(D) 7468.0−
5. For the given periodic function
=
62for4
20for2)(
t
tttf with a period 6=T . The
Fourier coefficient 1a can be computed as
(A) 2642.9−
(B) 1275.8−
(C) 9119.0−
(D) 5116.0−
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6. For the given periodic function
=
62for4
20for2)(
t
tttf with a period 6=T as
shown in Problem 5. The complex form of the Fourier series can be expressed as
−=
=k
tikw
k eCtf 0~
)( . The complex coefficient 1
~C can be expressed as
(A) i3734.04560.0 +
(B) i3734.04560.0 −
(C) i3734.04560.0 +−
(D) i4560.03734.0 −
7. Which of the following is the weak Dirichlets condition
a) Absolute Inerrability of the function.
b) Periodicity of the function.
c) Finite number of maximal’s and minima’s
d) None of the above.
8. With Respect to Fourier series “Multiplication in Time domain responds to
Convolution in frequency domain” is the propery of
a) Linearity
b) Time shifting
c) Convolution
d) None of the Above.
9. What is the magnitude of the exponential Fourier series coefficient of the fundamental term
(i.e., k = 1) of the periodic signal x(t) = 3 + 2 cos(3t) + sin(3t)
(a) 3/2
(b)√5/2
(c) 1/2
(d) 1
(e) None of the above.
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UNIT – 3
FOURIER TRANSFORMS
Contents:
3.1. Derivation of Fourier Transform from the Fourier Series.
3.2. Fourier Transform of Standard Signals.
3.3. Fourier Transform of Standard Signals.
3.4. Properties of Fourier Transform.
3.5. Hilbert Transform.
3.6. Parsevals Relations.
3.7. Inverse Fourier Transform.
3.8. Objective questions.
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Fourier Transforms
3.1. Derivation of Fourier Transform From Fourier Series:
a) a-periodic signal x(t)
b) Periodic signal.
x~ (t) constructed to be equal to x(t) over
one interval ‘T’.
- Let us consider a signal x(t) that is of finite duration.
i.e. x(t) = 0 if |t| > T1.
- Construct a periodic signal x~ (t) from the aperiodic signal x(t).
- As we choose the period ‘T’ to be larger x~ (t) is identical to x(t) over a larger
interval and as T →, x~ (t) and x(t1) are equal.
F.S. of x~ (t) carried out over interval –T/2 t T/2
We have x~ (t) =
−=n
tKJ
Koea
----- (1)
ak = T
1 dtetx
T
T
tJK o−
−
2/
2/
~
)(
----- Where o = T
2
x~ (t) = x(t) for | t | < T/2 and also since x(t) = 0
outside the interval
We can rewrite the above equation as
ak = T
1 dtetx
T
T
tJK o−
−
2/
2/
~
)(
=
− tJK oetxT
)(
1
Define envelope as X (J) of “Tak” as X(J) = dtetx tJ
−
− )(
So for coefficients ak
ak = )(1
oJkXT
------------- (2)
Combining (1) & (2) we can express x~ (t) in terms of X (J)
as x~ (t) =
−=k
tJk
ooeJkX
T
)(1
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We know that since
=→=
2
12 oo
TT
x~ (t) =
−= k
o
tJk
ooeJkX
.)(2
1 ----------- (3)
As T → ; x~ (t) approaches x(t) and in the limit eq. (3) becomes x(t).
As T → o → 0 and the Rt. Hand side of eq. (3) becomes an integral
i.e. x(t) =
−
deJX tJ)(2
1 ----------- (4)
X (J) =
−
− dtetx tJ)( ----------- (5)
Graphical interpretation of eq. (3)
think that t is a
constant
equations (4) & (5) are called “Fourier Transform Pair”.
(4) is called inverse F.T. equation.
(5) is called Fourier Transform of x(t)
3.2. Fourier Transform of standard signals :
3.2.1. Unit step fn :
u(t) = 0 for t < 0
1 for t > 0
We shall find the transform of an exponential e-at which starts at t = 0 and is O for t
0. Such a fn. is conveniently written as F [e-at u(t)] =
−
−− dtetue tJat )(
Since e-at u(t) = 0 for - < t < 0
We can write the above eq. as
F [e-at u(t)] =
+−−− =0 0
)( dtedtee tJatJat
=
JaJa
e tJa
+=
+−
+− 1
)(0
)(
i.e. F [e-at u(t)] = Ja +
1 for a > 0.
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F-1
+ )(
1
Ja = e-at u(t)
i.e. [e-at u(t)] =
−+
deJa
tJ
)(
1
2
1
+ )(
1
Ja =
−
−− dtetue tJat )(
To plot Ja +
1 on the graph.
Ja +
1 = a
TanJ
ea
1
22
1 −−
+
3.2.2. FT of Complex Exponentials:
f(t) = t e-at u(t)
F() = F [f(t)] =
−
−− dtetuet tJat )(
+−
0
)(. dtet tJa udv = uv - v du
Integrating by parts : Put t = u and e-(a+J)t = dv
To plot 2)(
1
Ja + t e-(a+Jt) dt =
dtJa
e
Ja
et tJatJa
.)()(
.
0
)()(
0 +−+−
+−−
+−
|F()| = 22
1
+a = 0 +
+−+−
+−=
+0
2
)()(
0)()(
Ja
edt
Ja
e tJatJa
a
F
1tan2)( −−= = 2)(
1
Ja + for a > 0
F() = aJ
ea
12 tan
22 )(
1 −−
+
3.2.3. : Fourier Transform of a gate fn:
GT (t) = 1 for |t| T/2 or – T/2 < t < T/2
0 for |t| > T/2
F() =
2/
2/
2/
2/
−
− −
−−−
−== tJtJtJ eJ
AdteAdteA
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=
−=
−−
−−2222
JJJJ
eeJ
Aee
J
A
= 2A
−
−
J
ee
JwJw
2
22
= 2A )2/(
)2/(sin
2
2sin
2
22
22
A
A
J
ee
JJ
==
−
−
= A Sa
2
3.2.4. :Fourier Transform of Impulse function:
Impulse fn. can be derived as a limiting case of step fn.
Say v = u(t) i(t) is undefined
We see that i = c dt
dv and
dt
dv = 0 because v is a constant. [u(t)]
dt
dv is ‘0’ for all values of t except at t = 0
derivative does not exist at t = 0 because the fn. is discontinuous at t = 0.
To circumvent this problem let us consider the step fn. as a limiting case.
ua(t) in the limit as a →0 because a step fn.
The derivative of ua(t) is a rectangular pulse of ht. a
1 and width ‘a’. As ‘a’ varies the pulse
shape varies but the area of the pulse remains unity.
Thus )(0
)( tudt
d
a
Ltt a
→=
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But the fn. )(tudt
da is a rectangular pulse of ht. 1/a and width a.
This can be describes as )]()([1
0)( atutu
aa
Ltt −−
→=
In the Lt. as a→0 the fn. )(t assumes the form of a pulse of infinite ht. and zero width. The
area under the pulse however remains constant and is equal to unity.
(t) = 1 at t = 0
−
− == 1)()]([ dtettF tJ = 0 at t 0
i.e. =)]([ tF 1
3.2.5. : Fourier Transform of a constant:
f(t) = A
Start with a gate fn. :
As → Gate fn. becomes a constant A
Thus F[A] = 2
SinA
Lt
→
=
→
222
Sin
LtA ------ (3)
= 2A ()
F[1] = 2 ()
Definition of (t) :
)()( tKtSaK
K
Lt=
→ ------ (1)
Sampling fn.
−
=
1)( dtKtSaK
------ (2)
Comparing (3) with (1) & (2) K = /2 ; and t = .
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So )(22
=
→
wSin
Lt.
Another way of finding F.T. of a constant :
Inverse F.T. of () =
−
−
− =
= 1)(;)(2
1)]([1 dedeF tJtJ
Because () exists at = 0, hence eJt at = 0 has value 1. Hence the integral 0
|=
tJe
hence 1.
Then F-1 [()] = ]1[2
1
F[1] = 2 ()
Simiarly F[A] = 2 A ()
3.2.6. Fourier Transform of Signum function:
e-a|t| = eat - < t < 0
Sgn (t) = 1 for t > 0 e-at 0 < t <
= 0 for t = 0
= -1 for t < 0
The above fn. is not absolutely integrable.
So let us consider the fn. e-a|t| Sgn (t) and substitute the Lt a → 0 to obtain the fn. sgn
(t).
F[Sgn(t)] =
−
−−
→dtetSgne
a
LttJta )(
0
||
=
+−
→
−−
−
−
0
0
0dteedtee
a
LttJattJat
=
+−+
−−
→
+−
−
−
||0
)(0)(
)()(0
Ja
e
Ja
e
a
Lt tJatJa
= JJJJaJaa
Lt 211
)(
1
)(
1
0=
+=
++
−−
→
i.e. F [Sgn(t)] = J
2
3.2.7. Fourier Transform of unit step function :
Start with Sgn(t)
Sgn(t) = -1 + 2 u(t)
F[Sgn(t)] = -F [1] + 2 F[u(t)]
Jw
2 = - 2 () + 2 F[u(t)]
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F u(t) = J
1 + ()
3.2.8. Fourier Transform of cos ot :
x(t) = cos ot
= tJtJ oo ee −
+2
1
F[cos ot] = F() =
+
−)(
2
1 tJtJ oo eeF
(1) = dteeeF tJtJtJ oo −
−
=2
1
2
1
=
−
−−
−
−−= dtedte
tJtJ oo )()()1(
2
1
2
1
This is F.T. of [1] with the operator as (-o)
Hence )(2 o
tJ oeF −=
)(2
1o
tJ oeF −=
Simiarly )(2
1o
tJ oeF +=
−
Thus F [cos ot] = (+o) + (-o)
Similarly F [sin ot] = J [ (+o) – (-o)]
F.T. of cos ot u(t) :
F [cos ot u(t)] = )(2
tuee
FtJtJ oo
+−
=
−
−
−−− + dttueedttuee tJtJtJtJ oo )(2
1)(
2
1
=
−
−
+−−−+ dttuedttue
tJtJ oo )(2
1)(
2
1 )()(
We know that F[u(t)] =
J
1)( +
Hence the above eq. becomes :
)()(2)(2
1
)(2
1)(
2)(
2 2222
oo
oo
oo
ooJJJJ
−
⊥=
−
−++=
++
−++
+−
= )(
)()(2
22
o
ooJ
−+−++
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= )(
)()(2 22
−+−++
o
oo
J
Similarly F [sin ot u(t)] = 22
)()(2
−++−−
o
ooo
J
3.2.9. Fourier Transform of e-a|t| for a > 0 :
X() =
−−
−
− +0
0
dteedtee tJattJat
=
+−
−
− +0
)(
0
)( dtedte tJatJa
= ||0
)(0)(
)()(
+−
−
−
+−+
−
Ja
e
Ja
e tJatJa
= 22
2
)(
1
)(
1
+=
++
− a
a
JaJa
22
|| 2
+=−
a
aeF ta
22
|| 2
+=−
a
aeF ta = F()
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3.2.10. Fourier Transform of a periodic fn. :
x(t) =
−=k
tJk
koea
expression for a periodic fn.
F[x(t)] = −=−=
=
k
tJk
k
k
tJk
koo eaFeaF by linearity property.
=
−=k
tJk
koeFa
------ (1) by ak being independent of t.
We know that F(1) = 2 ()
)(2 o
tJ oeF −=
)(2 o
tJkkeF o
−=
So the eq. (1) becomes
F [x(t)] = X() =
−=
−k
ok ka )(2
ak of f(t) i.e. F.S. of f(t)
F.T. of f(t)
F.T. of Real and even fns. and real & odd fns. :
F.T. of a real and even fn. is real & even.
F.T. of a real and odd fn. is imaginary & odd.
Ex : cos ot sin ot
Real & even Real & odd
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Transform of eternal exponential : F [cos ot] = (-o) + (-
o)
Transform of tJte oo
tJ o sincos += = F [J sin ot)= J[J{(+o) - (-
o)}
= - [ (+o)- (-
o)]
F[cos ot+J sin ot] = 2 (-
o)
−
−
−−−−−
−
=== dtedtedteeeFtJtJtJtJtJ oooo )()(
]1[.
F.T. [1] = 2 () hence F.T. )(2 o
tJ oe −=
3.3. Fourier Transform of periodic signals :
- We developed F.T. as a limiting case of F.S. by letting the period of the aperiodic
signal become “”.
- We show that F.S. is a limiting case of F.T. by proceeding in the opposite direction.
For a periodic fn. f(t)
−
dttf |)(|
- We found F.T. of cos ot and sin ot in the limit.
- We use here the same procedure assuming that the periodic signal exists only in a
finite interval (-T/2, T/2) and in the limit let → .
The F.T. of a periodic fn. is then the sum of the F.T’s of its individual components.
Periodic fn. f(t) with period can be expressed as
f(t) =
−=n
tJn
noeF
where o =
T
2
Taking F.T. of both sides.
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F[f(t)] =
−=
−=
=
n
tJn
n
n
tJn
noo eFFeFF
------ (1)
Substituting the transforms operation of tJn oe
.
)(2 o
tJ oeF −=
Equation (1) becomes
−=
−n
on nF )(2
=
−=
−n
on nF )(2
This means that the spectral density of a fn. or the F.T. of a periodic fn. consists of
impulses located at the harmonic frequencies of the signal and that the strength of each
impulse is the same as 2 times the value of the corresponding coefficients in exponential
F.S’s.
That means it is a sequence of equidistant impulses which is the same as the “periodic
fn. containing components only at discrete harmonic frequencies.
Find the F.T. of a periodic gate fn. :
f(t) =
−=n
tJn
koea
Where ak =
T
KSa
T
A
T = 4 /2
= T/2
Here ao = 2
2 A
T
TA=
For T = 4 /2 ; ak =
T
KSa
A
2
ak =
22
KSa
A
ak for T = 2
4
x(t) = 1 for |t| < /2
0 for /2 < |t| T/2
Fundamental frq. = T
2
x(t) is symmetric about Y-axis.
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Hence ak = −
−
2/
2/
dte
T
A tJk o
For k = 0 ; ao = T
A is the average value.
For k 0 ; ak = A −
−
2/
2/
1
dte
T
tJk o
= −
−
−
2/
2/
tJk
o
oeTkJ
A = ( ) 2/2/
oo kJkJ
o
eeTkJ
A−−
−
ak = A2 TK
kA
TkJ
ee
o
o
o
JkJk oo
2/sin2
2
2/2/
=
−−
-------- (1)
= k
kA o )2/(sin for k 0 o =
T
2
= k
kA o )2/(sin
For T = 4 /2 o =
=
2/4
2
Substituting for o in (1)
ak = 2
;)2/(2
AakSa
Ao =
ak = 2/2
2
)2/sin
2=
ow
k
kA
a1 = a-1 =
A a3 = a-3 =
−
3
1
a2 = a-2 = 0 a5 = a-5 = 5
1 a7 = a-7 =
−
7
1
Plot of ak =
22
kSa
A
for T = 4/2
Thus we can plot the ak for various values of
T = 8 /2 ; T = 16 /2 etc.
for T = 8 /2
for T = 16 /2
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An alternative way of interpreting this eq. is as samples of an envelope fn.
i.e. T ak =
2/sin2A
- i.e. with thought = ko of as a continuous variable.
- The fn.
2/sin2A represents the envelope of Tak.
- And the coefficients ak are equally spaced samples of this envelope.
ak = )2/(sin
T
A
Tak = A sin (/2)
= 2/
2/sin
A
=
)2/(sin2A
Tak = 2A
)2/(sin
i.e. ko =
Also for fixed values of /2 the envelope Tak is independent of T.
for T = 4 /2
T = 8 /2
T = 16 /2
- From the fig. we find that as T is increased i.e. as o is decreased the envelope is
sampled at closer intervals.
- As T becomes arbitrarily large the original periodic square wave approaches a
rectangular pulse (i.e. the given periodic signal in the time domain becomes a single
rectangular pulse i.e. an aperiodic signal corresponding to one period of the square
wave.
- The F.S. coefficients multiplied by T become more and more closely spaced samples
of the envelope.
- Ultimately the set of F.S. coefficients approaches the envelope of fn. as T → .
Find the F.T. of a periodic gate Fn. :
f(t) =
−=n
tJn
noeF
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Where Fn =
T
nSa
T
A = ( )2/
Sa
T
A
F[f(t)] =
−=
−
n
onT
nSa
T
A)(
2
The transform of f(t) therefore consists of impulses located at = o, = o, 2o etc.
The magnitude of these pulses is given by
T
nSa
T
A 2
This graph for 20
1= and T =
4
1 sec. then o = 8
F.T. of a sequence of equidistant impulses of unit strength
T (t) = (t) + (t-T) + ………… (t-nT) + (t+T) + (t+2T) + …….. (t+nT)
F.S. of the fn.
T (t) =
−=n
tJn
noeF
Fn = dtetT
T
T
tJn
To
−
−
2/
2/
)(1
Hence Fn = dteT
T
T
tJn
To
−
−
2/
2/
)(
1 consider the pulse is –T/2 to T/2
From the property of impulse fn.
Fn = T
1 i.e. Fn is a constant.
So the impulse train contains same impulse train in the F.T. spaced at = o, = o, 2o
….
So T (t) =
−=n
tJn oeT
1
F[T (t)] = 2
−=
−n
onT
)(1
=
−=
−
n
onT
)(2
= o .
−=
−n
on )(
F[T (t)] = )(. oo
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3.4. Properties of Fourier Transform:
3.4.1. Linearity Property :
f1 (t) F1 ()
f2 (t) F2 ()
Then for any arbitrary constants of a1 & a2
[a1 f1(t) + a2 f2(t)] a1F1 () + a2 F2 ()
3.4.2. Symmetry property :
f(t) F()
F(t) 2 f(-)
Proof :
F() =
−
− dtetf tJ)( ------ (1)
f(t) =
−
deF tJ)(2
1 ------- (2)
2 f(-t) =
−
− deF tJ)( From (2) by replacing t with – t
as is a dummy variable replace by x.
2f (-t) =
−
− dxexF Jxt)(
Put t = Hence
2 f(-) =
−
− dxexF Jx)(
As x is also a dummy variable replace it by another variable “t” we get
2 f(-) =
−
− = )]([)( tFFdtetF tJ
i.e. F[F(t)] = 2 f(-)
If f(t) is an even fn.
f(-t) = f(t)
Then F[F(t)] = 2 f()
F(t) 2 f()
3.4.3. Frequency differentiation :
If f(t) F()
Then -J t f(t) d
dF
Proof :
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F() =
−
− dtetf tJ)(
Differentiating both sides w.r. to :
−
−−= dteJttfd
dF tJ
)()(
)(
=
−
−− dtetfJt tJ))(( . This is F.T. of (-Jt f(t))
Here F [-Jt f(t)] =
d
dF )(
3.4.4. Frequency integration :
dFJt
tf)(
)(− F() =
−
− dtetf tJ)(
Integrating both sides w.r. to
−
−
−= dt
Jt
etfdF
tJ
)()(
−
−
−= dte
Jt
tfdF tJ
)()(
Thus dFJt
tf)(
)(
−
3.4.5. Time shifting :
f(t) F()
Then f(t – to) )(Fe otJ−
Proof : f(t) =
−
+
deF tJ)(
2
1
f(t-to) =
−
−+
deF ottJ )()(
2
1
=
−
−
−−
=
deFedeeF tJtJtJtJ oo )(2
1)(
2
1
Thus f(t – to) otJe
− F()
3.4.6. Conjugation and conjugate symmetry :
f(t) F()
Then f* (t) F* (-)
Proof :
−
−= dtetf tJ
F
)(
*
* )(
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= dtetf tJ
−
)(*
Replace by – .
Thus dtetfF tJ=−
−
− )()( ** . RHS is Fourier transform analysis eq. of f* (t)
This means )()( ** −Ftf
This conjugation property allows us to show that if f(t) is real then F() has conjugate
symmetry.
i.e. F(-) = )(* F because f(t) is real.
If f(t) is real then )()(* tftf =
then )()()(* FdtetfF tJ ==−
−
−
3.4.7. Time differentiation :
If f(t) F()
Then )()(
FJdt
tdfF =
f(t) = deF tJ
−
)(2
1
=
−
deFdt
d
dt
tfd tJ)(2
1)(
deJFdt
tdfF tJ
=
−
)(2
1)( =
−
deJF tJ)(2
1
= deFJ tJ
−
)]([2
1
)()(2
1)( 1 FJFdeFJdt
tfd tJ
−
−=
=
Thus )()(
FJdt
tdf )())((
)( 1 FJFJFFdt
tfdF ==
−
i.e. )()(
FJdt
tfdF =
Time and Frequency scaling :
If f(t) F()
f(at)
aF
a
||
1
Where a is a real constant
F[f(at)] =
−
− dteatf tJ)(
Put = at F[f(at)] = 0)(1
−
−
adefa
aJ
= 0)(1
+
−
−
afordefa
aJ
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Hence F[f(at)] =
−
−
defa
aJ
)(||
1 i.e. when a > 0
Hence f(at)
aF
a
||
1
=
aF
aatfF
1)(
For a < 0 When a < 0
Put at =
dt = a
d
−=−
aF
aatfF
1)(
F[f(at)] = F[f()] Thus
F[f(-at)] =
−
−−
−
defa
aJ )(
)(1
=
aF
aatfF
||
1)(
=
−
−
−
−
defa
aJ )(
)(1
=
−
−
−
−=−
defa
aFa
aJ )(
)(1
)/(1
=
−
−−
−
defa
aJ
)(1 = ( )aF
a/
1−
Hence F[f(at)] =
aF
a
||
1
3.4.8. Duality :
We have derived earlier
f1 (t) =
2/sin2)(
2/||0
2/||11 =
F
t
t
f2 (t) =
=
||0
||1)(
sin2F
t
t
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3.4.9. Convolution property :
f(t) =
−
− dtff )()( 21 convolution theorem
f1(t) F1()
f2(t) F2()
−
− )(.)()()( 2121 FFdtff
f1(t) * f2(t) F1() F2()
Proof :
F [f1(t) * f2(t)] =
−
−
−
− dtdtffe tJ )()( 21
=
−
−
−
− ddtetff tJ)()( 21
Put (t - ) = x ; t = (x + )
dt = dx
=
−
+−
−
ddxexff xJ )(
21 )()(
=
−
−
−
− ddxexfef xJJ )()( 21
F2() .
−
− = )()()( 121 FFdef J
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Thus f1(t) * f2(t) F1 () . F2 ()
3.4.10. Frequency convolution and modulation property :
If f1(t) F1()
f2(t) F2()
Then f1(t) . f2(t)
−
−
duuFuF )()(2
121
Put f3(t) = f1(t) . f2(t)
Then F[f3(t)] = F3 [] =
−
− dtetf tJ)(3
=
−
− dtetftf tJ)()( 21 ------ (1)
Inverse F.T. states that
f1 (t) =
−
deF tJ)(2
11
Substitute for f1(t) in (1)
= dtetfdeF tJtJ
−
−
)()(2
121
= ddtetfF tJ
−
−− )(
21 )()(2
1
F3() = dFF
−
−
)()(2
121
= )()(2
12*1 FF
Frequency convolution is defined as dFF
−
− )()( 21
3.4.11. Frequency shifting property :
f(t) F()
f(t) tJ oe
F( – o)
Proof :
−
−
−−− == dtetfdteetfetfFtJtJtJtJ ooo )(
)()()(
= F ( – o)
3.5. Hilbert Transform :
Hilbert transform fh(t) is a signal obtained by shifting the phase of every component is
f(t) by (-/2).
This represents Hilbert’s transform of f(t).
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It is defined as fh(t) =
−−
=
d
t
f
ttf
)(
)(11)(
1*
and inverse H.T. f(t) =
−−
−
d
t
fh
)(
)(1
By definition
)(
)(
2/
)(
)(
)(
h
h
F
tf
shifterphase
H
F
tf
−
To get fh(t) the signal f(t) is passed through a phase shifting network H() whose
output in fh (t).
The characteristics of such a system
(i) The magnitudes of all the frequency component present in f(t) should remain
unaltered when it is passed through the system.
i.e. |H()| = 1
(ii) The phase of the +ve frequency components should be shifted by -/2 and the
phase of –ve frequency components should be shifted by + /2.
Thus the transfer fn. H() = | H() | eJQ() ----- (1)
() =
−
02/
02/
for
for
So (1) can be written as
H() =
−
0
0
2/
2/
fore
fore
J
J
We know that JJeJ =+= 2/sin2/cos2/
JJe J −=−=− 2/sin2/cos2/
Hence H() becomes )(01
01)(
Sgn
for
for
J
H−=
−
=
i.e. H() = - J Sgn () ------- (2)
The response Fh() of the system is related to the input F() as
Fh() = F() . H()
Where f(t) F() and fh(t) Fh()
Substituting for H() from eq. (2)
Fh() = - J F() Sgn ()
Taking inverse F.T. of both sides
fh (t) = F-1 [-J F() . Sgn ()]
The time domain fn. of - J Sgn () = t
1
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i.e. t
1 -J Sgn ()
Using convolution theorem :
fh (t) =
ttf
1)(
1*
=
−−
d
t
f
)(
)(1
Which is Hilbert transform of f(t).
Applications of H.T. Properties of H.T.
(i) Generation of SSB Signals. (i) Signal f(t) and fh(t) have the same (EDS)
i.e. Energy density fn.
(ii) Design of minimum phase filters. (ii) f(t) & fh(t) have the same auto correlation
fn.
(iii) Representation of band pass signals. (iii) f(t) & fh(t) are mutually orthogonal.
(iv) H (ft) = fh (t)
H(fht) = -f (t)
Application of F.T. to Std. Problem :
(i)
−
=+ 1)()1( 2 dttt (ii) −
=−+
2
1
4 2)1()1( dttt
(iii) =−++
5
3
3 0)1()24( dtttt (iv)
−
−=−− 1)1()2( 4 dttt
f(t) =
−=
||
)(21 tfor
tt
F() = dtett
F tJ
−
=
= −
−−
−+
+
0
2/
2/
0
21
21
dte
tdte
t JwttJ
Ans.
42/ 2
Sat
F
3.6. Parseval’s relations :
E =
−
−
= dFdttf 22 |)(|2
1|)(|
E =
−
−
−
−
−
== dtdeFtfdttftfdttf tJ )(
2
1.)()(.)(|)(| **2
Reversing the order of integration
dFFddtetfF tJ )(.)(2
1)()(
2
1 **
−
−
−
−
=
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=
−
dF 2|)(|2
1
That is the total energy in the signal f(t) can be determined by computing the energy
per unit time |f(t)|2 and integrating over all time or by computing the energy per unit
frequency |F()|2 and integrating over all frequencies.
3.7. Inverse Fourier Transform:
1. Find Inverse F.T. of 2)3(
J
J
+ it is of the form J F().
2
3
2 )3(
1)(.
