THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS Osman Parlaktuna Osmangazi University Eskisehir, TURKEY.

THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS

Osman ParlaktunaOsmangazi University

Eskisehir, TURKEY

Circuit Analysis II Spring 2005 Osman Parlaktuna

A Resistor in the s Domain

R

+

viIn time domain v=Ri (Ohm’s Law).

Because R is constant, the Laplace transform of the equation

is V(s)=RI(s). This is the voltage-current relation for a

resistor in s-domain.

R

+

VI


An Inductor in the s Domain

+

v L i

Consider that the inductor carries an initial current of I0 v L

di

dtThe v-i relation in the time-domain is

The Laplace transform is

V(s)=L[sI(s)-i(0-)]=sLI(s)-LI0

+

V

sL I

+ LI0

I sV s

sL

I

s( )

( ) 0

+

V sL

I

I0s


A Capacitor in the s Domain

+

v C

iAssume that the capacitor is initially charged to V0 volts.

The voltage-current relation is i Cdv

dt

Taking the Laplace transform of the equation

I s C sV s v sCV s CV

V ssC

I sV

s

( ) [ ( ) ( )] ( )

( ) ( )

0

10

0I

+

V CV01/sC

+V0/s

1/sC+

V

I


The Natural Response of an RC Circuit

Rt=0

i

+

vC +

V0

+

VR

I1/sC

+V0/s

V

s sCI s RI s

I sCV

RCs s

iV

Re u t v Ri V e u t

VR

RC

tRC

tRC

0

0

1

00

1

1

0

( ) ( )

( )

( ) ( )


The Step Response of a Parallel Circuit

Opening the switch results in a step change in the current applied to the circuit

sCVV

R

V

sL

I

s

Vs s

Is s s

dc

IC

RC LC

L

ILC

RC LC

dc

dc

2 1 1

2 1 1[ ]


Is s s

Is s j s j

IK

s

K

s j

K

s j

K

Kj j

i t e t u t mA

L

L

L

Lt

384 10

64000 16 10

384 10

32000 24000 32000 24000

32000 24000 32000 24000

384 10

16 1024 10

384 10

32000 24000 4800020 10 126 87

24 40 24000 126 87

5

2 8

5

1 2 2

1

5

83

2

53 0

32000 0

( )

( )( )

( )( ).

( ) [ cos( . )] ( )

*


Transient Response of a Parallel RLC Circuit

Replacing the dc current source with a sinusoidal current source

)]()()[(

)()(

)]()()[()(

)()()(

)(

)(A cos

11222

11222

2

112

1

22

LCRC

LCI

L

LCRC

CI

gLCRC

C

mgmg

sss

s

sL

sVsI

sss

ssV

sIss

ssV

s

sIsItIi

m

m


mAti

mAtutetti

jjj

jK

jjj

jK

js

K

js

K

js

K

js

KsI

sss

ssI

sradmAI

Lss

tL

L

L

m

40000sin15

)()24000sin2540000sin15()(

90 105.12)48000)(6400032000)(1600032000(

)2400032000(10384

90 105.7)6400032000)(1600032000)(80000(

)40000(10384

240003200024000320004000040000)(

)101664000)(1016(

10384)(

/40000 ,24

32000

035

2

035

1

*22

*11

8282

5


Mesh Analysis336

42 8 4 42

0 42 90 10

40 9

2 12

15 14

2

1

12

168

2 12

7 8 4

2

14

12

1 2

1 2

1

2

ss I I

I s I

Is

s s s s s s

Is s s s s s

( . )

( )

( )

( )( )

( )( )

. .


i e e u t A

i e e u t A

i A

i A

t t

t t

12 12

22 12

1

2

15 14

7 8 4 14

336 90

42 4815

15 42

907

( ) ( ) ,

( . . ) ( ) .

( )( )

( )

( )( )

,


Thevenin’s Theorem

Use the Thevenin’s theorem to find vc(t).

Vs s

s s

Zs

s

s

s

Th

Th

( / )( . )

.. ( )

.

