System analysis in the time domain involves ( finding y(t)):

C

R

( )x t

( )y t

( ) ( ) ( )dy tRC y t x t

dt Solving the differential equation

Using the convolution integral

OR

Both Techniques can results in tedious (ممل )mathematical operation

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

Chapter 7 The Laplace Transform

Consider the following RC circuit ( System)

Fourier Transform provided an alternative approach

C

R

( )x t

( )y t

( ) ( ) ( )dy tRC y t x t

dt

Differential Equation

( ) ( ) ( ) ( )RC j Y Y X

Algebraic Equation

( )( ) ( ) 1

XYj RC

solve for ( )Y

Inverse Back( )y t

( ) ( ) ( )dy tRC y t x t

dt

C

R

( )x t

( )y t

( ) ( ) ( )dy tRC y t x t

dt

both side

( ) ( ) ( ) ( )RC j Y Y X

Fourier Transform

Solve for ( )

( )( ) ( ) 1

Y

XYj RC

( )y t

Unfortunately , there are many signals of interest that arise in systemanalysis for whi(not

ch theabsolutley integrable)

( ) x t dt

Example

Fourier Transform does not exists

2 ( ) , ( ) , ( ) tx t t x t t x t e

Inverse Back

( ) 1 2 ( )

1( ) ( ) ( ) +

( ) ( ) [ ( )+ ( )]

In fact signals such as

are not strictly integrable and their Fourier transforms

all contain some non conventional fu io

nct

o o o

x t

x t u tj

x t cos t

n such as .( )

A more general transform is needed

Suppose we have a function or signal that is not absolutely integrable as shown x t

Now multiply ( ) by tx et

The signal ( ) is absolutley integrable tex t The signal ( ) has Fourier Transform tex t

1( ) ( ) ( ) ( )2

j t j tx t X e d X x t e dt

Fourier Transform pairs was defined as

. ( ) ( )t t j te eF T x t x t e dt

( )( ) j tx t e dt

( )X j

Let us deined this ( ) as Complex Frequency jsj

0

. ( ) ( ) tt sF T x t x te e dt

( )X s L ( )b x t

were Denotes the operation of o Bilateral Laplace Transfobtaining the rm L b

Now multiply ( ) by and takes the Fourier Transform tx t e

Fourier Transform of ( )

The domain is not but

te x t

j

7.1 DEFINITIONS OF LAPLACE TRANSFORMS

The Laplace transform variable is complex

This show that the bilateral Laplace transform of a signal can be interpreted as the Fourier transform of that signal multiplied by an exponential function te

The Laplace transform of a time function is given by the integral

This definition is called the , or , Laplace transform — hence, the subscript bbilateral two-sided

Notice that the bilateral Laplace transform integral becomes the Fourier transform integral if      is replaced by  (i.e, 0)s j

The inverse Laplace transform

where

This transform is usually called, simply, the Laplace transform , the subscript b dropped

Laplace transform pairs

For most application , ( )  is zero for     0f t t

Unilateral (single sided) Laplace transform pairs

Forward transform

Inverse transform

Because of the difficulty in evaluating the complex inversion integral

Simpler procedure to find the inverse of Laplace Transform (i.e f(t) ) will be presented later

Complex domain

Using Properties and table of known transform ( similar to Fourier)

0

L ( ) ([ ] ) tst t e dt

0

L ( ) ([ ] ) tst t e dt

0

ts

t

e

1

The value of s make no difference

Assuming the lower limit is 0

Region of Convergence

j domains

0

Region of Convergence

We also can drive it from the shifting in time property that will be discussed later

Laplace Transform for an impulse (t)

( )1lim 1j t

te

s

1

lim 1t j t

te e

s

For this to converge then > 0lim 0t

te

region of convergence (ROC)

Re(s) > 0

is were the Laplace Transform Function ( )F s Pole

We derive the Laplace transform of the exponential function

A short table of Laplace transforms is constructed from Examples 7.1 and 7.2and is given as Table 7.2

