Chap 16 Applications of the Laplace Transform

7/30/2019 Chap 16 Applications of the Laplace Transform

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Chap 16 Applications of theLaplace Transform


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Outline

Introduction Circuit Element Models

Circuit Analysis

Transfer Functions

State Variables

Chap 16 Applications of the Laplace Transform 2


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Introduction

A system is a mathematical model of a physical

process relating the input to the output.


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Circuit Element Models

Steps in Applying the Laplace Transform1. Transformthe circuit from the time domain to

the s-domain.

2. Solvethe transformed circuit using nodalanalysis, mesh analysis, source transformation,

superposition, or any circuit analysis technique

with which we are familiar.

3. Take the inverse transform of the solution andthus obtain the solution in the time domain.


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Resistor Model


( ) ( ) ( ) ( )v t i t R V s RI s


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Inductor Model


( )

( ) ( ) ( ( ) (0 ))( ) (0 )

( ) (0 )) ( )

div t L

dt

V s V s L sV s i

sLI I s i I ss LisL s


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(

( )

( ) ( ) ( ( ) (0 )) () (0 )

( )) (0 )

dvi t C

dt

V s II s v

sCV s Cvs

s C sV s v I sC s

Capacitor Model



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Passive Elements with Zero Initials

Resistor

( ) ( )V s RI s

Capacitor

1( ) ( )V s I s

sC

Inductor

( ) ( )V s sLI s


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Impedance of Elements in s-domain

1 ( )( )

( ) ( )

I sY s

Z s V s


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Example 16.1

Q: Find vo(t) in the circuit, assuming zero initialconditions.


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Example 16.1 (cont.)

1 3

1

F

1

1

)

H

3

(

s

s

sL

C

us

s

t


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1 2

1 2

2

1 2

For mesh 1:

For mesh 2

1 3 31

3 30 5

1( 5 3)

3

:

I I

s s s

I s Is s

I s s I

2

2 2

3 2

2

2 3 2

1 3 1 31 ( 5 3)

3

3 ( 8 18 )

3

8 18I

s s s

s s I I s s s

s s s I

2 2 2 2

4

3 3 2( )

8 18 2 ( 4) ( 2)

He3

( ) sin n 2 V,ce, 02

t

o

ot

V s sI s s s

v e t t

2 2

2 2

sin( )

cos( )

at

at

e ts a

s ae t

s a

L

L


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Example 16.2

Q: Find vo(t) in the circuit. Assume vo(0)=5 V.


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10/( 1)2 0.5

10 10 101 1

2.5 ( 2

/2

10 10)

1 10

o o o

o o

o

s V V V

sV sV

V ss


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151

15

)1(

3525)()2(

101

10)2(

3525)()1(

where

21)2)(1(

3525

)2(251

10

22

11

s

so

s

so

o

o

s

ssVsB

sssVsA

s

B

s

A

ss

sV

sV

s

2(

10 15

Th

) (10 15 ) (

( )1

us

)

2

Vt

o

o

tv t

V ss s

e e u t


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Example 16.3

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


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/

( )( ) 0

/( )

( ) / ( / )

/ /( )

/ ( / )

( ) , 0 where /

o

o

o o o o

o o o

to o

o

VI s R sL LI

sLI V I V L

I sR sL s R sL s R L s s R L

I V R V RI s

s R L s s R L

V Vi t I e t R L

R R

(( )0 )o L

I i


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0 0

/ /

The final value ( ) / ,/

lim ( ) lim/ /

( ) (1 ), 0

o

o o o

s s

t to

o

i V RsI V L V

sI ss R L s R L R

Vi t I e e t

R

/

/ /

( ) / ( / )

( ) , 0 where /

o o o

to o

o

I V R V R

I s s R L s s R L

V Vi t I e t R L

R R

/

If the initial condition, 0,

( ) (1 ), 0

o

to

I

Vi t e t

R


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Circuit Analysis

Remember, equivalent circuits, withcapacitors and inductors, only exist in the s-

domain; they cannot be transformed back into

the time domain.


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Example 16.4

Q: Consider the circuit shown in (a). Find the value of thevoltage 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.


