applications of the laplace transform

APPLICATIONS OF THE LAPLACE

TRANSFORM

Chapter 16

Applications of LT

•ch16 Applications of the Laplace

Transform

•2

Introduction

Circuit Element Models

Circuit Analysis

Transfer Functions

State Variables

16.1 Introduction


Transform

•3

A system is a mathematical model of a physical

process relating the input to the output.

16.2 Circuit Element Models


Transform

•4

Steps in Applying the Laplace Transform:

1. Transform the circuit from the time domain to the s-

domain.

2. Solve the circuit using nodal analysis, mesh analysis,

source transformation, superposition, or any circuit

analysis technique.

3. Take the inverse transform of the solution and thus

obtain the solution in the time domain.

Current – voltage relationship


Transform

•5

s

vsI

sCsV

s

isV

sLsI

sRIsV

)0()(

1)(

)0()(

1)(

)()(

Resistor


Transform

•6

)()( )()(

:Resistor

sRIsVRtitv

Induktor


Transform

•7

s

i

SL

sVsI

LissLIsV

issILsVdt

diLtv

)0()()(

)0()()(

))0()(()( )(

:Inductors

Inductor


Transform

•8

Capacitor


Transform

•9

s

v

SC

sIsV

CvssCVsI

vssVCsIdt

dvCti

)0()()(

)0()()(

))0()(()( )(

:Capacitors

Capacitor


Transform

•10

Circuit with Zero Initials


Transform

•11

sCsZ

sLsZ

RsZ

sI

sVsZ

1)( :Capacitor

)( :Inductor

)( :Resistor

)(

)()(


Transform

•12

)()]([

)()]([

)(

)(

)(

1)(

saItai

saVtav

sV

sI

sZsY

L

L

Example 16.1


Transform

•13

Find vo(t) in the circuit of Fig. 16.4, assuming zero

initial conditions.

Example 16.1


Transform

•14

ssC

ssLs

tu

31F

3

1H 1

1)(

Example 16.1


Transform

•15

sssI

Isss

Is

Issss

IssIIs

sIs

Is

Iss

188

3

)188(3

3)35(

3

131

1

)35(3

135

30 2,mesh For

331

1 1,mesh For

232

223

222

22

121

21

Example 16.1


Transform

•16

0 V,2sin2

3)(

)2()4(

2

2

3

188

3)(

4

2222

ttetv

ssssIsV

to

o

Example 16.2


Transform

•17

Find vo(t) in the circuit of Fig. 16.7. Assume vo(0)=5 V.

Example 16.2


Transform

•18

)2(10

1

1010

25.2

1

1

/10105.02

10

1)10/(

sVsVV

s

s

VVVs

ooo

ooo

Example 16.2


Transform

•19

151

15

)1(

3525)()2(

101

10

)2(

3525)()1(

where

21)2)(1(

3525

)2(251

10

22

11

sso

sso

o

o

s

ssVsB

s

ssVsA

s

B

s

A

ss

sV

sVs

Example 16.2


Transform

•20

V )()1510()(

2

15

1

10)(

Thus

2 tueetv

sssV

tto

o

Example 16.3


Transform

•21

In the circuit of Fig. 16.10(a),

the switch moves from position

a to position b at t = 0. Find i(t)

for t > 0.

Example 16.3


Transform

•22

LRtR

Ve

R

VIti

LRs

RV

s

RV

LRs

IsI

LRss

LV

LRs

I

sLRs

V

sLR

LIsI

s

VLIsLRsI

otoo

ooo

oooo

oo

/ where0 ,)(

)/(

//

/)(

)/(

/

/)()(

0))((

/

Example 16.3


Transform

•23

0 ),1()(

,0 condition, initial In the

0 ),1()(

/

/

/lim)(lim

,/)( valueinitial The

/

//

00

teR

Vti

I

teR

VeIti

R

V

LRs

LV

LRs

sIssI

RVi

to

o

toto

ooo

ss

o

16.3 Circuit Analysis


Transform

•24

Remember, equivalent circuits, with capacitors and

inductors, only exist in the s-domain; they cannot be

transformed back into the time domain.

