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E C E 2 2 2 1 S I G N A L S A N D S Y S T E M S

ECE 2221 Signals and Systems, Sem 3 2010/2011, Dr. Sigit Jarot

1

Course Objectives

Dr. Sigit PW JarotECE 2221 Signals and Systems 2

To provide an analysis skill of the continuous‐time signals and systems as reflected to their roles in engineering

practice.

To expose students to both the time‐domain and frequency‐domain methods of analyzing signals and

systems.

To illustrate the potential applications of this course as a Pre‐requisite course to communication engineering and principles, digital signal processing and control system.

OBE (Outcome Based Education)Learning Outcomes


Classify, characterize and conduct basic of signals and systems.

Analyze continuous‐time signals and systems in time domain using convolution.

Analyze continuous‐time signals and systems in frequency domain using Laplace transform.

Analyze continuous‐time signals and systems in frequency domain using Fourier series and Fourier transform.

Acquire introductory‐level knowledge of discrete‐time signals and systems, and sampling theory.

Work in group to perform basic simulation of signals and systems analysis.

After completion of this course the students will be able to:

Course Synopsis


Introduction to Signals

Introduction to Systems

Time‐Domain Analysis of Continuous‐Time Systems

Frequency‐Domain System Analysis: the Laplace Transform

MID‐TERM Examination

Signals Analysis using the Fourier Series

Signals Analysis using the Fourier Transform

Introduction to Discrete Time Signals and Systems Analysis

FINAL Examination

Laplace Transform

The Laplace Transform Finding the Inverse Transform Pole‐Zero Region of Convergence (ROC)

Some Properties of the Laplace Transform Solution of Differential and Integro‐Differential Equations Zero‐State Response Stability Inverse System

Block Diagrams Bilateral Laplace Transform Filter Design by Placement of Poles and Zeros of H(s)

Dr. Sigit PW JarotECE 2221 Signals and Systems

5

Why frequency domain ?

We’ve been discussing about systems strictly in the time‐domain, all signals were represented as functions of time.

It is often much easier to analyze signals and systems when they are represented in the frequency domain.

Subject of Signals and Systems consists primarily on the following concepts:1. Writing signals as functions of frequency2. Looking at how systems respond to inputs of different frequencies3. Developing tools for switching between time‐domain and frequency‐

domain representations4. Learning how to determine which domain is best suited for particular

problem

Complex frequency

Frequency = the number of cycles per second of a period signal.

(this is correct definition, but now is the time to broaden your horizon)

For complete generality, we will allow this frequency to be a complex number.

Here we declare that any function written in the form of Kest (where K and s are in general complex constant) is characterized by the complex frequency s, which can be expanded as s=+ j .

Why Laplace Transform ?

Laplace transform is the dual (or complement) of the time‐domain analysis. In the time‐domain analysis, we break the input x(t) into impulsive

component, and sum the system response to all these components. In the frequency‐domain analysis, we break the input x(t) into exponentials

components of the form est, where s is the complex frequency:

Laplace transform is the tool to map signals and system behavior from the time‐domain into the frequency domain.

js

Time‐domain Analysis

h(t)x(t) y(t)

Frequency‐domain Analysis

H(s)X(s) Y(s)x(t) y(t)L L-1

Signals_and_Systems_Simon Haykin & Barry Van Veen 9

The Laplace Transform

6.1 Introduction Basic concepts:1. The Laplace transform (LT) provides a broader characterization of continuous‐time LTI

systems and their interaction with signals than is possible with Fourier transform.

Signal that is not absolutely integral Laplace transform

Fourier transform

2. Continuous‐time complex exponentials are eigenfunctions of LTI systems.The convolution of time signals becomes multiplication of the associated Laplace transform.

LTI system, h(t)x(t) y(t) = x(t) h(t)

x(t) = (t) y(t) = h(t)

Y(s) = X(s)H(s)

3. Two varieties of LT: (1) unilateral, or one‐sided, and (2) bilateral, or two‐sided. The bilateral Laplace transform (BLT) offers insight into the nature of system

characteristics such as stability, causality, and frequency response.



6.2 The Laplace Transform Complex exponential with complex frequency s = + j:

cos sin .st t te e t je t (6.1) Fig. 6.1

Figure 6.1 (p. 483)Real and imaginary parts of the complex exponential est, where s = + j.

p ; v ;s



6.2.1 Eigenfunction Property of est

1. LTI system:

LTI system, h(t)x(t) = est y(t) = x(t) h(t)

2. System output:

h(t) impulse response

*

y t H x t

h t x t

h x t d

s t

st s

y t h e d

e h e d

x(t) = est

3. Transfer function:

sH s h e d

(6.2)

st sty t H e H s e

The action of the LTI system on an input est is multiplication by the transfer

function H(s).

est Eigenfunction; H(s) eigenvalue



4. Polar form of complex‐valued transfer function:

( ) j sH s H s e H(s) amplitude of H(s); (s) phase of H(s)

j s sty t H s e e

We use s = + j to obtain

cos sin

j t jt

t t

y t H j e e

H j e t j j H j e t j

The LTI system changes the amplitude of the input by H( + j ) and shifts the phase of the sinusoidal components by ( + j ).