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1 1 1

1

1

2

1

2

1

10 / 0 [ (0) / ](0)0

10 / 3 5 1/ (0.1 )2 3 1

0.1 3 0.5

where (0) 5 V and (0) 1

( ) (35 30 ) ( )

A

( 3 2) 40 5

40 5 35 30

( 1)( 2) 1 2

So, Vt t

V s V V v si

s s s

s Vs s s

v i

s s V s

sV

v t e e u t

s s s s

KCL at node 1:

1


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Example 16.5: Superposition

Q: For the circuit in Example 16.4 , and the initial

conditions used in Example 16.4, use superposition tofind the value of the capacitor voltage. (i(0)=-1, v(0)=5)


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2

1

1 1 1

1

2

1

1

10 / 0 00 0

10 / 3 5 1/ (0.1 )

2 30.1 3

( 3 2) 30

30 30 30

( 1)( 2) 1 2

So, ( ) V(30 30 ) ( )t t

V s V V

s s

s V

s s

s s V

v t e e u

Vs s s

t

s

1)


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2)2 2 2

2

2

2

2

2

2

0 0 0(0)0

10 / 3 5 1/ (0.1 )

2 1

0.1 3

( 3 2) 10

10 10 10

( 1)(

( ) (10 10 ) (

2) 1 2

So, V)t t

V V Vi

s s s

s Vs s

s s V

Vs s s

t e e u

s

v t

i(0)=-1


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3)

3 3 3

3

2

3

2

3

3( ) (

0 0 00 0

10 / 3 5 1/ (0.1 )

20.1 3 0.5

( 3 2) 5

5 5 10( 1)( 2) 1 2

So, V5 10 ) ( )t t

V V V

s s

s Vs

s s V s

sVs s s s

v t e e u t

v(0)=5


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1

2

2 3

2

( ) ( ) ( )

{(30 10 5) ( 30 10 10) } ( )

( )

(35 30 V)V

( )

t t

t t

v t v t v t

e

v t

e e uu t

te

Example 16 6:


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Example 16.6:Thevenins Theorem

Q: Assume that there is no initial energy stored in the

circuit shown below at t= 0 and that is = 10 u(t). (a)

Find Vo(s) using Thevenins theorem. (b) Apply the

initial- and finial-value theorems to find vo(0+) and

vo(

). (c) Determine vo(t).


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(a)

1

oc TH

sc

1 1

1

Since 0,10 50

5

/ 2

( 2 ) 0 0100

5 2

100

2 3

x

x

x

I

s s

I V

V V

I s

V I V

s s

Vs


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1sc

oc

sc

T

TH

H

Hence,

100 /(2 3) 50

2 2 (2 3)

50/50 /[ (2 3)]

5 5 50 250

5 5 2

2 3

125

(( ) 4)3 2 8THo

V sI

s s s s

V s

I s s

V

Z s s

Z s

V

s ss s


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(b) Using the initial-value theorem we find

Using the finial-value theorem we find

01

0

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/125lim

4

125lim)(lim)0(0

s

s

sssVv

sso

s

V25.314

125

4

125lim)(lim)(

00

sssVv

so

so


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(c) By partial fraction,

00

47

4

By (a)

125

( ) 31.254

125( 4) ( ) 31.25

31.25 3

125

( 4) 4

( ) 31.

1.25

4

25(1 ( V) )

o s

s

o s

s

o

t

o

o

A

A sV s s

B

BV

s s s

s V ss

V

s

v t e

s

t

s

u


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Transfer Functions

)(

)()(

sX

sYsH

The transfer functionH(s) is the ratio of the output

response Y(s) to the input excitationX(s), assuming all

initial conditions are zero.


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Four Possible Transfer Functions

( )( ) Voltage gain( )

( )( ) Current gain

( )

( )( ) Impedance

( )

( )( ) Admittance

( )

o

i

o

i

V sH sV s

I sH s

I s

V sH s

I s

I sH s

V s

Unit Impulse Response


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Unit Impulse Responsev.s. Transfer Function

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

1

Let ( ) ( ) ( ) 1

Hence,

( ) ( ) or ( ) ( )

whe (r (e ) [ )]

x t t X s

Y s H s y t h

h t s

t

H

L


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Example 16.7

Q: The output of a linear system isy(t) = 10e-tcos4tu(t)

when input isx(t)=e-tu(t). Find the transfer function of

the system and its impulse response.