We know derive the Laplace Transform (LT) for a cosine function

Since

Proofat at st

0

[ ] [ ( )]e ( ) e eL f t f t dt

(s+a)t

0

( )ef t dt

Let s*= s+a at s*t

0

e ( ) ( )[ ] eL f t f t dt

( *) ( )F s F s a

0 20

2[ ] L cos t s

s

0 220

0

[ ] L sin ts

Example

0 L cos[ ] t 0 01 1

2 2 =L j t j te e

0 01 1

2 2[ ] = L + L[ ]j t j te e

1 L ( )[ ] = ( )

t

se u t

Since

Then 0

0

1L =(

[ ] )

j tej s

0

0

1L =( )

[ ] j tesj

00 0

1 11 1 L cos ( ) (2

2

])

[ s s

tj j

2 2

0

ss

0 220

0

[ ] L sin ts

Similarly

Real Shifting

Let

0.3( ) 5 tf t e 50.3s

0.3( ) 5 tf t e

2 0 0we need to put the function ( ) in a delay form ( ) ( ) in order to use

the Laplace Transform for time shift

f t f t t u t t

It is sometimes necessary to construct complex waveforms from simpler waveforms

(See example 7.3)

(Shift in time property)Since

Transform of Derivatives

( )L ( ) (0 )

df tf

dsF

ts

Proof

0

( ) ( )L

tsdf t df te dt

dt dt

Integrating by parts, ( ) ( )tsu e dv t df t

( ) ( )tsdu e v t ts f

b bb

aa a

udv uv vdu

0

( ) ( )L

tsdf t df te dt

dt dt

( ) ( )tsu e dv t df t ( ) ( )stdu e v t f ts

b bb

aa a

udv uv vdu

00

( ) + ( ) st stse f t f t e dt

( ) (0) ( ) (0 ) + ( ) s se f e f Fs s

0 1

( ) (0 ) F fs s

( ) ( ) (0 ) s sdf t F fdt

2( ) 2 ( )ti t e u t

=0

=0 =0 =0

2 1

2

1

2 3

1

( ) ( ) ( )

( ) ( ) ( )

t

t t t

n

n

n

n

nn n

n

s sd f t F f tdt

df t d f t d f tdt d t

sd

s

ts

2

2 =0

=0

2 ( )( ) ( ) ( ) t

t

s s sdf td f t F f t

dtdt

( ) ( ) (0 ) df t Fd

s fst

23

23 =0

=0 =0

3 2 ( ) ( )( ) ( ) ( ) t

tt

dx t d f td s s s sf t F f tdtdt dt

Proof

find the Laplace transforms of the following integrals:

Proof (See the book)

Since

Transfer Functions

Consider the following circuit

R( )x t ( ) i t

L

C

( )y t

input output

We want a relation (an equation) between the input x(t) and output y(t)

1( ) ' '( ) ( ) + ( )

t

di tx t L Ri t i t dtdt C

KVL

2

2

( ) ( ) ( )( ) + dx t di t i tdi tL Rdt dt Cdt

7.6 Response of LTI Systems

R( )x t ( ) i t

L

C

( )y t

input output

2

2

( ) ( ) ( )( ) + dx t di t i tdi tL Rdt dt Cdt

( )Since ( )

y ti t

R

2

2

( ) ( ) ( )( ) + dx t L dy t y tdy t Rdt R dt RCdt

Writing the differential equation as 2

2

( ) ( )( ) + ( ) dx t dy tdy tRC LC RC y tdt dtdt

R( )x t ( ) i t

L

C

( )y t

input output

2

2

( ) ( )( ) + { } { } { } { } ( ) 1RC dx t dy tdy t y tdt dtd

LCt

RC

Real coefficients, non negative which results from system components R, L, C

In general,

1 1

1 0 1 01 1

( ) ( ) ( ) ( ) + + + ( ) + + + ( )n n m m

n n m mn n m mdy t dy t dx t dx ty t y t

dt dt dt da a b b

ta b

' 'were , n ms ba s are real, non negative which results from system components R, L, C

Now if we take the Laplace Transform of both side (Assuming Zero initial Conditions)