)(
1)(
JtuetF
JatuetF tat
+=
+= −−
Let t e-3t u(t) = f1 (t)
Then F() = 2)3(
1
J+
We know that )()( FJtfdt
dF =
F-1 [J F()] = )(3 tuetdt
d t−
2. Find Inverse F.T. of F() = e-||
We know that 2
||
1
2
+=− teF
Using duality theorem
||
22
1
2 −=
+e
tF
)1(
12
||1
teF
+=−−
3. Inverse F.T. of ()
f(t) =
−
deF tJ)(2
1
= =
=
−
]1[2
1)(
2
1 de tJ
=−
2
1)]([1 F or F-1 [2 ()] = 1
Or F[1] = 2 ()
4. F-1 [ ( – o)]
f(t) =
−
−
−
=
dedeF tJ
o
tJ )(2
1)(
2
1
= tJ oe
2
1
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tJ
o
tJ
oo
o
eFe
F
=−
=− −− )(2;2
)( 11
F.T. of Complex and real fns. :
F.T. [f(t)] = F() = dtetf tJ−
−
)( ; Let f(t) = fR (t) + J fI(t)
= dttJttfJtf IR sincos)()( −+
−
= dtttfttfJdtttfttf RIIR
−
−
−++ sin)(cos)(sin)(cos)(
= FR () + J FI ()
Where FR() = dtttfttf IR
−
+ sin)(cos)( and
FI() = J dtttfttf RI
−
− sin)(cos)(
Inverse F.T. of F() is obtained from
f(t) =
−
deF tJ)(2
1
=
−
++
dtJtFJF IR sincos)()(2
1
fR (t) =
−
−
dtFtF IR sin)(cos)(2
1
fI (t) =
−
+
dtFtF IR cos)(sin)(2
1
Case : I f(t) is real :
Then fI (t) = 0 and F(-) = F* ()
So FR () =
−
−
−= dtttfFdtttf I sin)()(;cos)(
Case : II f(t) is even and real :
Let f(t) = fe (t)
FR () =
−
=0
cos)(2cos)( dtttfdtttfe
FI (w) = 0
So F() =
0
cos)(2 dtttf
Case : III f(t) is odd and real :
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Let f(t) = fo (t)
FR () = 0
FI () = -J F()
= - J
−
dtttfo sin)(
= - 2J
0
sin)( dtttfo
For non-symmetrical fns. :
F [f(t)] = F() = FR () + J FI ()
=
−
−
− dtttfJttf oe sin)(cos)(
F[f(t)] = Fe () + Fo ()
To understand duality property :
f(t) = 21
2
t+
Let us consider a signal whose F.T. is X() = 21
2
+
This is f(t) = e-1 | t | F() = 21
2
+
22
|| 2)(
+=−
a
aeofF ta
The synthesis eq. for this F.T. is :
e-|t| =
de tJ
+
−
21
2
2
1
Multiply by 2 and change t = -t
2 e-|t| =
de tJ
+
−
21
2
Interchange the variables t & we find that
2 e-|| = dtet
tJ−
−
+ 21
2
The RHS is analysis eq. of 21
2
t+
Thus ||
22
1
2 −=
+e
tF
Compare F e-|t| = )1(
22+
- Illustration of usefulness of the Fourier Transform linearity and time-shift properties
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f(t) = 2
1 f1 (t – 2.5) + f2 (t – 2.5)
F1() =
)2/(sin2 and F2() =
)2/3(sin2
Finally using the linearity and time shift properties of the F.T. yields
F() =
+−
)2/3(sin2)2/(sin2
5J
e
Problem of F.S. :
Find cosine representation of
o =
2
2 = 1
f(t) = 2
t 0 t T ao =
=
=
=
2
0
2
0
2
22
02
1
)2(222
1)(
2
1 tdt
tdttf
find ao = 2
1 an =
=
2
0
2
2
0)2(
2cos
22
2 dvu
dtntscotdtntt
an = 0 =
−
dtn
nt
n
ntt2
0
2
0
2
sinsin
)2(
2
bn =
−n
1 = 0
cos0
)2(
22
0
22=
+
n
nt
An =
=+n
ba nn
122 bn =
=
2
0
2
0
2)2(
2sin
22
2 dvu
dtntinstdtntt
n = 2/0
)/1(tan 1 =
−+− − n
=
−−
−
2
0
2
0
2
coscos
)2(
2dt
n
nt
n
ntt
=
+
−
2
0
222
sin
)2(
22
)2(
2
n
nt
n
f(t) =
=
+
+
1 2cos
1
2
1
n
ntn
=
−n
1
3.8. Objective questions:
1. Given two complex numbers: iCiC 41and,32 21 +=−= . The product 21 CCP =
can be computed as
(E) i52+
(F) i510+−
(G) i514+−
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(H) i514+
2. Given the complex number iC 431 += . In polar coordinates, the above complex
number can be expressed as iAeC =1 , where A and is called the amplitude and
phase angle of 1C , respectively. The amplitude A can be computed as
(I) 3
(J) 4
(K) 5
(L) 7
3. Given the complex number iC 431 += . In polar coordinates, the above complex
number can be expressed as iAeC =1 , where A and is called the amplitude and
phase angle of 1C , respectively. The phase angle in radians can be computed as
(M) 0.6435
(N) 0.9273
(O) 2.864
(P) 5.454
4. For the complex number ,43 iC +−= the phase angle in radians can be computed
as
(Q) 0.6435
(R) 0.9273
(S) 1.206
2.2143
5. Given the function .,0
,1)()(
=
=−=elsewhere
atifattf np The Fourier transform )(ˆ
0iwF
which will transform the function from time domain to frequency domain can be
computed as
(T) )( ta +
(U) afie )2( −
(V) 1 (W) )( at −
6. Given the function 1)(ˆ0 =iwF . The inverse Fourier transform )(tfnp which will
transform the function from frequency domain to time domain can be computed as
(X) ite
(Y) ite−
(Z) )0( −t
(AA) tfie )2( −
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UNIT – 4
SIGNAL TRANSMISSION THROUGH LINEAR
SYSTEMS Contents:
5.1 Classification of Systems.
5.2 Transmission through Linear Syatems.
5.3 LTI Systems.
5.4 Filter Characteristics of linear Systems.
5.5 Distortionless Transmission through System.
5.6 Ideal filter Characteristics.
5.7 Bandwidth and Rise time.
SIGNAL TRANSMISSION THROUGH LINEAR SYSTEMS
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A system is a set of elements connected together to perform a particular task.
Or
A system can be defined as a rational assembly of elements which produces a
required response when driven by a signal.
4.1. Classification of systems:
Systems can be classified as
1. Lumped parameter and distributed parameter systems.
a) Lumped parameter systems are the ones
constructed using lumped elements like L,
C & R etc. and they are described by
differential equations.
b) Distributed parameter systems are those
which are fns. of both space and time.
Ex : transmission lines.
2. Static and dynamic systems :
a) Static systems are those whose output
depends on the present input and not on
previous or the past ones.
b) Systems whose o/p depends on the
present and or past and previous inputs are
called dynamic systems or memory
systems.
3. Causal and non-causal systems :
a) A causal system in one whose o/p
depends on the present and the previous
inputs and not on the future ones and is
known as non-anticipatory system.
b) A non-causal systems o/p depends on
the future inputs. Such systems are called
anticipatory systems.
4. LTI & LTV :
a) A system is said to be time invariant if
a time delay or a time advance of the input
signal leads to an identical time shift in
the output signal.
b) Otherwise it is an LTV system.
5. Stable and unstable system :
a) Stable system is one which gives
bounded o/p for a bounded input
−
dtth |)(| .
b) Otherwise it is an unstable system.
Problems :
1. Given h(t) = e-t u(t) as the impulse response of the system. Find the transfer fn., test for
stability and causality.
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Transfer fn. L [h(t)] = 1
1
+S Re [S] > -1
For stability
−
dtth |)(|
−
−
− =0
1|)(| dtedttue tt
Hence the system is stable.
For causality h(t) must be zero for t < 0
Since h(t) = e-t u(t) and u(t) exists only.
for t > 0 ; So the gives system is causal.
2. H(S) = 1+S
eS
find h(t) and decide whether the system is causal or not
f(t) F(S) we know that e-t u(t) 1
1
+S
Using time shifting property of transform.
1
)1()1(
+++−
S
etue
St
h(t) = e-(t+1) u (t+1)
h(t) exists for t > -1 the system is non-causal i.e. between 0 & -1 hence non-causal.
3. dt
dy+ 3y = x find transfer fn. and impulse response.
Sy(S) + 3y(S) = x(S)
H(S) = 3
1
)(
)(
+=
SSX
SY Re [S] > - 3
Impulse response is h(t) = L-1 [H(S)] = e-3t u(t)
4. Given xydt
dy
dt
yd=++ 2
32
2
with conditions y(0+) = 3 and y1(0+) = -5.
x(t) = 2ut, Find y(t).
We know that L
dt
dy = Sy(S) – y(0+)
And L
2
2
dt
yd = S2y (S) – Sy (0+) – y1 (0+)
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Taking L1 transform for the given equation.
S2y (S) – 3S + 5 + 3Sy (S) – 9 + 2y (S) = X(S) = S
2 To find A
y(S) [S2 + 3S + 2] = S
2 + 3S + 4 Multiply by S and put S = 0
y(S) [(S+1) (S+2)] = S
2 + 3S + 4 A = 1
21
2=
Y(S) = )2()1(
43
)3()1(
2
++
++
++ SS
S
SSS To find B
21)2()1(
432 2
++
++=
++
++
S
C
S
B
S
A
SSS
SS Multiply by (S+1) and put S = 1
= )2()1(
][]2[]23[ 222
++
++++++
SSS
SSCSSBSSA
11
432
)21()1(
)1(4)1(32 2
−=−
−+=
+−−
−+−+
To find C
Multiply by (S+2) and put S = -2
32
8122
)12()2(
)2(4)2(32 2
=−+
=+−−
−+−+
A+B+C = 3 B+C = 2 From A = 1 ; B = -1 & C = 3
3A+2B+C = 4 2B+C = 1 Y(S) = 2
3
1
11
++
+−
SSS
2A = 2 2B+2-B = 1 y(t) = [1 – e-t + 3e-2t] u(t)
A = 1 C = 3
5. Find the impulse response of the system.
Vi = i R + C
1 idt
Vo = C
1 idt
Taking L.T. of both equations
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Vi [S] = RI [S] + CS
SI ][
Vo(S) = CS
SI )(
H(S) =
+
=+
=
+
=
RCSRC
RCS
CSR
CS
SV
SV
i
o
1
1
1
1
1
1
)(
)(
Impulse response = h(t) = L-1 )(1
1
1 / tueRC
RCSRC
RCt−=
+
6. Find the impulse response of the System.
x(t) = Ri + L dt
di
y(t) = L dt
di
X(S) = I(S) [R + LS]
Y[S] = LSI [S]
H[S] =
+
=
+
=+
=
L
RS
LS
L
RSL
LS
LSR
LS
SX
SY
)(
)(
)/(
/1
LRS
LR
L
RS
L
R
L
RS
+−=
+
−+
h(t) = -1 [H(S)] = (t) - t
L
R
eL
R −
= )()/(
1)( )/( tue
RLt RL
t−
−
Where L/R is time constant
= )(1
)( tuet
t
−
−
Find Linear or not
7. dt
dy + 3ty = t2 x(t)
dt
dy1 + 3ty1 = t2 x1 (t) ⎯⎯⎯ →⎯ abymultiply a
dt
dy1 + 3at y1 = at2 x1 (t) ------ (1)
dt
dy2 + 3ty2 = t2 x2 (t) ⎯⎯⎯⎯ →⎯ bbymultiply b
dt
dy2 + 3bt y2 = bt2 x2(t) ------ (2)
(1) + (2) = dt
d [ay1 + by2] + 3t [ay1 + by2] = at2 x1 (t) + bt2 x2(t)
= t2 [ax1(t) + bx2(t)]
x3 (t) = ax1 (t) + bx2(t)
dt
d [y3 (t)] + 3t y3 (t) = t2 [ax1 (t) + bx2 (t)] = t2 x3 (t)
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Hence Linear
8. dt
dy + 2y = x2 (t) → for an input of x(t)
For x1(t) dt
dy1 + 2y1 = x12 (t)
For x2(t) dt
dy2 + 2y2 = x22 (t)
dt
d [ay1 + by2] + 2 [ay1 + by2] = ax1
2 [t] + bx22 [t]
For ax1 (t) + bx2 (t) = x3(t)
dt
d [y3(t)] + 2 [y3(t)] = [ax1(t) + bx2(t)]
2 = a2x1(t)2 + b2x2
2 (t) + 2abx1(t) x2(t).
Hence not linear.
9. y(t) =
t
dx )( ; y1(t) = −
t
dx )(1
y2(t) = −
t
dx )(2
y3(t) = −
t
dx )(3 where x3 () = ax1 () + bx2 ()
= −
t
[ax1() + bx2 ()] d = ay1 (t) + by2(t)
y3(t) = ay1 (t) + by2 (t) hence linear.
10. dt
dxxy
dt
dy=+ 2
dt
dxxy
dt
dy 111
1 2 =+ ; dt
dxxy
dt
dy 222
2 2 =+
x3 (t) = ax1 (t) + bx2 (t) ;
++==+
dt
tdxb
dt
tdxatbxtax
dt
tdxtxy
dt
dy )()()()(
)()(2 21
213
333
= dt
tdxtxb
dt
tdxtabxtx
dt
tdxab
dt
tdxtxa
)()(
)()()(
)()()( 2
2
2121
211
2 +++
dt
dxbx
dt
dxaxbyaybyay
dt
d 22
112121 2 +=+++
Both are not equal hence non-linear.
11. Time invariance.
To check for time invariance we should determine the variance property for any input and
any time shift.
Thus let x1 (t) be an arbitrary input to the system.
and let y1(t) = Sin [x1(t)]
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Then consider a second input obtained by shifting x1(t) in time
x2(t) = x1 (t – t0)
Then o/p corresponding to this shift is
y2(t) = sin [x (t-t0)] ------ (1)
Similarly from y1 (t) = sin [x1(t)]
y1 (t – t0) = sin x (t – t0) ------ (2)
Comparing (1) & (2) we find that the system is time in variant.
12. Find time invariant or not
y(t) = t x(t) input is x(t)
y(t) = T [x(t)] = t x(t)
The o/p due to delayed input is
= x(t, T) = t [ x (t – T] = t [ x (t – T] = y (t, T)
If the o/p is delayed by T we set
y[t – T] = (t – T) x [t – T] Hence both are not equal
Hence it is a time variant system.
13. y(t) = x(t) cos (50 t)
Out put due to delayed in put
y(t, T) = x[t – T] cos 50 t
if o/p is delayed
y[t – T] = x[t – T] cos 50 (t – T)
y(t, T) y[t – T] Hence time variant.
14. y(t) = x(t2)
y(t, T) = x [t2 – T]
y[t – T] = x[(t – T)2]
y(t, T) y(t – T) Hence time variant.
15. y(t) = x(-t)
y(t, T) = x ( - t – T)
y(t – T) = x[-(t-T)] y(t, T) y(t – T)
= x[-t + T]
Hence time variant.
16. Causal or not :
y(t) = x(t) x (t – 1)
y(0) = x(0) x (-1) – Causal
y(1) = x(1) x(0) – Causal Hence Causal.
y(2) = x(2) x(1) – Causal
17. y(t) = x (t/2)
y(0) = x(0) Present causal
y(1) = y( ½ ) Previous causal
y(-1) = y (- 0.5) – Future input hence non-causal
18. y(t) = x( t )
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t = 0.49 t = 0.7
y(0.49) = x (0.7) Hence non-causal
19. x1(t) = eat u(t)
Causal because eat u(t) means x1 (t) = 0 for t < 0. Present
20. x(t) = e-2t u (-t)
u (-t) means fn. exists before t = 0. Hence causal.
21. y[n] = x[n] + ]1[
1
−nx
Solution :
n = 0 y [0] = x [0] + ]1[
1
−x - Causal present & previous.
n = 1 y [1] = x [1] + ]0[
1
x - Causal present & previous. Hence Causal
System
n = -1 y [-1] = x [-1] + ]2[
1
−x - Causal present & previous.
4.2.Signal Transmission through Linear system :
4.2.1 A linear system is one which satisfies both additive and homogeneity condition.
1) If two signals are x1 (t) & x2 (t)
And if the responses of the system are y1(t) & y2(t)
Then additive property states that
The response to x1(t) + x2(t) = y1(t) + y2(t)
2) Homogeneity property states that [or (scaling)]
for input ax1(t) response is ay1(t)
and for bx2(t) response is by2(t)
then ax1(t) + bx2(t) → ay1(t) + by2(t). Where a & b are constants.
Ex. 1 : A system S which is y(t) = t x(t) test whether linear.
x1(t) → y1(t) = t x1(t)
x2(t) → y2(t) = t x2(t)
Let x3(t) = ax1(t) + bx2(t) where a & b are constants.
If x3(t) is input to the system.
Then x3(t) → y3(t) = t x3(t) = t [ax1(t) + bx2(t)]
= a tx1(t) + b tx2(t) = ay1 (t) + by2 (t)
Hence ‘S’ Linear.
Ex. 2 : y(t) = x2 (t) Let x1(t) → y1(t) = x12 (t)
x2(t) → y2(t) = x22 (t)
Let x3(t) = a x1(t) + b x2(t) Then x3(t) → y3(t) – x32 (t)
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= [ax1(t) + b x2(t)]2 = a2 x1
2 (t) + b2 x22 (t) + 2ab x1(t) x2(t)
But ay1(t) + by2(t) = a x12 (t) + b x2
2 (t)
These two are not equal hence non linear.
Give a set of values :
If x1(t) = 1 x2 (t) = 0 a = 2 and b = 0
Then y1(t) = x12 (t) = 1 ; y2(t) = x2
2 (t) = 0
x3(t) = a x1(t) + b x2(t)
y3(t) = [ax1(t) + bx2(t)]2 = a2 x1
2(t) + b2 x22 (t) + 2ab x1(t) x2(t) ----- (1)
But for a linear system it should be y3(t) = ax12 (t) + bx2
2 (t) ----- (2)
i.e. By (1) y3(t) = 4 1 = 4
By (2) y3(t) = 2 1 = 2 Both are not equal. Hence the system is non
linear.
4.2.2. The impulse response of a linear system :
The transform of a unit impulse is unity
i.e.
−
== 1)([)( tFdtt
Hence the response of a system to a unit impulse input will be given by L-1 [H(S)]
where H(S) is the transfer fn. of the system.
)(
)(
)(
)(
)(
)(
SY
ty
SH
th
SX
tx
If a system is given an input signal x(t) whose LT is X(S) and the system output in the
form of Laplace transform is Y(S).
Then )(
)(
SX
SY = H(S) is called the transfer function of the system.
Thus H(S) = n
n
PS
a
PS
a
PS
a
SD
SN
−+
−+
−= .......
)(
)(
2
2
1
1
and h(t) = L-1 [H(S)] = tP
n
tPtPtP neaeaeaea ......321
321 +++
Then terms P1, P2 …….. Pn are the poles of the transfer fn. and are known as natural
frequencies of the system.
If the input is an impulse its L-transform is unity.
Hence the transfer fn. of the system itself is the response output. Thus for an impulse
input
H(S) is the response.
and inverse of H(S) is the impulse response in time domain.
Thus “The impulse response of a system consists of a linear combination of signals of
the natural frequencies of the system.
4.2.3. The response of a linear system :
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There are two components in the response of a system.
(i) The steady state response
(ii) The transient response
- The nature of the transient response is the characteristic of the system.
- The nature of the steady state response depends on the system and the driving fn.
Let the driving fn. be f(t) F(S)
Transfer fn. H(S)
and the response r(t) → R(S)
Then R(S) = H(S) F(S)
H(S) = )...()()(
)(
)(
)(
21
1
1
1
nPSPSPS
SN
SD
SN
−−−=
And F(S) = )...()()(
)(
)(
)(
21
2
2
2
mSSSSSS
SN
SD
SN
−−−=
Then R(S) = )...()()()...()()(
)()(
2121
21
mn SSSSSSPSPSPS
SNSN
−−−−−−
R(S) = )(
....)()()(
....)()( 2
2
1
1
2
2
1
1
m
m
n
n
SS
K
SS
K
SS
K
PS
C
PS
C
PS
C
−+
−+
−+
−+
−+
−
r(t) = tS
m
tStStP
n
tPtP mn ekekekececec .......... 2121
2121 +++++
= +j
tS
j
i
tP
iji eKeC
Transient Steady state
response response
4.3. Linear Time in variant system (LTI) :
A system is said to be time invariant if a time delay or a time advance of the input
signal leads to an identical time shift in the output signal.
Thus a time invariant system responds identically no matter when the input signal is
applied.
otS is time x(t) x(t-to) y1(t)
delay otS H For an LTI System
y2(t)
x(t) H otS Y1(t) = y2(t)
For an LTV i.e. Linear time variant system y1(t) y2(t)
4.3.1. Properties of impulse response of LTI Systems :
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1. Commutative property (interchange ability of input and impulse response)
x(t) y(t) h(t) y(t)
h(t) = x(t) x(t) * h(t) = h(t) * x(t)
2. Associative property :
[x(t) * h1(t)] * h2(t) = [x(t) * h2(t)] * h1(t)
x1(t) y(t) x(t) y(t)
h1(t) h2(t) = h2(t) h1(t)
z1(t) z2(t)
x(t) y(t)
h1(t) * h2(t)
3. Distributive Property :
4.4. Filter characteristics of linear system :
- A given system processes a signal in its characteristic way.
- The spectral density fn. of the input signal is given by F(S)
- The spectral density of the response fn. is given by H(S) F(S)
- The system therefore modifies the spectral density fn. of the input signal.
- It is evident from this that the system acts as a kind of filter to various frequency
components.
- Some frequency components are boosted in strength, some are attenuated and some
may remain unaltered.
- Thus each frequency component suffers a different amount of phase shift in the
process of transmission.
- i.e. the system modifies the spectral density according to its filter characteristics.
- The modification is carried out according to the transfer fn. H(S), which represents
the response of the system to various frequency components.
- So H(S) acts as a weighting fn. [as a filter] to various frequency components.
4.4.1. Low pass and High pass filter :
A sample RC low pass filter :
Let us take Vc(t) of the capacitor voltage as
output and Vs(t) as the system input.
The o/p voltage in this case is related to the
input voltage through the linear constant
coefficient differential equation.
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R i(t) + Vc(t) = Vs(t)
)()()(
tVtVdt
tVdRC sc
c =+ i(t) = C dt
tVd c )(
Hence )()()(
tVtVdt
tVdRC sc
c =+
The above system is an LTI system.
Let us assume input voltage is eJt = vs(t)
Hence O/P H(J) . eJt if we assume the transfer fn. as H(J)
i.e. Vc(t) = H(J) eJt
So RC dt
d [H(J) eJt] + H(J) eJt = VS (t) = eJt
i.e. H(J) eJt = JRC+1
1 eJt
H(Jw) = RCJ+1
1
RC dt
d H(J) eJt
= RC J H(J)
Hence
RC J H(J) eJt + H (J) eJt =
eJt
[RC J eJt + eJt] H(J) = eJt
(J RC + 1) eJt H(J) = eJt
For frequencies near , |H(J)| = 1.
For larger values of ‘’ |H (J)| is considerably smaller and infact steadily decreases
as ||→ increases. Thus this simple RC filter with VC (t) as o/p is a non ideal low pass filter.
Let us consider the time domain behaviour of this system.
The impulse response of the system is h(t) = )(1 / tue
RC
RCt− .
And the step response is :
S(t) = [1 – e-t/RC] u(t)
RC = , time constant of the cct.
h(t) =impulse response.
Step response of the RC low pass filter.
4.4.2. A simple RC High pass filter :
Suppose in the given RC network we choose VR(t) as the o/p.
Then the differential equation is
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dt
tdvRCtV
dt
tdvRC s
rr )(
)()(
=+= ------- (1)
VS(t) = VC(t)dt
tVdRC C )(
+
VS (t)=VC(t)+dt
tVdRC C )(
;dt
tVdRC C )(
+
=Vr(t)
)(1
)(tV
cJR
RtVr
S =
+
Or VS (t) R = Vr(t) R + cJ
tVr
)(
We can find the frequency response G(J) Or J RC VS (t) = RC J Vr(t) +
Vr(t)
of the system by putting VS(t) = eJt : )()()(
tVdt
tVdRC
dt
tdvRC r
rs +==
Then Vr(t) = G(J) eJt
G(J) =
JRC
JRC
+1
Then substituting these expressions in eq. (1) |G(J)| = RC J
+
−2221
1
CR
JRC
tJtJ eJGJeJGdt
d )()( = = 222
222
1
CR
CR
+
And dt
tVd s )( = J eJt.
+= −
CRJG
1tan)( 1
Hence RC J G(J) eJt + G (J) eJt = RC J eJt
RC J G(J) + G(J) = RC J.
G(J) [1 + RC J) = RC J
Magnitude of Frequency Response of HPF Phase diagram of a HPF
(Spectrum)
4.5. Distortion less transmission through a system :
- A system must attenuate all the frequency components equally to have a distortion
less transmission.
- i.e. H(J) should have a constant magnitude for all frequencies.
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- Not only that the phase shift of each component must also satisfy certain
relationships.
- Even if all the frequency components have been transmitted through the system with
same attenuation (or amplification) if they undergo different phase shifts then they
add up to form a different signal than the original one.
- Distortion less transmission means the o/p should be a true replica of the input.
- The replica may have a different magnitude but the preservation of the shape of the
wave form is important for distortion less transmission.
- We therefore conclude that a signal f(t) is transmitted w/o distortion if the response is
Kf(t-to) because the o/p may undergo some phase shift.
- This means that the response is an exact replica of the i/p with a magnitude K times
the original signal and delayed by to seconds.
If F(S) is the transform of f(t)
then by the time shifting theorem.
L [K f(t – to)] = K F(S) otSe−
F(S) H(S) = K F(S) otSe−
H(S) = K otSe−
So to achieve distortion less transmission through a system the transfer fn. of the
system must be of the form H(S) = K otSe−
.
For real frequencies
H(J) = K otJe
−.
It is evident that |H (J)|, the magnitude of the transfer fn. is K and that it is constant
for all values of . The phase shift for a components of frequency should be to.
In other words phase shift should be proportional to frequency.
() = n - to (n integer)
The magnitude and phase of the LTI system for distortion less transmission.
4.6. Ideal filters :
An ideal LPF transmits low frequency components w/o distortion.
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An ideal LPF characteristic and its impulse response :
Ideal HPF characteristics Ideal BPF characteristics
Paley Weiner criterion :
The Paley Weiner criterion is a frequency domain equivalent of the causality condition.
Causality is : (A Causal system is defined as the system which has no response to input
functions which are applied later) i.e. response does not begin before the input function is
applied in case of causal systems.
- A system violating the Paley Weiner criterion has a non causal impulse response. It
means that response is present prior to excitation. Obviously such systems can not be
realised physically as no physical system can produce response before excitation.
- On the other hand a system satisfying Paley Weiner criterion has a causal unit
impulse response.
h(t) = 1 for t < 0.
Such systems are physically realizable. These are time domain conditions for the
physical realisability of a system.
Frequency domain condition for physical realisability of a system can be interpreted
from the frequency domain Paley Weiner criterion of equation
+
d
H21
||)(|log| Paley Weiner criterion ----- (1)
(2) other requirement of Paley Weiner
criterion that |H(w)|2 should be
integrable
−
dH 2|)(| .
(a) Magnitude fn. |H(w)| of a physically realizable system may be zero for some
frequencies but it can never be zero for a finite band of frequencies. Otherwise the
integral will become infinite.
For example a system having a transfer fn. H() as shown in fig.
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is not physically realizable because its |H()| has zero value for a band of frequencies
beyond m.
Such systems are called ideal LPF’s.
On the other hand a system represented by
is physically realizable even for a small value of ‘’ as it has a non-zero value. Any where
for a band of frequencies although some discrete frequency have zero value.
(b) It also can be seen from the Paley Weiner criterion that the magnitude fn. |H()| for a
realizable system can not decay faster than a fn. of exponential order.
Thus |H()| = C e-|| is realizable.
and |H()| = 2−eC is not realizable.
4.7. Band width and Rise time :
- The system B.W. is referred to as the range of frequencies the system can handle.
- The system band width can be linked to a time fn. parameter called rise time as the
frequency domain computation of B.W. is cumbersome.
Let us consider a u(t) is applied to an ideal LPF.
- The response r(t) does not rise sharply.
- The time taken by the fn. to rise to its final value depends on the cut off frequency of
the filter i.e the system B.W.
We will find the relation between cut off frequency and the rise time.
The system response of an ideal LPF is a gate fn.
H() = G() e-Jd
F() = F[u(t)] = () + J
1
R() = [ () + J
1] H() = F() . H()
= () H() +
J
H )(
Since () exists at = 0 and |H()| = 1
= 0
Hence () H() = ().
So R() = () +
J
H )(
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r(t) = F-1
+
−
J
eG dJ)()(
F-1 [ ()] = 2
1 because 2 () = F[1]
Hence r(t) =
+
−−
J
eGF
dJ)(
2
1 1
=
deeJ
G tJdJ−
−
+
)(
2
1
2
1
Since G() = 1 for || m and also ‘0’ else where
r(t) =
dJ
dtem
m
J
−
−
+
)(
2
1
2
1
Putting the complex exponential fn. in terms of sinusoid.
r(t) =
ddt
dJ
dt m
m
m
m
)(sin
2
1)(cos
2
1
2
1 −
+
−
+
−−
(1) Vanishes on integration as it is an odd fn.
(2) Is an even fn. hence
r(t) =
ddtm
−
+
0
)(sin1
2
1 for an even fn.
−
=0
)(2)( dttfdttf
Put (t – d) = y so that d (t – d) = dy
d = )( dt
dy
−
So r(t) = dyy
ydtm
−
+
)(
0
sin1
2
1
= dyySa
dtm
−
+
)(
0
)(1
2
1
The integral of a sampling fn. is a standard integral known as sine integral Si(y) given by
Si (y) = y
dSa0
)(
Hence r(t) becomes r(t) = )(1
2
1dtSi m −
+
The Sine integral Si(t) :
Unit input fn. (t) and its response are Si(t) = t
dSa0
)(
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(a) Si (t) (b) u(t)
It is obvious from the r(t) fn. that
as cut off frequency m becomes less the response r(t) rises more slowly.
Assuming the minimum value as the initial value and maximum value as the final
value the rise time is defined as tr = Bm
12=
.
Where B is the band width and tr is the rise time.
B = 2
m obviously the rise time is inversely proportional to the B.W.
r(t) = )(1
2
1dtSi m −
+
UNIT – 5
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CONVOLUTION & CORRELATION OF
SIGNALS Contents:
5.1. Introduction.
5.2. Convolution Integral.
5.3. Graphical representation of convolution Integral.
5.4. Correlation and Convolution.
5.5. Autocorrelation.
5.6. Problems.
5.7. Properties of Autocorrelation.
5.8. Energy Spectral Density.
5.9. Power Spectral Density.
5.10. Differences between convolution and Correlation.
5.11. Detection of Periodic Signal in the presence of noisy by using
Correlation.
