03_The Laplace Transform

8/3/2019 03_The Laplace Transform

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1

The LaplaceTransform



2

Introduction Laplace transform is another method to

transform a signal from time domain to

frequency domain (s-domain) The basic idea of Laplace transform

comes from the Fourier transform

As we have seen in the previous chapter,not many functions have their Fouriertransform such as t , t 2, et etc.



3

The Fourier transform formula:

The Laplace transform formula is the

modification of the above formula, that is,

the term j is replaced by s

s is equal to + j , where is a large

positive real number

The Laplace transform formula:

However, the Laplace transform only

support the function f (t ) which domain t ≥ 0

dt et f t f F t j

)()}({)( F

0

)()}({)( dt et f t f sF st

L



4

Using definition, find the Laplace transform of

(a)

(b)

(c)

(d)

Example 1

)()( t ut f

)()( 5t uet g

t

)()( t ut t i

)(cos)( t ut t v



5

Solution(a)

ssse

se

dt edt et usF

st

st st

1101

)()(

0

00

(b)

5

1

)5(

10

)5(

1

)5(

)(

)5(

0

)5(

0

)5(

0

5

sss

e

s

e

dt edt eesG

st s

t sst t



6

(c)

22

0

2

000

11)00(

)(

ss

e

s

e

dt s

e

s

te

dt tes I

st

st st st



7

(d)

)(

cos)10(

))(cos()cos(

sin)00(

)(sinsincos)(

2

0

2

00

0

00

0

sV ss

tdt ess

dt set st es

tdt es

dt set t etdt esV

st

st st

st

st st st

1)(

2

s

ssV



8

Properties ofL

-transform Linearity

L{af (t ) ± bg(t )} = aL{ f (t )} ± bL{g(t )}

a

sF

aat f

1)(L

First shift theoremL{e−at f (t )} = F (s + a)

Second shift thoremL{ f (t − d ) u(t − d )} = e−dsF (s)

Time scaling



9

Properties ofL

-transform (cont.) Time derivatives

)0()()( f ssF t f L

)0()0()()( 2 f sf sF st f L

)0()0()()( )1(1)( nnnn f f ssF st f L

Time integral

)(1

)(0

sF s

d f

t

L



10

Example 2Determine the Laplace transform of

(a)

(b)

t et t 2sin3 43

4cos)1(

43 t

t et



11

Solutiont et

t 2sin3 43 L

4

2

4

3624

sss

t et t 2sin3 43LLL

4

coscos

4

3 t t et t L 43

41coscos t t et t LLL

522

6

1)3(

3

1 ss

s

s

s

(a)

(b)



12

Example 3Determine the Laplace transform of

(a)

(b) )2()2( 2 t ut

35 t e t



13

Solution(a) Let 3)( t t f then .

6)(

4s

sF

Therefore )(535 t f et e t t LL

)5( sF .)5(

64

s



14

(b) Let2)( t t f then .

2)(

3ssF

Therefore

)2()2( 2 t ut L

)(2 sF e s

.2

3

2

s

es

2)2()2( t t f

)2()2( t ut f L

Also



15

Inverse Laplace transform (IL

T) The inverse Laplace transform of F (s) is f (t ), i.e.

)()(1

sF t f

Lwhere L−1 is the inverse Laplace transform operator.



16

Example 4Find the inverse Laplace transform of

3

2

s(a) (b)

4

2

s

(d)9

652

s

s

(c)25

1

2s

(e)4)1(

12

s

s (f)4)1( 2 s

s



17

SolutionFrom the table of Laplace transform,

3

1 2

sL(a)

3

1 !2

sL

2t

4

1 2

sL(b)

4

1 !3

!3

2

sL

3

3

1t

25

12

1

sL(c)

22

1

5

5

5

1

sL t 5sin

5

1



18

9

652

1

s

sL

(d)22222 3

3

2359

65

ss

s

s

s

22

1

22

1

3

32

35

ss

sLL

t t 3sin23cos5

4)1(

12

1

s

s

L(e) t e t 2cos

Write



19

4)1( 2

1

s

sL

(f)