1 01

11

0( ) + ( ) + + ( ) ( ) + ( ) + + ( )n nn m

mn m

ma a a b bs s s s s s s s s sY Y bY X X X

We now define the transfer function H(s) ,

all initial conditions are zero

( )( )

( )

YH

X

ss

s 1 0

1

11 0

+ + +

+ + + m m

n

m m

n nn

b b b

a a a

s s

s s

1 1

1 0 1 01 1

( ) ( ) ( ) ( ) + + + ( ) + + + ( )n n m m

n n m mn n m mdy t dy t dx t dx ty t x t

dt dt dt da a b b

ta b

1 0

11 0

1 + + +

+ + +

( )( )

( )m

m

n

m

n n

m

n

b b b

sa

s

as a

sYH

ssX

s

' 'Sin ce , n ms ba s are real, non negative

( )

( )N sD s

The roots of the polynomials N(s) , D(s) are either real or occur in complex conjugate

The roots of N(s) are referred to as the zero of H(s) ( H(s) = 0 )

The roots of D(s) are referred to as the pole of H(s) ( H(s) = ± ∞ )

The Degree of N(s) ( which is related to input) must be less than orEqual of D(s) ( which is related to output) for the system to be Bounded-input, bounded-output (BIBO)

( )

( )N sD s

1 0

11 0

1 + + +

+ + +

( )( )

( )m

m

n

m

n n

m

n

b b b

sa

s

as a

sYH

ssX

s

Example : 3 2

2

4 + 2 1 6 8

( ) s ss

ss

sH

Using polynomial division , we obtain 2

19 174 + 2 + 6 8

( ) sss s

sH

Now assume the input x(t) = u(t) (bounded input) 1 ( ) s sX

2

2 1 19 174 + + 6 8

( ) ( ) ( ) ss s s s

Y Xs s sH

1

2

19 174 ( ) + 2 + ( 6 8)

( ) sts s s

y t L

unbounded

)

(

We see that for finite boundedInput (i.e x(t) =u(t) )

We get an infinite (unbounded)output

for m n BIBO

( )

( )N sD s

1 0

11 0

1 + + +

+ + +

( )( )

( )m

m

n

m

n n

m

n

b b b

sa

s

as a

sYH

ssX

s

The poles of H(s) must have real parts which are negative

The poles must lie in the left half of the s-plan

Proof (See the book)

( It is very similar to the Fourier Transform Property )

The loop equation (KVL) for this circuit is given by

If v(t) = 12u(t) ,find i(t) ?

The inverse transform, from Table 7.2

v(t) = 12u(t)

Now using the Laplace Table 7.2 to find i(t)

( ) ( )v t Ri t ( ) ( )V s RI s

( )( )L

di tv t Ldt

[ ]( ) ( ) (0 )L

V s L sI is s ( ) (0 )LI s Li

1( ) ( ) (s

0 ) L

I s V s isL

Source Transformation

R

Z R

L

Z sL

( )( ) Cdv ti t Cdt

( ) ([ ( ]0 ) )C

I s C sV vs

( ) (0 )C

sC s CvV

1( ) ( ) (0 ) C S

V s I s vsC

Source Transformation1

CZ

sC

10s

8 5s

1

2 s

2 2

0.5 2.5( )

1 1X

s ss

s s

( ) I s

1 R

( ) 0.5 cos 2.5 sin ( )x t t t u t

( ) i t

0 t 1 10

L H 5 8

C F

R1 Z 1 R

2 2

0.5 2.5( ) ( )

1 1

ssx t X

s

s s

L1 Z

10 10L H s

C 5 8 Z 58

Fs

C

(0 ) 2 Cv V

(0 ) 0 Li V

Example Using Laplace Transform Method find i(t)

10s

8 5s

1

2 s

2 2

0.5 2.5( )

1 1X

s ss

s s

( ) I s

KVL8 2( ) ( ) ( ) (1) ( ) 0

10 5X I

ss s s s

s sI I

2 2

8 20.5 2.5 1 ( )10 51 1

Iss s s

s ss s

2

2 2

15 25 20( )( 1)( 10 16)

s sss s s

I

10s

8 5s

1

2 s

2 2

0.5 2.5( )