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CONVOLUTION & CORRELATION OF SIGNALS
5.1. Introduction:
Convolution and correlation of signals :
Concept of convolution in time domain & frequency domain. Convolution is a
mathematical operation which is used to express the input output relationship in a linear time
invariant system. Mathematically the convolution of two functions x1(t) and x2(t) may be
defined as
x1(t) * x2(t) =
−
− dtxx )()( 21
Let us see how to interpret the equation :
A fn. x(t) can be represented as a summation of impulses.
under this point is equal to
x(t) (t-) i.e. x(t) (t-) = x() = A in the case.
Fig 1(a)
Fig 1(b)
An impulse fn. is used to represent the pt. A. of x(t). Fig. 1(a) shows the fn. x(t) which is
continuous in time.
- There is a point ‘A’ at t = in x(t) in fig. 1(a).
- This point can be represented as a weighted impulse as shown in fig. 1(b).
- We see that the amplitude of the impulse is A and is placed at t = .
- The fn. x(t) is continuous. Hence it can have infinite no. of such points.
- The infinite no. of weighted impulses means they are so close to each other such that
they completely represent x(t).
- Since the impulses are distinct from each other we have to add all of them to represent
x(t) i.e.
x(t) =
−
− dtx )()(
Here integration is used since is a continuous fn.
This equation represents any continuous time signal x(t) in terms of weighted
impulses. In the equation (t-) is the impulse at t = . Here we have used the shifting
property of the impulse fn.
5.2. Convolution integral (The convolution concept) :
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f
input signal output signal
x(t) y(t)
Let y(t) be the o/p to an input x[t]
y(t) = f [x(t)]
Putting x(t) from the eq. x(t) =
−
− dtx )()(
y(t) =
−
−
dtxf )()(
x() is the amplitude of the fn. x(t) at t = hence a constant.
Since the system is linear we can write the equation as
y(t) =
−
− dtfx )]([)(
If the unit impulse is applied as the input to the system.
It produces unit impulse response.
f
)(
)(
−t
t
)(
)(
−th
th because it is an LTI system.
Hence f [ (t-)] = h [t-]
Thus if the unit impulse is delayed by the impulse response also is delayed by the
same .
Thus y(t) =
−
− dthx )()(
i.e. the o/p y(t) is equal to the convolution of x(t) and h(t). Hence above equation is
also called convolution integral.
Convolution integral y(t) =
−
− dthx )()(
y(t) = x(t) * h(t)
5.2.1. Graphical representation of convolution :
LTI
system
LTI
system
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Shaded area corresponds to the value of convolution between x(t) and h(t) at “t1”.
Determine the o/p of the system if i/p is x(t) = e-at u(t) a > 0.
and h(t) = u(t)
Solution :
Case I t 0
Case I : The plot of h (t - ) is shown in the fig. for t > 0.
The value h(t - ) is obtained by shifting h (-) by t, towards right.
x() & h(t - ) over lap from 0 to t.
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Hence y(t) = −
t
dthx0
)()( for t > 0.
From the fig. h (t - ) = u (t - ) = 1 from 0 to t.
Hence the above equation becomes y(t) = −− =
t t
aa dede0 0
1.
= atat
t
a ea
ea
ea
−−− −=−−=
− 1
11
11
0
for t > 0
i.e. y(t) = )(11
tuea
at−−
Case 2 - t < 0 :
There is no over lap hence y(t) = 0.
Hence y(t) =
−
−−=− )1(1
)()( atea
dthx for t 0
= 0 for t < 0
Ex. 2 : The impulse response of the cct. is given by h(t) = e-t u(t)
This is driven by x(t) = e-3t [u(t) – u(t-2)] Find the o/p
x(t) = e-3t [u(t) – u(t-2)]
u(t) – u(t – 2) has the value of 1 from t = 0 to t = 2.
So x(t) = e-3t for 0 t 2
Similarly h(t) = e-t for t 0
The o/p is y(t) = x(t) * h(t) =
−
− dthx )()(
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Case 1 :
- < t < 0
Then −
−
0
)(.)( dthx
y(t) =
−
=00.)( dx for - < t < 0
Case 2 :
0 < t < 2 :
y(t) = 22
122
.3
2
00
2)(3
−=−
−=
−=
−−−
−−−−−−
taee
eee
edeett
tt
tt
tt
Case 3 :
2 < t < :
y(t) = −−−
2
0
)(3 dee t
= 4
2
0
2
122
. −−−
− −=
−e
eee
tt
Thus y(t) =
−=
−=
−=
−−
−−
212
2012
00
4
2
tforee
tforee
tfor
t
tt
5.2.2. Time convolution theorem of F.T. :
If f1(t) F1 () and f2(t) F2 ()
Then
−
− )(.)()(.)( 2121 FFdtff f1(t) * f2(t)
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Proof :
F[f1(t) * f2(t)] =
−
−
−
− dtdtffe tJ )()( 21
= ddtetff tJ−
−
−
− )()( 21
Put (t - ) = x
dt = dx
= ddxexff xJ )(
21 )(.)( +−
−
−
= ddxexfef xJJ
−
−−
−
)()( 21
−
−
−−
−
== )(.)()()()()( 121212 FFdefFdefF JJ
5.2.3. Frequency convolution :
If f1(t) F1 () and f2(t) F2 ()
Then f1(t) . f2(t)
−
−
duuFuF )()(2
121
Proof :
To prove F [f1(t) . f2(t)] = )()(2
12*1 FF
F-1 [F1() * F2()] = deduuFuF tJ
−
−
−
)()(
2
121
Interchange the integration.
−
−
− deuFduuF tJ)()( 21
Put ( - u) = y
Then d = dy
=
−
+
dyeyFduuF tuyJ )(
21 )()(2
1
=
−
+
dyeyFeduuF JytJut )()(
2
121
−
= dyeyFtf Jyt)(2
1)( 22
=
−
)(2.)(2
121 tfdueuF Jut So
−
= )(2)( 22 tfdyeyF Jyt
= 2 f1(t) . f2(t)
Thus F-1 [F1 () * F2 ()] = 2 f1 (t) . f2(t)
Or
f1(t) f2(t) 2
1 F1() * F2()
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5.3. Graphical representation of convolution : Graphical interpretation of convolution is very useful in system analysis as well as
communication theory.
Convolution integral f1(t) * f2(t) =
−
− dtff )()( 21
The independent variable in convolution is .
The functions f1() and f2 (-) are shown as
The term f2 (t - ) represents the fn. f2 (-) shifted t seconds along the +ve -axis.
- The value of convolution integral at t = t1 is given by the integral
f1(t) * f2(t) =
−
− dtff )()( 21 evaluated at t = t1
- This is clearly the area under the product curve of f1 () & f2 (t1-)
This is shown by the shaded area.
- We choose different values for t and shift the fn. f2 (-), accordingly and find the area
under the new product curve.
These areas represent the value of the convolution function f1(t) * f2(t) at the
respective values of “t”.
The plot of the area under the product curve as a fn. of t represents the desired
convolution for f1(t) * f2(t).
- The graphical mechanism of the convolution can be appreciated by visualizing a solid
frame represented by fn. f2 (-) which is being processed along the axis by t secs.
The fn. represented by this frame is multiplied by f1 () and the area under the product curve
is the value of the convolution fn. at t = t1.
So to find the value of f1(t) * f2(t) at any time “t0” we displace the rigid frame
representing f2 (-) by t0 secs, along the -axis and multiply the fn. with f1 (). The area under
the product curve is the desired value of f1(t) * f2(t) at t = t0.
Thus we summarise to evaluate convolution :
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1. Fold the fn. f2() about the vertical axis passing through the origin of the axis and
obtain f2(-).
2. Consider the folded fn. as a rigid frame and progress it along the -axis by an amount
say t0. The rigid frame now represents the fn. f2 (t0-).
3. The product of the fn. represented by displaced rigid frame with f1 () represents the
fn. f1() f2 (t0-) and the area under this curve is given by
−
− dtff )()( 021 = f1 (t) * f2 (t0-)
t = t0
4. Repeat this procedure for different values of t by successively progressing the frame
by different amounts and find the values of the convolution fn. f1(t) * f2(t) at those
values of t.
For +ve values of ‘t’ we get convolution by moving the frame along + axis, and the
convolution for –ve values of t. We get by moving the frame along the –ve -axis.
f1(t) * f2(t) = f2 (t) * f1(t)
5.4. Correlation and Convolution : There is a striking resemblance between correlation and convolution. Signals can be
compared on the basis of similarity of wave form. A comparison may be based on the amount
of component of one wave form contained in the other wave form.
If f1(t) and f2(t) are two wave forms then the wave form f1(t) contains an amount of
C12 f2(t) of that particular wave form f2(t) in the interval (t1 < t < t2).
i.e. C12 = 2
1
2
1
)()()(2
221
t
t
t
t
dttfdttftf ------ (1)
The magnitude of the integral ion the numerator above might be taken as an indication of the
similarity of the two signals.
If the integral vanishes i.e. if =2
1
0)()( 21
t
t
dttftf
Then two signals have no similarity in the interval (t1, t2).
If we consider a radar pulse as shown in the fig. both the pulses differ by a time delay
“td” for all other aspects they are identical. But still the equation (1) when applied to these
pulses in the interval (t1, t2) the integral is zero showing that they are not similar which is not
true.
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In order to search for a similarity between the two wave forms we must shift one
wave form w.r.t. the other by various amounts and see whether there exists a similarity.
The test signal
−
−= dttftf )()()( 2112
Where ‘’ is a searching parameter or a scanning parameter. It is obvious that this is a
fn. of and is known as cross correlation. Fn. between f1(t) and f2(t) and is denoted by 12 ().
It is same whether we shift f1(t) in the –ve direction by an amount of or f2(t) in the
+ve direction by an amount of .
i.e.
−
−
−=+= dttftfdttftf )()()()()( 212112
Graphically the cross correlation fn. can be obtained by shifting the fn. f2(t) by secs. And
multiplying it by f1(t). The area under the product curve gives the value of 12 ().
It is evident that the two wave forms are similar but are shifted w.r.to one another
then the cross correlation fn. will be finite over some range of indicating the measure of
similarity. If the cross correlation fn. vanishes every where then two wave forms are not
similar or they are not correlated.
5.5. Auto Correlation and Cross Correlation: Auto correlation is the comparison of the signal with itself when shifted by an
amount of . We define auto-correlation as
−
−
+=−= dttftfdttftf )()()()()( 111111
It is obvious that 11 () = 11 (-)
i.e. Auto correlation is an even fn. of .
There is a striking resemblance between correlation and convolution.
- In cross correlation of f1(t) and f2(t) we multiply f1(t) with f2(t) displaced by
seconds. The area under the product curve is the cross correlation between f1(t) and f2
(t).
- The convolution of f1(t) and f2(t) at t = is obtained by folding f2() backwards about
the vertical axis at the origin and taking the area of the product curve of f1() and the
folded fn. f2(-) displaced by t.
So the cross correlation of f1(t) and f2(t) is the same as the convolution of f1(t) and
f2(-t).
Analytically :
Convolution 12 (t) = f1(t) * f2 (-t)
= dtff )()( 21 −
−
Dummy variable may be replaced by another variable “x”.
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12 (t) = dxtxfxf )()( 21 −
−
Changing the variable t to
12 () = dxxfxf )()( 21 −
−
Change x to t :
12 () = dttftf )()( 21 −
−
Cross correlation between f1(t) & f2(t)
= Q21 () = dttftfdttftf )()()()( 1212
−
−
+=−
Convolution between f1(t) and f2(t)
5.6. Problems on Correlation:
5.6.1. The convolution of a fn. f(t) with unit impulse fn. produces the fn
itself :
f(t) * (t) =
−
− dtf )()(
() exists only at = 0
Hence the integrand f(t) * (t) = (t) * f(t) =
−
− dtf )()( is ‘0’ for all values of
except at = 0. Hence f (t - ) has value only at = 0. Hence f(t-) = f(t).
Thus f(t) * (t) =
−
−
==− )()()()()( tfdtfdtf
Hence f(t) * (t) = f(t)
Problems : (1) (2)
5.6.2. Correlation fn. for power signals :
R12 () = −
−→
/*
21 )()(2
1T
T
dttxtxTT
Lt
5.6.3. Correlation fn. for energy signal :
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R12 () = −
+→
2/
2/
*
21 )()(
T
T
dttxtxT
Lt
R12 () = −
−→
2/
2/
*
21 )()(
T
T
dttxtxT
Lt
- If the fundamental periods of two signals are different then “T” should be taken as the
common multiple of the two periods i.e. if T1 = 3 & T2 = 4 then T = 12.
- If one signal is energy signal and the other power signal then the correlation fn.
should be calculated using the definition of energy signals.
5.6.4. Auto correlation fn. of energy signal :
R() =
−
−
+=− dttxtxdttxtx )()()()( **
5.6.4. Auto correlation fn. for power signals (i.e. periodic signals) :
R() = −
−2/0
2/0
)()(1 *
0
T
T
dttxtxT
R() = −
+2/0
2/0
)()(1 *
0
T
T
dttxtxT
For any period R() = −
−→
2/
2/
* )()(1
T
T
dttxtxTT
Lt
Relation of Auto correlation with energy :
R() =
−
− dttxtx )(*)(
When the shift = 0.
Then R(0) =
−
−
== Edttxdttxtx 2|)(|)(*)(
That is when = 0 auto correlation is equal to total energy.
Relation of Auto correlation with power :
R() = −
−→
2/
2/
* )()(1
T
T
dttxtxTT
Lt
With ‘0’ shift i.e. = 0.
R(0) = −−
=→
=→
2/
2/
2
2/
2/
* |)(|1
)()(1
T
T
T
T
PdttxTT
Ltdttxtx
TT
Lt
Thus when = 0 Auto correlation is equal to the average power of the signal.
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5.7. Properties of Auto correlation :
Property 1 :
R() = R* (-) Auto correlation shows conjugate symmetry.
This statement means that the real part of R() is an even fn. of and the imaginary
part is an odd fn. of .
Proof :
R() =
−
− dttxtx )(*)(
R*() =
−
− dttxtx )()(*
R*(-) = )()()(* Rdttxtx =+
−
Thus R() = R* (-)
Property 2 :
The value of Auto correlation fn. at = 0 (i.e. the origin) is equal to the energy of the
signal.
i.e. R(0) = E =
−
dttx 2|)(|
Proof :
R() =
−
− dttxtx )(*)( =
Put = 0
R (0) =
−
−
== Edttxdttxtx 2|)(|)(*)(
Property 3 :
If is increased in either direction the auto correlation reduces. As decreases auto
correlation increases and it is maximum at = 0
|R ()| R(0) for all .
Consider the fn. x(t) and x(t+), |x(t) x(t + )|2 is always greater than or equal to
zero. It is equal to
x2 (t) + x2 (t + ) 2 x(t) x(t + ) 0
i.e. x2 (t) + x2 (t + ) 2 x(t) x (t + )
Integrating both sides.
−
−
−
+++ dttxtxdttxdttx )()(2|)(||)(| 22
E + E 2 R()
E R()
R(0) R() Since R(0) = E
Property 4 :
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The auto correlation fn. and energy spectral density fn. of energy signal form a F.T.
pair.
i.e. R() (f)
Definition and derivation of ESD (Energy Spectral Density) fn.
The ESD (Energy Spectral Density) fn. gives the distribution of energy of the signal in the
frequency domain.
We know that E =
−
−
= dXdttx 22 |)(|2
1|)(|
Let = 2f ; d = 2 df Parseval’s or Rayleigh’s theorem.
E = dffX
2|)2(|2
1 2
Normally X(2f) is written as X(f), then we have
E =
−
dffX 2|)(|
This equation gives the total energy of the signal from its frequency components
energy in frequency domain.
- In the equation |X(f)| is the amplitude spectrum of the signal. Let us denote the
squared amplitude spectrum |X(f)|2 of the signal by (f) i.e.
Energy spectral density (f) = |X(f)|2
- The ESD fn. of a periodic or non-periodic signal x(t) represents the distribution of
power or energy in the frequency domain.
- In other words for an energy signal or power signal the total area under the spectral
density curve, plotted as a fn. of frequency is equal to total energy or average power.
- For the energy and power signals two types of spectral density functions are normally
defined i.e. Energy Spectral Density (ESD) and Power Spectral Density (PSD).
Using the energy spectral density fn.
We can write E =
−
dff )(
This equation shows that the total energy of a signal is given by the total area under
the curve (f).
Let ESD of x(t) (the o/p to the system) be x(f)
And ESD of y(t) (the o/p of the system) be y (f)
Let LTI system have a pass band from fL to fH.
(i.e. an ideal filter) and all other frequencies will be blocked.
E =
−
dff )( The energy in the o/p is Ey =
−
dffy )( .
If )( fy is symmetric for +ve and –ve values of f then the above equation can be
written as Ey =
0
)(2 dffy
Since frequencies from fL to fH are only present (pass band).
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Ey = =H
L
H
L
f
f
f
f
y dffydff 2|)(|2)(2 since |y(f)|2 = y (f)
We know that y() = x() H(). This equation can also be written in term of ‘f’ as
y(f) = H(f) X(f). Hence the above equation
Ey = dffXfHH
L
f
f
2|)()(|2
= dffXfHH
L
f
f
22 |)(||)(|2
= dfffH x
f
f
H
L
|)(||)(|2 2 since x (f) = |X(f)|2
The filter passes all frequency set fL & fH. This means H(f) = 1 for fL f fH.
Hence the above equation will be Ey = dffx
f
f
H
L
)(2 .
This equation gives energy of the o/p signal in terms of ESD of i/p signal.
5.8. Energy Spectral Density: Relation of ESD to Auto correlation and the properties of ESD.
The auto correlation fn R() and ESD fn. (f) form a Fourier transform pair.
i.e. R() (f)
Proof :
Auto correlation is given as
R() =
−
− dttxtx )(*)(
Replace x* (t-) by its inverse transform.
We know that Inverse F.T. of X(f)
i.e. x(t) =
−
−
=
dfefXdfefX ftJftJ 22 )(2)(2
1
R() =
−
−
−
dtdfefXtx tfJ
*
)(2)()(
=
−
−
−−
dtdfefXtx tfJ
*
)(2* )()(
=
−
+
−
−
dfedtetxX fJftJ
f
22*)(
R() =
−
dfefXX fJ
f
2*)(
=
−
dfefX fJ 22|)(| This equation means that R() is IFT
of (f) since (f) = |X(f)|2.
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=
−
= )()( 2 Rdfef fJ
R() (f) i.e. R() is inverse F.T. of (f)
Thus R() and (f) are F.T. pair.
Ex. : Determine the auto correlation fn. and ESD of x(t) = e-at u(t)
To get auto correlation
R() =
−
− dttxtx )()( * This is energy signal Hence auto correlated fn.
energy signal is given by
If x(t) is real
Then R() =
−
− dttxtx )()(
Here e-at u(t) is a real signal. Putting the values in the above equation.
R() =
−
−−− − dttuetue taat )()( )(
We know that u(t) = 1 for t 0 and u(t - ) = 1 for t . Hence the limits of
integration will be to .
So R() =
−−−
dtee taat )(
= a
e
a
e
a
ee
a
ee
aaa
ata
2222
22
−−−
−
=
+
−=
−
This is auto correlation fn. for +ve values of .
This is because we have considered u(t - ) = 1 for +ve. We know that auto
correlation has conjugate symmetry.
R() = R* (-)
In this example R() is real
Hence R() = R(-)
Thus R() is an even fn. of .
Auto correlation fn. of e-at u(t)
To obtain ESD :
(f) = F.T. [R()]
= deR fJ
−
− 2)(
=
−
−−
dea
e fJa
2||
2
(f) = 22 )2(
2
2
1
fa
a
a +
i.e.
−−−
− −
−− +=0
22
0
2|| deedeedee fJafJafJa
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=
+−
−
− +0
)2(
0
)2( 2
2
1
2
1 de
ade
a
fafJa
=
+−+
−
+−−
− )2(2
1
)2(2
1 )2(
0
20
fJa
e
afJa
e
a
fJafa
= 2222 )2(
1
)2(
2
2
1
2(
1
2
1
2
1
fafa
a
afJafJaa +=
+=
++
−=
22
1
+a
This is the expression for ESD of e-at u(t).
For a periodic signal with period T Fourier coefficients X(k) are given as
X(k) = −
−
2/
2/
)( 0)(1
T
T
tkJdtetx
T
If we consider only one pulse/cycle of x(t) the limits of integration can be changed from - to
i.e. X(k) =
−
−dtetx
T
tkJ )( 0)(1
= )(1
0kXT
X(k) = ( ))2(1
0fkXT
since 0 = 2f0.
= )(1
0fkXT
here X(2f0) is written as X(kf0)
Substituting the value of X(k) is
P =
−=
−=
=kk
kfXT
kX 2
02
2 |)(|1
|)(|
Thus P =
−=k
fkXT
2
02|)(|
1
5.9. Power spectral density :
Periodic signals have infinite energy and are called power signals. So for a
power signal we find the average power.
We define the average power (simply power) of a signal f(t) as the average
power dissipated by a voltage f(t) applied a cross a 1 - resistor (or by the current
f(t) passing through a 1 - resistor).
Thus the average power of f(t) P = −
→
2/
2/
2 )(1
T
T
dttfTT
Lt.
Which is also the mean square value of f(t)
If we denote the mean square value of f(t) as )(2 tf .
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Then P = )(2 tf = −
→
2/
2/
2 )(1
T
T
dttfTT
Lt
To derive PSD :
Let us form a new fn. fT (t) by constructing f(t) outside the interval |t| > T/2.
i.e. fT (t) =
otherwise
Tttf
0
2/||)(
As long as T is finite fT (t) has finite energy. Let
FT (t) FT ()
Then energy ET of fT (t) is given by
ET =
−
−
= dfFdttf TT
22|)(|)(
But
− −
=
2/
2/
22)()(
T
T
T dttfdttf
Hence the average power P is given by
P = −
−→
=→
2/
2/
22 |)(|
)(1
T
T
T dfT
F
T
Ltdttf
TT
Lt
As T increases the energy of fT (t) also increases. Thus |FT ()|2 increases with
‘T’.
In the limit T
FT
2|)(| may approach a limit.
Assuming that such a limit exists we define.
Sf () is called the average power density spectrum or the power density
spectrum.
Hence average power P = −
→=
2/
2/
22 )(1
)(
T
T
dttfTT
Lttf
=
−
−
= dSdfS ff )(2
1)(
Note that |FT ()|2 = FT () FT (-)
Note that PSD fn. is an even fn. of .
So average power =
=0
2 )(2)( dfStf f
=
0
)(1
dS f
PSD of a signal retains only the information of its magnitude of FT (). The
phase information is lost.
That means all signals with identical PSD magnitude but different phase fns.
will have identical PSD’s.
Thus for a given signal there is a unique PDS but a PDS does not have unique
signal it may have infinite no. of signals.
For energy signals
If f(t) F()
f(t) cos ot )()(2
1oo FF −++
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and f(t) sin ot )()(2
oo FFJ
−−+
We can extend these results for power signals also.
Consider a power signal f(t) with PSD. Sf ()
Sf () = T
F
T
LtT
2|)(|
→
Consider the signal (t) given by
(t) = f(t) cos o (t)
S () = TT
LtT
2|)(|
→
FT (t) cos o (t) = T (t) T ()
T () = )()(2
1oToT FF −++
S () =
−++
→ T
FF
T
LtoToT
2|)()(|
4
1
=
−++
→ T
FF
T
LtoToT
22 |)(||)(|
4
1
FT ( + o) . FT ( - o) vanish because they are non-overlapping.
S () = )()(4
1ofof SS −++
S(f) = )(|)(|1
0
2
02fkffkX
T−
−
The definition shows that the PSD of the periodic signal is a discrete fn. of frequency
and it is defined at harmonics of fundamental frequency f0 i.e. why (f – kf0) appears in the
equation. The total area under the PSD curve gives the average power of the signal.
Hence
−=
−
−
−=k
dffkffkXT
dffS )(|)(|1
)( 0
2
02
Using shifting property of fn.
−=
−
−
−=k
dffkffkXT
dffS )(|)(|1
)( 0
2
02
= PfkXT k
=
−=
2
02|)(|
1
P =
−
dffS )(
Effects of system on PSD :
PSD of i/p and o/p are also related as that of ESD. Let PSD of o/p be Sy(f) and that of
input be Sx (f)
Sy (f) = |H(f)|2 Sx (f)
Here |H(f)| is the magnitude of the response fn.
Relation of PSD to auto correlation
R() (f)
Ex. : Find the average auto correlation fn. of the sine wave signal.
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x(t) = A sin (1t + ) Where 1 = T
2
We know that R() = −
−
2/
2/
*
0
0
0
)()(1
T
T
dttxtxT
for a periodic signal.
Since the signal is real valued and period T0 is not specified. We follow the definition
of auto correlation.
R() = −
+→
2/
2/
* )()(1
T
T
dttxtxTT
Lt
B A
= −
+++→
2/
2/
111 )(sin.)(sin1
T
T
dttAtATT
Lt
= −
++−→
2/
2/
111
2
)22[coscos2
1T
T
dttA
TT
Lt
R() = 1
2
1
2
cos22/
2/cos
2
1 At
T
TA
TT
Lt=
−→
Ex : Find the power of a signal x(t) = A + f(t) where A is a constant and the signal f(t) is a
power signal with zero mean value.
P = −
→
2/
2/
2|)(|1
T
T
dttxTT
Lt
= − − − −
++
→=+
→
2/
2/
2/
2/
2/
2/
2/
2/
222 )()(21
|)(|1
T
T
T
T
T
T
T
T
dttfdttAfdtATT
LtdttfA
TT
Lt
Since mean value of the signal is zero −
=→
2/
2/
0)(21
T
T
dttAfTT
Lt
Hence P = A2 + )(2 tf
)(2 tf is the mean square value of signal f(t).
Prove that :
R12 () = R21* (-)
R12 () =
−
− dttxtx )()(*
21
Put (t -) = n in the above equation.
Then R12 () =
−
+ dnnxnx )()(*
21 ------- (1)
R21 () =
−
− dttxtx )()(*
12
Let t = n
R21 () =
−
− dnnxnx )()(*
12
R21* () =
−
− dnnxnx )()( 1
*
2 -------- (2)
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R21* (-) =
−
+ dnnxnx )()( 1
*
2
Compare (1) and (2) both are equal
Hence R12 () = R21* (-)
Parseval’s power theorem :
The theorem defines the power of a signal in terms of its Fourier series coefficients
i.e. in terms of the amplitudes of the harmonic components present in the signal.
From f(t) → |f(t)|2 → f(t) f* (t)
The power of f(t) over one cycle is given by
P = − −
=
2/
2/
2/
2/
*2 )(.)(1
|)(|1
T
T
T
T
dttftfT
dttfT
Replacing f(t) by its exponential Fourier series.
P = −
−=
=
2/
2/
0
* 2)(
10
T
T n
tJn
nT
wheredteFtfT
With change in the order of integration and summation
P =
−= −n
T
T
tJn
n dtetfFT
2/
2/
* 0)(1
The integral is = TF*n
Hence P =
−=
−=
=n n
nnn FFF 2*||
This is Parseval’s power theorem.
Ex : Find the auto correlation of a gate fn. and determine the energy density spectrum of the
fn.
Shaded area is A2GT (T-
)
Shaded area AGT (T-)
That where is moved
in
either direction the
value
correlation is same.
To obtain 11 () we shift the gate fn. by -secs. And multiply the shifted fn.
A GT (t - ) with the original GT (t).
It is evident that the area under the product curve is given by
A2 [T - ||]
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As the gate fn. is an even fn. the auto
Correlation of the gate fn. is also an even fn.
The energy density spectrum of the gate fn. is given by (1/) times.
The F.T. of 11 () :
deTAde JJ −
−
−
−
−== |)|()()( 2
1111
The F.T. of a triangular pulse is given by
2
22
112
)(
=
SaTA
This ESD fn. S() = )(1
11
The energy density fn. S() =
2
22
2
1
SaTA
Directly from F.T. :
For a such fn. F() =
2
1 SaAT
i.e. S() = 2|)(|1
F
Hence S() =
2
22
2
1
SaTA
Correlation properties summarized :
−
= dttf |)(|)0(2
111
0)()0( 1111
)()( 1111 −=
)(|)(|)( 1
2
111 SF =
)(.)()( 2112 − FF
)()(*
2112 −
5.10. Differences between convolution & correlation :
1. In correlation physical time plays the roll of a dummy variable and it disappears after
the solution of the integral.
2. In convolution delay plays the roll of a dummy variable and disappear after the
integration.
3. Correlation is a fn. of delay parameter . Where as convolution is a fn. of time “t”.
- F.T. of Auto Correlation fn. is |F(|2 and hence it retains the information about the
magnitude of F1().
The phase information is absent.
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Hence a class of signals which have F.T. with the same magnitude fn. but different phase
fns. have the same auto correlation fn.
So for a given fn. there is a unique auto correlation fn. and for a given auto correlation fn.
there may be infinite no. of wave forms.
5.11. Detection of periodic signals in the presence of Noise : - Radar and sonar signals are full of noise. Here auto correlation techniques are used to
detect the signals.
- In meteorology one of the major problems is the detection of periodic components in
the presence of noise of random nature. We can successfully use cross and auto
correlation techniques to detect periodic signals in the presence of noise.
- The noise signal in practice is a signal with random amplitude variation such a signal
is un-correlated with any periodic signal.