6.2.2 Laplace Transform Representation1. A representation of arbitrary signals as a weighted superposition of eigenfunctions est:

Substituting s = + j into Eq.(2) and using t as the variable of integration, we obtain

j t

t j t

H j h t e dt

h t e e dt



H( + j ) is the Fourier transform of h(t)et.2. Inverse Fourier transform of H( + j ):

12

t j th t e H j e d

12

12

t j t

j t

h t e H j e d

H j e d

(6.3)

Substituting s = + j and d = ds/j into Eq.(6.3) we get

12

j st

jh t H s e ds

j

(6.4)

The limits on the integral are also a result of the substitution s = + j

H(s) is the Laplace transform of h(t) and h(t) is the inverse Laplace transform of H(s).

3. Laplace transform of x(t):

stX s x t e dt

(6.5)



4. Inverse Laplace transform of X(s):

12

j st

jx t X s e ds

j

(6.6)

Notation:

x t X sL

Eq. (6.6) represents the signal x(t) as a weighted superposition of complex

potentials est. The weights are proportional to X(s).

6.2.3 Convergence1. A necessary condition for convergence of the Laplace transform is the absolute

integrability of x(t)et:

tx t e dt

2. The range of for which the Laplace transform converges is termed the region of convergence (ROC).

Example:Fourier transform of x(t) = etu(t) does not exist, but if > 1, x(t)e t = e (1 )tu(t) is absolutely integrable, and so the Laplace transform, which is the Fourier transform of x(t)et, does exist. Fig. 6.2.



Figure 6.2 (p. 485)The Laplace transform applies to more general signals than the Fourier transform does. (a) Signal for which the Fourier transform does not exist. (b) Attenuating factor associated with Laplace transform. (c) The modified signal x(t)e‐t is absolutely integrable for > 1.

s

6.2.4 The s‐Plane1. Complex s‐plane: Fig. 6.3.2. Complex frequency: s = + j

Horizontal axis of s‐plane = real part of s; vertical axis of s‐plane = imaginary part of s.3. Relation between FT and LT:

0

X j X s

(6.7)

4. The j‐axis divides the s‐plane in half: left‐half and right‐half s‐plane.



Figure 6.3 (p. 486)The s‐plane. The horizontal axis is Re{s} and the vertical axis is Im{s}. Zeros are depicted at s = –1 and s = –4 2j, and poles are depicted at s = –3, s = 2 3j, and s = 4.

s ; v

6.2.5 Poles and Zeros

1. Laplace transform X(s) in terms of a ratio of two polynomials:

1

1 01

1 1 0

M MM M

N NN

b s b s bX ss a s a s a

1

1

MM kk

Nkk

b s cX s

s d

ck = zeros of X(s); dk = poles of X(s)

Fig. 6.3: pole; zero

Exponential Function est (1)

Exponential function is very important in signals & system, and the parameter s is a complex variable given by:

Therefore

Since the conjugate of s

Then

and

js

)sin(cos)( tjteeeee ttjttjst

js

)sin(cos)( tjteeeee ttjttjts

)(cos 21 tsstt eete


If =0, then we have the function e jt, which has a real frequency of

Therefore the complex variable s = + j is the complex frequency

The function est can be used to describe a very large class of signals and function. Here are a number of examples:1. A constant k = ke0t ( s = 0 )2. A monotonic exponential et ( s = , = 0 )3. A sinusoid cos t ( s = j, = 0 )4. An exponentially varying sinusoid et cos t ( s = j )

The Complex Frequency Plane s = + j

+ j

- j

s on x‐axis

s on y‐axiss on left of y‐axis s on right of y‐axis

s plane

The form est are Eigenfunction of CT

Why complex exponentials of the form est are indeed eigenfunctions of continuous‐time systems.

CT convolution formula

Substitute in x(t)=est input

Move est outside the integral

Use definition of Laplace transform

dtxhty )()()(

dehty ts )()()(

dehety sst

)()(

stesHty )()(

What is the meaning of the function of form est is Eigenfunction of CT ?

Function of the form est are indeed eigenfunctions of continuous‐time systems.

It means that when such a function is an input to a CT LTI system, the output is a function of the exact same complex frequency as the input, except it is multiplied by scaling factor.

