COURSE FILE OF SIGNALS & SYSTEMS - CMRCET · 2021. 6. 7. · 3. Signals & Systems-Simon Haykin and...

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.





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.





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.





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


24/7/2012

Unit-III: Fourier Transforms


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






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


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


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


24/9/2012

Unit-VII: Laplace Transforms


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





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


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





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.





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)





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.





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.





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.





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 + )





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

;





=

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





(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





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





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





=

+→=

+→

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





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





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.





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)(





(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 :





(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.





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]





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.





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.





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

.





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−





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 :





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.





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.





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





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





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.





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).





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)





+

=

++

=

+

++=

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





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





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





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





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





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 :





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)





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





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||





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.





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) :





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





2.8.4. Determine the Fourier series representations for the following signals:

(1) Each illustrated in the figures (a) - (f)





(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:





For ,

But, in

Substituting in x(t)

(b)

It is periodic with period 6. So, consider segment between .

The function in this interval is:





Let


For ,


(c)

It is periodic with period 3, so take the segment

The function in the interval is:





Let


For ,






(d)

It is periodic with period 2. So take segment

Here,

Let


For






(e)

It is periodic with period 6. So, take the segment

Here will be

Let


For






(f)

It is periodic with period 3. So, take the segment

Here will be

Let


For







(2) Let


Given .

For


(3) Let






Given .

For


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





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 .





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−


=

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−






=

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.





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.





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





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.





F-1

+ )(

1

Ja = e-at u(t)

i.e. [e-at u(t)] =

−+

deJa

tJ

)(

1

2

1

+ )(

1

Ja =

−


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





=

−=

−−

−−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

→=





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 = .





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)]





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

−+−++





= )(

)()(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()





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





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.





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.





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





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





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





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 :





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

)(

*

* )(





= 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





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





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





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).





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





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 **

−

−

−

−

=





=

−

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





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 :





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





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+−





(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( −





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





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.





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+)





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





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)





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)]





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 )





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)





= [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 :





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 :





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.





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





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.





- 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.





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.





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 )(





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

)(





(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





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.





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) :





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





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.





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 )()(





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)





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()





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 :





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.





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”.





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 :





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.





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 :





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).





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.





=

−

= )()( 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





=

+−

−

− +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 .





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 −++





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.





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)





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 - ||]





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.





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 .





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 (-)





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.





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.

\





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.





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





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.





- 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.





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.





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.





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).





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.





- 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





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)]





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.





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.





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.





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.





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)





F.T. → X(J) =

−

− dtetx tJ)(

=

−


=

+−−− +

==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)





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 :





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

→=

→





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’





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 :





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





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





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





= )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





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 −+





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.





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.





−

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)]





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





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 :





[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.





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 :






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 :


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 :


dt

tdx )( ⎯→L S X(S) with ROC containing R.

x(t) = +

−

J

J

st dseSXj

)(2

1





+

−

=

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 :





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

→=

→





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 :





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 :





(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





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





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





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





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.





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)





= 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





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





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)





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





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

−+

++

+





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





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 −

−−− −−−





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





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





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.





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.





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)





(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.





- 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.





- 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)





= 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





= ( )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.





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





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]





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 :





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)





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





= )()( **** 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





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

−−−

+++ −−−−





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)





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





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





6. JNTU PREVIOUS EXAMS QUESTION PAPERS:





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]





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]

********





Code No:A109210402 R09 SET-2


SIGNALS AND SYSTEMS


Time: 3 hours Max. Marks: 75

Answer any five questions


---

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





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]

********





Code No:A109210402 R09 SET-3


SIGNALS AND SYSTEMS


Time: 3 hours Answer any five questions

Max. Marks: 75


---

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.





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]

********





Code No:A109210402 R09 SET-4


SIGNALS AND SYSTEMS


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.





(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





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.





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





Code No: A109210402 R09 Set No. 4


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 transforms 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.





(b) Explain the Sampling theorem for Band Limited Signals with Graphical proof. [7+8] 4





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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Code No: A109210402 R09 Set No. 1


SIGNALS AND SYSTEMS Common to BME, ICE, ETM, EIE, ECE

Time: 3 hours Max Marks: 75 Answer any FIVE Questions


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





y(t) = R

e (t

)x( 2)d What is the impulse response h(t) for this system.

7





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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Code No: A109210402 R09 Set No. 3


SIGNALS AND SYSTEMS Common to BME, ICE, ETM, EIE, ECE

Time: 3 hours Max Marks: 75 Answer any FIVE Questions


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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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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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



Department of ECE


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.



Department of ECE


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.



Department of ECE


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.



Department of ECE


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.



Department of ECE


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.



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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.



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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).



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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



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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



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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 .



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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]



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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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UNIT – 3



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UNIT – 4



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UNIT – 5



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UNIT – 6



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MULTIPLE CHOICE QUESTIONS :

SET – 1



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COURSE FILE OF SIGNALS & SYSTEMS - CMRCET · 2021. 6. 7. · 3. Signals & Systems-Simon Haykin and...

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