Ifx(t)=e-tu(t) andy(t) = 10e-tcos4tu(t), then

172

)12(10

16)1(

)1(10

)(

)()(

Hence,

4)1(

)1(10)(and

1

1)(

2

2

2

2

22

ss

ss

s

s

sX

sYsH

s

ssY

ssX


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To find h(t),

2

2 2 2

10( 2 1) 4( ) 10 40

2 17 (

( ) 10 ( ) 40 sin

4

(

1)

4 )th t t e t

s sH s

s s s

u t


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Example 16.8

Q: Determine the transfer functionH(s) = Vo(s) /Io(s) ofthe circuit.


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2

2

2

( 4)

4 2 1/ 2

But

2( 4)2

5 1/ 2

Hence,

( ) 4 ( 4)( )2

12 1( )

o

o

o

o

o

s II

s s

s IV I

s sH ss s

s s

V sI s


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Example 16.9

Q: For the s-domain circuit, find: (a) the transferfunctionH(s) = Vo/Vi, (b) the impulse response, (c)

the response when vi(t) = u(t) V, (d) the response

when vi(t) = 8cos2tV.

E l 6 9 ( )


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(a)

(b)

1 , but1

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

1 1 ( 1) 1 ( 1) / ( 2) 2 3

So, . Thus,2

1( )

33 2

o ab

ab i i ab i

i o

o

i

V Vs

s s s sV V V V V

s s s s

V VHV

ss

V s

3 /

32

21( ) ( )

1 1)

2

(2

th t e u

H ss

t

E l 16 9 ( )


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(c)

3 32 2

0 32 0

3/23/2

( ) ( ), ( ) 1/

2( ) ( ) ( )

2 ( )

where

1 1( )2( ) 3

3 1 1( )

2 2 3

i i

o i

o s

s

o

ss

v t u t V s s

A BV s H s V s

s s s s

A sV ss

B s V ss

3 /2

0 32

Hence, for ( ) ( ),

1 1 1( ) V

1( ) (1 ) ( )

33o

i

t

v t u t

V ss s

v t e u t

3 /

3

2

21( ) ( )

1 1)

2

(2

th t e u

H ss

t

E l 16 9 ( )


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(d)

25

24

4

4)(

2

3

where

4)4)((

4)()()(

and,4

8)(then,2cos8)(When

2/32

2/3

2

232

23

2

sso

io

ii

s

ssVsA

s

CBs

s

A

ss

ssVsHsV

s

ssVttv

2 2

2 2

sin( )

cos( )

at

at

e ts a

s ae t

s a

L

L

E l 16 9


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Example 16.9

2 23 3

2 2

2 2

4 24 / 25( )

( )( 4) 424 3 3

4 ( 4)25 2 2

o

s Bs C V s

s s s s

s s B s s C s

2

Equating coefficients,

24 3 64

Constants: 0 4 25 2 25

3: 4

2

24 24: 0

25 25

C C

s B C

s B B

3 /

2425

2

2

232

Hence, for ( ) 8cos2 V,

24 3

24 4( ) cos2 sin 2 ( )

2

2 2( )

25 4 25 4

5

V

3

o

o

i

t

v

v t

t t

s

e t t u

V ss s s

t

St t V i bl


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State Variables

A linear system with m inputs andp outputs.

St t V i bl ( t )


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State Variables (cont.)

1

2

( )

( )( ) responsenting state vastate riav bleec s.

( )

tor

n

x t

x tt

x

n

t

x Ax Bz

x

A state variableis a physical property that characterizes

the state of a system, regardless of how the system got to

that state.

St t V i bl ( t )


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1 1

2 2

( ) ( ) / ( ) ( ) /

( ) ( ) / n n

x t dx t dt

x t dx t dt

x t dx t dt

x

1

2inpu

( )

( )( ) representint vector inpu .tsg

( )n

z t

z t

z

mt

t

z

St t V i bl ( t )


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x Ax Bz

y Cx Dz

1

2

( )

( )( ) the representing

( )

and are respectively and matrices.

and are respectively and mat

output vector

rices

output

.

s

p

y t

y tt

y t

n n n m

p n

p

p m

y

A B

C D

St t V i bl ( t )


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1

1

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

( ) ( )

( ) ( ) ( )

( )

( )

where system matrix

input coupling matrix

ou

(

t u

(

t

)

( ) )

p

s s s s s s s

s s

s s s

s

s

s

s s

I A B

X AX BZ I A X

H C I A B

BZ

X Z

Y CX DZ

Y

Z

A

B

C

D

matrix

feedforward matrixD

x Ax Bz

y Cx Dz

Steps to Apply the State Variable


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Steps to Apply the State VariableMethod 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. Apply KCL and KVL to the circuit and obtain

circuit variables (voltage and currents) in terms ofthe 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.