Example 16.4


Transform

•25

Consider the circuit in Fig.

16.12(a). Find the value

of the voltage 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.

Example 16.4


Transform

•26

V )()3035()( So,

2

30

1

35

)2)(1(

540

540)23(

A 1)0( and V 5)0( e wher

5.0132

31.0

0)1.0/(1

]/)0([)0(

5

0

3/10

/10

21

1

12

1

111

tueetv

ssss

sV

sVss

iv

ssV

ss

s

svV

s

i

s

VsV

tt

Example 16.5


Transform

•27

For the circuit in Fig. 16.12,

with the initial conditions used

in Example 16.4, use

superposition to find the

value of the capacitor

voltage.

Example 16.5


Transform

•28

For Fig. 16.13(a), we get

V )()3003()( So,

2

30

1

30

)2)(1(

30

30)23(

3231.0

0)1.0/(1

00

5

0

3/10

/10

21

1

12

1

111

tueetv

ssssV

Vss

sV

ss

s

V

s

VsV

tt

Example 16.5


Transform

•29

For Fig. 16.13(b), we get

V )()0110()( So,

2

10

1

10

)2)(1(

10

10)23(

1231.0

0)1.0/(1

01

5

0

3/10

0

22

2

22

2

222

tueetv

ssssV

Vss

sV

ss

s

V

ss

VV

tt

Example 16.5


Transform

•30

For Fig. 16.13(c), we get

V )()015()( So,

2

10

1

5

)2)(1(

5

5)23(

0.52

31.0

0)1.0/(1

00

5

0

3/10

0

23

3

32

3

333

tueetv

ssss

sV

sVss

Vs

s

s

V

s

VV

tt

Example 16.5


Transform

•31

V )()3035(

V )(})101030()51030{(

)()()()(

2

2

321

tuee

tuee

tvtvtvtv

tt

tt

Example 16.6


Transform

•32

Assume that there is no initial energy stored in the

circuit of Fig. 16.14 at t = 0 and that is = 10 u(t). (a)

Find vo(s) using Thevenin’s theorem. (b) Apply the

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

vo(). (c) Determine vo(t).

Example 16.6


Transform

•33

(a)

32

100

02

0

5

0)2(10

2/

50105

,0 Since

1

11

1sc

THoc

sV

s

VIV

s

sVII

ssVV

I

x

x

x

Example 16.6


Transform

•34

)4(

125

)82(

25050

325

5

5

5

32)]32(/[50

/50

)32(

50

2

)32/(100

2

Hence,

TH

sc

ocTH

1sc

ssssssV

ZV

sss

s

I

VZ

sss

s

s

VI

THo


Transform

•35

(b) Using the initial-value theorem we find

Using the finial-value theorem we find

01

0

/41

/125lim

4

125lim)(lim)0(0

s

s

sssVv

sso

s

V 25.314

125

4

125lim)(lim)(

00

sssVv

so

so

Example 16.6


Transform

•36

Example 16.6


Transform

•37

(c) By partial fraction,

V )()1(25.31)(

4

25.3125.31

25.31125

)()4(

25.314

125)(

4)4(

125

4

74

00

tuetv

ssV

ssVsB

sssVA

s

B

s

A

ssV

to

o

sso

sso

o

16.4 Transfer Functions


Transform

•38

The transfer function H(s) is the ratio of the output

response Y(s) to the input excitation X(s), assuming

all initial conditions are zero.

)(

)()(

sX

sYsH


Transform

•39

)(

)(Admittance)(

)(

)(Impedance)(

)(

)(gainCurrent )(

)(

)(gain Voltage)(

sV

sIsH

sI

sVsH

sI

sIsH

sV

sVsH

i

o

i

o


Transform

•40

)]([)( where

)()(or )()(

1)( that so ),()(

)()()(

1 sHth

thtysHsY

sXttx

sXsHsY

L

Example 16.7 •41

The output of a linear system is y(t) = 10e-t cos4t u(t)

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

the system and its impulse response.