Continuous‐timeLTI systemest H(s)est

Output has same frequency as input

The coefficient H(s) is in general a complex number.

Definition of Two‐sided (Bilateral) Laplace Transform

For a signal x(t), its Laplace transform is defined by:

The signal x(t) is said to be the inverse Laplace transform of X(s). It can be shown that

where C is a constant chosen to ensure the convergence of the first integral.

This general definition is known as two‐sided (or bilateral) Laplace Transform

dtetxsX st)()(

jC

jC

stdsesXj

tx )(21)(

Definition of One‐sided (Unilateral) Laplace Transform

Symbolically, it can be shown as X(s)= L [x(t)] and x(t)= L-1 [X(s)] . It is also a common practice to use a bidirectional arrow to indicate a Laplace transform pair x(t) X(s) .

Note that: L-1 {L [x(t)]}=x(t) and L {L -1 [X(s)]}=X(s) .

If all the signals are causal its Laplace transform is defined by:

One‐sided (unilateral) vs. Two‐sided (bilateral):• Unilateral is used in the solution of differential

equation with initial condition and analysis of causal LTI CT systems

• Bilateral is used in the steady‐state (zero initial condition) analysis of LTI CT system and non‐causal system

0

)()( dtetxsX st

Linearity of the Laplace Transform

The principle of superposition holds in Laplace transform

Dr. Sigit PW JarotECE 3211 Communication Engineering

25

)()(

)()(

)]()([

)]()([

2211

2211

2211

2211

sXasXa

dtetxadtetxa

dtetxatxa

txatxa

stst

st

L

Unilateral Laplace Transform Pairs (1)

Finding inverse Laplace transform requires integration in the complex plane – beyond scope of this course.

So use a Laplace transform table (analogous to the convolution table)

Region of Convergence (ROC)

Also called the region of existence. The region of convergence (ROC) of a Laplace transform is defined as the set of values of s for which the Laplace transform integral can be evaluated.

There are several rules that can be used to quickly determine the ROC without performing any integration. All that is needed is the type of signals and the location of the poles of the transform.

A pole is defined as a value of s that causes the denominator of the transform to become zero.

Types of Signals

0

. . .

left‐sided signal(starts somewhere, ends at ‐∞)

0

right‐sided signal(starts somewhere, ends at +∞)

0

two‐sided signal(starts at ‐∞, ends at +∞)

0

finite‐duration signal(starts somewhere, ends somewhere)

. . .

. . . . . .

Rules of the ROC

The ROC is always a region of the s‐plane to the left or right of a vertical line, or a strip between two vertical line.

The ROC never contains any poles. If x(t) is right‐sided, then the ROC is right‐sided. i.e. Re{s}>a, where a is the Re{rightmost pole}

If x(t) is left‐sided, then the ROC is left‐sided. i.e. Re{s}<a, where a is the Re{leftmost pole}

If x(t) is two‐sided or the sum of a left and right sided signal, then the ROC is either a strip (a<Re{s}<b)or else the individual ROC will not overlap, producing the null set.

If x(t) is finite duration, then the ROC is the entire s‐plane.

ROC

For right-sided signal, ROC is always to the right of s plane

j

ROC, >

0

right‐sided signal(starts somewhere, ends at +∞)

. . .

ROC

j

For left-sided signal, ROC is always to the left of s plane

ROC, < - 0

. . .

left‐sided signal(starts somewhere, ends at ‐∞)

ROC

j

For two sided signal, ROC is two-sided

ROC, ‐ < <

0

two‐sided signal(starts at ‐∞, ends at +∞)

. . . . . .

Inverse Laplace Transform

In principle we can recover x(t) from X(s) via

Rather to perform the inverse Laplace transform, we will merely manipulate the given expression until we see patterns we recognize from the Laplace transform table.

This is basically a heuristic scheme and is one that will become more obvious with practice.

PFE (partial fraction expansion) is very helpful for this purpose.

jC

jC

stdsesXj

tx )(21)(

Do not use this formula !!

Properties of the Laplace Transform

Time Shifting Property Frequency Shifting Property Time‐Differentiation Property Time‐Integration Property Time‐Convolution & Frequency Convolution Property Initial‐Final Value

Linearity Property

Linearity

)()()()( sYsXtytxL

Time‐Shifting Property

Delaying x(t) by t0 (i.e. time shifting) amount to multiply its transform X(s) by e-sto.

The more accurate statement of the time‐shifting property is

0)()( 000 tesXttx st

L

0)()()(

)()()(

0000

tesXttuttx

sXtutx

stL

L

Frequency‐Shifting Property

Frequency shifting transform X(s) by s0 amounts to multiplying its time signal x(t) by

Observe symmetry (or duality) between frequency‐shift and time‐shift properties.