If s(t) is a periodic signal and n(t) represents the noise signal.
−
=−→
2/
2/
0)()(1
T
T
allfordttntsTT
Lt .
If we denote )( sn as the cross correlation of s(t) & n(t) then 0)( = sn .
Hence it is not possible to detect a signal by cross correlation.
5.11.1. Detection by auto correlation :
Let s(t) be a periodic signal mixed with n(t) then the recd. Signal f(t) is [s(t) + n(t)].
Let )(),( ssff and )( nn be the auto correlation fns. of f(t), s(t) and n(t)
respectively.
Then −
−→
=
2/
2/
)()(1
)(
T
T
ff dttftfTT
Lt
= −
−+−+→
2/
2/
)]()([)]()([1
T
T
dttntstntsTT
Lt
= )()()()( nssnnnss +++
Since periodic signal s(t) and noise signal n(t) are uncorrelated
0)()( == tt nssn
So )()()(~
nnssff +=
Thus )( ff has two components )(~
&)( nnss .
The auto correlation fn. of a periodic signal is also periodic fn. of the same frequency
and auto correlation fn. of non-periodic fn. )( nn tends to zero for large value of .
Therefore for sufficiently large values of . )( ff is equal to )( ss .
So )( ff will exhibit a periodic nature at sufficiently large value of .
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If )( ff exhibits a periodic nature at sufficiently large values of it follows that f(t)
contains a periodic signal of frequency displayed by )( ff . Because )( ff exhibits a
periodic nature it is possible to separate )( ss (periodic component) and )(~
nn as shown
above.
In practice it is difficult to compute auto correlation fn. of f(t) over an infinite interval.
They are done over a finite interval of course with large values of . To that extent error is
introduced in the detected signal. If the signal s(t) is very weak then )( ss has a small
amplitude and under such conditions it is necessary to make very large in order to reduce
the error well below.
Auto correlation fn. of a given f(t) may be calculated by numerical techniques used
for correlation. Using these numerical techniques it is possible to evaluate the fn. on digital
computers. Automatic electronic correlaters are available in the market to carry out these fn.
automatically.
5.11.2. Extraction of a signal from Noise by filtering :
A signal masked by noise can be detected either by correlation techniques or by
filtering. Actually the two techniques are equivalent.
The correlation technique is a means of extraction of a given signal in the time
domain where as filtering achieves exactly the same results in the frequency domain.
Correlation in the time domain corresponds to filtering in the frequency domain.
Relation between correlation and filtering :
Consider cross correlation fn. 12 () of signals f1(t) & f2(t).
If f1t F1() and f2(t) F2 ()
12 () F1 () . F2 (-)
The fn. 12 () may be obtained in time domain by evaluating the integral by a cross
correlation.
Alternatively if the signal f1(t) is applied to a system with the transfer Fn. F2 (-) the
output will be12 ().
It is thus seen that the cross correlation between f1 (t) and f2 (t) may be effected by
applying f1(t) to the input terminals of a linear system whose transfer fn. is F2 (-). This
operation represents a filter.
f1(t) f2 (t) 12() f1(t) 12()
Cross correlation F2 (-)
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F1() F1() . F2 (-)
The impulse response h(t) of the system with transfer fn. F2 (-)
h(t) = F-1 [F2 (-)]
f2(t) F2 (), f2(t) F2 ()
f2(-t) F2 (-)
h(t) = f2 (-t)
i.e. the cross correlation of signals f1(t) and f2(t) is the response of a system with
transfer fn. F2 (-) where the driving fn. is f1 (t).
The result can be applied for detecting a periodic signal from a random noise signal.
Lt s(t) be periodic signal component and n(t) random noise component.
Then the recd signal is f(t) = s(t) + n(t)
to perform cross correlation we need a periodic signal c(t) of the same period as that
of s(t) i.e. we need a system with an impulse response c(-t) or which has a transfer fn. c(-).
c(t) c()
Where c(-t) c(-)
Since c(t) is a periodic signal of period T0, c() the F.T. of the c(t) will consist of
impulses at = o, o , 2o, ……… where o = 0
2
T
So c(-) also consists of impulses located at the same frequencies.
c(t) = on
o
tJn
nT
whereec o
=
−=
2
F.T. of c(t) → c() = −n
on nc )(2
Since c(-) = c* ()
c(-) = −n
on nc )(2*
Thus c(-) consists of impulses at = o, o , 2o, …… it amounts to filtering
of all frequency except fundamental frequency of the signal and its harmonics.
It is evident that the operation of cross correlation is equivalent to filtering all the
noise frequency components except the fundamental frequency of s(t) and its harmonics.
If the nature of the filter is such that the relative attenuation of all the frequency
components were uniform then the output will be an exact replica of s(t).
Detection by cross correlation Detection by auto correlation
In cross correlation method the cross
correlation between two fns. f(t) and c(t)
gives )(~
sc directly without any additional
noise terms. Here it is possible to conclude
whether the signal is periodic (is present or
absent) in f(t) over any value of ‘’.
This fn. of f(t) contains two terms
)(~
)(~
nnss and . To arrive at the proper
conclusion we have to observe whether
)(~
ff is periodic over a very large value of
. Where )(~
nn becomes negligible.
Here the knowledge of the frequency is
essential.
There is no such necessity.
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UNIT – 6
SAMPLING
Contents:
6.1. Sampling Theorem.
6.2. Siognal Recovery from its sampled version.
6.3. Sampling of Band pass Signals.
6.4. Types of Sampling.
6.5. Problems.
\
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SAMPLING
6.1. Sampling Theorem :
Sampling theorem is significant in communication systems because it provides the
basis for transmitting analog signals by use of digital techniques. The sampling theorem can
be stated in two different ways.
1. A band limited signal having no frequency components higher than fm Hz. is
completely described by its sample values at uniform intervals less than or equal to
mf2
1sec apart. This is frequency domain statement. Time interval between samples is
mf2
1sec → “Nyqluist period”.
2. A band limited signal having no frequency components beyond fm Hz. may be
completely recovered from its samples taken at a rate of at least 2fm samples/sec. 2fm
is “Nyqluist rate”.
This is called uniform sampling theorem as the samples are taken at uniform intervals.
Proof :
- A band limited has zero value of F.T. beyond the frequency fm Hz.
- No useful signal is band limited in the mathematical sense because its F.T. extends
from - to . But after a particular frequency the transform diminishes to such as
extent that it can be neglected and the F.T. provides a definite band width.
Such signals are considered band limited signals for all practical purposes.
The above statement can be proved by using convolution theorem.
Proof :
- Let us consider band limited signal f(t) having no frequency components beyond fm
Hz. i.e. F() = 0 for || > m where m = 2fm.
- Where the signal is multiplied by a periodic impulse train T (t) the product yields a
sequence of impulse located at uniform intervals of T secs.
- The strength of the resulting impulses is equal to the value of f(t) at the corresponding
instants.
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Figure shows :
f(t) F()
T (t) o wo ()
FS (t) = f(t) . T (t) T
1 [F() * o ()]
The F.T. of fS (t) can be obtained by using frequency convolution of
F() & o o ()
Thus F.T. of f(t) fS (t) = 2
1 [F() * o o ()]
Putting T
o 1
2=
We get fS (t) = T
1 [F() * o ()]
- Thus the spectrum of fS (t) can be obtained by convolving F() & o o ().
- Convolution can be obtained by folding (flipping) T (t) around the vertical axis.
Since T (t) is an even fn. of time folding yields the same T (t).
- The operation of convolution yields F() repeating itself every o radians/sec.
- F() repeating periodically forms the spectrum of fS (t) and is denoted by FS ().
- The spectrum FS () can be achieved analytically also the periodic fn. 0 0 () can
be written as the sum of impulses located at = 0, 0, 20, ….. n0.
Thus 0 () =
−=
−m
m )( 0 where m = 1, 2, 3, …..
Thus fS (t)
−
−=m
mFT
)()(1
0*
FS () =
−=
−m
mFT
)()(1
0*
By using sampling property of “” function
FS () =
−=
−m
mFT
)(1
0
The summation represents F() repeating every 0 radians / sec. as the one obtained
by graphical convolution.
From the fig. above we can see that F() repeats periodically w/o overlapping
provided
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0 2 m Or mfT
22
Or T mf2
1 sec.
Where T is the uniform sampling interval.
Sampling rate f0 = T
1 ; i.e. f0 = 2 fm samples / sec.
Thus the sampling theorem is proved.
6.1.1. Recovery of the original spectrum :
- We find that as long as the signal is sampled at interval ‘T’ or at a sampling rate “f0”
the spectrum will repeat w/o overlap.
- The spectrum extends upto and the ideal B.W. of the sampled signal is infinite.
- However the desired spectrum F() centered at = 0 can be recovered by passing the
sampled signal with spectrum FS () through an LPF with cut off frequency m. The
transfer fn. of the ideal LPF is rectangular i.e. a gate fn.
The sampled signal after filtering
yields all the frequency components
present in the desired signal. f(t)
having spectrum F().
if (a) 0 - m = m they touch each other
(b) 0 - m > m they do not touch.
(c) 0 - m < m they overlap.
6.1.2. Nyqluist interval :
- The maximum sampling interval is
T = mf2
1 secs. for complete recovery of the sampled signal.
- This is known as the “Nyqluist sampling interval”.
- Similarly minimum sampling rate f0 = 2fm is called “Nyqluist rate”.
When the band limited signal is sampled at Nyqluist sampling interval (i.e. Nyqluist
sampling rate) the spectrum FS () will contain non-overlapping F() repeating
periodically but each spectrum F() will be touching the neighbouring ones in the FS ().
6.1.3. Aliasing :
When a band limited signal is sampled at a rate lower than Nyqluist rate i.e. f0 < 2fm
(or T > mf2
1) then periodically repeating F() in the spectrum FS () overlap with the
neighbouring over as shown in the picture.
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- The signal is under sampled and is said to produce aliasing in the under sampled
process.
- Because of the overlapping due to aliasing phenomenon, it is no longer possible to
recover f(t) from fS (t) by LPF.
- Since the spectral components in the overlap regions, add up the signal is distorted.
- To combat the aliasing effect we may use prior to sampling an LPF called anti-
aliasing filter and then the filtered signal is sampled at a rate slightly higher than the
Nyqluist rate.
fS = T
1 number of samples / sec.
fS 2 fm ; fS = 2fm is the minimum rate of sampling is called “Nyqluist sampling
rate”.
To explain Aliasing let us take an examples :
x(t) = cos ot We want to sample this are reconstruct by passing the
sampled o/p through an LPF of B.W. “o”.
F.T. of cos ot [ ( + o) + ( - o)]
Case 1 – When s = 4o In this case s - o = 3o
LPF characteristics allows o to
pass through. Hence we get the
reconstructed o/p as x(t) =
cos ot.
Case 2 – When s = 3o
Signal at s - o = 1.5 o
by missing of the signals o &
s.
In this case we still get
x(t) = cosot at the o/p.
Case 3 – When s = 1.5 o We get in this case when s <
2o.
x(t) = cos ot + cos (s - o) t
Which is not same as the original
signal hence distorted.
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6.2. Signal recovery from its sampled version :
We have seen that the original signal f(t) can be recovered in frequency domain by
passing its sampled version through an LPF. With a cut off frequency of m (fm).
Let us see now that we can reconstruct the f(t) in time domain from its sampled
version.
For fn. f(t) sampled at Nyqluist rate 0 we have 0 = 2m.
And FS () =
−=
−n
onFT
)(1
because o = 2m FS () =
−=
−n
mmFT
)2(1
FS () F()
LPF
fS (t) f(t)
H() = T m
G 2 ()
Recovery of a fn. from its sampled version :
The LPF has a cut off frequency of m radians.
The transfer fn. of an LPF is a Gate fn.
Hence the base band spectrum can be recovered from FS () by multiplying FS ()
with m
G 2 ().
FS () . m
G 2 () = )(1
FT
F() = FS () . T )(2 mG
- The gate fn. )(2 mG is representing an LPF with a cut off frequency m. In other
words the action of the LPF is equivalent to multiplying the sampled signal FS ()
with a gate fn. T )(2 mG and the action yields the base band spectrum F().
The gate fn. has ht. T and width 2m.
- The fn. f(t) can be obtained by evaluating the time domain equation of (Inverse F.T.
of) F().
- The time domain equation of RHS can be evaluated by using time convolution
theorem.
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f(t) = fS (t) *
mT
Sa (mt) i.e. if the frequency
domain
= fS (t) * Sa (m t) as T = o
2 is a gate fn. its time
domain
Equation is a Sa fn.
The sampled fn. fS (t) can be considered as a
sum of impulses located at sampling instants (nT) 022
fff
mmm ==
=
having strength equal to fn at that instant. 10 ==
fTT m
Hence the sampled fn. fS (t) can be expressed as
fS (t) = −n
n nTtf )( where fn is the nth sample of f(t).
Substituting for fS (t) the fn. f(t) is given as [i.e. fS (t) = Sa (m t)
f(t) = tSanTtf m
n
n *)( −
Using sampling property of “” fn we have.
f(t) = −n
mn nTtSaf )(
Putting m T = we get ; 2fm T = mT
=
=
m
mm
f
f
f
f
2
2
2
2
0
because T = 0
1
f
We get f(t) = −n
mn ntSaf )(
- This equation represents that the fn. f(t) can be constructed by multiplying the sample
fn by a sampling fn. Sa(mt - n) and adding the multiplied values.
- Here Sa(mt - n) represents the sampling fn. at the sampling instants T = nT.
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6.3. Sampling of Band pass signals :
- We have discussed so far band limited signals centered around the origin i.e. low pass
signals.
- Let us consider band pass signals which is a more general case. Let the band be
represented by H upper cut off and L lower cut off frequencies.
The case we have considered so far was C = 0. i.e. the sampling of band limited
signals.
Case I : If either H or L is a harmonic of the sampling frequency S.
i.e. S = 2(H - L) = 2 2m = 4 m.
- Let us consider L is an integral multiple of (harmonic of) S i.e. (L = n S).
Then the spectrum of the corresponding band pass signal contains the entire pass band
and the original pass band signal can be recovered from its sampled version by passing the
signal through a band pass filter with sharp cut off having pass band L to H.
Case II :
If L or H not harmonic of S.
Then a more general sampling condition is given as follows.
m
mcS
)(2
+= ; where m is the largest integer not exceeding
m
mc
2
+
Actually case I is a special case of Case II.
Where the frequency band occupied by the signal is located between adjacent
multiples of 2 m i.e. m =
+
m
mc
f
ff
2 here the minimum sampling rate is twice the band
width of the signal i.e. 2 2m = 4 m. This is same as Case I.
- In the case of BP sampling x(t) can be recovered.
With out any error from its samples x(KTS) taken at regular intervals of TS if the sampling
rate fS is such that
fS = m
f
TS
221=
Where m is the largest integer not exceeding B
f2
Where f2 is the highest frequency contained in the Band Pass Signal and B is (f2 – f1).
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If must be noted that sampling frequencies higher than m
f22 may not always permit recovery
of x(t) without distortion (i.e. we may not be able to avoid aliasing) unless fS > 2f2. If x(t) is
ideally sampled using impulses the signal x(t) can be recovered from its samples by an ideal
BPF whose transfer fn. H(f) is given by
H(f) =
otherwise
ffffor
0
||1 21
In fact as stated above in the sampling theorem, the required sampling rate for a Band
Pass Signal depends on ‘m’ i.e. on “f2/B”.
fS = m
f22
m = largest integer of (f2/B).
6.4. Types of Sampling:
Flat Top Sampling :
- The circuitry needed for natural sampling is very complicated because the pulse shape
at the top has to be maintained. These complications can be reduced by Flat Top
Sampling.
- Here the pulses have a constant amplitude with in the pulse interval.
- The constant amplitude of the pulse can be chosen at any value of f(t) within the pulse
interval. Say the value at the beginning of the pulse is chosen in the diagram below.
- The Flat Top Sampled Signal fm (t) is shown in the figure (e) and it is the convolution
of the impulse sampled signal fS (t) of (a) and the non-periodic pulse P(t) of width
and of ht. 1 shown in figure (c).
- The spectrum of fS (t) and P(t) are shown in figs. (b) & (d).
- The spectrum of fm (t) is shown at (f) which is obtained by multiplying FS () and
P(). As the P() value is different at different frequencies the shape of Fm () is not
similar to FS () which shows that a distortion will be introduced if the signal is
recovered by an ideal LPF of cut off frequency m.
- To ward of this difficulty the signal is passed through a filter whose transfer fn. is
)(
1
P.
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- Thus when the Flat Top Sampled Signal is passed through an LPF its o/p is P().
F() and when this is passed through equalizer whose transfer fn. is )(
1
P we get the
original signal f(t) as F().
- The amplitude of the sampling pulse train is adjusted to 1.
- The duration of the pulses is .
- They are separated by TS.
- The F.S. of a aperiodic train of pulses is given by
V(t) =
=
+
1
2cos
2
n o
n
oo T
ntC
T
A
T
A
Where A = Amplitude of the pulse
= Duration of the pulse
To = Period of the pulse train and
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Cn is given by Cn =
o
o
T
n
T
n
2
2sin
Here
So V(t) = fC (t) ; A = 1 & To = Ts
Hence
fc (t) =
+
+
+ .....
42cos
221
S
n
SSS T
tCC
T
tC
TT
The constant Cn is given by
Cn = )/2(
/2sin
S
S
Tn
Tn
for n = 1, 2, 3, ……
The o/p of the multiplier is then
f(t) . fc(t) =
+
+
+ .......
22cos)(
2cos)(
2)( 21
ssss T
tCtf
T
tCtf
Ttf
T
Let us choose Ts = mf2
1
Where fm = maximum frequency comp. in f(t).
We then get
f(t) . fc(t) = .......)2(2cos)(2
)( 1 ++ tfCtfT
tfT
m
ss
If we don’t consider the multiplying factor of the first term in the RHS it is the f(t)
itself.
If f(t) . fc(t) is passed through an LPF, f(t) is obtained at the o/p, the cut off frequency
of the LPF = fm.
Ex : The signal x(t) = 2 cos 200 t + 6 cos 180 t is ideally sampled at a frequency of 150
samples / sec. The sampled version X0(t) is passed through an ideal LPF with a cut off
frequency of 110 Hz.
What frequency components will be present in the o/p of the LPF ?
Write down the expressions for its o/p signal.
x(t) = 2 cos 200 t + 6 cos 180 t
= 2 cos 2 (100) t + 6 cos 2 (90)t
Hence taking F.T. of both side
X(f) = [ (f + 100) + (f – 100)] + 3 [ (f + 90) + (f – 90)]
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Spectrum of sampled version of x(t)
fs =
−=
++−+++−n
ffffffffT
)()()()(1
02020101
f1 = 100 f1 = -100 f2 = 90 f2 = -90
250 = 100 + f0 -100 + f0 = 50 f2 + f0 = 240 f2 + f0 = 60 = (-90+150)
-50 = 100 – f0 -100 – f0 = -250 f2 – f0 = -60 f2 – f0 = -90 – 150
= - 240
The output of the filter will be
x(t) = 2 [cos 2 (50) t + cos 2(100) t]
+ 6 [cos 2(60) t + cos 2 (90)t]
6.5. Problems:
Ex. 1 :
The signal x(t) = 10 cos 150 t is ideally sampled at a frequency.
fs = 200 samples/sec. Sketch the spectrum of x (t)
x(t) = 10 cos 2 (75)t
X(f) = 5 [(f-75) + (f+75)]
Since the spectrum X (f) of x(t) is given by
X (f) =
−=
−n
snffXT
)(1
.5 sfT=
1
The sketch is as follows :
f = 75 f = -75
f + fs f + fs = 75 + 20 =
125
75+200=275 f-fs = -75–200 = -
275
f – fs = 75-200
= -125
Ex. 2 : For the above probability if sampling is done at 100 SPs. Find the spectrum of
sampled o/p.
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fs = 100 SPs
f = 75 f = -75
f – fs = 75-100 = -25 f + fs = -75+100 =
25
f + fs = 75+100 = 175 f–fs =-75–100 = -
175
T (t) =
−=n
tJn
koea
ax = T
dtetT
T
T
tJ o1
)(1
2/
2/
=−
−
= T
1
T (t) = tJn
n
oeT
−=
1
F.T. [T (t)] = dteeT
tJtJn
n
o −
−
−= .
1
= dteT
tnJ
n
o )(]1[
1 −−
−=
−
= )(2
o
n
nT
−
−=
Ex. 3 : How many minimum no. of samples are required to exactly describe the
following signal.
(a) (b)
x(t) = 10 cos (6 t) + 4 sin (8t)
= 10 cos 2 [3] t + 4 sin 2 [4] t
T1 of (a) = 3
1 & T2 of (b) =
4
1
3
4
1
4
3
1
2
1 ==T
T
T = 3T1 = 4T2 = 1 Period = 1 of the combined signal.
The highest frequency in the combined signal is 4 Hz.
Hence the required sampling frequency = 8 CPS.
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Ex. 4 : Determine the minimum sampling frequency to be used to sample the signal
x(t) =100 sin2c (100 t) if the signal x(t) is to be recovered w/o any distortion.
Solution x(t) = 100 sin2 c (100 t) = 10 sinc (+100 t). 10 sinc (100 t)
We know that 10 sin c (100 t) 0.1 [f/100]
F.T. [100 sin c2 100 t] = 0.1 [f/100] * 0.1 [f/100]
Convolution of two identical rectangular pulses results in a triangular pulse whose
base width in twice that of each rectangular pulse.
F [100 sinc2 100 t] = [0.1 (f/100)] * 0.1 [ (f/100)]
= 0.01 100 (f/200)
Where [f/200] denotes a triangular pulse as shown.
Thus the signal x(t) = 100 sin c2 100 t
is a low pass signal, band limited to 100 Hz.
Hence the Nyquist rate is 200 CPS.
Triangular spectrum of
the x[t] = 100 sinc2 100t.
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UNIT – 7
LAPLACE TRANSFORMS
Contents:
7.1. Introduction.
7.2. Laplace Transform.
7.3. Properies of Laplace Transform.
7.4. Problems.
7.5. Inverse Laplace Transforms.
7.6. Region of Convergence.
7.7. Properties of ROC.
7.8. Laplace Transforms of some useful Signals.
7.9. Laplace Transforms of periodic Functions using Wave Synthesis.
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LAPLACE TRANSFORMS
7.1. Introduction:
Fourier analysis is extremely useful in the study of problem of practical importance
involving signals and LTI systems.
This has been possible because
(i) Broad classes of signals can be represented as linear combination of periodic complex
exponentials and those complex exponentials are eigen fns. of LTI systems.
(ii) Laplace transforms while having all the advantages of Fourier transform they have
additional capabilities like analysis of stability of LTI systems.
7.2. The Laplace transform : When the impulse response of an LTI system is h(t) to a complex exponential input of
the form est.
Y(t) = H(s) est
Where H(s) =
−
− dteth st)( for [S = J] this equation is the Fourier transform of
h(t).
LAPLACE TRANSFORM OF A GENERAL SIGNAL IS
X(S) =
−
− dtetx st)(
The complex variable S → ( + J] = real part
= impart
General denotion :
)()( sXtx ⎯→⎯
When S = J
X(J) =
−
− dtetx tJ)(
Which corresponds to F.transform of x(t)
JS
SX=
|)( = F [x(t)]
When S is not purely imaginary also, Laplace transform bears a relationship to F.T.
Consider X(S) with S = ( + J)
X( + J) =
−
+− tJetx )()( dt
X( + J) = ( )
−
−− dteetx tJt )(
RHS in F.T. of ( )tetx −)(
Ex. 1 : Let a signal x(t) = e-at u(t)
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F.T. → X(J) =
−
− dtetx tJ)(
=
−
−− dtetue tJat )(
=
+−−− +
==0 0
)( 01
aJa
dtedtee tJatJat
L.T. is X(S) =
−
+−−− =0
)()( dtedtetue tastsat
with S = ( + J)
X ( + J) =
−+−
0
)( dtee tJta
This is F.T. of e-(+a)t u(t).
X( + J) = Ja ++ )(
1 for ( + a) > 0.
If S = + J and = Re[s] > -a
X(S) = as +
1 ; Re[s] > - a (for the series to converge).
That is e-at u(t) as
TL
+⎯⎯→
1. ; Re[s] > - a
7.3. Properties of L.T.
7.3.1. The linearity properties :
If x1(t) X1(S) & x2(t) X2(S)
Then [ax1 (t) + bx2 (t)] aX1 (S) + bX2 (S)
L.T. of [ax1 (t) + bx2 (t)] =
−
−
+0
21 ))()(( dtetbxtax st
=
−
−
−−
+=+0
212
0
1 )()()()( sbXsaXdtetxbdtetxa stst
Ex. : L[x(t)] = L [cos t] u(t) = L 2
1 [eJt + e -Jt] u(t)
Using linearity property we can write L cos (t) u(t) = )(2
1)(
2
1tueLtueL tJtJ −+
= 22)(
1
2
1
)(
1
2
1
+=
++
− S
S
JSJS Re [s] > 0
Similarly L [sin (t) u(t)] = 22
+S
7.3.2. Differentiation property (with respect to S) :
If f(t) F(S), then (-t)n f(t) n
n
ds
d F(S)
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We know that F(S) =
−
−0
)( dtetf st
Differentiate both sides with ‘S’ 2
11
SdS
dF−=
+
−−
−
−==0 0
)()()()(
dtetftdtetfdS
d
dS
SFd stst
32
2 1)2(1
SdS
Fd−−=
= L [(-t)1 F(t)]
43
3 1)3()2()1(
SdS
Fd−−−=
Similarly )()()( 2
2
2
tftLdS
SFd−= then
n
nn
dS
SFdtftL
)()()( =−
1
!)1(+
−=
n
n
n
n
S
n
dS
Fd
(-t)n f(t) )(.. SFdt
dn
nTL⎯⎯→ ---------- (2)
Let f(t) = 1 ; Then F(S) = 11 −= SS
From (2) F(S) = S
1
1)()(!
)1()(1
=−=−=+
tftLS
nSF
dS
d n
n
n
n
n
2
1)1(
SdS
dF−=
Thus L [tn] = 1
!+nS
n
)2(1
)1(3
2
2
2
−−=SdS
Fd
i.e. tn 0][Re!
1
.. ⎯⎯→+
SS
nn
TL 1
1)1(
+−=
n
n
n
n
SdS
Fd
7.3.3. Differentiation property with respect to ‘t’ :
If f(t) ⎯⎯→ ..TL F(S) then L[f1(t)] = SF (S) – f (0-)
Where f(t) = )]([ tfdt
d
Proof : L[f1 (t)] =
−
−−−
− − −
−−=0 0 0
1 )()()()(
vdu
stuv
stvst
du
dtetfsdtetfdtetf
L [f1 (t)] = - f(0-) + SF(S)
= SF (S) – f (0-)
Thus L[f1 (t)] = SF (S) – f(0-)
7.3.4. Complex shift property :
If f(t) )(.. SFTL⎯⎯→ then )()]([ .. aSFtfe TLat +⎯⎯→− where a is a complex constant.
Proof : e-at f(t) =
+−−−
− −
−+
==0 0
)( ]Re[)(
1aS
asdtedtee tasstat
= F(S+a) L Re[S] > -a
Thus e-at f(t) u(t) )(.. aSFTL +⎯⎯→ Re[S] > -a .
7.3.5. Integration theorem :
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If f(t) = −
t
dx )( then L[f(t)] = F(S) = −
+
0
)(1)(
dxSS
SX
Given f(t) = −
t
dx )( then f1 (t) = x(t)
L[f1(t)] = S F(S) – f(0-) = X(S) = SF(S) -
−
−
0
)( dx .
Then SF (S) - )()(
0
SXdx =
−
−
F(S) = −
+
0
)(1)(
dxSS
SX
7.3.6. Initial value theorem :
It states that if L.T. of f(t) = F(s) i.e. f(t) ⎯⎯→ ..TL F(S)
Then )()(0
SSFS
Lttf
t
Lt
→=
→ if f(t) is continuous at t = 0.
Proof : We know that L[f1(t)] = SF (S) – f(0-)
−−
−
−=0
1 )0()()( fSSFdtetf st
Consider −
−
→
0
1 )( dtetfS
Ltst
as f(t) is continuous at t = 0
f1(t) does not have an impulse at
t= 0
So 0)(0
1 =→
−
−
dtetfS
Ltst
i.e. as S→
−
−0
1 )( dtetf st
converges i.e. it becomes ‘0’.