E ample 16 10


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Example 16.10

Q: Find the state-space representation of the circuit.Determine the transfer function of the circuit when vs

is the input and ixis the output. TakeR = 1, C=

0.25 F, andL = 0.5 H.

Example 16 10 (cont )


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Applying KCL at node 1:

Applying KVL around the outer loop:

,

L

C

x C

s L s

s

div L

dtdv

i Cdt

dv V

i i i C idt R

div v v L v

v iv

RC C

vv

L L

vdt

i

1 1

11

0

01

0

RC C

s

LL

x

v vv

i iv

iiR



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1 14 2

1 1

11

If 1, , and

04 4 0, ,

0 2 0 2

10 1 0

0 4 4 4 40 2 0 2

RC C

LL

R C L

R

s sss s

A B

C

I A

x Ax Bz

y Cx Dz1

1

( )

( ) ( )

( ) ( )

( )

( )

s

s s

s s

s

s

I A B

H C I A

X Z

ZB

YD



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Taking the inverse of this gives

1

2

2

1

2 2

4

2 4adjoint of( )

determinant of 4 8

( )

4 0 8

1 0 1 02 4 2 2 8

( )

8

4 8 4 8 4 8

s

ss

s s

s

s

s

s

s s

s s s s s

AI A

A

C I A BH

x Ax Bz

y Cx Dz1

1

( )

( ) ( )

( ) ( )

( )

( )

s

s s

s s

s

s

I A B

H C I A

X Z

ZB

YD

Example 16 11


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Example 16.11

Q: Consider the circuit, which may be regarded as atwo-input, two-output. Determine the state variable

model and find the transfer function of the system.



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1 1

1

K1

0VL 6 6

6

:s s

s o

v i i i v i

v i v v

01 0 1

12 4

2( )

22

3 23

KCL at node 1

4

:

s

s s

vi i v i i

i vi v ii v

vv i v i

1 3 32

KCL

3 2 3

2 32 ( )

3 2

at node 2:

o io o o o

s

o s

v v vv i v v i i

i v vv i v i v

State variables: andv i

2s i

v v i v v



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2 1 1 1

2 4 4 0

s

i

vv v

vi i

2 4 4

2

s

s i

i v i v

v v i v v

3

3( )

2

i

o

o s

v vi

v v i v

32 23 3 2

1 13 3

0

0 0

o s

o i

v vv

i vi

Example 16 12


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Example 16.12

Q: Assume we have a system where the output isy(t)and the input isz(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.

2

2( ) ( )3 2 ( ) 5 ( )d y t dy t y t z t

dt dt



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1

2 1

2

1 1

2 2

1

2

1 2

( ), ( )Two state variables:

( )

0 1 0( )

2 3 5

(

( ) 2 ( ) 3

) 1 0

1 0

(

0 1

0 1 2

) 5 ( ) )

3

2 3 5 (x x z ty

x y t x y t

x x y t

x

x xz t

x x

x

t

tx

s

y t y t z t

s

y

I A1

2 3

s

s

2

2

( ) ( )

3 2 ( ) 5 ( )

d y t dy t

y t z tdt dt



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)2)(1(

5

2)3(

5

501

2)3(

5

0

2

13)01(

)()(

isfunctiontransferThe

2)3(

2

13

)(isinverseThe

1

1

ssss

s

ss

s

s

ss

ss

s

s

s

BAICH

AI



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23

5

)(

)()(

)(5)(]23[

2

2

sssZ

sYsH

sZsYss

HW Ch16 (Due Day: 8/24/2012)


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HW Ch16 (Due Day: 8/24/2012)

Chap 16 Applications of the Laplace Transform

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Transcript of Chap 16 Applications of the Laplace Transform