22

1

22

1

2)1(

2

2

1

2)1(

1

ss

sLL

t et et t sin

2

1cos

Since the ILT of the term cannot be found

directly from the table, we need to rewriteit as the following

2222

2222

2)1(

2

2

1

2)1(

1

4)1(

1

4)1(

1

4)1(

1)1(

4)1(

ss

s

ss

s

s

s

s

s



20

Example 5Find the inverse Laplace transform of

8

( 2)

s

s s

(a) (b)

2

9

2 7 4s s

(d)2

7 20

( 4 20)

s

s s s

(c)3 2

4 1

2

s

s s s

(e)2

25 6

s

s s



21

SolutionWe use the partial fractions technique:

1 8

( 2)

s

s s

(a)

1

2

9

2 7 4s s

(b)

1 4 3

2s s

24 3 t e

1 2 1

2 1 4s s

/ 2 4t t e e

1

12

1 1

4s s

=LL

L =L

=L



22

1

3 2

4 1

2

s

s s s

(c) 1

2

4 1

( 1)

s

s s

1

2

1 3 1

( 1) 1s s s

1 3 t t e t e

where, if we let2

1( )F s

s , then ( ) . f t t Hence,

1 1

2

1

( 1) ( )( 1)

t t

F s e f t e t s

=LL

=L

L =L



23

1

2

7 20

( 4 20)

s

s s s

(d) 1

2

1 3

4 20

s

s s s

1

2 2

1 2 5

( 2) 16 ( 2) 16

s

s s s

1

2

1 3

( 2) 16

s

s s

1

2

1 ( 2) 5( 2) 16

s

s s

2 254

1 cos 4 sin 4t t

e t e t

=LL

=L

=L

=L



24

21

2 5 6

s

s s

(e) 1

2

5 61

5 6

s

s s

1 5 61

( 2)( 3)

s

s s

1 4 912 3s s

2 3( ) 4 9t t t e e

L =L

=L

=L=L





26

Example 6

(a) )2)(1(

1

ss

Use the convolution theorem to find the inverseLaplace transforms of the following:

(b))9(

122 ss

(c))5(

72 ss



27

Solution

)2)(1(

11

ssL(a)

2

1

1

1 11

ssLL

t t ee 2

t

t d ee0

2

t

t d e0

3

t t

e

0

3

3

33

22 t t t t eeee



28

)9(

122

1

ssL(b)

9

314 2

1

ssL

9

314

2

11

ssLL

)3sin1(4 t

t

d 0

3sin14

t

033cos4

)3cos1(

34 t



29

)5(

7

2

1

ssL(c)

5

117

2

1

ssL

5

117 1

2

1

ssLL

t et 57

t

t d e0

)(57

t

t d e0

)(57

t t t

t

d ee

0

)(5

0

)(5

57

57

t t ete

0

)(50

257

5

07

25

)1(7

5

7 5t et )15(

25

7 5 t et



30

Circuit applications1. Transfer functions

2. Convolution integrals

3. RLC circuit with initial conditions

sC C

sL L

R R

1



31

Transfer function

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

)()()( t xt ht y

)(Input

)(Output

)(

)()(,functionTransfer

s

s

s X

sY s H

In s-domain, )()()( s X s H sY

In time domain,

Network

System



32

Example 7 (Pb.13.64, pg.751)For the following circuit, find H (s)=V o(s)/ V i(s).

Assume zero initial conditions.



33

SolutionTransform the circuit into s-domain withzero i.c.:

)(sV s )(sV o

s

s

10



34

ss

sso

V ss

V ss

V

s

s

sV

s

s

sV

3092

20

)52)(2(20

20

2

52

2052

20

210

// 4

10 // 4

2

Using voltage divider

3092

20

)(

)()(

2

sssV

sV s H

s

o



35

Example 8 (Pb.13.65, pg.751)Obtain the transfer function H (s)=V o(s)/ V i(s),

for the following circuit.