)0()(0
)( −=→
=→
ftft
LtSSF
S
Lt
7.3.7. Final value theorem :
If f(t) ⎯⎯→ ..TL F(S) then )(0
)( SSFS
Lttf
t
Lt
→=
→
Proof : dtetfS
LtfSSF
S
Ltst−
−
−→
=−→
)(0
)0()(0
0
1
= )0()()(0
)( |00
1 −
−
−→
==→ −−
ftft
Lttfdte
S
Lttf st
i.e. )0()()0()(0
−− −→
=−→
ftft
LtfSSF
S
Lt
i.e. )()(0
tft
LtSSF
S
Lt
→=
→
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7.3.8. Time shift theorem :
i.e. f(t) F(S)
then f(t – t0) u (t – t0) ⎯⎯→ ..TL 0ste−
F(S) to > 0
L.T. [f(t – t0) u (t – t0)] =
−−
−
−=−−0
000
0
)()()(t
stst dtettfdtettuttf
Put (t – t0) = v ; t = v + t0 dt = dv
When t = t0 v = 0 & When t = ; v =
So
−−−+−
− −
==0 0
)()()()( 000 SFedvevfedvevf
stsvsttvS
Thus f(t – t0) u (t – t0) ⎯→LT 0ste−
F(S)
7.3.8. Convolution theorem :
If f1(t) ⎯→LT F1(S)
& f2(t) ⎯→LT F2(S)
Then F1(S) . F2(S) = LT
−
−
t
dtff0
21 )(.)(
Proof : F1(S) =
−
−0
1 )( def s
F1(S) F2(S) =
−
−0
21 )(.)( deSFf s
We know that 0ste−
F(S) = L.T. [f(t – t0) u (t – t0)]
=
−
− −
−−0 0
21 ])()()( ddtetutff st
= dtedtutff st−
− −
−−
0 0
21 )()()(
=
−
−0
)( dtetf st where f(t) =
−
−−0
21 )()()( dtutff
f(t) = −
−
t
dtff0
21 )()(
Selection of Limits :
f(t) =
−
−−0
21 )()()( dtutff
u(t - ) = 1 for (t - ) > 0 or t > or < t
Hence upper Lt. of the integral is ‘t’
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Thus f(t) =
−
−−0
21 )()()( dtutfTf
f(t) = −
−
t
dtff0
21 )()( Hence F1(S) . F2(S) = L.T. −
t
dtff0
21 )()(
7.4. Problems:
1. Find the L.T. of x(t) = e-3t cos (2 100 t) u(t)
We know that e-at f(t) F(s+a) Here a = 3
Here f(t) = cos 200 t F(S) = 22 )200( +S
S
F(S+a) = X(S) = 22 )200()3(
3
++
+
S
S Re (S) > -3
2. F(S) = 136
82 ++
+
SS
S find f(0) and f1(0) using the initial value theorem.
→S
LtSF(S) =
2
2 1361
/81
136
8
)136(
)8(
SS
S
SS
S
SS
SS
++
+=
++
+=
++
+ = 1
1)(0
)( =→
=→
tft
LtSSF
S
Lt initial value i.e. f(0) = 1
Now to find f1 (0)
From the initial value theorem
We may write )0()()0()(0
11 −−→
==→
fSSFSS
Ltftf
t
Lt
Since f(0) = 1, and SF(S) = 136
82
2
++
+
SS
SS
)0()()(0
1 fSSFtft
Lt Lt
S−=
→→ ; 1)(
136
1321
136
822
2
−=++
−=−
++
+SSF
SS
S
SS
SS
2
2
2
2 1361
132
136
132
136
)132(1)(
SS
S
SS
SS
SS
SSSSFS
++
−
=++
−=
++
−=−
21
2)0()()(
0
1 ==−=→
−
→ fSSFStft
LtLt
S
7.5. Inverse Laplace Transform:
(a) Inverse Laplace transforms :
All the poles are distinct :
F(S) = n
n
SS
K
SS
K
SS
K
−+
−+
−........
2
2
1
1
Kj = jSS
j SFSS=
− |)()(
(b) Poles are not distinct :
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Poles are not distinct means that some of the poles repeat. Let the ith pole have a
multiplicity of m. then if all the other poles are distinct.
F(S) =
)()()(....
)()()()( 2
2
2
1
1 21
n
n
j
i
m
i
i
i
i
i
i
SS
K
SS
K
SS
K
SS
K
SS
K
SS
K
SS
Km
−+
−+
−++
−+
−+
−+
−
i
mSS
m
ii SFSSK=
−= |)()(
To find )1( −miK multiply both sides by (S – Si)
m and differentiate before evaluating the
resultant S=Si.
F(S) (S – Si)m = .....)()(.....
)(
)(
)(
)( 2
1
1
2
2
1
1
11+−+−+
−
−+
−
− −− m
i
m
ii
m
i
m
i SSKSSKSS
SSK
SS
SSK
+ n
m
in
j
m
ij
iiiSS
SSK
SS
SSKKSSK
mm −
−+
−
−++−
−
)(.......
)(
)()(
)1(
Differentiate both sides w.r. to s :
)1(
|)()(−
=−=
m
i
i
SS
m
i KSFSSds
d
i.e.
−
=
=−
)()(!
1)(
SFSSds
d mir
r
rK
i
rm
SS
i
Ex. : Using partial fraction expansion find f(t).
(i) F(S) = 23
32 ++
+
SS
S (ii)
12
122 ++
−
SS
S
K1 = (S + 1) F(S) = 2)2(
)3(
)2()1(
3|
1
=+
+=
++
+
−=SS
S
SS
S
(i) F(S) = 21)2()1(
)3( 21
++
+=
++
+
S
K
S
K
SS
S
K1 = (S+1) F(S) 21
2
)2(
)3(|
1
==+
+=
−=S
S
S
K2 = (S+2) F(S) 112
32
)5(
)3(|
2
−=+−
+−=
+
+=
−=S
S
S
F(S) = )2(
1
1
2
+−
+ SS ; f(t) = [2 e-t – e-2t] u(t)
K2 = F(S) (S + 2) = 1)1(
)3(|
2
−=+
+
−=SS
S
F(S) = 2
1
1
2
+−
+ SS
f(t) = 2 e-t – e-2t for t 0.
(ii) F(S) = 22 )1(
12
12
12
+
−=
++
−
S
S
SS
S
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F(S) = 2
21
)1(1 ++
+ S
K
S
K
K2 = 312)()1( ||
11
2 −=−=+−=−= SS
SSFS
K1 = 22)12()()1( ||11
2 ==−=+−=−= SS
Sds
dSFS
ds
d
F(S) = tt eteSS
−− −=+
−+
32)1(
3
1
22
for t 0.
System transfer function:
)(3)(
5)(2411)(
2
2
txdt
tdxty
dt
dy
dt
tyd+=++
with f(0) = 0 ; f1 (0) = 0
L.T. [f1(t)] = SF (S) – f(0)
L.T. [f11(t)] = S[SF(S) – f1(0)] – f(0)
1. Find the transfer fn. of the system described by the differential eqn.
)(3)(
5)(2411)(
2
2
txdt
tdxty
dt
dy
dt
tyd+=++
with f(0) = 0 ; f1 (0) = 0 = 0 & L.T. f1(t) = SF (S) – f(0)
L.T. f11(t) = S[SF(S) – f1(0)] – f(0)
2. Table of C – T’s
3. Properties of L.T’s
4. Problem : F(S) = 136
82 ++
+
SS
S find f(0) & f1 (0) using initial value theorem.
(a) SF(S) = 11
1
1361
81
136
8
136
)8(|
2
2==
++
+
=
++
+=
++
+
=S
SS
S
SS
S
SS
SS
1)(0
)( =→
=→
tft
LtSSF
S
Lt initial values = f(0)
(b) To find f1 (0) :
f1(0) = )0()()(0
1 fSSFSS
Lttf
t
Lt−
→=
→
)0()( fSSFSS
Lt−
→
We know that f(0) = 1 ------------- (1)
And SF(S) = 136
)8(2 ++
+
SS
SS
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SF(S) – 1 = 136
132
136
13681
136
)8(22
22
2 ++
−=
++
−−−+=−
++
+
SS
S
SS
SSSS
SS
SS
S[SF(S) – 1] = 21/2136
1
132
136
)132(|
2
2
2==
++
−
=++
−
=s
SS
S
SS
SS
i.e. )0()()(0
1 fSSFSS
Lttf
t
Lt−
→=
→ = 2
i.e. f1 (0) = 2 -------------- (2)
Problem on multiple roots :
Problem : Let X(S) = 3
2
2
1
3 )2()2()2()2()3(
)12(
++
++
+=
++
+
S
K
S
K
S
K
SS
S
X1(S) = (S+2)3 X(S) = (2S+1)
K2 = (S+2)3 F(S) |2−=S
= 1
3
32
14
)3(
12|
2
−=
+−
+−=
+
+
−=SS
S -------- (3)
K1 = )3()12(3
12)()2(
!1
1||
22
3 ++=
+
+=+
−=−=
SSds
d
S
S
ds
dSFS
ds
d
SS
= -1 (2S+1) (S+3)-2 + 2 (S+3)-1
= |2
2 )3(
2
)3(
)12(
−=+
++
+−
SSS
S
K0 = |2
3
2
2
)()2(!2
1
−=
+S
SFSds
d
Let X(S) = 3
31
2
2111
3 )2()2()2()2(
12
++
++
+=
+
+
S
C
S
C
S
C
S
S
C31 = (S+2)3 F(S) |2−=S
= 31)2(2)2(
)12()2(|
2
3
3
−=+−=+
++
−=SS
SS
C21 = 2)12()()2( ||22
3 =+=+−=−= SS
Sds
dSFS
ds
d
C11 = 0)12()()2(!2
12
2
2
3
2
2
| =+=+−=
Sds
dSFS
ds
d
S
Thus X(S) = 32 )2(
3
)2(
2
+−
+ SS ; x(t) = 2t e-2t – 6t2 e-2t
= 2t e-2t (1 – 3t) u(t)
X(S) = 3)2()3(
)12(
++
+
SS
S
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= )3()2()2()2(
4
3
3
2
21
++
++
++
+ S
K
S
K
S
K
S
K
K4 = (S+3) X(S) |
3−=S
= 51
16
)2(
12|
3
3=
−
+−=
+
+
−=SS
S
K3 = (S+2)3 X(S) |2−=S
= |2
)3(
12
−=+
+
SS
S
= 332
14−=
+−
+−
K2 =
+
+=+
−=)3(
12)()2( |
2
3
S
S
ds
dSXS
ds
d
S
= 1
)14()1(2
)3(
)12()3(2|
2
2
+−−=
+
+−+
−=SS
SS = 5
2
3
)3(
5)()2(
+=+
SSXS
ds
d
K1 = ||2
3
2
3
2
2
)3(
52
2
1)()2(
!2
1
−=−=+
−=+
SSS
SXSds
d
= 101
10−=
−
X(S) = )3(
5
)2(
3
)2(
5
2
532 ++
+−
++
+
−
SSSS
= )3()2(
103 ++
−
SS
= [-5e-2t + 5t e-2t + 5e-3t – 6t2 e-2t] u(t)
Ex. 1 : x(t) = e-at u(t)
X(S) =
+
−−
−0
dtee stat
=
+−
0
)( dte tas
= as +
1 for Re [s] > -a
Ex. 2 : Let us consider a signal.
x(t) = -e-at u(-t)
X(S) =
−
−− −− dtetue stat )(
= −
+−
+=−
0
)( 1
asdte tas
For convergence in the example we need Re [s+a] < 0 or Re [s] < -a
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i.e. –e-at u(-t) ⎯⎯→ ..TL as +
1 Re[s] < - a.
From Ex. (1) & (2) we see that X(S) is same through x(t)1s are different.
i.e. In specifying the Laplace Transform of a signal both the algebraic expressions and
the range of values of s for which the expression is valid are required. [i.e. the Region of
convergence has to be specified].
ROC can be shown conveniently as follows :
S → a complex no.
Fig. of display is a complex plane. [S-plane] →
axisyalongpart
axisxalongpart
''Im
''Re
ROC for Ex. 1 ROC for Ex. 2
Ex. 3 :
x(t) = 3e-2t u(t) – 2e-t u(t)
X(S) =
−
−
−−−− − dttueedtetue sttstt )(2)(3 2
= 1
2
2
3
+−
+ SS
7.6. To Determine ROC : x(t) is a sum of two real exponentials.
And X(S) is the sum of the Laplace Transforms of each of the individual terms.
e-t u(t) ⎯→L 1
1
+S ; Re [S+1] > 0 → Re {S} > -1
e-2t u(t) 2
1
+⎯→
S
L ; Re {S+2] > 0 → Re {S} > -2
For combined convergence Re{S} > -1
3 e-2t u(t) – 2e-t u(t) 23
12 ++
−⎯→
SS
SL , Re{S} > - 1
ROC for Ex. 3
Ex. 4 :
x(t) = e-2t u(t) + e-t (cos 3t) u(t) cos 3t = 2
33 tJtJ ee −+
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Using Euler’s formula we get
x(t) = )(2
1
2
1 )31()31(2 tueee tJtJt
++ +−−−−
X(S) =
−
−
−
−
+−−−−−
++ dtetuetuedtetue sttJtJstt )(
2
1)(
2
1)( )31()31(2
e-2t u(t) ⎯→L 2
1
+S Re {S} > - 2
e-(1-3J)t u(t) ⎯→L )31(
1
JS −+ {Re (S)} > -1.
e-(1+3J)t u(t) ⎯→L )31(
1
JS ++, Re {S} > -1
For all the three components to converge Re {S} > -1 Or
e-2t u(t) + e-t (cos 3t) u(t) ⎯→L )2()102(
1222
2
+++
++
SSS
SSS Re [S] > -1
7.7. Properties of ROC:
ROC for L.T’S
We know that for one LT, there may be more than one signal when we take inverse L.T. This
difficulty can be obviated by specifying ROC. There are many constraints on ROC.
Property 1 : The ROC of X(S) consists of strips parallel to JW-axis in the S-plane.
ROC of X(S) consists of those values of S = + J for which the F.T. of x(t) e-t converges.
That is the ROC of LT of x(t) consists of those values of S for which x(t) e-t is absolutely
integrable.
−
− dtetx t|)(|
Property 1, thus follows, since this condition depends only on the real part of S.
Property 2 : For rational L.T. the ROC does not contain any poles.
At a pole, X(S) is infinite. Hence X(S) does not converge at a pole. Thus the ROC can
not contain values of S that are poles.
Property 3 : If x(t) is of finite duration and is absolutely integrable then ROC is entire S-
plane.
Signals are of finite duration which means outside certain range their values are zero
and any fn. which weighs them and is integrated over has finite value because the duration is
limited. Hence the total S-plane must be the ROC for such signals.
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f(t) Finite duration f(t) multiplied by decaying f(t) multiplied by growing
signal exponential exponential
Mathematically :
(i) 2
1
|)(|
T
T
dttx (ii) −2
1
|)(|
T
T
t dtetx (iii) 2
1
|)(|
T
T
tdtetx
S = ( + J) to be in the ROC x(t) e-t should be integrable.
(i) Verifies that S is the ROC for = 0
(ii) Verifies that for > 0, the maximum value of e-t over the interval on which x(t) is non
zero
is 1Te
−. So we can write
−− 2
1
2
1
1 |)(||)(|
T
T
T
T
Tt dttxedtetx . Thus RHS is bounded so the
LHS.
S-plane for Re {S} > 0 must be in ROC :
Similarly for < 0 −−
2
1
2
1
2 |)(||)(|
T
T
T
T
Tt dttxedtetx is integrable. Hence the ROC
includes entire S-plane.
Property 4 :
If x(t) is right sided and if the line Re[S] = 0 is in ROC, Then all values of S for
which Re[S] > 0 will also be in the ROC.
Rt. Sided signal means x(t) = 0 prior to some finite time
T1.
If for values of say 0 if the L.T. converges, then
−
−dtetx
t0|)(|
.
Since x(t) is right sided.
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−
1
0|)(|T
tdtetx
If > 0 say 1 then te 1− decays faster than
te 0− .
i.e. if 1 > 0 it must also be true that x(t) te 1− is absolutely integrable, since t
e 1− decays
faster than t
e 0− as t → i.e. dteetxdtetxtt
TT
t )( 010
11
1 .|)(||)(| −−−
−
=
−−−
1
0101 |)(|)(
T
tTdtetxe
Since T1 is finite the RHS of the inequality is finite hence x(t) te 1− is absolutely
integrable. Hence the result.
Property 5 :
If x(t) is left sided and if the line Re{S} = 0 is in the ROC, then all values of S for
which Re{S} < 0 will also be in the ROC.
Similar argument as for the Rt-sided.
Property 6 :
If x(t) is double sided and if the line Re{S} = 0 is in the ROC then the ROC will
consists of a strip in the S-plane that includes the line Re{S} = 0.
- A two sided signal is of infinite extent for
both t > 0 and t < 0.
- Divide the signal into two parts by taking as
arbitrary time T0 as shown in (b) & (c)
- The x(t)2 L.T. converges for which both xR(t)
and xL(t) converge.
- From property 4, the ROC of L[xR(t)] consists
of half-plane Re|S| > R for some value of R.
- From property 5 the ROC of L[xL(t)] consists
of a half plane for which Re(S) < L for some value of L.
Thus the ROC of L[x(t)] is the over lap of these two half planes.
ROC for L[xR(t)] ROC of x(t) is the overlap ROC for L[xL(t)]
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of ROC’s of L[xR(t)] & L[xL(t)]
The inverse Laplace Transform :
We know that X( + J) = F [x(t) e-t] =
−
−− dteetx tJt )(
For values of S = ( + J) in the ROC, we can invert this relationship using the inverse
F.T. as
x(t) e-t = F-1
−
+
=+ deJXJX tJ)(2
1)(
- Multiplying both sides by et
x(t) =
−
++
deJX tJ )()(2
1
- That is fixed and varying from - to , we can recover x(t) from its L.T. for a set
of values of S = + J in the ROC.
- We can thus recover x(t) by replacing J by S as is constant and ds = Jd ; S = +
J
ds = J
d
as is
constant
i.e. x(t) = +
−
J
J
st dseSXJ
)(2
1 ------ (1)
J
dd =
- The contour of integration in equation is the st.
line with S-plane corresponding to all points of
“S” satisfying [R(S)] = .
- This line is parallel to J-axis.
- Further more we can choose any line i.e. any
value of such that X( + J) converges.
- The equation of (1) involves complex integration.
But for the transform of rational nature the inverse
Laplace transform can be evaluated by partial fraction method.
Ex :
X(S) = )2()1(
1
++ SS Re [S] > -1
Partial fraction expansion.
X(S) = )2()1()2()1(
1
++
+=
++ S
B
S
A
SS
A = (S+1) X(S) |1−=S
= 21
1
+− = 1
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B = (S+2) X(S) |2−=S
= -1 = )12(
1
+−
Thus X(S) = )2(
1
)1(
1
+−
+ SS
There are two possibilities for the inverse transform.
Case – I : Transform of the from )(
1
aS +
ROC is towards Rt of Re[S] > -a
In this case we have to determine which ROC is to associate with each of the
individual 1st order terms.
Since in this case ROC for X(S) is Re(S) > -1.
Pole Zero plot for 1
1
+S i.e. pole at S = -1 Pole Zero plot of X(S) =
)2()1(
1
++ SS
and its ROC and its ROC
Pole zero plot for 2
1
+S i.e. pole at S = -2
and its ROC
Pole zero plot for 2
1
+S i.e. pole at -2 & its ROC.
Data shows that Re(S) > -1 hence it is RHS, ROC.
So )(1
1tue
S
t−+
Re (S) > -1
e-2t u(t) 2
1
+S Re (S) > -2
We thus obtain :
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[e-t – e-2t] u(t) )2()1(
1
++ SS Re (S) > -1.
Example 2 : Let us assume that the expression for
X(S) = 2
1
1
1
+−
+ SS the ROC is in the L.H. plane of Re(S) < -2.
With this new ROC
ROC for 1
1
+S ROC of
2
1
+S
for Re[S] < -1 for Re[S] < -2
With this new ROC
ROC is towards left of both the poles.
This must be true for each of the poles.
That is the ROC for the term corresponding to the pole S = -1
Re(S) < 1, while ROC for the term with pole at S = -2 in Re(S) < -2.
X(S) =
−
−− dtetu st)(e- t- Then e-t u(-t) 1
1
+S Re {S} < -1
−
−−
0
t-e- dte st e-2t u(-t) 2
1
+S Re {S} < -2
−
+−
0
t1)(s-e- dt Thus )2(
1
)1(
1
+−
+ SS
1)(S
1e
1)(S
1-
0
t1)(s-
+=
+−
−
+ So that [-e-t + e-2t] u (-t) )2()1(
1
++ SS Re (S) < -2
Suppose in the above problem the ROC is given as
ROC for X(S) = )2()1(
1
++ SS is -2 < Re[S] < -1
i.e. ROC is strip shows in the figure.
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So x(t) = e-t u(-t) + e-2t u(t) )2()1(
1
++ SS -2 < Re[S] < -1
Properties of Laplace Transform :
1. Linearity property :
x1(t) ⎯→L X1(S) with ROC R1
and
x2(t) ⎯→L X2(S) with ROC R2
Then
ax1(t) + bx2(t) ⎯→L aX1 (S) + bX2(S) with ROC R1 R2.
2. Time shifting :
x(t) ⎯→L X(S) with ROC = R,
Then
x(t – t0) ⎯→L 0ste−
X(S) with ROC = R
3. Shifting in the S-domain :
x(t) ⎯→L X(S) with ROC = R
Then ts
e 0 x(t) ⎯→L X(S – S0) with ROC R + Re {S0} ------ (1)
Here the ROC is associated with X(S – S0) is that of X(S), shifted by Re {S0}.
Thus for any value S that is in R, the value of S + Re [S0] will be in R1.
In the fig. if X(S) has a pole or zero at S = a, then X(S – S0) has a pole or zero at S – S0 = a -
….. i.e. S = a + S0.
i.e. as a special case if S0 = Jo.
Then equation (1) becomes tJ oe
x(t) ⎯→L X(S - Jo) with ROC = R.
RHS can be interpreted as a shift in S-plane parallel to J-axis.
i.e. the L.T. of x(t) has a pole or zero at S = a, then the L.T. of tJ oe
x(t) has a pole or
zero at S = a + Jo.
4. Time scaling :
x(t) ⎯→L X(S) with ROC = R
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then x(at) ⎯→L
a
S
a ||
1 with ROC R1 =
a
R.
i.e. for any value S in R, the value of S/a will be in R1.
ROC of X(S) ROC for
a
S
a ||
1 For a > 1 there is a
compression
for 0 > a > -1 in the size of the ROC of
Expansion in the size X(S) as shown in fig. (c)
for
of the ROC of X(S) by
a
S
a ||
1 by a factor
a
1
a factor a
1
- For 0 > a > -1, there is a expansion in the size of the ROC of X(S) by a factor ‘a’ as
shown in (b).
- For ‘a’ greater than 1, there is a compression in the size of the ROC of X(S) as shown
in fig. (c).
x(-t) ⎯⎯→ ..TL X(-S) with ROC = -R as shown in (b) there is reversal of ROC
in addition to a
scaling.
5. Conjugation property :
x(t) ⎯→L X(S) with ROC = R
Then x* (t) ⎯→L X* (S*) with ROC = R
Therefore X(S) = X* (S*) when x(t) is real.
6. Convolution property :
If x1(t) ⎯→L X1(S) with ROC = R1.
x2(t) X2(S) with ROC = R2
Then
x1(t) * x2(t) ⎯→L X1(S) X2(S) with ROC containing R1R2.
7. Differentiation in Time domain :
x(t) ⎯→L X(S) with ROC = R
dt
tdx )( ⎯→L S X(S) with ROC containing R.
x(t) = +
−
J
J
st dseSXj
)(2
1
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+
−
=
J
J
st dsesSXjdt
txd
.)(2
1)(
= +
−
J
J
st dseSSXj
)(2
1
)()(
SSXdt
txd
Ex. : L.T. of x(t) = t e-at u(t) L.T. [t e-at] =
+−
0
)( dtet tasv du
e-at u(t) ⎯→L aS +
1 Re {S} > -a =
dtas
e
as
et tastas
+−−
+− +−+−
0
)()(
0)()(
.
t e-at u(t) 2)(
1
)(
1
aSaSds
dL
+=
+−⎯→ Re {S} > -a =
+−
+−
2
)(
0)( as
e tas
i.e. 3
2
)(
1)(
2 aStue
t at
+− Re {S} > -a =
2)(
1
as +
And so on n
Latn
aStue
n
t
)(
1)(
)!1(
1
+⎯→
−
−−
Re {S} > -a
9. Integration in Time domain :
If x(t) ⎯→L X(S) with ROC = R
Then −
⎯→
t
L SXS
dx )(1
)( with ROC containing R {Re{S}>0}
10. Initial and Final value theorem :
- Under x(t) = 0 for t < 0 and x(t) contains no impulses we can calculate from the
Laplace transform the initial value of x(t).
X(0+) = )(SXSS
Lt
→
- Also if x(t) = 0 for t < 0 and x(t) has a finite limit as t →, then the final value
theorem states that
)(0
)( SXSS
Lttx
t
Lt
→=
→
Table of properties :
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Property Signal
x(t)
x1(t)
x2(t)
Laplace
Transform
X(S)
X1(S)
X2(S)
ROC
R
R1
R2
1 Linearity ax1(t) + bx2(t) ax1(S) + bX2(S) At least R1R2
2 Time shifting x(t – t0) 0ste−
X(S) R
3 Shifting in ‘S’
domain
tse 0 x(t) X(S – S0) Shifted version of R[i.e. S is
in the ROC if (S-S0) is in R]
4 Time scaling x(at)
a
SX
a ||
1
Scaled ROC (i.e. S is in the
ROC if S/a is in R)
5 Conjugation x* (t) X* (S*) R
6 Convolution x1(t) * x2(t) X1(S) . X2(S) At least R1R2
7 Differentiations
in time domain )(tx
dt
d
S X(S) At least R
8 Differentiation in
S-domain
-t x(t) )(SX
dS
d
R
9 Integration in the
time domain −
t
dx )( )(1
SXS
At least R {Re{S} > 0}
Initial & Final value theorem
10. If x(t) = 0 for t < 0 and x(t) contains no impulses at t = 0 then x(0+) = )(SXSS
Lt
→
If x(t) = 0 for t < 0 and x(t) has a finite limit as t → then )()( SXSS
Lttx
t
Lt
→=
→
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7.8. Laplace Transforms of some useful signals
Transform
pair
Signal Transform ROC
1 (t) 1 All S.
2 u(t) 1/S Re {S} > 0
3 - u (-t) 1/S Re {S} < 0
4 )()!1(
1
tun
t n
−
−
1/Sn Re {S} > 0
5 )()!1(
1
tun
t n
−−
− −
1/Sn Re {S} < 0
6 e-at u(t)
)(
1
aS +
Re {S} > -a
7 -e-at u(-t)
)(
1
aS +
Re {S} < -a
8 )()!1(
1
tun
et atn
−
−−
naS )(
1
+
Re {S} > -a
9 )()!1(
1
tuen
t atn
−−
− −−
naS )(
1
+
Re {S} < -a
10 (t – T) e-sT All S.
11 [cos ot] u(t)
22
oS
S
+
Re {S} > 0
12 [sin ot] u(t)
22
o
o
S
+
Re {S} > 0
13 [e-at cos ot] u(t)
22)( oaS
aS
++
+
Re {S} > -a
14 [e-at sin ot] u(t)
22)( o
o
aS
++
Re {S} > -a
15 un (t) = n
n
dt
td )(
Sn All S.
16 u-n (t) = timesn
tutu )(........)( ** 1/Sn Re {S} > 0
Laplace Transform of signals :
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x(t) = - eat u(-t) determine L.T. & ROC for the signal
X(S) =
−
− dtetx st)(
=
−
− − dttue at )(
u(-t) =
01
00
tfor
tfor
Hence the limits :
X(S) = −
−+−
0
dtee stat
= −
−−−
0
)( dte tas
=
0)(
)(−
−−
− as
e tas
)()(0
)()(
as
e
t
Lt
as
e
t
Lt tastas
−−→−
−→
−−−−
The second term will converge if power of exponent is –ve.
Now that t → - (-ve infinity) hence (s-a) should be –ve.
So we can write X(S) = )()(
)()(0)(
as
e
as
e asas
−−
−
−−−−−
= )(
0
)(
1
asas −−
− for (s-a) < 0
= )(
1
as − for (s-a) < 0 or s < a.
Laplace transform pair can be written as
-e-at u(-t) )(
1
as
L
−⎯→ , ROC S < a
Ex. 2 :
(i) x(t) = e-at u(t) (ii) x(t) = -e-at u(-t)
We know eat u(t) )(
1
as −, ROC S > a
Hence L.T. of e-at u(t) )(
1
as +, ROC S > -a
(ii) x(t) = -e-at u(-t)
We know that L.T. of
-eat u(-t) )(
1
as
L
+⎯→ , ROC S < - a
Ex. 3 :
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(i) x(t) = e-2t u(t) – e2t u(-t)
(ii) x(t) = 3e-2t u(t) – 2e-t u(t)
X(S) of x(t) = e-2t u(t) –e2t u(-t)
X(S) =
−
− dtetx ts)(
=
−
−− −− dtetuetue tstt )()( 22
=
−
−
−
−− −−+ dtetuedtetue tsttst )()( 22
)2(
1
+s for ROC S > -2
)2(
1
−s for ROC S < 2
ROC for X(S) = )2(
1
)2(
1
−+
+ ss, ROC S > -2 and S < 2
Re[S] = -2 < S < 2
ROC of L.T. of x(t) = e-2t u(t) – e2t u(-t)
(ii) x(t) = 3 e-2t u(t) – 2 e-t u(t)
L.T. → X(S) =
−
− dtetx st)(
=
−
−−
−
−− − dtetuedtetue sttstt )(2)(3 2
3 e-2t u(t) ⎯⎯→ ..TL )2(
3
+s, ROC S > -2
2 e-t u(t) ⎯⎯→ ..TL
)1(
2
+s, ROC = S > -1
X(S) = )1(
2
)2(
3
+−
+ ss ROC S > -2, and S > -1
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Re{S} > -2 &
Re {S} > -1
ROC of 3 e-2t u(t) - -2 e-t u(t)
Proof of time scaling :
Let x(t) ⎯⎯→ ..TL X(S) with ROC R.