1 1X

s ss

s s

( ) I s

2

2 2

15 25 20( )( 1)( 10 16)

s sss s s

I

( Imaginary Roots)

2

2 2

15 25 20( )

( 1)( 10 16)s s

ss s s

I

2(15 25 20)( )( )( 2)( 8)j

s ss js s s

1 2 3 4

( ) ( ) ( 2) ( 8)s s sA A A A

j sj

2 8( ) cos( ) sin( ) ( )t ti t t t e e u t

Inversion of Rational Function ( Inverse Laplace Transform)

Let Y(s) be Laplace Transform of some function y(t) .

We want to find y(t) without using the inversion formula .

We want to find y(t) using the Laplace Transform known table and properties

Objective : Put Y(s) in a form or a sum of forms that we know it is in the Laplace Transform Table

Y(s) in general is a ratio of two polynomials Rational Function

2

3 2

2 Example ( ) = 2

2 5 3

s sss s

Ys

When the degree of the numerator of rational function is less theDegree of the dominator

Proper Rational Function

2

3 2

2 Example ( ) = 2

2 5 3

s sss s

Ys

Highest Degree is 2

Highest Degree is 3

Examples of proper rational Functions

1 1( ) =1

Y ss

2

22 2

2 6 6( ) =( 2)(

2 2)

sss s

Y ss

Examples of not proper rational Functions

32( ) = 1

Y sss

However we can obtain a proper rational Function through long division

32( ) = 1

Y sss

1= 1 +1s

We will discuss different techniques of factoring Y(s) into simpleknown forms

Simple Factors

2

10Let ( ) =( 0 1

1 6)

ss

Ys

If we check the Table , we see there is no form similar to Y(s)

However if we expand Y(s) in partial fractions:

2

10( 10 16)s s

( 2) ( 8)s s

A B

and ( 2) ( 8)

A Bs s Are available on the Table

Next we develop Techniques of finding A and B

Heavisdie’s Expansion Theorem

2

10 10( ) ( 10 16) ( 2)( 8) ( 2) ( 8)

ss s s s s s

A BY

Multiply ( ) both side by (s+2) and set s =2

2 2 2

10 ( 2) ( 2) ( 2)( 2)( 8) ( 2) (

X X8)

Xss s

s s ss s s s

A B

10 ( 2) ( 8) 2(

282 )

BA

10 0(6)

A 5 3

A

X

Similarly Multiply both side by (s +8) and set s =8

8 8 8

10 ( 8) ( 8) ( 8)( 2)( 8) ( 2) (

X X8)

Xss s

s s ss s s s

A B

5 3

B

2 85 5(t) = ( ) ( )3 3

t ty e u t e u t 2 85= ( )3

t te e u t

2

2 2

(15 25 20)Let ( )

( 1)( 10 16)s s

ss s s

Y

Imaginary Roots

2(15 25 20)( )( )( 2)( 8)j

s ss js s s

1 2 3 4

( ) ( ) ( 2) ( 8)s s sA A A A

j sj

Using Heavisdie’s Expansions, by multiplying the left hand side andRight hand side by the factors

( ) , ( ) , ( 2) , ( 8)s sj s sj , , 2 , 8s s s sj j and substitute respectively

We obtain 1 2 3 4

1 1(1 ) , (1 ) , 1 , 22 2

A j A j A A

(1/2)(1 ) (1/2)(1 ) 1 2( )( ) ( ) ( 2) ( 8)

ss

j jYj sjs s

2 ( )te u t 8 ( )te u t

From Table

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

j jj js s

Can be inverted in two methods:

(a)(1/2)(1 )

