ltm laplace transform 2011a mk

8/4/2019 LTM Laplace Transform 2011a Mk

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N u ễn Côn Phươn

The Laplace Transform



Contents

1. Basic Elements Of Electrical Circuits

2. Basic Laws

3. Electrical Circuit Analysis

4. Circuit Theorems

5. Active Circuits

6. Capacitor And Inductor

7. First Order Circuits. econ r er rcu s

9. Sinusoidal Steady State Analysis

10. AC Power Analysis

. -12. Magnetically Coupled Circuits

13. Frequency Response

15. Two-port Networks

The Laplace Transform 2




f (t ) = 0

(integrodifferential)i (t ), v (t ), …Circuit

F (s ) = 0

(algebraic)I (s ), V (s ), …





• Definition

• Two Important Singularity Functions• Transform Pairs

• Properties of the Transform

• Inverse Transform• Initial-Value & Final-Value Theorems

• Laplace Circuit Solutions

• Analysis Techniques

•

• Transfer FunctionThe Laplace Transform 4



Definition ( ) f t

t 0 0( ) ( ) ( )st

s L f t f t e dt

s j

0( ) t f t e dt

1

1

1 1( ) ( ) ( )

2

j st

j f t L F s F s e ds

j





• Definition






•

• Transfer FunctionThe Laplace Transform 6



Two Important Singularity Functions (1)

( )u t 10 0

( )t

u t

t 0

( )u t a

t 0 a

( )1

u t at a




Two Important Singularity Functions (2)Ex. 1

Determine the Laplace transform for the waveform?

( )u t 1

( ) ( ) st F s u t e dt

t 0

0 1

st

e dt

0

1 st e s

1

s






( ) ( ) st F s u t a e dt

( )u t a

1

0 0 1

a st

adt e dt

t 0 a

1 st

a

e s

ase

s






( ) [ ( ) ( )] st F s u t u t a e dt

t 0 a

0

1

( )

st

u t e dt s

( )u t

1

0( )

st st e

u t a e dt s

t 0

1 1as ase e F s

t

( )u t a

0


s s s 1a



Two Important Singularity Functions (5)( )t

0 0t t

t 0

( ) 1 0t dt

( )t a

0t a t a

t 0 a( ) 1 0

a

at a dt

2 1 2( )t f a t a t t t a dt


1 1 20 ,t a t a t




Determine the Laplace transform of an impulse function?

0( ) ( )

st

F s t a e dt

2 1 2( )

( ) ( )t

t

f a t a t f t t a dt

( ) as F s e





• Definition






•

• Transfer Function




Transform Pairs (1)Ex. 1

Find the Laplace transform of f (t ) = t ?

0( ) st F s te dt

1Let & & st st st u t dv e dt du dt v e dt e s

200 0

1( ) 0t st st t e e F s e dt

s s s s




Transform Pairs (2)Ex. 2

Find the Laplace transform of f (t ) =cosωt ?

0( ) cos st F s te dt

0 2

j t j t st e e

e dt

0 2

e edt

2 s j s j

s


2 2 s



Transform Pairs (3)

f (t ) ( )t ( )u t at e t at te sin at cosat

F ( s) 1 s s a 2 s 2( ) s a 2 2

a

s a 2 2

s

s a





• Definition






•





Properties of the Transform (1) Property f (t ) F ( s)

1. Magnitude scaling

2. Addition/subtraction

( )t ( ) F s

1 2( ) ( ) f t f t 1 2( ) ( ) F s F s

1 s .

4. Time shifting

a a ( ) ( ), 0 f t a u t a a ( )

ase F s

( ) ( ), 0 f t u t a a [ ( )]

as

e L f t a

. requency s t ng

6. Differentiation

7. Multiplication by t

( )e t s a

( ) / n nd f t dt

1 2 1 1( ) (0) (0) ... (0)

n n n o n s F s s f s f s f

( )nt f t ( 1) ( ) / n n nd F s ds

8. Division by t

9. Integration

( ) / f t t ( ) s

F d

0( )

t

f d ( ) / F s s


10. Convolution1 2 1 2

0( ) * ( ) ( ) ( )

t

f t f t f f t d 1 2( ) ( ) F s F s



Properties of the Transform (2)Ex. 1

Find the Laplace transform of 10( ) 5 cos20 ?