Solution:

If x(t)=e-tu(t) and y(t) = 10e-t cos4t u(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

Example 16.7


Transform

•42

To find h(t),

)(4sin40)(10)(

4)1(

44010)(

22

ttuetth

ssH

t

Example 16.8


Transform

•43

Determine the transfer function H(s) = Vo(s) / Io(s) of

the circuit in Fig.3

Example 16.8


Transform

•44

1122

)4(4

)(

)()(

Hence,

2/15

)4(22

But

2/124

)4(

2

2

2

ss

ss

sI

sVsH

ss

IsIV

ss

IsI

o

o

oo

o

Example 16.9


Transform

•45

For the s-domain circuit in Fig. 16.19, find: (a) the

transfer function H(s) = Vo/Vi, (b) the impulse

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

response when vi(t) = 8cos2t V.

Example 16.9


Transform

•46

(a)

(b)

32

1)( Thus, .

32 So,

32

1

)2/()1(1

)2/()1(

)1(11

)1(1

but ,1

1

sV

VsH

s

VV

Vs

sVV

ss

ssV

s

sV

Vs

V

i

oio

iabiiab

abo

)(2

1)(

1

2

1)(

2/3

23

tuesh

ssH

t

Example 16.9


Transform

•47

(c)

3

1

2

1)(

2

3

3

1

)(2

1)(

where

)(2

2)()()(

/1)( ),()(

2/32/3

0230

23

23

ss

o

s

so

io

ii

ssVsB

sssVA

s

B

s

A

sssVsHsV

ssVtutv

Example 16.9


Transform

•48

(d)

V )()1(3

1)(

11

3

1)(

),()(for Hence,

2/3

230 tuetv

sssV

tutv

to

i

25

24

4

4)(

2

3

where

4)4)((

4)()()(

and ,4

8)( then ,2cos8)(When

2/32

2/3

2

232

23

2

ss

o

io

ii

s

ssVsA

s

CBs

s

A

ss

ssVsHsV

s

ssVttv

Example 16.9


Transform

•49

2

3

2

3)4(4 22 sCssBsAs

ABBAs

CBs

ACCA

0:2

34:

3

8

2

340:Constants

ts,coefficien Equating

2

Example 16.9


Transform

•50

V )(2sin3

42cos

25

24)(

4

2

25

32

425

24)(

V, 2cos8)(for Hence,

64/25.24/25, ,25/24 gives theseSolving

2/3

22

23

2524

tuttetv

ss

s

ssV

ttv

CBA

to

o

i

16.5 State Variables


Transform

•51


Transform

•52

A state variable is a physical property that

characterizes the state of a system, regardless of

how the system got to that state.

vectors.state ingresponsent vector state

)(

)(

)(

)(

where

2

1

n

tx

tx

tx

t

n

x

BzAxx


Transform

•53

input. ngrepresentior input vect

)(

)(

)(

)(

and

)(

)(

)(

2

1

2

1

m

tz

tz

tz

t

tx

tx

tx

n

n

z

x


Transform

•54

DzCxy

BzAxx

matrices. and ly respective are and

matrices. andly respective are and

outputs ngrepresentitor output vec the

)(

)(

)(

)(

where

2

1

m pnp

mnn n

p

ty

ty

ty

ty

p

DC

BA


Transform

•55

matrix dfeedforwar

matrixoutput

matrix couplinginput

matrix system re whe

)()(

)()(

)()()(

matrix.identity theis e wher

)()()(

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

1

1

D

C

B

A

DBAICZ

YH

DZCXY

I

BZAIX

BZXAIBZAXX

ss

ss

sss

sss

sssssss

Steps to Apply the State Variable Method to

Circuit Analysis:


Transform

•56

1. Select the inductor current i and capacitor voltage v as the state variables, making sure they are consistent with the passive sign convention.

2. Applying KCL and KVL to the circuit and obtain circuit variables (voltage and currents) in terms of the 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.

Example 16.10


Transform

•57

Find the state-space representation of the circuit in

Fig. 16.22. Determine the transfer function of the

circuit when vs is the input and ix is the output. Take R

= 1Ω, C = 0.25 F, and L = 0.5 H.