( )j

s j

(1/2)(1 ) ( )jtj e u t

(1/2)(1 ) ( )

js j

(1/2)(1 ) ( )jtj e u t

(1/2)(1 ) ( )

js j

(1/2)(1 ) ( )jtj e u t

(1/2)(1 ) ( )

js j

(1/2)(1 ) ( )jtj e u t

combine

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

js s

jj j

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

jt jtj e u t j e u t

1( ) ( ) ( ) ( )2 2

jt jt jt jtje e u t e e u t

cos( ) ( ) sin( ) ( )t u t t u t

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

jt jt jt jte e u t e e u tj

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

j jj js s

Can be combined as

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

j jj js s

(1/2)(1 )( ) (1/2)(1 )( )( )( )s ss

j j j jj js

2

11

ss

2 2

1 1 1

ss s

(b)

cos( ) ( ) sin( ) ( )t u t t u t

(1/2)(1 ) (1/2)(1 ) 1 2( )( ) ( ) ( 2) ( 8)

ss

j jYj sjs s

2 ( )te u t 8 ( )te u tcos( ) ( ) sin( ) ( )t u t t u t

2 8( ) cos( ) ( ) sin( ) ( ) ( ) ( )t ty t t u t t u t e u t e u t

2 8( ) cos( ) sin( ) ( )t ty t t t e e u t

Repeated Linear Factor

( )( ) ( ) ( )n

sss

Ps

YQ

If

Then its partial fraction

1 2 2 n2 3

( )( ) + + + ( ) ( )( ) ( ) ( )n

sss ss s sA A A A RY

Q

Were( )

m ( )

1 ( ) ( )( )!

n mn

n ms

s sdA Yn m ds

2

10 ( )

( 2) (exam

)e

8pl

ss

s sY

Repeated Linear Factor

2

10( )

( 2) ( 8)

ss

sY

s

Let

1 22

( )( 2) ( 8)( 2)

A A BY ss ss

Then

(2 1)2

1 2(2 1)

2

101 ( 2)(2 1)! ( 2) ( 8) s

ssd

s s sA

d

2

10

( 8)s

sdsd

s

2

2

10( 8) 10( 8)

s

s ss

2

2 210( 8) 10( )( 8)2

8036

209

( )

m ( )

1 ( ) ( )( )!

n mn

n ms

s sdA Yn m ds

2

1

n

m

2

2

n

m

Also A2 can be found using Heavisdie’s

by multiplying both sides by

B can be found using Heavisdie’s

by multiplying both sides by

(2 2)2

2 2(2 2)

2

101 ( 2)(2 2)! ( 2) ( 8) s

ssd

s s sA

d

2

10

( 8)s

ss

22

10( )( 8)

103

206

1 22

( )( 2) ( 8)( 2)

A A BY ss ss

209

103

1 22

( )( 2) ( 8)( 2)

A A BY ss ss

209

103

To find B , we use Heaviside

1 22

2

888 8

10( 8) ( 8) ( 8) ( 8)( 2) ( 8)( 2) ( 8) (

X2)

X Xsss s

ss s s ssA A

s sB

ss

0 0

20 9

B

209

2

(20/9) (10/3) (20/9)( )

( 2) ( 8)( 2)s

s ssY

2

20 1 1 (3/2)( ) ( 8) ( 2)9 ( 2)

ss

Ys s

8 2 220 3( ) ( )9 2

t t ty t e e te u t

Repeated Linear Factor

3

10( )

( 2) ( 8)

ss

sY

s

Let

1 2 32 3

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

ss ss sA A A BY

Then

Can be found using Heaviside’s expansion

8( 8) ( )

ssB s Y

10

27

3 3

2( 2) ( )

ssA Ys

10

3

(3 2)3

2 (3 2) 2

1 ( 2) ( )(3 2)! s

ss

dA Yd

s

2

10

( 8)s

sdsd

s

209

1 2 32 3

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

ss ss sA A A BY

209

103

1027

A1 Can be found using Heaviside differentiation techniques

(3 1)3

1 (3 1) 2

1 ( 2) ( )(3 1)! s

ss

dA Yd

s

2

2

2

101 ( 8)2

s

ssds

d

3

2

1601 ( 8)2

ss

2027

2027

2 3

1 110 20 101 1( )( 8) ( 2)27 9 3( 2) ( 2)

ss s

Ys s

8 2 210 5 4

27 3 3( ) ( )t t tt ty t e e e u t