t f t e t

1 2 1 2( ) ( ) ( ) ( ) f t f t F s F s

10( ) [5] [ ] [cos20 ]t F s L L e L t

( ) ( ) f t AF s

[5] 5 [1] L L 5

[5] L s [1] L

s

10 1[ ]t L e

2 2 2[cos20 ]

20 400

s s L t

s s


2 2

5 1 5 2400 4000( )

10 400 ( 10)( 400)

s s s F s

s s s s s s




Find the Laplace transform of the waveform? 5

t 0 1 2 3

5

( ) 5 ( 1) 5 ( 2) f t u t u t

t 0 1 2 32

25 s s se e

5 s s s


t 0 1 2 3





t 0 1 2 3( ) ( 5 10)[ ( 1) ( 2)] f t t u t u t

5 ( 1) 10 ( 1)tu t u t

5 ( 2) 10 ( 2)tu t u t

5 ( 1) 5( 1 1) ( 1)tu t t u t

t 0 1 2 3

5 ( 2) 5( 2 2) ( 2)tu t t u t

5( 2) ( 2) 10 ( 2)t u t u t

1( ) 5( 1) ( 1) 5 ( 1)

10 ( 1)

f t t u t u t

u t

t 0 1 2 3

The Laplace Transform 2110 ( 2)

t u t u t

u t





t 0 1 2 3( ) ( 5 10)[ ( 1) ( 2)] f t t u t u t

5( 1) ( 1) 5 ( 1)t u t u t

10 ( 1)

5( 2) ( 2) 10 ( 2)

u t

t u t u t

t 0 1 2 3

u t

5( 1) ( 1) 5 ( 1)t u t u t

12

2 2

5( ) 5 5

s s se e

F s e s s s


t 0 1 2 3

2

5

(1 )

s se

s e s





t 0 1 2 3





• Definition






•






f (t ) = 0


F (s ) = 0

(algebraic)I (s ), V (s ), …




Inverse Transform (1)1

1 1 0( ) ...m m

m m P s a s a s a s a

1

1 1 0( ) ...n n

n nQ s b s b s b s b

1 2 n K K K

1 2

...n

s p s p s p

1( )Complex- conjugate poles : ( ) P s F s 1

*

1 1 ...

s s s

K K

s j s j

1

1 1

( )Multiple poles : ( )

( )( )n

P s F s

Q s s p


11 12 1

2

1 1

... ...( ) ( ) ( )

n

n s p s p s p



Inverse Transform (2)

1 2

1 2

Simple poles : ( ) ...

( )

n

n

F s

Q s s p s p s p

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

( )i

i i

s p

P s s p K

Q s

1 i p t ii

K L K e

i

1 2 n p t p t p t


n



Inverse Transform (3)Ex. 1

Find the inverse La lace transform of 225 300 640 s s

F s

( 4)( 8) s s s

1 2 3 ( )( ) ; ( ) 0 ... 0 0 ... 0i i

K K K P s F s s p K

i s p

2 2

25 300 640 25 300 640 640 s s s s 1 0

0 0( 4)( 8) ( 4)( 8) 4 8 s

s s s s s s s

2 2

2 4

4 4

2

( 4) ( ) ( 4) ( 4)( 8) ( 8)

25( 4) 300( 4) 640

s

s s K s F s s s s s s s


( 4)( 4 8)




Find the inverse La lace transform of 225 300 640 s s

F s

( 4)( 8) s s s

1 2 3 ( )( ) ; ( ) 0 ... 0 0 ... 0i i

K K K P s F s s p K

1 220; 10 K K

2 225 300 640 25 300 640 s s s s

i s p

3 8

8 8

2

( 4)( 8) ( 4)

25( 8) 300( 8) 6405

s

s s s s s s s

( 8)( 8 4)

20 10 5 4 8t t


4 8 s s s




Find the inverse La lace transform of 100( 6) s

F s

( 1)( 3) s s




Inverse Transform (6)*

1 1 1( )Com lex- con u ate oles : ...

P s K K F s

1( )( )( )Q s s j s j s j s j

1 1

( )( )

P s s j K K

s j

*

1 1 K K

1( ) F s 1

s j

1 1... ...

s j s j s j

( ) ( ) ( ) ( )... ... j j t j j t t j t j t t K e e K e e K e e e

cos sin je j

cos sin cos sin ...t t K e t t t t


12 cos( ) ...t K e t




Find the inverse La lace transform of 24 76 s s

F s

( 2)( 6 25) s s s

1 2 3( )K K K

F s

1 1

( )( ) ; ( ) 2 cos( ) ...