)()( 00 ssXetx

Lts

0)()( 000 tesXttx st

L

tse 0

Time Scaling – Frequency Scaling

Time scaling

01)(

a

asX

aatx

L

Frequency scaling

0)(1

aasX

atx

a

L

Time‐differentiation Property

First order time differentiation

Second order time differentiation

)0()()( xssXtxdtd L

)0()0()()( 22

2 x

dtdsxsXstx

dtd L

Frequency‐differentiation Property

Frequency differentiation property

)()( sXdsdttx

L

Time‐integration PropertyFrequency‐integration Property

Time‐integration Property

ssXdx

Lt )()(0

Frequency‐integration Property

dzzXttx t

s

L

)()(

Time Convolution & Frequency Convolution Property

Time‐convolution property

Convolution in time domain is equivalent to multiplication in s (frequency) domain)

)()()()( sHsXthtxL

Frequency‐convolution property

Convolution in frequency s domain is equivalent to multiplication in time domain

)()(21)()( sHsX

jthtx

L

Initial and Final Value Theorem

How to find the initial and final value of a function x(t) if we know its Laplace Transform X(s)?

Conditions : Laplace transforms of x(t) and dx/dt

exist. X(s) numerator power (M) is less than

denominator power (N), i.e. M<N.

Conditions: Laplace transforms of x(t) and

dx/dt exist. sX(s) poles are all on the LFP or

origin.

)(lim)0()(lim0

ssXxtxst

Initial Value Theorem

)(lim)()(lim0

ssXxtxst

Final Value Theorem

Introduction to the Laplace Transform

(REV I EW )

49

Definition of the Laplace Transform

The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, given F(s).

js

dtetfsFtf st

where

)()()]([ :LT sided-One0

L

50

)( )( :Pair LT

)(21)()]([ :LT Inverse

)Re( )(

:Region eConvergenc

1

1

1

0

sFtf

dsesFj

tfsF

sdttfe

j

j

st

ct

L

51

Region of Convergence

52

Example

Determine the Laplace transform of each of the following functions:

(a) u(t), (b) e‐atu(t), a0, (c) (t). Solution:

(a)

53

Example

sss

es

dtetu stst

1)1(1)0(1

11)]([0

0

L

54

Example

ase

as

steetue

tas

statat

11

)]([

0

)(

0L

55

Example

1)()]([ 00

edtett stL

56

Example

Determine the Laplace transform of function: sin t u(t). Solution:

22

0)()(

00

1121

)(21

2)(sin][sin)(

sjsjsj

dteej

dtejeedtettsF

tjstjs

sttjtj

stL

57

Properties of the Laplace Transform

Linearity

Scaling

Time shift

)()()]()([ 22112211 sfasfatfatfa L

asF

aatf 1)]([L

)()]()([ sFeatuatf saL

58

Frequency Shift

)()]([ asFtfe at L

59

Time Differentiation

)0()0(

)0()(

)0()0()()]([

)0()()]([

)1(02

1

2

nn

nnn

n

fsfs

fssFsdt

fd

fsfsFstf

fssFtf

L

L

L

60

Time Integration

Frequency Differentiation

)(1)(0

sFs

dttft

L

dssdFttf )()]([ L

n

nnn

dssFdtft )()1()]([ L

61

Time Periodicity

Initial and Final ValuesTse

sFsF

1)()( 1

)(lim)0( ssFfs

)(lim)(0

ssFfs

ch15 Introduction to the Laplace Transform

64

65

Example

Obtain the Laplace transform of f(t) = (t) + 2u(t) – 3e–2t,t 0.

Solution:

)2(4

213121

)]([3)]([2)]([)(2

2

ssss

ss

tuetutsF tLLL

66

Example

Determine the Laplace transform of f(t) = t2sin2t u(t). Solution:

32

2

22

22

222

22

)4(1612

)4(4

42)1(]2sin[)(

22]2[sin

ss

ss

dsd

sdsdttsF

st

L

L

67

Example

Find the Laplace transform of the gate function

68

Example

)(1010)(

)]3()2([10)(

3232

ssss

eess

es

esG

tututg

69

Example

Find the Laplace transform of the periodic function in this figure.