36

SolutionTransform the circuit into s-domain (We canassume zero i.c. unless stated in the question)

)(sV s )(sV o

)(s I s2

)(2 s I

s



37

I ss

I s I s

V

I I I V

s

o

932

3)3(2

9)2(3

We found that

293

9

932

9

)(

)()(

2

ss

s

s

s

sV

sV s H

s

o



38

Example 9 (P.P.15.14, pg.705)Use convolution to find vo(t ) in the circuit of

Fig.(a) when the excitation (input) is the

signal shown in Fig.(b).





40

Step 4: Find vo(t )

d vt ht vt ht v

sV s H sV

s

t

so

so

)()()()()(

)()()(

0

)(20)1(20

2020

102)(

22

0

2

0

2

0

)(2

t t t t

t t t t

t t

o

eeee

eed ee

d eet v

For t < 0 0)( t vo

For t > 0





42

Therefore, it is important to consider the

initial current of an inductor and the initialvoltage of a capacitor

For an inductor

Taking the Laplace transform on bothsides of eqn gives

or

dt

t di Lt v L

L

)()(

)a1.....()0()()()]0()([)( L L L L L Lis I sLissI LsV

)b1.....()0()(

)(s

i

sL

sV s I L L

L



43

)0()()()( L L L Lis I sLsV s

i

sL

sV

s I L L

L

)0()(

)(



44

For a capacitor

Taking the Laplace transform on bothsides of eqn gives

or

dt

t dvC t i C

C

)()(

)a2.....()0( / 1

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

C C C C Cv

sC

sV vssV C s I

)b2.....()0(

)(1

)(s

vs I

sC sV C

C C



45

)0( / 1

)()(

C

C

C Cv

sC

sV s I

s

vs I

sC sV C

C C

)0()(

1)(



46

Example 10 (Pb.16.23, pg.750)Consider the parallel RLC circuit of thefollowing. Find v(t ) and i(t ) given that

v(0) = 5 V and i(0) = −2 A.



47

SolutionTransform the circuit into s-domain (use thegiven i.c. to get the equivalents of L and C)

)(s I

)(sV s

4

161

s

80s4

8

10

)(sV



48

Then, using nodal analysis

208

)96(516

96

16

16

80

)208(

16

14

80

)8(

4

8

016

14

80 // 10

2

2

ss

sV

s

s

ss

V ss

s

V s

s

V

s

s

V I



49

Since the denominator cannot be factorized,

we may write it as a completion of square:

22222 2)4(

)2(230

2)4(

)4(5

4)4(

)96(5)(

ss

s

s

ssV

V)()2sin2302cos5()( 4 t uet t t v t

Finding i(t),

sssss

sV I 2

)208()96(25.1

48

2



50

A)(])2sin375.112cos6(4[)( 4 t uet t t i t

Using partial fractions,

sss

C Bs

s

A

ssss

ss I

2

208

2

)208(

)96(25.1)(

22

It can be shown that 75.46,6,6 C B AHence,

22222

2)4(

)2(375.11

2)4(

)4(64

208

75.4664)(

ss

s

sss

s

s

s I





52

SolutionThe i.c. are not given directly. Hence, at firstwe need to find the i.c. by analyzing the circuitwhen t ≤ 0:

V24 )0( L

i

)0(C

v

5

V0)0(,A8.45

24)0( C L vi



53

Then, we can analyze the circuit for t > 0 by

considering the i.c.

1025.6625.0

)10(3

625.0

3

1 // 625.0

32

101010

ss

s

ss I

ss

3)0( L Li

s625.0

s

101

)(s I o

Let

I



54

Using current divider rule, we find that

)8)(2(

48

1610

48

1025.6625.0

30

10

10

1

2

210

10

ssss

ss I

s I I

s

so

Using partial fraction we have

2

8

8

8)( sss I o

A)()(8)(28

t ueet it t

o

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03_The Laplace Transform