( )

t P s K s j f t K e t

Q s

2 2

3 2 2

4 76 4 76( 2) 8

( 2)( 6 25) 6 25

s s s s K s

s s s s s

2

1 2

3 4

4 76( 3 4) 6 8 10

( 2)( 6 25) s

s s K s j j

s s s

o53.1


3 o 2 3 o 2( ) 2 10 cos(4 53.1 ) 8 20 cos(4 53.1 ) 8t t t t f t e t e e t e





F s

( 4 5) s s s




Inverse Transform (9)1 11 12 1

2

( )Multiple poles : ( ) ... ...n

n n

P s K K K F s

1 1 1 1

11 1( ) ( )n

n s p s p F s K

1

1 1 1[( ) ( )]n

n

s p

d s p F s K

ds

2

1 1 22[( ) ( )] (2!)n

n

d s p F s K

ds

1

1 1

1[( ) ( )]

n jn

j n

d K s p F s


1 s p




Find the inverse La lace transform of

210 34 27( )

s s F s

s s

11 12 21 12

1( ) ; [( ) ( )]

n jn

j n j

K K K d F s K s p F s

2 22 2

12 23

10 34 27 10 34 27( 3) ( ) ( 3) 5

3 s

s s s s K s F s s

s s s

1 s p

s s

22

11

3 3

10 34 27[( 3) ( )]

s s

d d s s K s F s

ds ds s

2

2

3

(20 34) (10 34 27)7

s

s s s s

s


2 20

0

( ) 3( 3) s

s

s s K sF s s

s s




Find the inverse La lace transform of

210 34 27( )

s s F s

s s

11 12 21 12

1( ) ; [( ) ( )]

n jn

j n j

K K K d F s K s p F s

7 5 3 K K K

1 s p

7 5 3 F s

3 ( 3) s s s

3 3t t






F s

( 1)( 2) s s





• Definition






•





Initial-Value & Final-Value Theorems (1)

0t s s s

0Final value theorem : lim ( ) lim ( )

t s f t sF s




Initial-Value & Final-Value Theorems (2)Ex.

Find the initial and final values of 5( 1) s

F s

( 2 2) s s s

2

5( 1)

(0) lim ( ) lim 0

s

f sF s

20 0

5( 1)( ) lim ( ) lim 2.52 2 s s

s f sF s s s





• Definition



• Inverse Transform

• Initial-Value & Final-Value Theorems



•





Ex.

Find the current i(t )?

t = 0

+

200

Method 1

Laplace Circuit Solutions (1)

–

1V ( )i t 1 L R

div v e L Ridt

di n

dt

n e e e

3

2002000

100 10

R

2000t

ni Ke

10.005A

200 f

ei

R

2000

0.005

t

f ni i i Ke

2000 0(0) 0.005 0.005 0 0.005i Ke K K


2000( ) 0.005(1 ) At i t e



Laplace Circuit Solutions (2)

f (t ) = 0


F (s ) = 0(algebraic)

I (s ), V (s ), …




Laplace Circuit Solutions (3)Ex.


t = 0

+

200

Method 2 –

1V( )i t

0.1 200 1 L R

div v e idt

0.1 200 [1] 0.1 [200 ] L i L L L idt dt

1

[1] L s

[200 ] 200 ( ) L i I s

1 2 1 1( )nn n n o nd f t ...

n

dt

0.1 0.1[ ( ) (0)] 0.1 ( )di

L sI s i sI s


10.1 ( ) 200 ( ) sI s I s

s



f (t ) = 0


F (s ) = 0(algebraic)

I (s ), V (s ), …Circuitin s -domain





• Definition







•





Circuit Element Models (1)

Rv t

RV s

v Ri

( ) ( ) Af t AF s ( ) ( )V s RI s




Circuit Element Models (2)( )i t ( ) I s

L

sL

(0)i

+ –

(0) Li

v L dt

( )df t

( ) [ ( ) (0)]V s L sI s i


dt




C

1

sC

–+ (0)v

s

0

1( ) (0)