70

Example

)1()1(

21

)()(

)1(2222)(

)1(2)1()1(2)(2)1()11(2)(2

)1(2)(2)]1()([2)(

221

222

1

sssTs

ssss

seeese

sFsF

sees

ess

es

sF

tututttututttu

ttuttutututtf

71

Example

Find the initial and final values of the function whose Laplace transform is

Solution:)258)(3(

20)( 2

ssssH

0)001)(01(

0)/25/81)(/31(

/20lim

)258)(3(20lim)(lim)0(

2

2

2

ssss

ssssssHh

s

ss

72

Example

0)2500)(30(

0)258)(3(

20lim

)(lim)(

20

0

ssss

ssHh

s

s

73

The Inverse Laplace Transform

Steps to Find the Inverse Laplace Transform:1. Decompose F(s) into simple terms using partial fraction

expansion.2. Find the inverse of each term by matching entries in Table

74

Simple Poles

)()()(

)()(

)(

2121

2

2

1

1

tuekekektf

SFpsk

psk

psk

psksF

tpn

tptp

psii

n

n

n

i

75

Repeated Poles

)(

)()!1(!2

)(

)()()()(

)(

1

12321

11

21

2

1

1

tf

tuetn

ketktekektf

sFps

kps

kps

ksF

ptnnptptpt

nn

76

Complex Poles

)()()coscos()(

)()()(

)(

)()()(

)()(

)()(

111

1221

221

12211

12221

1221

tftuteBteAtf

sFs

Bs

sA

sFs

BsA

sFs

AsA

sFbass

AsAsF

tt

77

Example

Find the inverse Laplace transform of

Solution:4

61

53)( 2

ssssF

0 ,)()2sin353(4

61

53)]([)( 21111

ttutesss

sFtf

t

LLLL

78

Example

Find f(t) given that

Solution:

)3)(2(12)(

2

sss

ssF

2)3)(2(

12)3)(2(

12)(

32)3)(2(12

0

2

0

2

ss ss

sssFA

sC

sB

sA

ssss

79

Example

7)1)(3(

129)2(

12)()3(

8)1)(2(

124)3(

12)()2(

3

2

3

2

2

2

ss

ss

ssssFsC

ssssFsB

80

Example

Find v(t) given that

Solution:2

2

)2)(1(410)(sss

ssV

2)2(1

)2)(1(410)(

2

2

2

sD

sC

sB

sA

sssssV

81

Example

134

52)(

)12)(410()20)((

410)]()2[(

22)1)(2(

44)1(410)()2(

14)1)(1(

14)2(410)()1(

1)2)(1(

4)2)(1(

410)(

222

222

2

2

2

2

2

2

22

21

2

2

1

20

2

2

0

s

ss

ss

ss

ss

sssssss

sss

dsdsVs

dsdD

ssssVsC

ssssVsB

sssssVA

82

Example

Find the inverse transform of the frequency‐domain function in Example 15.7:

Solution:)258)(3(

20)( 2

ssssH

)258(3)258)(3(20)(

let We340258

22

2

ssCBs

sA

ssssH

jsss

83

Example

.2 that so ,10 ,2But

443420344)34)(4(

201

.102 Since

3252025375

20

21020

25820)()3(

323

BCA

CBACBAs

CA

CACAss

sHsAs

s

84

The Convolution Integral

The convolution of two signals consists of time‐reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product.

t

dthxtxthty0

)()()(*)()(

)](*)([)()( 2121 tftfsFsF L

85

Application to Integro‐differential Equations

Example Use the Laplace transform to solve the differential equation

subject to Solution:

)(2)(8)(6)(2

2

tutvdt

tdvdt

tvd

.2)0(,1)0( vv

ssVvssVvsvsVs 2)(8)]0()([6])0()0()([ 2

86

Example

42)4)(2(24)(

Hence,

2424)()86(

2)(86)(62)(

.2)0( ,1)0(

2

22

2

sC

sB

sA

ssssssV

sss

sssVss

ssVssVssVs

vv

87

Example

41

)2)(4(2

)2(24)()4(

21

)2)(2(2

)4(24)()2(

41

)4)(2(2

)4)(2(24)(

where

4

2

4

2

2

2

0

2

0

ss

ss

ss

sssssVsC

sssssVsB

ssssssVA

88

Example

)()21(41)(

42)(

Hence,

42

41

21

41

tueetv

ssssV

tt

89

Example

Solve for the response y(t) in the following integrodifferential equation.

Solution:2)0( ),()(6)(5

0 ytudyty

dtdy t

ssY

ssYyssY 1)(6)(5)]0()([

90

Example

515

212)()3(

313

312)()2(

where32)3)(2(

12)(

21)65)((

2)0(

33

22

2

ss

ss

sssYsB

sssYsA

sB

sA

ssssY

ssssY

y

91

Example

)()53()(3

52

3)(

Thus

32 tueetyss

sY

tt

System Stability

EC E 2 2 2 1

What is a Stable System ?