( )

480 0 002

20 0 002

480

10

600 002 20

20 0 002

80 7500

10

4

4


Is

s s s

Is

s s

s

s

Is s

i t te e u t A

C

C

C

ct t

480 10

80 7500 10 2 10

6

10000 25 10

6

5000

30000

5000

6

5000

30000 6

4

4 5

2 6 2

2

5000 5000

/ ( )

[ ( ) / ( )] [( ) / ]

( )

( )

( ) ( ) ( )

VsC

Is

s

s s

v t te u t V

c C

ct

1 2 10 6

5000

12 10

5000

12 10

5

2

5

2

5 5000

( ) ( )

( ) ( )


MUTUAL INDUCTANCE EXAMPLE

i A i1 2060

125 0 0( ) , ( )

The switch has been in position a for a long time. At t=0,the switch moves instantaneously to position b. Find i2(t)

Using the T-equivalent of the inductors, and s-domain equivalent gives the following circuit


Is s s s

i t e e u t At t

2

23

2 5

1 3

125

1

125

3

125

.

( )( )

. .

( ) . ( ) ( )

( )

( )

3 2 2 10

2 12 8 101 2

1 2

s I sI

sI s I


THE TRANSFER FUNCTIONThe transfer function is defined as the ratio of the Laplacetransform of the output to the Laplace transform of the inputwhen all the initial conditions are zero.

H sY s

X s( )

( )

( ) Where Y(s) is the Laplace transform of the output,

and X(s) is the Laplace transform of the input.

H sI s

V s R sL sC

sC

s LC RCs

H sV s

V s s LC RCs

g

g

1

2

2 2

1

1

11

1

( )( )

( ) /

( )( )

( )


EXAMPLE

V V V

s

V s

Vs

s sV

H sV

V

s

s s

p j p j

z

g

g

g

0 0 06

0 2 6

02 6

1 2

1

1000 250 0 05 100

1000 5000

6000 25 101000 5000

6000 25 10

3000 4000 3000 4000

5000

.( )

( )( )

,

Find the transfer function V0/Vg

and determine the poles and zeros of H(s).


Assume that vg(t)=50tu(t). Find v0(t). Identify the transient and steady-state components of v0(t).

V s H s V ss

s s s

K

s j

K

s j

K

s

K

s

K K K

v e t

t u t V

g

t

0 2 6 2

1 1 22

3

14 0

2 34

04 3000 0

4

1000 5000

6000 25 10

50

3000 4000 3000 4000

5 5 10 79 7 10 4 10

10 5 10 4000 79 7

10 4 10

( ) ( ) ( )( )

( )

. , ,

[ cos( . )

] ( )

*


The transient component is generated by the poles of the transfer function and it is

10 5 10 4000 79 74 3000 0 e tt cos( . )

The steady-state components are generated by the polesof the driving function (input).

( ) ( )10 4 10 4t u t


Time Invariant SystemsIf the input delayed by a seconds, then

L x t a u t a e X s

Y s H s X s e

y t L Y s y t a u t a

as

as

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

1

Therefore, delaying the input by a seconds simply delays theresponse function by a seconds. A circuit that exhibits thischaracteristic is said to be time invariant.


Impulse Response

If a unit impulse source drives the circuit, the response of the circuit equals the inverse transform of the transfer function.

x t t X s

Y s H s

y t L H s h t

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

1

1

Note that this is also the natural response of the circuitbecause the application of an impulsive source is equivalentto instantaneously storing energy in the circuit.


CONVOLUTION INTEGRAL

)()( ttx

y(t)

a b t

x(t)

a bt

Circuit N is linear with no initial stored energy. If we know the form of x(t), then how is y(t) described? To answer this question, we need to know something about N. Suppose we know the impulse response of the system.

Nx(t) y(t)

Nx(t) y(t)(1)

t

)()( thty

t


)( t

Instead of applying the unit impulse at t=0, let us suppose that it is applied at t=λ. The only change in the output is a time delay.