t

v i x dx vC

0( )( )t s f d s

1 (0)( ) ( ) vV s I s sC s


(0)v s




++ + sM +––

1 1 2

1(0)i2(0)i

1 2 –2

–1 1 s 2 s

–2

–1

1 2( ) ( )di t di t

1 2 1 2

1 1

dt dt 1 1 1 1 1 2 2

2 1( ) ( )di t di t


2 2dt dt

2 2 2 2 2 1 1




i t I s

R( )v t R( )V s

L

sL

( )v t (0)i

( )V s + –

(0) Li





C ( )v t

1

sC ( )V s

–+ (0)v

s

++ + sM +––

1 1 2(0) (0) L i Mi 2 2 1(0) (0) L i Mi

1 L 2 L

+

–2( )v t

+

–1( )v t

1 sL 2 sL

+

–2( )V s

+

–1( )V s


1( )i t 2( )i t 1( ) I s 2( ) I s




• Definition

• Two Important Singularity Functions

• Transform Pairs






•





Analysis Techniques (1)

KVL/KCL : ( ) ( ) ... ( ) 0 x t x t x t

1 2KVL/KCL : ( ) ( ) ... ( ) 0n X s X s X s

i (t ), v (t ), …Circuit

Inverse TransformCircuit Element Models DC circuit analysis techniques

, , ,

mesh analysis, sourcetransformation, superposition,

Thevenin/Norton equivalent, …)


I (s ), V (s ), …in s -domain



Analysis Techniques (2)Ex. 1


t = 0

+

200

–

1V( )i t (0) 0i

1200 ( ) 0.1 ( ) 0.1 (0) 200 ( ) 0.1 ( ) I s sI s i I s sI s

+ –

0.1 s

1 21 10( )

(0.1 200) ( 2000) 2000

K K I s

s s s s s s

1

s( ) s

+–0.1 (0)i

1

0

100.005

2000 s

K s

2

2000

100.005

s K s


. .( )

2000 I s

s s

( ) 0.005(1 ) At i t e





=

+

200

100mH2000( ) 0.005(1 ) At i t e

1. Solve for initial capacitor

voltages & inductor currents2. Draw an s-domain circuit

–

1V ( )i t 3. Use one of DC circuit

analysis techniques to solve

for voltages or/and currents

-(0) 0i Circuit Element Models

200

4. Find the inverse Laplace

transform to convert them

back to the time domain

+ – 0.1 s

1

1200 ( ) 0.1 ( ) I s sI s

10( ) I s


s( ) s

+–




Find the voltage v(t )?

t = 0

10k

+

4

6

4

110

1 525 10( ) // ( )1 2

sV s R J s sC s

(0) 0v 25 At e 25 F –

v

625 10 s 10k

5A

2 s

–

( )V s4

1 24 10 K K

1. Solve for initial capacitor voltages &

6

1

25 10 s

4

4

1

2

4 102 10

4 s

K s

n uc or curren s

2. Draw an s-domain circuit3. Use one of DC circuit analysis

techniques to solve for voltages or/and

-

4

4

2

4

4 10 2 102

s

K s


-

4. Find the inverse Laplace transform to

convert them back to the time domain4 2 4( ) 2 10 ( ) Vt t v t e e




Find the current i(t )? ( )i t ( ) s + 4

t

8 +8

(0) 1A

8

i ( )v t

(0)i

sL

( )V s + –

–

2 H

8V

( )i t –

12V

2( )

2 4

s I s s

(0) Li

+– 42 + –

1. Solve for initial capacitor voltages &1 2

6

( 2) 2

s K K

s s s s

s s

n uc or curren s



-

1

0

632 s

s K s

2

( ) 3 2 At

i t e


-


convert them back to the time domain2

2

62

s

s K

s




Find the voltage v(t )? + 4

t

8 +( )i t 1

( ) I s

12 81 s s

(0) 8 Vv –

2 F8V

( )v t –

12V –

C ( )v t

sC

( )V s

+

12 42

s

s

0.5 K K

–+ 4

1

8

s + –+

–

(0)v

s

1. Solve for initial capacitor voltages &( 0.125) 0.125 s s s s

0.54 K

2 s s

s –

n uc or curren s



-

00.125 s s

2

0.54 K


-


convert them back to the time domain. s

0.125( ) 4(1 ) Vt v t e




Write the mesh equations in the s-domain? 1C 1

R 22 L

+ –

+ –

1( )e t 2 ( )e t 3

L3(0)i2 (0)i

– +

+

1(0)v( )i t

sL

( ) I s

3 –3

1 R 2 2

sL1(0)v

s +

(0)i

+ –

+ –

+ –

1

1

sC 3 sL

2 2(0) L i

+ – +–

( )i t ( ) I s

1( ) E s2( ) E s

3 3

(0) L i

3(0)v+ –

+–

C ( )v t

1

sC ( )V s

(0)v


3

1

sC

s

–

s




Write the mesh equations in the s-domain? 1C 1

R 22

L

+ –

+ –

1( )e t 2 ( )e t 3

L3(0)i2(0)i

– +

+

1(0)v1

1

1: ( ) A A R I s

sC

3 –3

3

3

1 3

[ ( ) ( )]