Ball sitting in valley with rough sides.An impulse input will set ball in motion, but the ball returns to the resting state

Ball sitting in valley on frictionless surface.An impulse input will cause the ball to be set in motion forever

Ball sitting on top of a hill.One small push and the ball keeps on falling forever

What is a Stable System ?

BIBO Stable (a.k.a. stable in BIBO sense) If bounded inputs (i.e. finite amplitude, no matter how large) produce bounded

outputs (i.e. finite output, never go to infinity)

If the input stops and the output slowly decays away, or else the input continues and the output signal behaves itself (doesn’t blow up)

If the output remains at a constant size forever even though the input signal has long since stopped

If the input stops or remains finite and the output signal continues to grow larger and larger until reaching the infinite proportions.

Stable

Marginally Stable

Unstable

Conditions for Stability

A system is BIBO stable if and only if its impulse response is absolutely summable or integrable.

A system is BIBO stable if and only of the ROC includes the j‐axis(continuous‐time system or the unit circle (discrete‐time system)

(1) Impulse Response

(2) ROC of System Function

kkhordhStabilityBIBO )()(

Im

Re

Im

Re1

For causal system, stability means that all poles are in the shaded area.(locations of zeros do no affect system stability

Conditions for Stability

All poles are in the left‐half plane. This implies that the roots of the characteristic equation (denominator of

the system function H(s)) are all negative, or at least have negative real parts.

(3) Characteristic Equations

0 as 0a102 bass 0,0 ba2

023 cbsass cabandcba 0,,3

System Order

Characteristic Equation

Stability Criteria

Causal, stable systems

The system function of an LTI system, H(s), can be used to categorize systems. Causal system ⇒ ROC is to the right of the rightmost pole of H(s). Causal, stable system ⇒ ROC is to the right of the rightmost pole of H(s) and all the poles are in the left‐half of the s plane.

More on the definition of stable systems at a later time.

Block Diagram

ECE 2 2 2 1 S I GNA L S AND SYST EMS

Block Diagram

The transfer function is given inside the block The input in this case is E(s) The output in this case is C(s) C(s) = G(s) E(s)

Cascaded System

Parallel System

Feedback System

Block Diagram Algebra

Example 1

collapse summing junctions

form equivalent cascaded system in the forward path and equivalent parallel system in the feedback path

form equivalent feedback system and multiply by cascaded G1(s)

Example 2

Applications of the Laplace Transform

108

Applications of LT

Introduction Circuit Element Models Circuit Analysis Transfer Functions State Variables

109

Introduction

A system is a mathematical model of a physical process relating the input to the output.

110

Circuit Element Models

Steps in Applying the Laplace Transform:1. Transform the circuit from the time domain to the s‐domian.2. Solve the circuit using nodal analysis, mesh analysis, source

transformation, superposition, or any circuit analysis technique.3. Take the inverse transform of the solution and thus obtain the

solution in the time domain.

111

sisI

sCsV

sisV

sLsI

sRIsV

)0()(1)(

)0()(1)(

)()(

112

)()( )()( :Resistor

sRIsVRtitv

113

si

SLsVsI

LissLIsV

issILsVdtdiLtv

)0()()(

)0()()(

))0()(()( )(

:Inductors

114

Inductor

115

Capacitor

sv

SCsIsV

CvssCVsI

vssVCsIdtdvCti

)0()()(

)0()()(

))0()(()( )(

:Capacitors

116

Capacitor

117

Circuit with Zero Initials

sCsZ

sLsZRsZ

sIsVsZ

1)( :Capacitor

)( :Inductor)( :Resistor

)()()(

118

)()]([)()]([

)()(

)(1)(

saItaisaVtav

sVsI

sZsY

LL

119

Example

Find vo(t), assuming zero initial conditions.

120

Example

ssC

ssLs

tu

31F 31

H 1

1)(

121

Example

sssI

Isss

Is

Issss

IssIIs

sIs

Is

Iss

1883

)188(3

3)35(31311

)35(313530 2,mesh For

3311 1,mesh For

232

223

222

22

121

21

122

Example

0 V,2sin2

3)(

)2()4(2

23

1883)(

4

2222

ttetv

ssssIsV

to

o

123

Find vo(t). Assume vo(0)=5 V.

124

Example

)2(101

101025.2

11

/10105.02

101)10/(

sVsVVs

sVVVs

ooo

ooo

125

Example

151

15)1(3525)()2(

101

10)2(

3525)()1(

where21)2)(1(

3525

)2(251

10

22

11

sso

sso

o

o

sssVsB

sssVsA

sB

sA

sssV

sVs

126

Example

V )()1510()(2

151

10)(

Thus

2 tueetvss

sV

tto

o

127

Example

In the circuit, the switch moves from position a to position b at t = 0. Find i(t) for t > 0.