Then x(at) ⎯⎯→ ..TL
a
SX
a ||
1 ROC
a
R
Meaning expansion in Time domain is compression in frequency domain and vice
versa.
Proof :
L[x(at)] =
−
− dteatx st)(
Let at = t = /a and dt = a
1 d. Limit will be
So L x(at) =
−
−
−
− = dexa
da
ex asas // )(11
)(
=
a
SX
a
1 ROC R/a
- Let us consider –ve value of a i.e.
L [x(-at)] =
−
−− dteatx st)(
Let -at = ; t = -/a ; dt = - a
1 d. Limits of integration will interchange
i.e. L [x(-at)] = −
+
−−
− d
aex as 1
)( )/(
= ( )
−
−−
−−=
a
RROC
aSX
adex
a
as 1)(
1 )/(
From this we can write
L [x(at)] =
a
SX
a ||
1, ROC =
a
R
Convolution of x(t) with u(t) and its transform :
- x(t) * u(t) =
−
− dtux )()(
- u(t - ) = −
whereelse
teitortfor
0
..0)(1 i.e. the limit is from - to
t.
- Hence convolution becomes
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x(t) * u(t) = −−
=
tt
dxdx )()(.1
i.e. −
t
dx )( = x(t) * u(t)
Taking Laplace transform both sides
)()()( * tutxLdxL
t
=
−
Using convolution property we can write RHS as
)(.)()( tuLtxLdxL
t
=
−
= X(S) . S
1 =
S
SX )(, ROC : R [Re(S) > 0}
2. Integration in the S-domain :
If x(t) ⎯→L X(S) : ROC : R
Then ⎯→
S
L dSSXt
tx)(
)( ROC : R
Meaning : Frequency domain integration corresponds to dividing the time domain signal by
t.
Proof : Consider RHS of above equation
−
−
=
S
st
S
dsdtetxdsSX )()(
Change the order of integration
dtt
etxdtdsetx
S s
stst
−
−
−−
−=
.)()(
−
−−
−
−
−−
−→=
−dt
t
e
t
e
S
Lttxdt
t
etx
stst
s
st
)(.)(
i.e.
−
−
−−
== dtet
txdt
t
etxdSSX st
st
S
)()()( i.e. R.T. of
t
tx )(
i.e. t
tx )(
S
dSSX )(
7.9. Laplace transform of a periodic fn. : If the L.T. of the first cycle of a periodic fn. is F1(S) then the L.T. of the periodic fn.
with period T is given by
F(S) = )(1
11 SF
e ST−−
Proof : Consider a sine wave with cycles
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A periodic sine wave can be obtained by adding individual cycles.
Here u(t – T) = 1 for t T Let f1(t) → F1(S)
(1) So f2(t) = 0 for t < T
= 1 for t T
(2) Similarly f3(t) = 0 for t < 2T
= 1 for t 2T and so on.
So f(t) = f1(t) + f1(t) [u(t-T)] + f1(t) . u(t-2T) + f1(t) . u(t-3T) + ……..
L[f(t)] = F1(S) + e-TS F1(S) + e-2TS F1 (S) + ……
F(S) = F1(S) . STe−−1
1
Determine the L.T. of :
(i) x(t) = A sin ot u(t)
(ii) x(t) = A cos ot u(t)
(i) x(t) = A sin ot u(t)
X(S) =
−
− dtetx st)(
=
−
A sin ot u(t) e-st dt
Sin ot = tJtJ oo eeJ
−−
2
1
X(S) =
−
−−
tJtJ oo eeJ
A
2 u(t) e-st dt
= )()(2
tueLtueLJ
A tJtJ oo −−
We know that
aS
tueL at
−=
1)( ROC Re[S] > a or > a
o
tJ
JStueL o
−=
1)( ROC : Re[S] > - Jo i.e. Re [S] > 0
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Here ROC : S > Jo means
ROC : ( + J) > 0 + Jo. But we consider only real part.
So ROC : > 0 i.e. Re [S] > 0
Hence X(S) =
+−
− )(
11
2 oo JSJSJ
A
ROC : Re(S) > 0
= 2222
2
2o
o
o
o
S
A
S
J
J
A
+=
+
i.e. A sin ot u(t) 22
o
oL
S
A
+⎯→ ROC : Re (S) > 0
Similarly A cos ot u(t) 22
o
L
S
AS
+⎯→ ROC : Re (S) > 0
Finding L.T. by wave form synthesis :
1. f(t) To find the L.T. of this pulse.
f1(t) [1–e-s] = 1 – 2e-s + e-2s
f2(t)
f(t) = f1(t) + f2(t) Thus the L.T. can be found.
2.
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Representation of triangular pulse
L.T. f(t) = L.T. [f1(t) + f2(t) + f3(t)]
f(t) = f1(t) + f2(t) + f3(t)
= At u(t) – 2A (t-1) u (t-1) + A (t-2) u(t-2)
L. [f(t)] = ( )SSSS
eeS
Ae
S
A
S
eA
S
A 2
2
2
22221
2 −−−−
+−=+−
L.T. of a ramp function :
Determine the L.T. of the ramp fn.
Def. of ramp fn. r(t) =
whereelse
tfort
0
0 Or
r(t) = t u(t)
L[r(t)] =
0
r(t) e-st dt
= dts
e
s
et stst
−−
−
−−
00
.1
= 0 - dts
e st
−
−
0
for S > 0
=
−
−0
1
s
e
S
st
= 0 -
− s
e
S
01 for S > 0
= 2
1
S
Find the L.T. of
f1 (t) = T
A2− (t – T/2) [ Since ramp is shifted to t = T/2]
f2 (t) = u(t) – u (t – T)
So f(t) = f1(t) . f2(t)
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= T
A2− (t – T/2) [u(t) – u(t–T)] [t – T/2 = t – T + T/2]
= T
A2− (t – T/2) u(t) +
T
A2 [t – T + T/2] u (t – T)
By rearranging the second term.
f(t) = ( ) )()(2
)(2/2
TtuATtT
AtuTt
T
A−
+−+−−
= )()()(2
)()(2
TtuATtuTtT
AtuAtut
T
A−+−−++−
L[f(t)] = F(S) = S
eA
S
e
T
A
S
A
ST
A STst −−
+++−22
21.
2
= ( ) ( )
−−+ −− STST e
Se
T
TS
A1
11
2
2
L.T. of
We know that L.T. For L.T. is
( ) ( )
−−+ −− STST e
Se
T
TS
A1
11
2
2
from the previous prob.
for a periodic fn. F(S) = STe
SF−−1
)(1
= ( ) ( )
−−+
−
−−
−
STST
STe
Se
T
TS
A
e1
11
2
2
1
1
=
−−
+−
−
Se
eT
TS
AST
ST 1
1
1
2
2
Ex. L.T. of a square pulse :
(a) A unit step fn.
starting at t = 0
(b) A unit step fn.
Starting at t = T
Thus the gate fn. is given by
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f(t) = u(t) – u(t–T)
(c) A gate fn. obtained by
Adding unit step at
(a)&(b)
L f(t) = L [u(t) – u(t–T)
= STeSS
−−11
= ( )STeS
−−11
Inverse L.T. using partial fractions expansions :
Case I : Simple and real roots.
F(S) = n
n
SS
K
SS
K
SS
K
SS
K
−+
−+
−+
−.......
2
2
1
1
0
0
K0 = (S – S0) F(S) |0SS =
; K1 = (S – S1) F(S) |1SS=
; K2 = (S-S2) F(S) |2SS =
and
so on
Ki = (S – Si) F(S) |iSS=
Ex. 1 : X(S) = )3()2(
222
−+
−+
SSS
SS, ROC Re {S} > 3
X(S) = )3()2(
210
−+
++
S
K
S
K
S
K
K0 = S X (S) |0=S
= 3
1
)3()20(
200=
−+
−+
K1 = (S + 2) X(S) |2−=S
= 5
1
10
244
)32()2(
)2(2)2( 22
−=−−
=−−−
−+− −
K2 = (S-3) X(S) |3=S
= 15
13
15
269
)23(3
)3(2)3( 22
=−+
=+
+ −
Thus X(S) = 3
15/13
2
5/13/1
−+
+−
SSS
x(t) = L-1
−+
+−
3
15/13
2
5/13/1
SSS
=
−+
+−
−−−
3
15/13
2
5/13/1 111
SL
SL
SL
−
SL
3/11 = 1/3 u(t)
(1) u(t) ⎯→L 0)(:1
→ SRROCS
e
(2) Parallelly e-2t u(t) )2(
1
+⎯→
S
L ROC, Re [S] > -2
(3) e+3t u(t) 3
1
−⎯→
S
L ROC : Re {S} > 3
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ROC for Re {S} > 0, Re {S} > -2, Re{S} > 3.
The given ROC of X(S) is Re {S} > 3
Thus x(t) = )(15
13)(
5
1)(
3
1 32 tuetuetu tt +− −
Ex. 2 : Given ROC : -2 < Re {S} < -1
X(S) = )2()1(
1
++ SS
= )2()1(
10
++
+ S
K
S
K
K0 = (S+1) X(S) |1−=S
= 1
K1 = (S+2) X(S) |2−=S
= -1
X(S) = )2(
1
)1(
1
+−
+ SS
x(t) = L-1 X(S) = L-1
+1
1
S - L-1
+ 2
1
S
Given, ROC of X(S) is given as -2 < Re {S} < -1
i.e. Re {S} > -2 and Re {S} < -1
- From this we observe that the region for Re {S} > -2 is right sided.
- Hence the time domain term corresponding to this ROC will also be right sided.
i.e. L-1
+ )2(
1
S = e-2t u(t) for ROC : Re {S} > -2
- Parallelly the region for Re {S} < -1 is left sided. Hence the time domain term will be
left sided.
Consider L-1 )1(
1
+S = - e-t u(-t) for ROC : Re{S} < -1
Thus x(t) = - e-t u(-t) – e-2t u(t)
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Ex. 3 : Find the inverse L.T. of F(S) = )1()2(
3
−+
−
SS
If the ROC is (i) -2 < Re (S) < +1
(ii) Re (S) > 1
(iii) Re (S) < -2
The given F(S) = )1()2(
3
−+
−
SS
= 12
10
−+
+ S
K
S
K
K0 = (S+2) F(S) |2−=S
= 121
3−=
+
−
F(S) = 1
1
2
1
−−
+ SS
(i) Inverse L.T. for -2 < Re (S) < 1
The given fn. F(S) has two poles at S = -2 and S = 1.
- Consider the pole at – 2. It lies on the left side of the ROC.
So e-2t u(t) )2(
1
+⎯→
S
L ROC Re(S) > -2.
- Consider the pole at S = 1. It lies on the Rt. of the ROC. Hence the time domain
signal is left sided i.e.
-e+t u(-t) 1
1
−⎯→
S
L , ROC : Re{S} < 1.
So f(t) = L-1 F(S) =
−−
+
−−
1
1
2
1 11
SL
SL
= e-2t u(t) + et u(-t)
(ii) To obtain L-1 for Re (S) > 1
ROC for Re {S} > 1
)1(
1
2
1
−−
+ SS
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We observe that the poles are towards left of the ROC and the ROC Rt. sided. Hence the time
domain signals will also be Rt. sided.
e-2t u(t) )2(
1
+⎯→
S
L ROC Re {S} > -2
and e+t u(t) 1
1
−⎯→
S
L ROC Re {S} > 1
f(t) = e-2t u(t) + et u(t) = e-2t [u(t)] + et u(t)
= [e-2t + et] u(t)
(iii) To obtain L-1 for Re (S) < -2
1
1
2
1
−−
+ SS
The ROC here is left sided. Hence time
domain signals also will be left
sided. i.e.
-e-2t u(-t) 2
1
+⎯→
S
L ROC Re {S} < -2
-et u(-t) 1
1
−⎯→
S
L ROC Re {S} < 1
Thus f(t) = -e-2t u (-t) – [-et u(-t)]
= [et – e-2t] [u(-t)]
Case II : Complex Roots :
If there are complex roots then F(S) can be written as
SDJSJS
SNSF
SD
SN
1)()(
)()(
)(
)(
+−−−==
= )(
)(
)()( 1
121
SD
SN
JS
K
JS
K+
+−+
−−
)(
)(
1
1
SD
SN is void of
complex roots.
Here K1 = F(S) (S - - J) |)( JS +=
& K2 = F(S) (S - + J) |)( JS −=
Ex. :
F(S) = )4()2(
122
2
++
++
SS
SS
F(S) = )2()2()2(
122
JSJSS
SS
−++
++
= )2()2(2
210
JS
K
JS
K
S
K
−+
++
+
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K0 = (S+2) F(S) |2−=S
= 8
1
]4)2[(
1)2(2)2(2
2
=+−
+−+−
K1 = (S + J2) F(S) |2JS −=
= 0
002
2
2
4531.11
13.535
88
43
)22()22(
1)2(2)2(
)2()2(
12|
=
+
+=
−−+−
++−
−+
++
−=J
J
JJJ
JJ
JSS
SS
JS
Similarly
K2 = (S-J2) F(S) |2JS=
= 0
02
2
2
4531.11
13.335
88
43
)22()22(
1)2(2)2(
)2()2(
12|
−
−=
−
−=
++
++
++
++
=J
J
JJJ
JJ
JSS
SS
JS
K1 = 0.442 8.130 = 0.437 + J 0.0625
K2 = 0.442 -8.13 = 0.437 – J 0.0625
So F(S) = 2
8/1
+S + 0.437
−+
+ )2(
1
)2(
1
JSJS + J 0.0625
−−
+ 2
1
2
1
JSJS
= 4
40625.0
4
2437.0
2
8/122 +
−+
++
+ S
JJ
S
S
S
f(t) = 22
1
22
11
2
2125.0
2874.0
2
1
8
1
++
++
+
−−−
SL
S
SL
SL
L-1 2
1
+S = e-2t u(t)
L-1 0.874 22 2+S
S = 0.874 cos 2t u(t)
L-1 0.125 22 2
2
+S = 0.125 sin 2t u(t)
f(t) = )(2sin125.02cos874.08
1 2 tutte t
++−
Case III : Multiple Roots :
F(S) = )()(
)(
10 SDSS
SNn−
F(S) = SD
SN
SS
K
SS
K
SS
K n
nn
1
1
0
1
1
0
1
0
0 )(
)(.........
)()(+
−+
−+
−
−
−
Here SD
SN
1
1 )( are devoid of multiple roots.
For this Kj is given by = |0
)(!
11
SS
jSF
ds
dj
J=
Where J = 0, 1, 2 ….
Ex. : F(S) = S
A
S
K
S
K
S
K
SS
S+
++
++
+=
+
−
)1()1()1()1(
2 2
2
1
3
0
3
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A = SF(S) |0=S
= 3)1(
2
+
−
S
S |
0=S
= 31
20− = -2.
F1(S) can be written as
F1 (S) = (S+1)3 F(S) = S
S 2−
K0 = 32
)()(!0
1|||
11
1
1
10
0
=−
==−=−=−= SSS
S
SSFSF
ds
d
K1 = 222
)(!1
1|||
1
2
11
1 ==
−=
−=−=−= SSSSS
S
ds
dSF
ds
d
K2 = 2)1(
4
2
14
2
1)(
!2
13
1
3
1
12
2
|| =
−−=
−=
−=−= SSS
SFds
d
So F(S) = SSSS
2
)1(
2
)1(
2
)1(
323
−+
++
++
Given ROC
From the table of Transform :
n
Latn
aStue
n
t
)(
1)(
)!1(
1
+⎯→−
−
−−
ROC Re[S] < - a
i.e. - e-t u(-t) 1
1
+⎯→
S
L
- t e-t u(-t) 2)1(
1
+S ROC : Re [S] < -1
- 2
2t e-t u(-t)
3)1(
1
+S
Similarly - u (-t) S
L 1⎯→ ; ROC Re [S] < 0
Here note that ROC of Re [S] < 0 includes ROC of Re [S] < -1.
Thus inverse L.T. of F(S) becomes
f(t) = L-1 F(S)
= utuetuettuet ttt 2)(2)(2)(2
3 2
+−−−−−− −−−
= )(222
32
2
tueetet ttt −
−−− −−−
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L.T. inverse by convolution integral :
If L f1(t) = F1(S) and L f2(t) = F2(S) then the convolution theorem states that
L[f1 (t) * f2(t)] = F1(S) . F2(S)
i.e. if a fn. can be expressed as a product of F1(S) and F2(S) then
L-1 F(S) = L-1 [F1(S) . F2(S)] = f1(t) * f2(t)
i.e. L-1 F(S) = −
t
dftf0
21 )(.)(
Ex. : Using convolution integral find out L-1 )2(
12 +SS
F(S) = )2(
12 +SS
Lt F1(S) = )2(
1
+S & F2(S) =
2
1
S
F(S) = F1(S) . F2(S) and f1(t) = L-1 F1(S) = L-1 )2(
1
+S = e-2t.
And f2(t) = L-1 F2(S) = L-1 2
1
S = t
The convolution of f1(t) and f2(t) gives the fn. f(t) i.e.
f(t) = f1(t) * f2(t) = −
t
dftf0
21 )(.)(
= −−− =
t t
tt deede0 0
22)(2 .
f(t) =
t
t eee
0
222
41
2.
−−
= ( ) ( )
−−−− 0222
4
10
2
1eetee ttt
= 44
1
2
2tet −
+−
This is the required inverse Laplace Transform. Here we have dropped u(t) after every
term including u(t) we can write the equation as
f(t) = )(4
1)(
4
1)(
2
2 tuetutut t−+−
Ex. 2 :
Determine the inverse L.T. of )(
1222 aSS −
using convolution theorem.
F(S) = )(
1222 aSS −
= F1(S) . F2(S)
Let F1(S) = 2
1
S and F2(S) =
)(
122 aS −
F2(S) =
+−
− )(
1
)(
1
2
1
aSaSa
f1(t) = L-1 F1(S) = L-1 2
1
S = t
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f2(t) = L-1 F2(S) = L-1
+−
− aSaSa
11
2
1 =
atat eeaaS
LaS
La
−−− −=
+−
− 2
1
)(
11
2
1 11
By convolution f(t) = f1(t) * f2(t)
= −
t
dtff0
21 )(.)(
= −−− −
t
tata deea
0
)()(
2
1.
= −
− −
t
aatt
aat
dea
ede
a
e
0022
Integrating by parts we get :
f(t) =
−−
−−
− −
−− t aaat
t aaat d
a
e
a
eed
a
e
a
ee
a00
112
1
=
−−
−
− −−−
taa
at
taa
at
a
e
a
ee
a
e
a
ee
a0
2
0
22
1
=
−+
−−
−−−
−− −
−−
2222
10
10
2
1
ae
a
e
a
ete
ae
a
e
a
ete
a
atatat
atatatat
at
=
−+−+−−
−
2222
11
2
1
a
e
aa
t
a
e
aa
t
a
atat
= ( )
−+− −atat ee
aa
t
a 2
12
2
1
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UNIT – 8
Z-TRANSFORMS
CONTENTS:
8.1. Introduction.
8.2. Z-Transform.
8.3. Properties od ROC.
8.4. Properties of Z-Transforms.
8.5. Inverse Z-Transforms.
8.6. Z-Transform Pairs.
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Z-TRANSFORMS
8.1. Introduction:
Discrete time signal :
- The discrete time signals are defined at
discrete instants of time and
are represented by x(n).
Where n is an index.
Amount deposited in bank month-wise
(A discrete time signal)
- In this case the independent variable is discrete.
- The other type of discrete time signal is obtained by sampling a continuous time
signal at regular intervals. This class of discrete signals are used in computers.
The main / fundamental difference between the continuous and discrete time signals
is that in the continuous time signal, the signal is present all the time where as in the discrete
time signals, the signal is present only at particular instants of time called discrete intervals of
time.
The discrete time signal is of two types.
(a) The signal for which the independent variable itself is discrete.
(b) The signal for which the independent variable is continuous but the values are
sampled only at discrete time instances.
- A discrete time signal is represented as x[n] where ‘n’ is a discrete time independent
variable and it takes only integral values.
Discrete time complex exponential & sinusoidal signals :
As in continuous time, an important signal in discrete time is the complex exponential
signal or sequence, defined by x[n] = C n.
Where C and are in general complex numbers.
- This could alternatively be expressed in the form x[n] = C en
Where = e.
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Real exponential signals :
- If C and are real, they behave as detailed below.
- If || > 1 the magnitude of the signal grows exponentially with ‘n’.
- If || < 1 we get a decaying exponential.
- Further if is +ve all the values of C n are of the same sign.
- But if is –ve then the sign of x[n] alternates.
- If =1, then x[n] is constant. Where as if = -1 then x[n] alternates between +C and
–C.
- Real valued discrete time exponentials are often used to describe population growth
as a fn. of generation and total return on investment as a fn. of day, month or a
quarter.
x[n] = C n
Sinusoidal signals :
- Another important complex exponential is obtained by using the form x[n] = C en
and by constraining to be purely imaginary (so that || = 1). Specifically let us
consider.
x[n] = nJ oe
---------- (1)
This signal is closely related to
x[n] = A cos (on + ) ---------- (2)
With o and with units of radians and n-dimensionless.
nJ oe
= cos on + J sin on ----------- (3)
A cos (on + ) = nJJnJJ oo ee
Aee
A −−+22
--------- (4)
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(1) & (2) are general examples of discrete time signals with infinite total energy but
finite power.
For example 1|| 2 =nJ oe
i.e. every sample of the signal contributes 1 & hence total energy is
.
General complex exponential signals :
- The general discrete time complex exponential can be written and interpreted in term
of real exponentials and sinusoidal signals.
- Specifically if we write C and in polar form.
x[n] = C n.
C = |C| eJ and = || oJe
; C n = |C| ||n eJ (+Jn)
Then x[n] = C n = |C| ||n cos (on + ) + J |C| ||n Sin (on + )
- Thus for || = 1, the real and imaginary parts of a complex exponential sequence are
sinusoidal.
- For || < 1 they correspond to sinusoidal sequences multiplied by a decaying
exponential.
- For || > 1 they correspond to sinusoidal sequences multiplied by a growing
exponential.
Examples of these signals are shown :
Growing discrete time sinusoidal signal Decaying discrete time sinusoidal
signal
Periodicity of discrete time using complex exponential signals :
There are a no. of differences between continuous time and discrete time signals as
there are many similarities.
- One of the differences in the discrete time exponential nJ oe
.
In the case of continuous time signal
tJ oe
has two main properties.
(1) The larger the magnitude of o the higher in the rate of oscillation of the signal.
(2) tJ oe
is periodic for any value of o.
Now let us describe the discrete versions of both these properties.
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- The first of these properties is different in discrete time as a consequence of an
important distinction between discrete time and continuous time complex
exponentials namely.
Consider the discrete time complex exponential with frequency o + 2.
nJnJnJnJ ooo eeee
== + 2)2(
- We see here that the exponential at frequency o + 2 is the same as that at
frequency o.
- Thus we have a very different situation from the continuous time case in which the
signals tJ oe
are all distinct for distinct values of o.
- In discrete time these signals are not distinct as the signal with frequency o is
identical to the signals with frequencies o 2, o 4 and so on.
- So to consider the discrete time complex exponential we need consider only a
frequency interval of length 2 in which to choose o.
So in most occasions we use 0 o 2 or - o< .
The signal nJ oe
does not have a continually increasing rate of oscillation as o is
increased in magnitude.
- (Slide). As we increase o from 0, we obtain signals that oscillate more and more
rapidly until we reach o = .
- As we continue to increase o, we decrease rate of oscillation until we reach o = 2,
which produces the same constant sequence as at o = 0.
- So the low frequency discrete – time exponentials have values of o near 0 or 2.
And any other even multiple of .
While the high frequencies (corresponding to rapid variation) are located near o =
and other odd multiples of .
In particular of o = or any other odd multiple of .
nnJJn ee )1()( −==
So the signal changes sign at each point in time thus oscillating more rapidly.
- The second property is about the periodicity of discrete time complex
exponential :
For nJ oe
to be periodic with period N > 0
We must have nJNnJ oo ee
=
+ )(
Or equivalently NJ oe
= 1
oN must be a multiple of 2. i.e. there must be an integer m such that
oN = 2m
Or equivalently
N
mo =2
- i.e. the signal nJ oe
is periodic if o / 2 is a rational number otherwise it is not
periodic.
- Same observations hold for discrete time sinusoids.
- From above we can also determine the fundamental period and frequency of discrete-
time complex exponentials.
- Here we define the fundamental frequency of a discrete time periodic signal as we did
in continuous time.
That is if x[n] is periodic with fundamental period N.
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- Its fundamental frequency is 2/N.
- Consider then a periodic complex exponential
x[n] = nJ oe
with o 0.
For periodicity, o must satisfy the condition N
mo =2
------- (1)
for some pair of integers m & N. With N > 0.
- The fundamental frequency of the periodic signal nJ oe
is
mN
o=2
------- (2)
The fundamental period can also be written as N = m
o
2
Eq. (1) & (2) are different from their continuous time counter parts.
Comparison of signals tJ oe
and
nJ oe
. tJ oe
nJ oe
Distinct signals for distinct values of o. Identical signals for values separated by
multiples of 2.
Periodic for any choice of o. Periodic only if o = 2 m/N for some
integers N>0 and m.
Fundamental frequency o. fundamental frequency o/m.
Fundamental period
o = 0 undefined
o 0, o
2
Fundamental period.
o = 0, undefined
o 0,
o
m
2
8.2. Z-Transforms :
The Z-transform of a sequence x(n) is defined as
- X(Z) = n
n
Znx −
−=
)( -------- (1)
Where Z is a complex variable, in polar form Z can be expressed as
Z = r eJ -------- (2)
Where r is radius of a circle.
- If the sequence x(n) exists for n in the range - to then eq. (1) represents two sided
or a bilateral Z-transform.
- On the other had if the sequence exists only for n 0 then eq. (1) changes to
X+ (Z) = n
n
Znx −
=
)(0
--------- (3)
Which is called one sided Z-transform.
Substituting (2) in (1) we have
X (r eJ) = nJ
n
renx −
−=
)()( --------- (4)
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= nJn
n
ernx −−
−=
)( ---------- (5)
- The above equation represents F.T. of a signal x(n) r-n. Hence inverse DTFT of X(r
eJ) must be x(n) r-n.
So we can write x(n) r-n =
−
dererX nJJ )()(2
1 Z = r eJ
Substituting Z = reJ and d = JZ
dZ dZ = r . eJ Jd
x(n) = −
C
n dZZZXj
1)(2
1 dZ = JZ d
d = JZ
dZ
Where C
denotes integration around the circle of radius |Z| = r in a counter-clock
wise direction.
8.2.1. Definition :
The Z-transform of an arbitrary signal x(n) is
X(Z) = n
n
Znx −
−=
)( ------- (7)
And inverse transform of X(Z) is
x(n) = −
C
n dZZZXj
1)(2
1 -------- (8)
The Z-transform X(Z) is a ratio of polynomials on Z-1 given by
X(Z) = N
N
M
M
ZaZaZa
ZbZbZbb−−−
−−−
+++
+++
........1
........2
2
1
1
2
2
1
10
The roots of Nr. are the values for which X(Z) = 0 and are called zeros of X(Z) and
the roots of Dr are values for which X(Z) becomes and are called the poles of X(Z).
- The Z-transform exists when the “” converges.
- The sum may not converge for all values of Z.
- The values of Z for which the sum of eq. (7) converges is called the region of
converge (ROC).
8.2.2. Examples:
Ex. 1 : Find the Z-transform and ROC for the signal x(n) = an u(n).
X(Z) = n
n
Znx −
−=
)(
= nn
n
Znua −
−=
)( u(n) =
01
00
nfor
nfor
= nn
n
Za −
=
0
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= ( )nn
aZ 1
0
−
=
---------- (9)
This is a geometric series of finite length.
Eq. ‘9’ converges when (aZ-1) < 1
Or a < Z or Z > a
ROC of example 1
We know that
a + ar + ar-1 + ……… = r
a
−1 if |r| < 1 ------- (10)
The summation at eq. (9) converges if |aZ-1| or |Z| > |a|
By using (10) we can write
X(Z) = 11
1−− aZ
; ROC |Z| > |a|
Which implies that the ROC is exterior to the circle of radius in the fig.
i.e. if ( )n
n
aZ
=
−
0
1 has to converge |(aZ-1)| < 1 or 1Z
a or |Z| > |a|
So the area of ROC is exterior to the circle a as shown in the fit.