(0) (0)

A B sL I s I s sC

v v

1 R 2 2

sL1(0)v

s +1 3 3 s s

+ –

+ –

1

1

sC 3 sL

2 2 (0) L i

+ – +–

2 2: ( ) B B R sL I s

( ) A I s ( ) B I s1( ) E s

2( ) E s

3 3

(0) L i

3(0)v+ –

+–

3

3

1 [ ( ) ( )] B A sL I s I s sC


3

1

sC

s3

3 3 2 2 2(0) (0) ( )v

L i L i E s s




Write the node equations in the s-domain? +–

2C

2(0)v

–1

L

1(0)i2 (0)i

2 L

3(0)v

3C

+1( ) j t 1 R 3 R

3( ) j t





Solve for v(t ) ?8

(0) 0; (0) 0; L C i v

+

1H

2

+

–

( )v t

15( )

5 ( ): 012

a

a

V s

V s sa s s

–

5 ( ) Au t 15 ( ) Vu t s

2

3 2

10 35 15

( )a

s s

V s

8

s

1

s

a

2

3 2

( ) 10 35 15( ) 2 2

1aV s s s s

V s

+ –5 15

s

2

+

–

( )V s s

11 1210( 3) s K K


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



Analysis Techniques (11)Solve for v(t ) ?

8 1

a

Method 1

Ex. 7

+5 15

s

2

s +( )V s

11 12

2 2

10( 3)( )

( 1) 1 ( 1)

s K K V s

s s s

2

12 2 1

1

10( 3)( 1) 10( 3) 20

( 1) s

s

s K s s

s

–

s s

2

11 2

11

10( 3)( 1) 10 3 10

( 1) s s

d s d K s s

ds s ds

10 20( ) ( ) 10(2 1) Vt V s v t t e


s s




8 1

a

Method 2

Ex. 7

+5 15

s

2

s +( )V s

( ) A I s

Suppose the current source flows via the inductor.

5 1 15( ) 2 ( ) A A s I s I s

s s s

–

s s

2

3( ) 5

( 1) A

s I s

s

2

3( ) 10

sV s





8 1

a

Method 3

Ex. 7

+5 15

s

2

s +( )V s

52

25 10( 0.5)

( )1 ( 1)

2ab

s

s s s

V s s s s

–

s s

10( 0.5) 2 10 s s8

s

1

s

a

52 21

( 1) ( 1)2 s s s s

52

+

–

5( ) s

V s


b




8 1

a

Method 3

Ex. 7

+5 15

s

2

s +( )V s

152

15( )

1 ( 1)2

C s

s I s s s

s

–

s s

8

s

1

s

a

15

2

( ) 2( 1) s

V s s

+ –

15

s

2

+

–

15( ) s

V s





8 1

a

Method 3

Ex. 7

+5 15

s

2

s +( )V s8 1

s

a

–

s s5

2

10( )

( 1) s

sV s

s

5

2

+

–

5( ) s

V s

s

b

a 15 15( ) ( ) ( ) s s

V s V s V s

1515

( ) 2V s

8

s

1

s + 2 2

10 30

( 1) ( 1)

10 3

s

s s

s


s s –

15

s

– s 2( 1) s




8 1

a

Method 4

Ex. 7

+5 15

s

2

s +( )V s( ) Z s s

–

s s

2

5 15 3( ) 5

s E s s

s s s

Z s

1

s2

35

3( ) 5

1 ( 1)2

s s s I s s s

+ –( ) E s

2

+

–

( )V s

s

2 2

3 3( ) 2 5 10

s sV s





8 1

a

Method 5

Ex. 7

+5 15

s

2

s +( )V s8

s1 s

a

–

s s( )eq Z s ( )eq Z s s s

+( )e E s

eq s

2

+

–

( )V sa

–

+ s s

( )eq E s

5 15

( )eq E s s s s

3 s

( )( ) 2

eq E s

V s


+ –5

s

15

s –

eq s s

2

310

( 1)

eq

s

s

0.5v



0.5v


Solve for i(t ) ?