Example 16.10


Transform

•58

C

i

RC

vv

R

Vi

dt

dvCiii

dt

dvCi

dt

diLv

Cx

C

L

gives 1 nodeat KCL Applying

i

v

Ri

vi

v

i

v

R

vi

L

v

L

vi

vvdt

diLvvv

x

s

LL

CRC

s

sLs

01

0

0

,

11

11

Example 16.10


Transform

•59

s

s

s

ss

R

LCR

LL

CRC

2

44

02

44

0

0

0101

,2

00 ,

02

44

0

obtain we, and , ,1 If

11

11

21

41

AI

C

B A

Example 16.10


Transform

•60

Taking the inverse of this gives

84

8

84

82

801

84

2

0

42

401

)()(

84

42

4

)-(sI oft determinan

)-(sI ofadjoint )(

222

1

2

1

ssss

s

ss

s

s

ss

ss

s

s

s

BAICH

A

AAI

Example 16.11


Transform

•61

Consider the circuit in Fig. 16.24, which may be

regarded as a two-input, two-output. Determine the

state variable model and find the transfer function of

the system.


Transform

•62

KVL di loop kiri

Karena i1 harus dibuang,

dengan KVL pada loop vs,

R1Ω, R2Ω, dan C 1/3 F,

1

1

66

06

1

ivi

iiv

s

s

vviv os 1

(1)

(2)

s

ss

vivi

vviiiviv

iivv

ii

442

3

223

)(22

11

100

1

KCL di node 1

Substitusikan ke (2), selanjutnya ke (1):

(3)

(4)

(5)

Example 16.11


Transform

•63

Di node 2,

Substitusikan (4) ke (3) untuk memperoleh

Kemudian substitusikan (7) dan (8) ke persamaan (6)

)(3

2

3

22

32

3

3

1

2

ss

o

oooo

vivivvi

v

ivvivv

(6) (7)

(8)

3

io

vvi

KVL di loop sebelah kanan:

is vvivv 2

Example 16.11


Transform

•64

i

s

o

o

i

s

v

v

i

v

i

v

v

v

i

v

i

v

31

23

31

32

32

0

0

0

04

11

42

12

State equations yang diperoleh:

Example 16.12


Transform

•65

Assume we have a system where the output is y(t) and

the input is z(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.

)(5)(2)(

3)(

2

2

tztydt

tdy

dt

tyd

Example 16.12


Transform

•66

32

1

32

10

10

01

01)(

)(5

0

32

10

)(532)(5)(3)(2)(

)(let Now

)( ),(Let

2

1

2

1

2

1

212

12

1

s

sss

x

xt

tzx

x

x

x

tzxxtztytytyx

tyxx

tyxtyx

AI

y

Example 16.12


Transform

•67

)2)(1(

5

2)3(

5

501

2)3(

5

0

2

13)01(

)()(

isfunction transfer The

2)3(

2

13

)( is inverse The

1

1

ssss

s

ss

s

s

ss

ss

s

s

s

BAICH

AI

Example 16.12


Transform

•68

23

5

)(

)()(

)(5)(]23[

2

2

sssZ

sYsH

sZsYss

Application: Network Stability


Transform

•69

Example 16.13


Transform

•70

Determine k, so that the

Circuit is stable!

Applying mesh analysis:


Transform

•71

In matrix form

And the determinant is

Network Synthesis


Transform

•72


Transform

•73

Example i


Transform

•75

Jika kapasitor tidak bermuatan saat t=0,

tentukan vo(t) saat t > 0


Transform

•76

KCL:

I + 4I = s.Vo

5I = sVo

Dimana

I = (5-Vo)/2 sehingga

5(5-Vo) = 2sVo Vo = 12.5/(s+5/2)

vo(t) = 12.5 e -2.5t V

Example ii


Transform

•77

Convolution of 2 signals


Transform

•78


Transform

•79


Transform

•80


Transform

•81


Transform

•82

applications of the laplace transform

Documents

laplace transform chapter

introduction ch16 applications

initials ch16 applications

resistorch16 applications

circuit of

circuit analysis technique

mesh analysis

time domain