128

Example

LRtRVe

RVIti

LRsRV

sRV

LRsIsI

LRssLV

LRsI

sLRsV

sLRLIsI

sVLIsLRsI

otoo

ooo

oooo

oo

/ where0 ,)(

)/(//

/)(

)/(/

/)()(

0))((

/

129

Example

0 ),1()(

,0 condition, initial In the

0 ),1()(

//

/lim)(lim

,/)(valueinitial The

/

//

00

teRVti

I

teRVeIti

RV

LRsLV

LRssIssI

RVi

to

o

toto

oooss

o

130

Circuit Analysis

Remember, equivalent circuits, with capacitors and inductors, only exist in the s‐domain; they cannot be transformed back into the time domain.

131

Example

Find the value of the voltage across the capacitor assuming that the value of vs(t) = 10u(t) V and assume that at t = 0, ‐1 A flows through the inductor and +5 V is across the capacitor.

132

Example

V )()3035()( So,

230

135

)2)(1(540

540)23(

A 1)0( and V 5)0( e wher

5.013231.0

0)1.0/(1

]/)0([)0(5

03/10/10

21

1

12

1

111

tueetv

sssssV

sVss

ivss

Vs

s

ssvV

si

sVsV

tt

133

Example

Assume that there is no initial energy stored In the circuit at t = 0 and that is = 10 u(t). (a) Find vo(s) using Thevenin’s theorem. (b) Apply the initial‐ and finial‐value theorems to find vo(0+) and vo(). (c) Determinevo(t).

134

Example

(a)

32100

02

05

0)2(102/

50105

,0 Since

1

11

1sc

THoc

sV

sVIV

s

sVIIss

VV

I

x

x

x

135

Example

)4(125

)82(25050

3255

55

32)]32(/[50

/50)32(

502

)32/(1002

Hence,

TH

sc

ocTH

1sc

ssssssV

ZV

ssss

IVZ

ssss

sVI

THo

136

(b) Using the initial‐value theorem we find

Using the finial‐value theorem we find

010

/41/125lim

4125lim)(lim)0(0

ss

sssVv

ssos

V 25.314

1254

125lim)(lim)(00

s

ssVvsoso

137

Example

138

Example

(c) By partial fraction,

V )()1(25.31)(4

25.3125.31

25.31125)()4(

25.314

125)(

4)4(125

4

74

00

tuetvss

V

ssVsB

sssVA

sB

sA

ssV

to

o

sso

sso

o

139

Transfer Functions

The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero.

)()()(

sXsYsH

140

)()(Admittance)(

)()(Impedance)(

)()(gainCurrent )(

)()(gain Voltage)(

sVsIsH

sIsVsH

sIsIsH

sVsVsH

i

o

i

o

141

)]([)( where

)()(or )()(1)( that so ),()(

)()()(

1 sHth

thtysHsYsXttx

sXsHsY

L

142

Example

The output of a linear system is y(t) = 10e‐t cos4t u(t) when input is x(t)=e‐tu(t). Find the transfer function of the system and its impulse response.

Solution: If x(t)=e‐tu(t) and y(t) = 10e‐t cos4t u(t), then

172)12(10

16)1()1(10

)()()(

Hence,4)1()1(10)( and

11)(

2

2

2

2

22

ssss

ss

sXsYsH

sssY

ssX

143

Example

To find h(t),

)(4sin40)(10)(

4)1(44010)( 22

ttuetth

ssH

t

144

Example

Determine the transfer function H(s) = Vo(s) / Io(s) of the circuit

145

Example

1122)4(4

)()()(

Hence,2/15

)4(22

But2/124

)4(

2

2

2

ssss

sIsVsH

ssIsIV

ssIsI

o

o

oo

o

146

Example

For the s‐domain circuit in Fig. 16.19, find: (a) the transfer function H(s) = Vo/Vi, (b) the impulse response, (c) the response when vi(t) = u(t) C, (d) the response when vi(t) = 8cos2t V.

147

Example

(a)

(b)

321)( Thus, .