x

v

+

–

2F

6

i

(0) 0; (0) 0C Lv i

6( ) 2 ( ) ( ) ( 2) ( ) 0 x A c A

V s I s I s s I s 1H

6

( ) 2 ( ) 0.5 ( ) ( 2) ( ) 0 x A x AV s I s V s s I s

s

.c x

2 4( ) 2 ( ) 2 ( ) x A AV s I s I s

s s

( ) xV s

+2

2 2

6

s

( ) I s I s

2 ( ) 2 ( ) 0.5 2 ( )

( 2) ( ) 0

A A A

A

I s I s I s s s s

s I s


s

–

s8 12( ) ( )

( 2)( 6) A

s I s I s

s s s

0.5v




Solve for i(t ) ?

x

v

+

–

2F

6

i

1 2 38 12( )

( 2)( 6) 2 6

s K K K I s

s s s s s s

1H

1

0

8 121

( 2)( 6) s

s K

s s

.c x

2

2

8 120.5

( 6) s

s K

s s

( ) xV s

+2

2 2

6

s

( ) I s

36

1.5( 2) s

s K

s s


s

–

s( ) 1 0. 1. Ai t e e




Find the current i(t )?0t

2

152

8(0) 1A

8

i + –

4

2 H8V ( )i t

8 + –

2

.9( )2 4 ( 2)( 9)

s s I s s s s

*

1 2 2 K K K

2 3 3 s s j s j 2

1 2

16.51.58

s K

+– 42

( ) I s

+ –

152 s

2

2

16.50.35

( 2)( 3)

s K

s s j

o146.3

2 9 s


2 o( ) 1.58 0.70cos(3 146.3 ) At i t e t



Analysis Techniques (21)

i (t ), v (t ), …Circuit

Inverse TransformCircuit Element Models DC circuit analysis techniques

(KVL, KCL, nodal analysis,

Circuit

mesh analysis, source

transformation, superposition,

Thevenin/Norton equivalent, …)

, , …in s -domain





• Definition


• Transform Pairs






•





Convolution Integral (1)1 2 1 2 1 2

0 0( ) ( )* ( ) ( ) ( ) ( ) ( )

t t

f t f t f t f t f d f f t d

2( ) f t

t 0

t

t 0

1( ) f t 2( ) f t


0 0

2



Convolution Integral (2)Ex. 1

Find the convolution of the two signal?1

1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 4

22( ) f

1

1

( ) f t

22( ) f

1

0 1 2 3 4

0 1 2 3 4

1( ) f t 1 2

0 1: 1; 0t f f


1 2( )* ( ) 0 f t f t





1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 4

22( ) f

1

1

( ) f t

0 1 2 3 41 2

0 1: ( )* ( ) 0t f t f t

1 21 2 : 1; 2t f f


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

t t t f t f t f t f d d t





1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 4

22( ) f

1

1

( ) f t

0 1 2 3 41 2

0 1: ( )* ( ) 0t f t f t

1 2 : ( )* ( ) 2( 1)t f t f t t

1 22 3: 1; 2t f f


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

t t t

t t t f t f t f t f d d





1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 4

22( ) f 1

1

( ) f t

0 1 2 3 41 2

0 1: ( )* ( ) 0t f t f t

1 2 : ( )* ( ) 2( 1)t f t f t t

1 23 4 : 1; 2t f f 1 22 3: ( )* ( ) 2t f t f t


3 3 3

1 2 1 2 11 1( )* ( ) ( ) ( ) 1 2 2 8 2

t t t f t f t f t f d d t





1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 4

22( ) f

1

1

( ) f t

0 1 2 3 41 2

0 1: ( )* ( ) 0t f t f t

1 2 : ( )* ( ) 2( 1)t f t f t t

1 24 : 1; 0t f f 1 22 3: ( )* ( ) 2t f t f t

1 23 4 : ( )* ( ) 8 2t f t f t t


1 2( )* ( ) 0 f t f t





1( ) f t

t 0 1 2 3 4

2

1 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t


t 0

2

1 2 3 41 2

0 1: ( )* ( ) 0t f t f t

1 21 2 : ( )* ( ) 2( 1)t f t f t t

1 22 3: ( )* ( ) 2t f t f t

2

1 2( )* ( ) f t f t

1 24 : ( )* ( ) 0t f t f t

1 2:t t t t






( )w t

t 0 1 2 3 41 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t

f t f t f t f t f d f f t d 2 e

2

2e

( )w t

t 0 1 2 3 4

2( )

