Lapplace Transform

8/11/2019 Lapplace Transform

http://slidepdf.com/reader/full/lapplace-transform 1/22

Lapplace Transform

Pekik Argo Dahono



Definition

j s

dt et f s F t f L st 0

)()()(

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



Examples

sdt et u L s F

t ut f st /11)]([)(

)()(

0

a sdt ee s F

t uet f st at

at

1)(

)()(

0

1)()(

)()(

0 dt t s F

t t f



Examples

22

220

00

)(

cos)(

21

2sin)(

)(sin)(

s s s F

t ut t f s

dt ee j

dt e jee

dt et s F

t ut t f

t j st j s

st t jt j

st



Properties

)()()()( 2121 sbF saF t bf t af L Linearity:

)/(1

)( a s F a

at f L Scaling:

)()()( s F eat uat f L as

)()()( a s F t ut f e L at

Time shift:

Frequency shift:



Examples

22

2222

22

)(

cos)(

)]([

/1)]([

4

2

2/21

)2/(21

2sin

sin

a s

a s s F

t et f

se

at u L

st u L

s s s F t L

st L

at

as



Properties

)0()0(')0()(])(

[

)0()()('

121 nnnnn

n

f f s f s s F sdt

t f d L

f s sF t f L

)(1

)(0

s F s

dt t f Lt

Time differentiation:

Time integration:



Example

221

1

22

)0()(1

)(

)('1

cos)(

)(

sin)(

s s

f s sF s F

t f t t f

s s F

t t f



Properties

ds

sdF t tf L

)()(

)()0( lim s sF f s

)()( lim0

s sF f s

Frequency diferentiation:

Initial value theorem:

Final value theorem:



Inverse Laplace Transform

)()(

s D s N

s F

Steps:1. Decompose F(s) into simple terms using partial fraction expansion.2. Find the inverse of each term by matching entries in Laplace transform table.



Simple poles

i p s

ii

n

n

n

s F p sk

p sk

p sk

p sk

p s p s p s s N s F

)(

)())(()()(

2

2

1

1

21

t pn

t pt p nek ek ek t f 2121)(

:Solution



Repeated Poles

p s

n

m

m

mn

p s

nn

nn

nn

n

s F p sdsd

mk

s F p sk

p sk

p sk

p sk

p s s N s F

)(!1

)(

)()( 11

1

pt nn pt pt et

mnk tek ek t f 1

21 !)(

:Solution



Complex Poles

t jt j

j s

j s

e K e K t f

s F j s K

s F j s K j s

K

j s

K

j s j s s N

s

s N bas s

s N s F

21

2

1

21

222

)(

)(

)()()()(



Example

t t s

s

s

t

eet f

s F sC

s F s B

s sF A s

C s

B s A

s F

s s s s s F

t et f

s s s s F

323

2

0

2

2

782)(

7)(3

8)(2

2)(32

)(

3212

2sin353)(

46

153)(

t t t

s s

s

s

s

eteet f

s s

s s s s s s s F s

ds

d D

s F sC

s F s B

s sF A

s D

s

C s

B s A

s s s

s s s F

3

122

32

1

2

1

2

3

0

22

3

25.25.125.32)(4

133

3262323)(1

23

)(1

1227

)(3

2)(

11313

62)(



The Convolution Integral

t ht xt y

d t h xt y

)(

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 d t h xt xt ht y

t t x

0)(

thencausalissystemtheand

0for0)(If



Properties

s F s F t f t f L

d f t ut f

t f t t f

t t f t t t f

t f d t f t t f

t yt xt f t yt xt f

t yt f t xt f t yt xt f

t xt ht ht x

t

oo

2121

''

)(



Steps to evaluate the convolutionintegral

• Folding: Take the mirror image of h( λ ) aboutthe ordinate axis to obtain h(- λ ).

• Displacement: Shift or delay h(- λ ) to obtainh(t- λ ).

• Multiplication: Find the product of h(t- λ ) and x( λ ).

•

Integration: For a given time t, calculate thearea under the product h(t- λ )x(λ ) for 0< λ<t toget y(t) at t .



Example



Laplace Applications

t t eet v

C B A

sC

s B

s A

s s s s s

sV

s s s

sV s s

s sV v s sV v sv sV

vv

vdt dv

dt

vd

42

2

22

2

2

2

41

21

41

)(

4/12/14/1

424224

)(

24)(86

2)(8)0()(6)0(')0()(s

:Jawab

2)0('1)0(

286



Example

t t

s

s

s

t

teet v

sV sdsd

C

sV s B sV s A

sC

s

B s

A

s s

s s sV

s s s

s s

sV s s

s sV v s sV v sv sV s

vv

evdt dv

dt vd

2

2

2

2

2

1

22

2

22

2

2

2

2)(

0)(2

2)(211

22121

66)(

166

51

1)(44

11

)(4)0()(4)0(')0()(

1)0(')0(

44



Example

t t

t

eet y

B A

s B

s A

s s s

s s s sY

s s

sY s s s

s sY s sY y s sY

y

t u ydt ydt dy

32

2

2

0

53)(

53

323221

6521)(

21)(

65

1)(

6)(5)0()(

2)0(

)(65



End

Lapplace Transform

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