32 So,

321

)2/()1(1)2/()1(

)1(11)1(1

but ,1

1

sVVsH

sVV

VssVV

ssssV

ssV

Vs

V

i

oio

iabiiab

abo

)(21)(

121)(

2/3

23

tuesh

ssH

t

148

Example

(c)

31

21)(

23

31

)(21)(

where)(2

2)()()(

/1)(),()(

2/32/3

0230

23

23

sso

sso

io

ii

ssVsB

sssVA

sB

sA

sssVsHsV

ssVtutv

149

Example

(d)

V )()1(31)(11

31)(

),()(for Hence,

2/3

230 tuetv

sssV

tutv

to

i

2524

44)(

23

where4)4)((

4)()()(

and ,4

8)( then ,2cos8)(When

2/32

2/3

2232

23

2

sso

io

ii

sssVsA

sCBs

sA

ssssVsHsV

sssVttv

150

Example

23

23)4(4 22 sCssBsAs

ABBAs

CBs

ACCA

0:234:

38

2340:Constants

ts,coefficien Equating

2

151

Example

V )(2sin342cos

2524)(

42

2532

42524)(

V, 2cos8)(for Hence,64/25.24/25,,25/24gives theseSolving

2/3

2223

2524

tuttetv

sss

ssV

ttvCBA

to

o

i

152

State Variables

153

A state variable is a physical property that characterizes the state of a system, regardless of how the system got to that state.

vectors.state ingresponsent vector state

)(

)()(

)(

where

2

1

n

tx

txtx

t

n

x

BzAxx

154

input. ngrepresentior input vect

)(

)()(

)(

and)(

)()(

2

1

2

1

m

tz

tztz

t

tx

txtx

n

n

z

x

155

DzCxyBzAxx

matrices. and ly respective are and matrices. andly respective are and

outputs ngrepresentitor output vec the

)(

)()(

)(

where

2

1

m pnpmnn n

p

ty

tyty

ty

p

DC BA

156

matrixdfeedforwar matrixoutput

matrix couplinginput matrix system re whe

)()()()(

)()()(matrix.identity theis e wher)()()(

)()()( )()()(

1

1

DCBA

DBAICZYH

DZCXYI

BZAIX

BZXAIBZXX

ssss

sss

sss

sssssAss

157

Steps to Apply the State Variable Method to Circuit Analysis:

1. Select the inductor current i and capacitor voltage vas the state variables, making sure they are consistent with the passive sign convention.

2. Applying KCL and KVL to the circuit and obtain circuit variables (voltage and currents) in terms of the state variables. This should lead to a set of first‐order differential equations necessary and sufficient to determine all state variables.

3. Obtain the output equation and put the final result in state‐space representation.

158

Example 16.10

Find the state‐space representation of the circuit. Determine the transfer function of the circuit when vs is the input and ix is the output. Take R = 1Ω, C = 0.25 F, and L = 0.5 H.

159

Example

Ci

RCvv

RVi

dtdvCiii

dtdvCi

dtdiLv

Cx

C

L

gives 1 nodeat KCL Applying

iv

Ri

viv

iv

Rvi

Lv

Lvi

vvdtdiLvvv

x

sLL

CRC

s

sLs

01

00

,

11

11

160

Example

ss

ss

s

R

LCR

LL

CRC

244

0244

00

0101

,200

,0244

0

obtainwe,and, ,1 If

11

1121

41

AI

C

B A

161

Example

Taking the inverse of this gives

848

8482

801

8420

424

01

)()(8442

4

oft determinan ofadjoint )(

222

1

21

sssss

sss

sss

sss

s

s

BAICHA

AAI

162

Example

Consider the circuit in Fig. 16.24, which may be regarded as a two‐input, two‐output. Determine the state variable model and find the transfer function of the system.

163

Example

At node 1, KCL gives

At node 2,

vviv

iviiiv

os

ss

1

1 66061

ss

s vivivviiiviv

iivvii

4423

223

)(22

1

100

1

)(23

322

33

23

31

2

ss

o

ioooo

o

vivivviv

vviivvivv

164

Example

i

s

o

o

i

s

s

vv

iv

iv

vv

iv

iv

vvivv

31

23

31

32

32

1

00

0

0411

4212

2

165

Example

Assume we have a system where the output is y(t) and the input is z(t). Let the following differential equation describe the relationship between the input and output

Obtain the state model and the transfer function of the system.

)(5)(2)(3)(2

2

tztydt

tdydt

tyd

166

Example

321

3210

1001

01)(

)(50

3210

)(532)(5)(3)(2)()(let Now)(),(Let

2

1

2

1

2

1

212

12

1

ss

ss

xx

t

tzxx

xx

tzxxtztytytyxtyxxtyxtyx

AI

y

167

Example

)2)(1(5

2)3(55

01

2)3(50

213

)01()()(

isfunction transfer The2)3(

213

)( is inverse The

1

1

sssss

sss

s

ss

sss

s

s

BAICH

AI

168

Example

235

)()()(

)(5)(]23[

2

2

sssZsYsH

sZsYss

ECE 2221 SIGNALS AND SYSTEMS - International …staff.iium.edu.my/sigit/ece2221/pdf/04 Laplace...

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Transcript of ECE 2221 SIGNALS AND SYSTEMS - International …staff.iium.edu.my/sigit/ece2221/pdf/04 Laplace...