N )( th

Next, suppose that the unit impulse has some strength other than unity. Let the strength be equal to the value of x(t) when t= λ. Since the circuit is linear, the response should be multiplied by the same constant x(λ)

)()( tx N )()( thx


Now let us sum this latest input over all possible values of λ and use the result as a forcing function for N. From the linearity, the response is the sum of the responses resulting from the use of all possible values of λ

dtx )()(

N dthx )()(

From the sifting property of the unit impulse, we see that the input is simply x(t)

N dthx )()(

X(t)


dzzhztxdzzhztxty )()()()()(

)(*)()( thtxty

dthxty )()()(

Our question is now answered. When x(t) is known, and h(t), the unit impulse response of N is known, the response is expressed by

This important relation is known as the convolution integral. It is often abbreviated by means of

Where the asterisk is read “convolved with”. If we let z=t-λ, then dλ=-dz, and the expression for y(t) becomes

dzzhztxdzzthzxthtxty )()()()()(*)()(


Convolution and Realizable Systems

For a physically realizable system, the response of the system cannot begin before the forcing function is applied. Since h(t) is the response of the system when the unit impulse is applied at t=0, h(t) cannot exist for t<0. It follows that, in the second integral, the integrand is zero when z<0; in the first integral, the integrand is zero when (t-z) is negative, or when z>t. Therefore, for realizable systems the convolution integral becomes

tdzzhztxdzzthzxthtxty

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


EXAMPLE

)(2)( tueth th(t)

x(t)

t

y(t)

0

0

)](2)][1()([

)()()(*)()(

)1()()(

dzzueztuztu

dzzhztxthtxty

tututx

z

1


Graphical Method of Convolution


t tz tedzety0

10 )1(22)(

Since h(z) does not exist prior to t=0 and vi(t-z) does not exist for z>t, product of these functions has nonzero values only in the interval of 0<z<t for the case shown where t<1.

When t>1, the nonzero values for the product are obtained in the interval (t-1)<z<t.

1 )1(22)(1

teedzety tzt

t


EXAMPLE

Apply a unit-step function, x(t)=u(t), as the input to a system whose impulse response is h(t)=u(t)-2u(t-1)+u(t-2), and determine the corresponding output y(t)=x(t)*h(t).


10 1)(0

ttdztyt

When t<0, there is no overlap and y(t)=0 for t<0

For 0<t<1, the curves overlap from z=0 to z=t and product is 1. Thus,

When 1<t<2, h(t-z) has slid far enough to the right to bring under the step function that part of the negative square extending from 0 to z=t-1. Thus,

21 ,211)(1

1

0

1

0 1

ttzzdzdztyt

t

tt t

t


Finally, when t>2, h(t-z) has slid far enough to the right so that it lies entirely to the right of z=0

2 011)(1

1

2

tdzdzty

t

t

t

t


Convolution and the Laplace Transform

dtffLtftfL )()()(*)( 2121

0 0 2121 ])()([)(*)( dtdtffetftfL st

Let F1(s) and F2(s) be the Laplace transforms of f1(t) and f2(t), respectively. Now, consider the laplace transform of f1(t)*f2(t),

Since we are dealing with the time functions that do not exist prior to t=0-, the lower limit can be changed to 0-

Since e-st does not depend on λ, we can move this inside the inner integral. Reversing the order of integration

0 21021 ])()([)(*)( ddttffetftfL st


f1(λ) does not depend on t, and it can be moved outside the inner integral

)()(

)()(

)]([)(

])([)(

])()[(

])()[()(*)(

21

0 12

0 21

0 0 21

0 0 2)(

1

0 0 2121

sFsF

defsF

dsFef

ddxxfeef

ddxxfef

ddttfeftftfL

s

s

sxs

xs

st


STEADY-STATE SINUSOIDAL RESPONSE

If the input of a circuit is a sinusoidal functionx t A t( ) cos( )

x t A t A t

X sA s

s

A

sA(s

s

Y s H sA(s

s

( ) cos cos sin sin

( )( cos ) ( sin )

cos sin )

( ) ( )cos sin )

2 2 2 2

2 2

2 2

=


The partial fraction expansion of Y(s) is

Y sK

s j

K

s j( )

*

1 1

terms generated by the poles of H(s)

If the poles of H(s) lie in the left half of the s plane, the correspondingtime-domain terms approach zero as t increases and they do notcontribute to the steady-state response. Thus only the first two termsdetermine the steady-state response.