Ex. 2 : Determine the Z-transform of the signal –bn u (-n-1). Find ROC
X(Z) = n
n
Znx −
−=
)(
= n
n
n Znub −
−=
−−− )1( u(-n-1) = 1 for (-n-1) > 0
= n
n
n Zb −−
−=
−1
- n > 1
Put n = -1 = ( )
=
−−1
1
n
nZb n -1
The limits are the highest limit is
-1.
The above series is an infinite series and converges if (b-1Z) < 1 Or |Z| < |b|. Hence
X(Z) = 11
1
1
1
1 −−
−
−=
−=
−−
bZbZ
Z
Zb
Zb ; |Z| < |b|
That is ROC is interior to the circle of Radius B.
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ROC of example 2
Ex. 3 : Find Z-transform of x(n) = an u(n) – bn u(-n-1) & its ROC.
The given signal is a two sided infinite duration series having values of n from - to
.
So X(S) =
−=
−
n
nZnx )( u(-n-1) = 1 for (-n-1) > 0
=
−=
−−−−n
nnn Znubnua )1()( -n-1 > 0
=
=
−
−
−− −0
1
n
nnnn ZbZa n + 1 < 0
=
=
−− −
0 1
11 )()(n
nn ZbaZ n < -1
- The first series converges if |aZ-1| < 1 or |a| < |Z| or |Z| > |a|
- The second series converges if |b-1Z| < 1 or |Z| < |b|
- The two ROC’s do not over lap if |b| < |a| and hence X(Z) does not exist.
- If |b| > |a| the two ROC’s over lap and X(Z) exists.
Therefore the ROC for X(Z) is |a| < |Z| < |b|. That is for an infinite duration two sided
signal the ROC is a ring in the Z plane.
X(Z) = bZ
Z
aZ
Z
−+
− ROC |a| < |Z| < |b|
|Z| < |b|
ROC of a two sided sequence for |b| > |a| ROC’s of both over lap.
|b| < |a|
No over lap
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8.3. Properties of Region of convergence ROC :
1. The ROC is a concentric ring or a circular disc in the Z-plane centered at the origin.
2. The ROC can not contain any poles.
3. If x(n) is a causal sequence then the ROC is the entire Z-plane except at Z = 0.
4. If x(n) is an anti-causal sequence then the ROC is the entire Z-plane except Z = .
5. If x(n) is a finite duration two sided sequence then the ROC is entire Z-plane except at Z
= 0 and Z =.
6. If x(n) is an infinite duration two sided sequence then the ROC will consist of a circular
ring in the Z-plane, bounded on the interior and exterior by a ring not containing any
poles.
7. The ROC of an LTI stable system contains the unit circle.
8. The ROC must be a connected region.
Ex. Find the Z-transform and ROC of the causal sequence.
x(n) = (2, -1, 3, 2, 01, )
X(Z) = .x (-1) . Z + x(0) + x(1) . Z-1 + x(2) . Z-2 + x(3) . Z-3 + x(4) . Z-4 + x(5) . Z-5
+…….
The given sequence values are x(0) = 2, x(1) = -1, x(2) = 3, x(3) = 2, x(4) = 0 & x(5)
= 1
Substituting these values :
X(Z) = 2 – Z-1 + 3Z-2 + 2Z-3 + 1.Z-5
The X(Z) converges for all values of Z except at Z = 0.
Left Hand Sequence :
Find the Z-transform and ROC of the anti causal sequence.
x(n) = (3, 2, -1, -4, 1)
X(Z) =
=
−
n
nZnx )( x(0) = 1 x(-1) = -4
x(-2) = -1 x(-3) = 2
x(-4) = 3
= + x(-4) Z4 + x(-3) Z3 + x(-2) Z2 + x(-1) Z + x(0) Z0 + …..
X(Z) = 3 Z4 + 2Z3 – Z2 – 4Z + 1
The X(Z) converges for all values of Z, except at Z = .
Two sided sequence :
x(n) = [1, 2, 0, -4, 3, 2, 1, 6, 5]
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x(0) = 3 ; x(1) = 2, x(2) = 1, x(3) = 6, x(4) = 5
x(-1) = -4, x(-2) = 0, x(-3) = 2, x(-4) = 1
X(Z) =
−=
−
n
nZnx )( we get
X(Z) = Z4 + 2Z3 – 4Z + 3 + 2Z-1 + Z-2 + 6Z-3 + 5Z-4
The X(Z) converges for all values except at Z = 0 and Z = .
Z-transform pairs :
Sequence Z [SLIDE] Slide of Page No. 10.13, Table-1
8.4. Properties of Z-transform : 8.4.1. Linearity Property:
If X1 (Z) = Z{x1 (n)} and X2 (Z) = Z [x2(n)}
Then Z[ax1 (n) + b x2 (n)} = aX1 (Z) + b X2 (Z)
Proof :
Z[ax1 (n) + bx2(n)] =
−=
−+n
nZnxbnxa )]()([ 21
=
−=
−=
−− +=+n n
nn ZXbZXaZnxbZnxa )()()()( 2121
The ROC of the sum sequence transform is the intersection of the individual
transforms.
R1 R2
8.4.2. Time Shifting :
If X(Z) = Z{x(n)} and the initial condition for x(n) are “0” then
Z[x(n-m)] = Z-m X(Z)
Proof :
Z{x (n – m)} =
−=
−−n
nZmnx )(
Let (n –m) = P then n = P + m, Now
Z {x (n – m)} =
−=
+−
n
pmZPx )()(
= Z-m
−=
−− =n
mP ZXZZPx )()(
Z {x (n –m)} = Z-m X(Z)
If m > 0 the ROC of Z-m X(Z) is the same as that of X(Z) except Z = 0.
8.4.3. Multiplication by an exponential sequence :
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If X(Z) = Z{x(n)} then Z[an x(n)] = X (a-1Z).
Proof :
Z[an x(n)] =
−=
−=
−−− =n n
nnn ZanxZnxa )()()( 1
= X (a-1 Z)
The ROC is R1 < |(a-1Z)| < R2 Or |a| R1 < |Z| < |a| R2
8.4.4. Time reversal :
If X(Z) = Z{x(n)} then Z{x(-n)} = X (Z-1)
Proof :
Z{x(-n)} =
−=
−−n
nZnx )(
Let -n = l then
Z {x(-n)} =
−=
−−− =n
l ZXZlx )()()( 11
The ROC is |R1| < |Z-1| < R2
Or 12
1||
1
RZ
R
8.4.5. Multiplication by n :
If Z{x(n)} = X(Z) then Z{n x(n)} = -Z dZ
d X(Z)
Proof : We know X(Z) =
−=
−
n
nZnx )(
Z{n x(n)} =
−=
−=
−−− =n n
nn ZnxnZZnxn 1)(.)(
=
−=
+−
n
nZnnxZ ][)( )1(
= + Z ( )
−=
−=
−−
−=
−
n n
nn ZdZ
dnxZZ
dZ
dnx )()(
= - Z )()( ZXdZ
dZZnx
dZ
d
n
n −=
−=
−
In the same way we can define
Z{nk x(n)} = )(ZXdZ
dZ
K
−
8.4.6. Convolution :
- A very important property in the analysis of discrete time system is convolution
property.
- According to this property the Z-transform of convolution of two signals is equal to
the multiplication of their Z-transform.
Z{x(n) * h(n)} = X(Z) . H(Z)
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Where x(n) * h(n) denotes the linear convolution of the sequences.
Proof : Let y(n) = x(n) * h(n) then we have
y(n) =
−=
−k
knhkx )()(
Taking Z-transform of both sides ; we have
Y(Z) =
−=
−
−=
−
n
n
k
Zknhkx )()(
On interchanging the order of summation, we have
Y(Z) =
−=
−−
−=
− −n
kn
n
k ZknhZkx )()()(
On replacing (n-k) by l, we have
Y(Z) =
−=
−
−=
−
n
l
l
k ZlhZkx )()(
= X(Z) H(Z)
8.4.7. Time expansion :
The signal xK (n) =
kofmultipleanotisnif
kofmultipleaisnifknx
0
)/(
Then Z[xk (n)] = X(ZK)
has (k-1) zeros inserted between successive values of the original signal.
If Z{x(n)} = X(Z)
Then Z{xk(n)} = X(ZK)
Proof : Z[xk (n)] =
−=
−
n
nZknx )/(
Let n/k = l then
Z{xk (n)} =
−=
−=
−− ==l l
klklk ZXZlxZlx )()()()(
8.4.8. Conjugation :
If Z {x(n)} = X(Z)
Then Z{x*(n)} = X* (Z*)
Proof : Z {x*(n)} =
−=
−
n
nZnx )(*
=
*
*)()(
−=
−
n
nZnx
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= )()( **** ZXZX =
8.5. Inverse Z-transform : The process of finding x(n) from its Z-transform X(Z) is called the inverse Z-
transform and is denoted as follows.
x(n) = Z-1 {X(Z)}
There are 4 methods often used to find the inverse Z-transform.
1. Long division method.
2. Partial fraction method.
3. Residue method
4. Convolution method
8.5.1. Long division method :
X(Z) =
−=
−
n
nZnx )( ------- (1)
Eq. (1) has both +ve and –ve powers of Z.
- If the sequence is x(n) is causal then
X(Z) =
=
−
0
)(n
nZnx have only –ve powers of Z.
- If the sequence is non-causal
Then X(Z) has +ve powers of Z. with ROC |Z| < r
- If the X(Z) is ratio of two polynomials
X(Z) = )(
)(
ZD
ZN =
N
N
M
M
ZaZa
ZbZbb−−
−−
++
++
......1
.....1
1
1
10
We can generate a series in Z by dividing the Nr. by the Dr.
- If X(Z) converges for |Z| > r then we can obtain series as
X(Z) = x(0) + x(1) Z-1 + x(2) Z-2 + ……..
- If X(Z) converges for |Z| < r then we can get a sums
X(Z) = x(0) + x(-1)Z + x(-2) Z2 + ……
Which is non-causal.
Ex. 1 :
If x[n] is causal X(Z) = 21
1
21
21−−
−
+−
+
ZZ
Z
1-2Z-1+Z-2 1 + 2Z-1 1+4Z-1 + 7Z-2 + 10Z-3 + 13Z-4 + 16Z-5 and so on 16Z-5-13Z-
6
1–2Z-1+Z2
4Z-1-Z-2
4Z1-8Z-2+4Z-3 Thus X(Z) = 1+4Z-1+7Z-2+10Z-3+13Z-4+16Z-5+and
so on
7Z-2-4Z-3 So [x(n) = 1, 4, 7, 10, 13, 16, …..]
7Z-2-14Z-3+7Z-4
10Z-3-7Z-4
10Z-3-20Z-4+10Z-5
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13Z-4 – 10Z-5 13Z-4-26Z-5+13Z-6
Other method : If x[n] is non-causal
Z-2 -2Z-1 + 1 2Z-1 + 1 2Z+1 + 5Z2 + 8Z3 + 11Z4 + 14Z5
2Z-1 – 4 + 2Z
5-2Z
5-10Z+15Z2 Since x(n) is non-causal
8Z-5Z2 We begin the division
with
8Z-16Z2+8Z3 higher –ve power of Z.
11Z2-8Z3
11Z2-22Z3+11Z4
14Z3-11Z4
14Z3-28Z4+14Z5
17Z4-14Z5
X(Z) = 2Z + 5Z2 + 8Z3 + 11Z4 + 14Z5 + ……….
x(n) = 14, 11, 8, 5, 2, 0
x(-5) x(-4) x(-3) x(-2) x(-1) x(0)
8.5.2. Partial fraction expansion method :
X(Z) = N
N
M
M
ZaZa
ZbZbb
ZD
ZN−−
−−
++
++=
......1
.....
)(
)(1
1
1
10
- If M is < N and aN 0 then the rational function.
X(Z), is said to be proper.
- On the other hand if M N then it is an improper fraction.
i.e. X(Z) =
.Pr
1
1)(1
10)(
)(.....
)(
)(
fnrationaloper
Polynomial
NM
NMZD
ZNZbZcc
ZD
ZN+++= −−
−
−
- The inverse Z-transform of polynomial X(Z) of the above can be easily found.
- So let us focus on inversion of rational function.
If X(Z) = N
N
M
M
ZaZa
ZbZbb−−
−−
++
++
......1
.....1
1
1
10
- Multiply Nr. and Dr. ZN to eliminate –ve power of Z.
Thus X(Z) = N
NN
MN
M
NN
aZaZ
ZbZbZb
......
.....1
1
1
10
++
++−
−−
Z
ZX )( =
N
NN
MN
M
NN
aaZZ
ZbZbZb
......
.....1
12
1
1
0
++
++−
−−−−
= )(........)()(
.....
21
12
1
1
0
N
MN
M
NN
PZPZPZ
ZbZbZb
−−−
+++ −−−−
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The poles of Equation :
)(
........)()(
)(
2
2
1
1
N
N
PZ
C
PZ
C
PZ
C
Z
ZX
−++
−+
−=
The CK = |)(
)(
KPZ
KZ
ZXPZ
=
− K = 1, 2, 3, ……..N
If X(Z) has a pole of multiplicity l, then the denominator (Z – PK)l and the expansion
of the form is no longer valid. Let us assume l = 2 and
2
21 )()(
1)(
PZPZZ
ZX
−−=
Then 2
2
3
2
2
1
1
)()(
)(
PZ
C
PZ
C
PZ
C
Z
ZX
−+
−+
−=
C1 = |1
)()( 1
PZZ
ZXPZ
=
− ; C2 = |2
)()( 2
2
PZZ
ZXPZ
dZ
d
=
−
C3 = |2
)()( 2
2
PZZ
ZXPZ
=
−
Cik =
kPZ
l
kil
il
Z
ZXPZ
dz
d
il =
−
−
−
−
)()(
)!(
1
8.5.3. Convolution method :
- In this method the given X(Z) is split into X1 (Z) and X2 (Z)
Such that X(Z) = X1(Z) . X2(Z)
Next we have to find x1(n) and x2(n) by taking inverse Z-transform.
From convolution property of Z-transform, we know
Z[x1(n) * x2(n)] = X1(Z) . X2(Z) = X(Z)
We can find x(n) by convoluting x1(n) and x2(n).
Given X(Z) = 21
1
231
31−−
−
++
+
ZZ
Z ; |Z| > 2
= )2()1(
)3(
++
+
ZZ
ZZ
Let X(Z) = X1(Z) . X2(Z)
Where X1(Z) = 1+Z
Z ; and X2(Z) =
)2(
3
22
3 1
++
+=
+
+ −
Z
ZZ
Z
Z
Z
Z
x1(n) = (-1)n u(n)
And x2(n) = (-2)n u(n) + 3(-2)n-1 u(n-1)
x(n) = x1(n) * x2(n)
= (-1)n u(n) * [(-2)n u(n) + 3(-2)n-1 u(n-1)]
= (-1)n u(n) * (-2)n u(n) + 3(-1)n u(n) * (-2)n-1 u(n-1)
= −=
− −−−n
nk
knk knuku )()2()()1( )(
x1(n) = (-1)n u(n)
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x2(n) = (-2)n u(n) + 3(-2)n-1 u(n-1)
x(n) = x1(n) * x2(n)
= (-1)n u(n) * [(-2)n u(n) + 3(-2)n-1 u(n-1)]
= (-1)n u(n) * (-2)n u(n) + 3 (-1)n u(n) * (-2)n-1 u(n-1)
= −=
−−
−=
− −−−−+−−−n
nk
knkn
nk
knk knukuknuku )1()2()()1(3)()2()()1( 1)(
u(k) = 1 for k 0 u(k) = 1 for k 0
u(n-k) = 1 for n – k > 0 u(n-1-k) = 1 for n–1–k > 0
n > k n–1 > k
k < n k < n–1
i.e. =
n
k 0
So −
=
1
0
n
k
n – k > 0
n > k → = −
=
−−
=
− −−+−−1
0
1
0
)2()1(3)2()1(n
k
knkn
k
knk
k < n
= (-2)n −
=
−
=
− −−−+1
00
1 )2()1()2(3)5.0(n
k
kkn
k
nk
= (-2)n −
==
−+−+1
00
)1( )5.0()2(3)5.0(n
k
kn
k
nk
= (-2)n
−
−=→
−
−−+
−
−=
+−
+ n
k
nk
nn
n
a
aa
0
11
1
1
1
5.01
)5.0(1)2(3
5.01
)5.0(1
= (-2)n 2 . [1 – (0.5)n+1] + 3 (-2)n-1 2 [1 – (0.5)n]
= (-2)n 2 [1 – 0.5 (0.5)n] – 3(-2)n [1 – (0.5)n]
= 2(-2)n – (-2)n (0.5)n – 3(-2)n + 3(-2)n (0.5)n
= 2(-2)n – (-1)n – 3(-2)n + 3(-1)n
= +2 (-1)n – (-2)n for n 0
x[n] = [2(-1)n – (-2)n] for n 0
+ 3 −
−=
−− −−−−1
1 )1()2()()1(n
nk
knk knuku
= (-2)n =
−
=
−−− −−−+−−n
k
n
k
kknkk
0
1
0
1 )2()1()2(3)2()1(
= (-2)n =
−
=
−− −−−+n
k
n
k
kknk
0
1
0
1 )2()1()2(3)5.0(
= (-2)n =
−
=
−−+n
k
n
k
knk
0
1
0
1 )5.0()2(3)5.0(
= (-2)n
−
−=
−
−−+
−
−=
+−
+ n
k
nk
nn
n
a
aa
0
11
1
1
1
5.01
)5.0(1)2(3
5.01
)5.0(1
= (-2)n . 2 [1 – 0.5 (0.5)n] + 3 (-2)n-1 . 2 [1 – (0.5)n]
= (-2)n 2 [1 – 0.5 (0.5)n] – 3(-2)n [1 – (0.5)n]
= 2 (-2)n – (-1)n – 3 (-2)n + 3(-1)n
= 2(-1)n – (-2)n for n 0
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x(n) = [2(-1)n – (-2)n] u(n)
8.6. Z-Transform Pairs:
S.No. Sequence
x[n]
Z-transform
X(Z)
ROC
1 (n) 1 All Z
2 u(n) 11
1−− Z
|Z| > 1
3 -u(-n-1) 11
1−− Z
|z| < 1
4 (n-m) Z-m All Z except Z = 0 if m > 0
All Z except Z = 0 if m < 0
5 an u(n) 11
1−− aZ
|Z| > (a)
6 -an u(-n-1) 11
1−− aZ
|Z| < (a)
7 nan u(n) 21
1
)1( −
−
− aZ
aZ |Z| > (a)
8 -nan u(-n-1) 21
1
)1( −
−
− aZ
aZ |Z| < (a)
9 [cos on] u[n] 21
1
)cos2(1
cos1−−
−
+−
−
ZZ
Z
o
o
|Z| > 1
10 [sin on] u[n] 21
1
)cos2(1
sin−−
−
+− ZZ
Z
o
o
|Z| > 1
11 rn cos o n u(n) 221
1
cos21
cos1−−
−
+−
−
ZrZr
Zr
o
o
|Z| > (r)
12 rn sin o n u(n) 221
1
cos21
sin−−
−
+− ZrZr
Zr
o
o
|Z| > (r)
13 n
1 ; n > 0
- log (1-Z-1) 1 < (Z)
14 nk an, k < 0 11
1−−
−−
aZdZ
dZ
n
|Z| < (a)
15 an for all n )1()1(
11
2
−−−
−
aZaZ
a |a| < |Z| <
a
1
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6. JNTU PREVIOUS EXAMS QUESTION PAPERS:
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Code No:A109210402 R09 SET-1
B.Tech II Year - I Semester Examinations, December 2011
SIGNALS AND SYSTEMS
(COMMON TO ECE, EIE, BME, ETM, ICE)
Time: 3 hours Max. Marks: 75
Answer any five questions
All questions carry equal marks
- - -
1.a) Derive the expression for component vector of approximating the function f1(t)
over f2(t) and also prove that the component vector becomes zero if the f1(t) and
f2(t) are orthogonal. 1 0 t π
b) A rectangular function f(t) is defined by f (t) π t 2π
−1
Approximate this function by a waveform sint over the interval (o, 2π) such that the
mean square error is minimum. [15]
2.a) List out all the properties of Fourier Series
b) Obtain the trigonometric Fourier series for the half wave rectified sine wave shown in
Figure.1. [15]
Figure.1
3. Determine the Fourier transform for the double exponential pulse shown in Figure.2. [15]
Figure.2
4.a) Define Linearity and Time-Invariant properties of a system.
b) Show that the output of an LTI system is given by the linear convolution of input
signal and impulse response of the system. [15]
5.a) State and prove Parseval’s Theorem.
b) Find the convolution of two signals x(n) = { 1, 1, 0, 1, 1} and h(n) = { 1, -2, -3, 4}
and represent them graphically. [15]
6.a) State and Prove the sampling theorem for Band limited signals.
b) Discuss the effect of aliasing due to under sampling. [15]
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7.a) Define Laplace Transform and Its inverse. b) Define Region of convergence and state its properties.
c) Find the Laplace transform of f(t) = sin at cos bt and f(t) = t sin at [15]
8.a) Find the two sided Z-transform of the signal x(n) = (1/3)n n ≥ 0
= (-2)n n ≤ -1
b) Determine the inverse Z-Transform of X(z) = z /(3z2 – 4z +1), if the region of
convergence are i) z > 1 ii) z < 1/3 iii) 1/3 < z < 1 [15]
********
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Code No:A109210402 R09 SET-2
B.Tech II Year - I Semester Examinations, December 2011
SIGNALS AND SYSTEMS
(COMMON TO ECE, EIE, BME, ETM, ICE)
Time: 3 hours Max. Marks: 75
Answer any five questions
All questions carry equal marks
---
1.a) Define a complete set and hence show that the error can be minimized when the
function f(t) is approximated using n set of orthogonal functions.
b) A rectangular function f(t) is defined by
Approximate this function by a waveform single term sint, two terms sint and sin3t,
three terms sint, sin3t and sin5t over the interval (o, 2π) and show that the mean
square error is minimum when the function is approximated by three terms rather than
single term. [15]
2.a) Derive the necessary expression to represent the function f(t) using Trigonometric
Fourier Series.
b) Bring out the relationship between Trigonometric and Exponential Fourier series. [15]
3.a) Prove that the time shift in time domain is equal to phase shift in frequency domain.
b) Find the Fourier transform of the function
i) f(t) = e-a|t|
sin(t) ii) f(t) = cos at2 iii) f(t) = sin at
2 [15]
4.a) What are the requirements to be satisfied by an LTI system to provide distortionless
transmission of a signal?
b) Bring out the relation between bandwidth and rise time? [15]
5.a) Show that autocorrelation and power spectral density form a Fourier Transform Pair.
b) Discuss the process of extraction of a signal from noise in frequency domain. [15]
6. Define Sampling Theorem and discuss the way of performing sampling using impulse
sampling technique. [15]
7.a) State and Prove Initial value and Final value theorem w.r.to Laplace transform.
b) Find the Laplace transform of the periodic rectangular wave shown in Figure.1. [15]
Figure.1
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8.a) Determine the impulse and unit step response of the systems described by the difference equation y(n) = 0.6y(n-1)-0.08y(n-2)+x(n)
b) Define Region of Convergence and state its properties w.r.to Z- Transform. [15]
********
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COURSE FILE OF “SIGNALS AND SYSTEMS”
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Code No:A109210402 R09 SET-3
B.Tech II Year - I Semester Examinations, December 2011
SIGNALS AND SYSTEMS
(COMMON TO ECE, EIE, BME, ETM, ICE)
Time: 3 hours Answer any five questions
Max. Marks: 75
All questions carry equal marks
---
1.a) Discuss the concept of orthogonality in complex functions and derive the expression
for component vector of approximating the function f1(t) over f2(t) in case of complex
functions.
b) Derive the expression for Mean square Error in approximating a function f(t) by a set
of n orthogonal functions. [15]
2.a) State the necessary and sufficient conditions for the existence of Fourier series
representation of a Periodic Signal.
b) Obtain the trigonometric Fourier series for the signal shown in Figure.1. [15]
Figure.1
3.a) State and prove any Four Properties of Fourier Transform. b) Find the Fourier Transform of
i) f(t) = e-at
Cos(bt) ii) f(t) = t cosat. [15]
4.a) Define the terms:
i) Signal Bandwidth ii) System bandwidth
iii) Linear time Variant system iv)Paley-wiener criteria for physical realizability. b) Test the linearity, causality, time-variance, stability of the system governed by the
equation
i) y(n) = ax(n) + b ii) y(n) = n cos[x(n)] [15]
5.a) Explain the process of detection of periodic signals by the process of correlation.
b) Define autocorrelation and state its properties. [15]
6. Define Sampling Theorem and discuss the way of performing sampling using Natural
sampling technique and compare it with impulse sampling. [15]
7.a) State any four properties of Laplace transform.
b) Find the Laplace transform of the wave form shown in Figure.2.
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Figure.2
c) Find the inverse Laplace transform of (S-1) / (S) (S+1). [15]
8.a) Using scaling property determine the Z-transform of an cosωn and find its ROC.
b) Using differentiation property find the Z-transform of x(n) = n2 u(n).
c) Obtain the Z-transform of x(n) = -anu(-n-1) and find its ROC. [15]
********
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Code No:A109210402 R09 SET-4
B.Tech II Year - I Semester Examinations, December 2011
SIGNALS AND SYSTEMS
(COMMON TO ECE, EIE, BME, ETM, ICE)
Time: 3 hours Answer any five questions Max. Marks: 75
R09 Set No. 2
II B.TECH – I SEM EXAMINATIONS, NOVEMBER - 2010
SIGNALS AND SYSTEMS Common
to BME, ICE, ETM, EIE, ECE Time: 3 hours Max Marks: 75 Answer any FIVE Questions
All Questions carry equal marks ? ? ? ?
c) (a) Write short notes on "Ideal BPF".
In the following network, determine the relationship between R's and C's in order to have a distortion less attenuation while signal is transmitted through the network shown in gure 1b. [8+7]
1
Figure 1b
2. (a) State the three important spectral properties of periodic power signals.
(b) Determine the Fourier series of the function shown in gure 2b. [5+10]
Figure 2b 3. (a) With the help of graphical example explain sampling theorem for Band limited
signals.
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(b) Explain brie y Band pass sampling. [8+7] 4. (a) Find the Z-transform and ROC of the signal
x(n) = [4(5n) 3(4
n)] u(n)
1
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Code No: A109210402 R09 Set No. 2
(b) Find the Z-transform as well as ROC for the following sequences: [7+8]
i. 13 n u( n)
ii. 13 n [u( n) u (n 8)]
5. (a) State the properties of the ROC of Laplace transforms.
(b) Determine the function of time x(t) for each of the following laplace transforms
and their associated regions of convergence. [7+8]
i. (s+1)2/ s
2-s+1 Re fSg > 1/2
ii. s2- s+1/ (s+1)
2 Re fSg > -1
6. (a) The rectangular function f(t) in gure 6a is approximated by the signal 4 Sin t.
Figure 6a
show that the error function fe(t) = f(t)-4/ Sin t is orthogonal to the function Sin t over the interval (0,2 ).
(b) Determine the given functions are periodic or non periodic.
i. a Sin 5t + b cos 8t
ii. a Sin (3t/2) + bpcos (16t/15) + c Sin (t/29)
iii. a cos t + b Sin 2t
Where a, b, c are real integers. [10+5]
7. (a) Determine the Fourier Transform of a trapezoidal function and triangular RF pulse f(t) shown in gure 7a. Draw its spectrum.
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KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 2
Figure 7a
(b) Using Parsevals theorem for power signals, Evaluate R
e 2t
u(t)dt. [10+5] 8. (a) Consider an input x[n] and an impulse response h[n] given by
hx[
[n
n]] == u2
1[n
n+
2 2]u
:[n 2];
Determine and plot the output y[n] = x[n] h[n]. (b) Bring out the relation between Correlation and Convolution.
(c) Explain the properties of Correlation function. [7+4+4]
? ? ? ? ?
3
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 4
II B.TECH – I SEM EXAMINATIONS, NOVEMBER - 2010
SIGNALS AND SYSTEMS Common
to BME, ICE, ETM, EIE, ECE Time: 3 hours Max Marks: 75 Answer any FIVE Questions
All Questions carry equal marks ? ? ? ? ?
1. (a) State the properties of the ROC of Laplace Transforms.
(c) Determine the function of time x(t) for each of the following Laplace trans- forms and their associated regions of convergence. [7+8] i
. ii.
(s + 1)
2 s2
s + 1 s2
s + 1 (s + 1)
2
RefSg > 1=2
RefSg > 1
2. (a) Explain the conditions under which any periodic waveform can be expressed using
the Fourier series.
(b) Find the Trigonometric Fourier series for a periodic square form shown in gure 2b, which is Symmetrical with respect to the vertical axis? [5+10]
Figure 2b 3. (a) What is an LTI system? Derive an expression for the transfer function of an LTI
system.
(b) The signal v(t) = cos !0t + 3 Sin 3!0t + 0.5 Sin 4!0t is ltered by an RC low pass
lter with a 3dB frequency fc = 2f0. Find the output power S0.[8+7] 4. (a) Impulse train sampling of x[n] is used to obtain
1 P
g[n] = x[n] [n kn] k= 1
if X(ej!
) for 3 =7 j!j , determine the largest value for the sampling interval N which ensures that no aliasing takes place.