000 2 : ( ) 2 1 2 2(1 )t t t t f t e d e e

2 t 0 1 2 3 4


002 : ( ) 2 1 2 2( 1)t t t t f t e d e e e





( )w t

t 0 1 2 3 41 2 1 2 1 2

0 0

( ) ( )* ( ) ( ) ( ) ( ) ( )t t

f t f t f t f t f d f f t d 2 e

22 t e

( )w

t 0 1 2 3 4

2

000 2 : ( ) 1 2 2 2(1 )t t f t e d e e

t t 0 1 2 3 4


222 : ( ) 1 2 2 2( 1) t

t t t f t e d e e e



Convolution Integral (10) Property f (t ) F ( s)

1. Magnitude scaling

2. Addition/subtraction

( )t ( ) F s

1 2( ) ( ) f t f t 1 2( ) ( ) F s F s

1 s .

4. Time shifting

a a ( ) ( ), 0 f t a u t a a ( )

ase F s

( ) ( ), 0 f t u t a a [ ( )]ase L f t a

. requency s t ng

6. Differentiation

7. Multiplication by t

( )e t s a

( ) / n nd f t dt

1 2 1 1( ) (0) (0) ... (0)

n n n o n s F s s f s f s f

( )nt f t ( 1) ( ) / n n nd F s ds

8. Division by t

9. Integration

( ) / f t t ( ) s F d

0

( )t

f d ( ) / F s s


10. Convolution1 2 1 2

0( ) * ( ) ( ) ( )

t

f t f t f f t d 1 2( ) ( ) F s F s

C l i I l (11)E 3




Find vo(t )?+

+ v

–

( )iv t

0.2F

( )ov t e

( ) 5 5 50.2( ) ( )1( ) 1 5 11

C o i

C

Z s sV s V s Z s s s s

Method 1: 5( ) ( ) 6.25( ) Vt t

o oV s v t e e

Method 2: ( ) ( ) ( ) ( ) ( )* ( )o i o iV s H s V s v t h t v t

55

( ) ( ) 55

t

H s h t e s

5 5 4 5 4

t t t t

t t t


00 0 0

5

.

6.25( ) V

o i

t t e e

Th L l T f



The Laplace Transform• Definition


• Transform Pairs






•



T f F ti (1)



Transfer Function (1)( )in I s ( )out I s

+

( )inV s

+

( )out V s( ) s – –

( )Out s

( ) In s

If ( ) ( ) ( ) 1 ( ) ( )in t t In s H s Out s


T f F ti (2)E 1



Transfer Function (2)Linear

++Ex. 1

Find the transfer function h(t ) of the filter?an pass

filter – –

iv t ov

( ) 10 ( )iv t u t

10( ) ( ) ( ) ( )o iV s H s V s H s

1( ) ( )o s sV s

1 ( )odv t


10 dt

T f F ti (3)Ex 2



Transfer Function (3)Ex. 2

Find the transfer function H ( s)? +

–

( )iv t 1

0.2F

( )ov t

1( ) 50.2C Z s s

1( ) 51

0.2

o i i i iC Z s s

s

( ) 5( )( ) 5

o

i

V s H sV s s


Transfer F nction (4)



Transfer Function (4)

A circuit is stable if : lim ( ) finite x

h t

j

1 2

( )( )

( )( )...( )n

N s H s

s p s p s p

1 2 ... n p t p t p t h t k e k e k e u t

n


lie in the left half of the s-plane

Transfer Function (5)Ex 3




An active filter has the transfer functionk

2 (4 ) 1 s k s

For what values of k is the filter stable?

j A circuit is stable when all the poles of its transfer

function H ( s) lie in the left half of the s-plane

2

1,2

(4 ) (4 ) 4

2

k k p

4 0k


4k

Transfer Function (6)Ex 4




+Given the transfer function5

–

( )iv t 1

C

( )ov t

5 s

Find C ?

( ) 1( ) ( ) ( ) ( )

1( ) 11

C o i i i

C

Z s sC V s V s V s V s R Z s Cs

++

( ) 1 5( ) oV s

H s –

i s 1

sC

( )oV s

i

5 1C


0.2FC

ltm laplace transform 2011a mk

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