KH s A s

s j

H j A j

j

H j A jH j Ae

s j

j

1

2

2

1

2

( ) ( cos sin )

( ) ( cos sin )

( ) (cos sin )( )

H j H j e

KAH j e

y t A H j t

j

j

ss

( ) ( )

( )

( ) ( ) cos[ ( )]

( )

[ ( )]

1 2


EXAMPLE

If the input is 120 cos(5000t+300)V, find the steady-state expression for v0

H ss

s s

H jj

j

j

j

v t

t V

ss

( )( )

( )( )

( )

cos( )

cos( )

1000 5000

6000 25 10

50001000 5000 5000

25 10 5000 6000 25 10

1 1

6

2

645

120 2

65000 30 45

20 2 5000 15

2 6

6 6

0

00 0

0

=


THE IMPULSE FUNCTION IN CIRCUIT ANALYSIS

The capacitor is charged to an initial voltageV0 at the time the switch is closed. Find the expression for i(t) as R 0

IR s

CCC

C C

Vs

sC sC

VR

RC

e

e

0

1 2

0

1 1 1

1 2

1 2

( ) ( ) ( )

i tV

Re u tt RCe( ) ( )/

0

As R decreases, the initial current (V0/R)increases and the time constant (RCe)decreases. Apparently i is approaching animpulse function as R approaches zero.


The total area under the i versus t curve represents the total chargetransferred to C2 after the switch is closed.

Area = qV

Re dt V Ct RC

ee

000

/

Thus, as R approaches zero, the current approaches an impulse

strength V0Ce. i V C te 0 ( )


Series Inductor CircuitFind v0. Note that opening the switch forcesan instantaneous change in the current L2.

i A i1 20 10 0 0( ) , ( )

V

s

V

s

Vs

s s

s

s

Vs s

v t t e u t V

s

t

0 0100

0

0

05

2 15

30]

3 100

40 7 5)

5)

12 7 5)

5

6012

10

5

12 60 10

[( )

( .

(

( .

( ) ( ) ( ) ( )


Does this solution make sense? To answer this question, first letus determine the expression for the current.

Atueti

sssI

t

s

)()24()(

5

24

255

30)(

5

100

Before the switch is opened, current through L1 is 10A and in L2 is 0 A,after the switch is opened both currents are 6A. Then the current in L1

changes instantaneously from 10 A to 6 A, while the current in L2 changesinstantaneously from 0 to 6 A. How can we verify that these instantaneousjumps in the inductor current make sense in terms of the physical behaviorof the circuit?


Switching operation places two inductors in series. Any impulsive voltageappearing across the 3H inductor must be balanced by an impulsive voltage across the 2H inductor. Faraday’s law states that the induced voltage is proportional to the change in flux linkagebefore switching

After switching

( )v ddt

L i L i1 1 2 2 3 10 2 0 30( ) ( ) Wb - turns

( ) ( ) ( )

( )

L L i i

i A

1 2 0 5 0

030

56

Thus the solution agrees with the principle of the conservationof flux linkage.


Impulsive SourcesWhen the voltage source is applied, the initialenergy in the inductor is zero; therefore the initialcurrent is zero. There is no voltage drop across R,so the impulsive source appears directly across L

iL

V x dx iV

LA

t

1

000

0 ( ) ( )

Thus, in an infinitesimal moment, the impulsive voltage source has stored

w LV

L

V

LJ

1

2

1

20

202

iV

Le u t

RL t 0 ( ) ( )

Current in the circuit decays tozero in accordance with thenatural response of the circuit


EXAMPLE

I

s

s s

i t e u t A

V s Is s

v t t e u t V

s

t

t

50 30

25 512

5

4

12 4

15 2 3260

5

60

32 60 60

100

5

0

05

( ) ( ) ( )

( )

( ) ( ) ( ) ( )

Find i(t) and v0(t) for t>0

THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS Osman Parlaktuna Osmangazi University Eskisehir, TURKEY.

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Transcript of THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS Osman Parlaktuna Osmangazi University Eskisehir, TURKEY.