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
(b) Explain the Sampling theorem for Band Limited Signals with Graphical proof. [7+8] 4
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 4
5. Find the power of periodic signal g(t) shown in gure5. Find also the powers of
(a) -g(t)
(b) 2g(t)
(c) g(-t)
(d) g(t)/2. [15]
Figure 5
6. (a) An AM signal is given by
f(t) = 15 Sin (2 106t) + [5 Cos 2 10
3t + 3 Sin2 10
2t] Sin 2 10
6t
Find the Fourier Transform and draw its spectrum.
(b) Signal x(t) has Fourier Transform x(f) = j2 f
3+j=10 .
i. What is total net area under the signal x(t).
t
ii. Let y(t) = R
x( )d what is the total net area under y(t). [8+7]
7. (a) Find the inverse Z-transform of the following X(z).
i. X(Z) = log ( 1 / (1-az 1)), z > a
ii. X(Z) = log ( 1 / (1-a 1z)), j j j j
z < a
j j j j
(b) Find the Z-transform X(n), x[n] = (1/2)n u[n] + (1/3)
n u[-n-1] [8+7]
8. (a) Which of the following signals or functions are periodic and if what is its
fundamental period.
i. g(t) = e j60
t
ii. g(t) = 10 Sin (12 t) + 4 Cos (18 t)
(c) Let two functions be de ned by:
x1(t) = 1 , Sin (20 t) 0 -1 , Sin (20 t) < 0
X2(t) = t, Sin (2 t) 0 -t Sin (2 t) < 0
Graph the product of these two functions vs time over the time interval -2 < t < 2. [8+7]
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CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 1
II B.TECH – I SEM EXAMINATIONS, NOVEMBER - 2010
SIGNALS AND SYSTEMS Common to BME, ICE, ETM, EIE, ECE
Time: 3 hours Max Marks: 75 Answer any FIVE Questions
All Questions carry equal marks ? ? ? ? ?
2. (a) Evaluate the following integrals:
8 R
i. [u(t + 3) 2 (t):u(t)]dt 1
5 2
R ii. (3t)dt
1 2
(b) A even function g(t) is described by g(t) = 8 15 2t 3t 30 tt << 7
3
< : 2 7 t < 10
ii. What is the value of g(t) at time t = 5
ii. What is the value of 1st
derivative of g(t) at time t = 6. [8+7]
2. (a) Distinguish between Energy and Power signals.
(b) Derive the expression for Energy density spectrum function of a energy signal f(t) from fundamentals and interpret why it is called Energy density spectrum.
[5+10]
3. (a) Explain the concept of generalized Fourier series representation of signal f(t).
(b) State the properties of Fourier series. [8+7]
4. (a) Explain the properties of the ROC of Z transforms. 1+ z
1
(b) Z transform of a signal x(n) if X(z) =
1 + 13 z 1 .
Use long division method to determine the values of
i. x[0], x[1], and x[2], assuming the ROC to be jzj > 1
3
ii. x[0], x[-1], and x[-2] , assuming the ROC to be jzj < 1
3 . [7+8] 1
5. (a) correlationAsignaly(t)
andgiven
PSDby
ofy
y(t)(t)
.= C0 + n
P=1 CnCos(n!0t + n). Find the auto-
(b) Explain the Graphical representation of convolution with an example. [8+7]
6. (a) Consider an LTI system with input and output related through the equation. t
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
y(t) = R
e (t
)x( 2)d What is the impulse response h(t) for this system.
7
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 1
(b) Determine the response of this system when the input x(t) is as shown in gure 6b.
Figure 6b (c) Consider the inter connection of LTI system depicted in gure 6c.
Figure 6c Here h(t) is an in part (a). Determine the output y(t) when input x(t) is again given gure above, using the convolution integral. [5+5+5]
7. (a) Consider the signal x(t) = (sin 50 t / t)2 which to be sampled with a sampling
frequency of !s = 150 to obtain a signal g(t) with Fourier transform G(j! ). Determine
the maximum value of !0 for which it is guaranteed that
G(j!) = 75 X(j!) for j!j < !0 where X(j!) is the Fourier transform of x(t).
(b) The signal x(t) = u(t + T0) - u(t - T0) can undergo impulse train sampling without
aliasing, provided that the sampling period T< 2T0. Justify. [7+8]
8. (a) Explain the method of determining the inverse Laplace transforms using Par-tial
fraction method, for the following cases
i. Simple and real roots ii. Complex roots
iii. Multiple or repeated roots.
(b) Find the Laplace transform of the function
f(t) = A Sin !0t for 0 < t < T/2. [3+3+4+5]
? ? ? ? ?
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CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 3
II B.TECH – I SEM EXAMINATIONS, NOVEMBER - 2010
SIGNALS AND SYSTEMS Common to BME, ICE, ETM, EIE, ECE
Time: 3 hours Max Marks: 75 Answer any FIVE Questions
All Questions carry equal marks ? ? ? ? ?
2. (a) Derive polar Fourier series from the exponential Fourier series representation and
hence prove that Dn= 2jCnj. (b) Determine the trigonemetric and exponential Fourier series of the function
shown in gure 1b. [5+10]
Figure 1b
2. (a) Write short notes on \orthogonal vector space".
(b) A rectangular function f(t) is de ned by:
1 0 < t < 1 t < 2
Approximate above function by a nite series of Sinusoidal functions. [8+7]
3. (a) Using the Power Series expansion technique, nd the inverse Z-transform of the following X(Z):
i. X(Z) =
ii. X(Z) =
Z
2Z2 3Z+1
Z 2Z
2 3Z+1
jZj < 1
2
jZj > 1
(b) Find the inverse Z-transform of
(Z+1)(
jZj > 2.
X(Z) = Z(Z 1)(Z 2) [8+7]
4. (a) Determine the inverse Laplace transform for the following Laplace transform and their associated ROC. i
. ii.
s + 1 (s
2
+5s + 6) (s
2 +5s + 6)
(s + 1 )2
3 < Refsg < 2
Refsg > 1
(b) Explain the constraints on ROC for various classes of signals, with an example. [9+6]
f(t) =
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9
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
Code No: A109210402 R09 Set No. 3
5. (a) Find the Fourier Transform for the following functions shown in gure 5a.
Figure 5a
(b) Find the total area under the function g(t) = 100 Sin c ((t-8)/30). [10+5]
6. (a) Explain brie y detection of periodic signals in the presence of noise by corre-lation.
(b) Explain brie y extraction of a signal from noise by ltering. [8+7]
7. (a) Find the transfer function of Lattice network shown in gure 7a.
Figure 7a
(b) Sketch the magnitude and phase characteristic of H(j!). [8+7]
8. Determine the Nyquist sampling rate and Nyquist sampling interval for the signals.
(a) sinc(100 t).
(b) sinc2(100 t).
(c) sinc(100 t) + sinc(50 t).
(d) sinc(100 t) + 3 sinc2(60 t). [3+4+4+4]
? ? ? ? ?
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10
CMR COLLEGE OF ENGINEERING AND TECHNOLOGY
KANDLAKOYA(VILLAGE), MEDCHAL.
COURSE FILE OF “SIGNALS AND SYSTEMS”
PREPARED BY B.CHAKRADHAR- Asst. Professor.
7. DESCRIPTIVE QUESTION BANK:
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
1ST UNIT
1. a) Write short notes on “Orthogonal Vector Space”.
b) A rectangular function f(t) is defined by
f(t) =
−
)20(1
)0(1
t
t
Approximate the above function by a finite series of Sinusoidal functions.
2. a) Define and sketch the following elementary signals
i. Unit impulse signal
ii. Unit step signal
iii. Signum function
b) Explain the Analogy of vectors and signals in terms of orthogonality and evaluation of
constant.
3. a) Sketch the single sided and double sided spectra of the following signal
x(t) = 2 sin (10t - /6).
b) Show that the functions Sin n0t and Sin m0t are orthogonal over any interval
(t0, t0 + 2 / 0) for integer values of n and m.
4. a) Write short notes on “Orthogonal functions”.
b) Define the following Elementary signals.
i. Real Exponential Signal
ii. Continuous time version of sinusoidal signal and Bring out the relation between
Sinusoidal and complex exponential signals.
5. a) Define Orthogonal signal space and bring out clearly its application in representing a
signal.
b) Obtain the condition under which two signals f1 (t) & f2 (t) are said to be orthogonal to
each other. Hence, prove that sin n0t and cos m0t are orthogonal to each other for all
integer values of m, n.
6. a) A rectangular function defined as
f(t) =
−
2)2/3(
)2/3()2/(
)2/(0
tA
tA
tA
Approximate the above function by A cos (t) between the intervals (0, 2) such that the
mean square error is minimum.
b) Prove the following.
i. (n) = u(n) – u(n – 1) ii. u(n) = =
n
k
K )(
7. a) Sketch the single sided and double sided spectra of the following signal
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
x(t) = 2 Sin (10t - /6)
b) Derive polar Fourier series from the exponential Fourier series representation and hence
prove that Dn = 2 |Cn|.
8. a) Distinguish between Orthogonal vectors and Orthogonal functions.
b) Consider the complex valued exponential signal
x(t) = A ea+Jt, a > 0. Evaluate the real and imaginary components of x(t) for the
following cases.
i. a real, a = a1.
ii. a imaginary, a = j1
iii. a complex, a = a1 + j
c) Consider the rectangular pulse x(t) as shown in the below figure.
x(t)
A
-0.5 0.5 t
Repeat the above rectangular pulse in terms of weighted sum of two step functions.
9. a) Sketch the following signals.
i. [(t – 1) / 2] + (t – 1)
ii. f(t) = 3u(t) + tu(t) – (t – 1) u (t – 1) – 5u(t – 2)
b) Evaluate the following integrals.
i.
5
0
2sin)( dttt
ii.
−
− − dtte at )10(2
10. a) Define and sketch the following signals.
i. Real exponential signals for C 0, a > 0
ii. Even continuous time signal
iii. Unit doublet
iv. Real part of damped complex exponential for a = 0
b) Evaluate the following integrals.
i.
−+ edtt t)3(
ii.
−+ dttttt sin)1(cos)(
c) Discuss the signal with a neat sketch.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
2ND UNIT
1. a) Prove that Sine(0) = 1 and plot Sine function.
b) Determine the Fourier series representation of that Signal x(t) = 3 Cos (t/2 + /4)
using the method of inspection.
2. Prove the following properties.
a) The FS symmetry properties for
i. Real valued time signals
ii. Real and even time signals
b) Obtain the Fourier series representation of an impulse train given by
x(t) =
−=
−n
nTt )( 0
3. a) Explain about even and odd functions.
b) Obtain the trigonometric Fourier series for the periodic waveform as shown in the below
figure.
4. a) Prove that the normalized is given by p =
−=n
nC 2|| , where |Cn| are complex Fourier
coefficients for the periodic wave form.
b) Determine the Fourier series expansion for the signal x(t) shown in the below figure.
5. a) State the three important spectral properties of periodic power signals.
b) Assuming T0 = 2, determine the Fourier series expansion of the signal shown in the
below figure.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
6. a) Show that the magnitude spectrum of every periodic function is Symmetrical about the
vertical axis passing through the origin.
b) With regard to Fourier series representation, justify the following statements.
i. Odd functions have only Sine term.
ii. Even functions have no sine terms.
iii. Functions with half wave Symmetry have only odd harmonics.
7. Show that the Fourier series for a real valued signal can be written as
x(t) =
=
++1
)(sin)()(cos)()(n
oo tnnAtnnBoB
Where B(n) and A(n) are real valued coefficients and express cn in terms of b(n) and A(n).
8. a) Write short notes on “Exponential Fourier Spectrum”.
b) Find the Fourier series expansion of the periodic triangular wave shown in the below
figure.
9. a) State the properties of Complex Fourier series.
b) Determine the Fourier series of the function shown in the below figure.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
3RD UNIT
1. a) Find the Fourier Transform of the signal shown in the below figure.
b) Find the Fourier Transform of the signal given below.
y(t) = −
otherwise
tt
0
2210cos
2. a) Obtain the Fourier transform of the following functions :
i. Impulse function (t)
ii. DC Signal
iii. Unit step function
b) State and prove time differentiation property of Fourier Transform.
3. a) State and prove properties of correlation function.
b) If V(f) = AT sin 2 fT / 2 fT find the energy contained in V(t).
c) Obtain the Fourier Transform of the following :
i. x(t) = A sin (2 fct) u(t)
ii. x(t) = f(t) Cos (2 fct) + )
d) State and prove the following properties of Fourier Transform.
i. Multiplication in time domain
ii. Convolution in time domain
a) Find Fourier Transform of the following time function.
x(t) = e-3t [u(t+2) – u(t – 3)]
b) State and prove frequency and time shifting properties of Fourier Transform.
4. a) Explain the concept of Fourier transform for periodic signals.
b) Find out the Fourier Transform of the periodic pulse train shown in the below figure.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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5. a) State and prove time convolution and time differentiation properties of Fourier
transform.
b) Find and sketch the Inverse Fourier transform of the Waveform shown in the below
figure.
6. a) Find the Fourier transform of the signal x(t) = 21
2
t+ .
b) Explain how Fourier transform can be derived from Fourier series.
7. Find the Fourier Transform of the following functions.
a) A single symmetrical Triangle Pulse.
b) A single symmetrical Gate Pulse.
c) A single cosine wave at t = 0.
8. a) Distinguish between Fourier series and Fourier transform.
b) State the conditions for the existence of Fourier transform of signal.
c) Find the Fourier transform of the signum function and plot it’s amplitude and phase
spectra.
9. a) Find and sketch the Inverse Fourier transform of the waveform shown in the below
figure.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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4TH UNIT
1. a) Explain how input and output signals are related to impulse response of a LTI system.
b) Let the system function of a LTI system be 2
1
+j. What is the output of the system
for an input (0.8)t u(t) ?
2. a) Explain the difference between the following systems.
i. Time invariant and time variant systems.
ii. Causal and non-causal systems.
b) Consider a stable LTI system characterized by the differential equation :
)(2)(
)(3)(
4)(
2
2
txdt
txdty
dt
tdy
dt
tyd+=++
Find its impulse response and transfer function.
3. a) Explain the difference between the following systems.
i. Linear and Non-linear systems.
ii. Causal and Non-causal systems.
b) Consider a stable LTI system characterized by the differential equation :
)()(2)(
txtydt
tdy=+
Find its impulse response.
4. a) There are several possible ways of estimating an essential bandwidth of non-band
limited signal. For a low pass signal, for example, the essential bandwidth may be
chosen as a frequency where the amplitude spectrum of the signal decays to k percent
of its value. The choice of k depends on the nature of application. Choosing k = 5
determine the essential bandwidth of g(t) = exp (-at) u(t).
b) Differentiate between signal bandwidth and system bandwidth.
5. a) Find the impulse response of the system shown in the below figure. Find the transfer
function. What would be its frequency response ? Sketch the response.
b) Explain the characteristics of an ideal LPF, Explain why it can’t be realized.
6. a) What is a LTI system ? Explain its properties. Derive an expression for the transfer
function of a LTI system.
b) Obtain the conditions for the distortion less transmission through a system. What do
you understand by the term signal bandwidth ?
7. a) Find the impulse response to the RL filter shown in the below figure.
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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5TH UNIT
1. a) State and prove properties of auto correlation function ?
b) A filter has an impulse response h(t) as shown in figure 5b. The input to the network is a
pulse of unit amplitude extending from t = 0 to t = 2. By graphical means determine the
output of the filter.
1 h(t)
1 t
Figure 5b
2. a) Find the energies of the signals shown in figures 1, 2.
b) Determine the power of the following signals.
g4 (t)
g3 (t) 1
e-t/5
-1 t
5
Figure 1 Figure 2
i. (10 + 2 sin 3t) Cos 10 t.
ii. 10 Cos 5t cos 10t.
3. a) A signal y(t) given by y(t) = C0 +
=
+1
)(cosn
non tnC . Find the auto correlation and
PSD of y(t).
b) Find the mean square value (or power) of the output voltage y(t) of the system shown in
figure 2. If the input voltage PSD. S2 () = rect (/2). Calculate the power (mean square
value) of input signal x(t).
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Figure 2
4. a) A waveform m(t) has a Fourier transform M(f) whose magnitude is as shown in figure 2.
Find the normalized energy content of the waveform.
1
|M(f)|
-1 1 f
Figure 2
b) The signal V(t) = cos 0t + 2sin 3 0t + 0.5 sin 40t is filtered by an RC low pass filter
with a 3 dB frequency.
fc = 2f0. Find the output power So.
c) State parseval’s theorem for energy X power signals.
5. a) The signal v(t) = cos 0t + 2 sin 30t + 0.5 sin 40t is filtered by an RC low pass filter
with 3 dB frequency fC = 2fO.
i. Find Gi (f)
ii. Find G0 (f)
b) Let G(f) denote the Fourier transform of real valued energy signal g(t), and Rg () its
autocorrelation function, show that
−
−
=
dffGfd
d
dRg 422
2
|)(|4)(
6. a) State and prove properties of cross correlation function.
b) If V(f) = AT sin 2 fT / 2 fT find the energy contained in V(t).
7. a) Explain briefly detection of periodic signals in the presence of noise by correlation.
b) Explain briefly extraction of a signal from noise by filtering.
8. a) For the signal g(t) = 2a / (t2+a2). Determine the essential band width B Hz of g(t) such
that the energy contained in the spectral components of g(t) of frequencies below B Hz is
99% of signal energy Eg.
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b) Show that the auto correlation function of g(t) = C cos (0t + 0) is given by
Rg() = (c2/2) cos 0 , and the corresponding PSD is
Sg () = (c2/2) [(-0) + (+0)].
9. a) Using frequency domain convolution, find X(f) for x(t) = A Sinc2 2wt.
b) Show that correlation can be written in terms of convolution as
R() = )](*)([/1
txtxT
TLt−
→
c) The input to an RC Low pass filter is x(t) = sinc2 wt. Find output energy Ey.
10. Find the power of periodic signal g(t) shown in figure 5c. Find also the powers of
(a) –g(t) b) 2g(t) c) g(t)
Figure 5c
11. a) Determine an expression for the correlation function of a square wave having the values
1 or 0 and a period T.
b) The energy of a non periodic wave form v(t) is E =
−
dttv )(2 .
i. Show that this can be written as E =
−
−
dfefvtvdt ftj 2)()( .
ii. Show that by interchanging the order by integration we have E =
−
*)( vfv
(f) df =
−
dffv 2|)(| .
12. a) Derive Parsavel’s theorem from the frequency convolution property.
b) Find the cross correlation between [u(t) + u(t - )] and e-t u(t).
13. a) Find the Z-transform X(n).
i. x[n] = (1/2)n u[n] + (1/3)n u[n]
ii. x[n] = (1/3)n u[n] + (1/2)n u[-n-1]
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b) Find inverse z transform of x(z) using long division method
x(z) = (2 + 3z)-1 / (1 + z-1) (1 + 0.25 z-1 – (z-2) / 8)
14. The complex exponential representation of a signal f(t) over the interval (0, T)
f(t) = −=
+
n
ntjen 2)(4
3
(a) What is the numerical value of T
(b) One of the components of f(t) is A Cos 3t. Determine the value of A.
(c) Determine the minimum no. of terms which must be maintained in representation
of f(t) in order to include 99.9% of the energy in the interval (0, T).
6TH UNIT
1. a) Consider the signal x(t) =
250sin
t
twhich to be sampled with a sampling frequency
of s = 150 to obtain a signal g(t) with Fourier transform G(j). Determine the
maximum value of 0 for which it is guaranteed that G(j) = 75 (j) for || 0
where X(j) is the Fourier transform of x(t).
b) The signal x(t) = u(t + T0) – u(t – T0) can undergo impulse train sampling without
aliasing, provided that the sampling period T < 2T0. Justify.
c) The signal x(t) with Fourier transform X(j) = u( + 0) – u( - 0) can undergo
impulse train sampling without aliasing, provided that the sampling period T < /0.
Justify.
2. a) With the help of graphical example explain sampling theorem for Band limited signals.
b) Explain briefly band pass sampling.
3. a) Explain briefly impulse sampling.
b) Define sampling theorem for time limited signal and find the nyquist rate for the
following signals.
i. rect300t
ii. -10 sin 40t cos 300t
4. a) Determine the Nyquist rate corresponding to each of the following signals.
i. x(t) = 1 + cos 200 pt + sin 4000 t
ii. x(t) = t
t
4000sin
b) The signal Y(t) is generated by convolving a band limited signal x1(t) with another band
limited signal x2(t) that is
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y(t) = x1(t) * x2(t)
where
x1(j) = 0 for || > 1000
x2 (j) = 0 for || > 2000
Impulse train sampling is performed on y(t) to obtain
yp(t) =
−=
−n
nTtnTy )()(
Specify the range of values for sampling period T which ensures that y(t) is recoverable
from yp(t).
5. Determine the Nyquist sampling rate and Nyquist sampling interval for the signals.
(a) sin c(100t) (b) sin (100t) (c) sin c (100t) + sin c(50t)
(d) sin c(100t) + 3 sin c2 (60t)
6. a) Explain Flat top sampling.
b) A Band pass signal with a spectrum shown in figure 6b is ideally sampled. Sketch the
spectrum of the sampled signal when fs = 20, 30 and 40 Hz. Indicate if and how the
signal can be recovered.
Figure 6b
7. a) A signal x(t) = 2 cos 400 t + 6 cos 640 t is ideally sampled at fs = 500Hz. If the
sampled signal is passed through an ideal low pass filter with a cut off frequency of 400
Hz, what frequency components will appear in the output.
b) A rectangular pulse waveform shown in figure 6b is sampled once every TS seconds and
reconstructed using an ideal LPF with a cutoff frequency of fs/2. Sketch the
reconstructed waveform for Ts = 6
1sec and Ts =
12
1 sec.
Figure 6b
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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8. a) A periodic waveform is formed by eliminating the alternate cycle of a Sinusoidal
waveform as shown in figure 6a.
Figure 6a
i. Find the Fourier series (exponentially) by direct evaluation of the coefficients.
ii. If the waveform is shifted to the left by seconds, the new waveform f(t + ) is
odd function of the time whose Fourier series contains only sine terms. Find the
Fourier series of f(t + ). From this series, write down the Fourier series for f(t).
9. a) Let x(t) be a signal with Nyquist rate 0. Also let y(t) = x(t) p(t-1), where p(t) =
−=
−n
nTt )(1 , and T < 0
2
Specify the constraints on the magnitude and phase of the frequency response of a filter
that gives x(t) as its output when y(t) is the input.
b) Explain the Sampling theorem for Band Limited Signals with analytical proof.
10. a) Find the output voltage v(t) of the network shown in figure 6a when the voltage applied
to the terminals a b is given by f(t) = e-t/4 u(t) + e-t/2 u(-t)
Figure 6a
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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b) Show that the impulse response of ideal low pass filter is h(t) = 2
)(
2
0ttSa
−
for a
distortion transmission plot the impulse response of h(t).
7TH UNIT
1. a) Obtain the inverse laplace transform of F(s) = )2(
12 +ss
by convolution integral.
b) Using convolution theorem find inverse laplace transform of 222 )( as
s
+.
c) Define laplace transform of signal f(t) and its region of convergence.
2. a) When a function f(t) is said to be laplace transformable.
b) What do you mean by region of convergence.
c) List the advantages of Laplace transform.
d) If (t) is a unit impulse function find the laplace transform of )]([2
2
tdt
d .
3. a) State and prove the properties of Laplace transforms.
b) Derive the relation between Laplace transform and Fourier transform of signal.
4. a) State the properties of the ROC of L.T.
b) Determine the function of time x(t) for each of the following laplace transforms and their
associated regions of convergence.
i) 1
)1(2
2
+−
+
ss
s Re {S} > ½
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ii) 2
2
)1(
1
+
+−
s
ss Re {S} > -1
5. a) Determine the Laplace transform and the associate region convergence for each of the
following functions of time.
i) x(t) = 1 0 t 1
ii) x(t) = 21
10
2
− t
t
t
t
b) State and prove initial value theorem of L.T.
6. a) The system function of a causal LTI system is H(s) = 22
12 ++
+
ss
s. Determine the
response y(t) when the input is x(t) = e-|t|.
b) State and prove initial value and final value theorems.
7. Consider the following signals, find laplace transform and region of convergence for each
signal.
(a) e-2t u(t) + e-3t u(t)
(b) e-4t u(t) + e-5t sin 5t u(t)
(c) State properties of laplace transform.
8. Determine the function of time x(t) for each of the following Laplace transforms and their
associated regions of convergence.
(a) 9
12 +s
Re {S} > 0
(b) 92 +S
S Re {S} < 0
(c) 9)1(
12 ++
+
s
s Re{S} < -1
9. a) The two periodic functions f1(t) and f2(t) with zero dc components have arbitrary
waveforms with periods T and T2 respectively. Show that the component in f1(t) of
waveform f2(t) is zero in the interval (- < t < a).
b) Complex Sinusoidal signal x(t) has the following components.
Re{x(t)} = xR (t) = ACos (t + )
Im {x(t)} = xI (t) = ASin (t + )
The amplitude of x(t) is given by the square root of )()( 22 txtx IR + . Show that this
amplitude equals A and is therefore independent of the phase angle .
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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10. a) Find the initial values and final values of the function F(s) = sssss
sss
2453
67172345
23
++++
+++.
b) Explain the Step and Impulse responses of Series R-C circuit using Laplace transforms.
11. a) Consider the signal x(t) = (sin 50 t / t)2 which to be sampled with a sampling
frequency of s = 150 to obtain a signal g(t) with Fourier transform G(j). Determine
the maximum value of 0 for which it is guaranteed that G(j) = 75 j() for || - 0.
where X(j) is the Fourier transform of x(t).
b) The signal x(t) = u(t + T0) – u(t – T0) can undergo impulse train sampling without
aliasing, provided that the sampling period T < 2T0. Justify.
8TH UNIT
1. a) A finite sequence x[n] is defined as x[n] = {5, 3, -2, 0, 4, -3} Find X[Z] and its ROC.
b) Consider the sequence x[n] = −
otherwise
aNnan
0
0,10 Find X[Z].
c) Find the Z-transform of x(n) = cos (n) u(n).
2. a) Find the inverse Z-transform of
X(Z) = )2()1(
352 23
−−
++−
ZZ
ZZZ |Z| < 1
b) Find the inverse Z-transform of X(Z) = 2
3
−Z |Z| > 2
c) Find the Z-transform of an sin (n) u(n).
3. a) State and Prove the properties of the z-transform.
b) Find the Z-transform of the following Sequence.
x[n] = an u[n]
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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4. a) Find the Z-transform and ROC of the signal x[n] = [4. (5n) – 3 (4n)] u(n)
b) Find the Z-transform as well as ROC for the following sequences.
i) )(3
1nu
n
−
and ii) )]8()([
3
1−−−
nunu
n
5. a) Using the Power Series expansion technique, find the inverse Z-transform of the
following X(Z) :
i. X(Z) = 2
1||
132 2
+−Z
ZZ
Z
ii. X(Z) = 1||132 2
+−
ZZZ
Z
b) Find the inverse Z-transform of
X(Z) = 2)2()1( −− ZZZ
Z |Z| > 2
6. a) Find the Z-transform of the following Sequences.
i. x[n] = a-n u[-n-1]
ii. x[n] = u[-n]
iii. x[n] = - an u [-n-1]
b) Derive relationship between z and Laplace Transform.
7. Using the method indicated, determine the sequence that goes with each of the following Z
transforms :
a) Partial fractions :
X(z) =
−+
−
−−
−
21
1
2
51
21
zz
z , and x[n] is absolutely summable.
b) Long division :
X(z) =
1
1
2
11
2
11
−
−
+
−
z
z , and x[n] is right sided.
c) Partial fractions :
X(z) = 1
8
1
4
1
3
−−− zz
, and x[n] is absolutely summable.
8. a) Determine and sketch the auto correlation function of the following exponential pulses
i. f(t) = e-at u(t)
ii. f(t) = e-a|t|
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b) Determine the cross correlation function R12 () of the two signals g1(t) and g2(t) denoted
by
g1(t) = A Cos(2 f1t + 1), 0 t T
= 0 , elsewhere
g2(t) = A Cos(2 f2t + 2), 0 t T
= 0 , elsewhere
9. a) Which of the following signals or functions are periodic and if what is its fundamental
period.
i. g(t) = e-j60t
ii. g(t) = 10 Sin (12t) + 4 Cos (18t)
b) Let two functions be defined by :
x1(t) = 1, Sin (20t) 0
-1, Sin (20t) < 0
x2(t) = t, Sin (2t) 0
-t Sin (2t) < 0
Graph the product of these two functions vs time over the time interval -2 < t < 2.
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8. OBJECTIVE QUESTION BANK
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UNIT – 1
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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UNIT – 3
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
UNIT – 4
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
UNIT – 5
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
UNIT – 6
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
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MULTIPLE CHOICE QUESTIONS :
SET – 1
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Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni
CMR COLLEGE OF ENGINEERING & TECHNOLOGY
Kandlakoya (V), Medchal (M), RR Dist. A.P.
Department of ECE
Course File of Signals and Systems by B.Chakradhar/ S.Krishnaveni