THE LAPLACE TRANSFORM THE LAPLACE TRANSFORM IN CIRCUIT ANALYSISIN CIRCUIT ANALYSIS

A Resistor in the s DomainA Resistor in the s Domain

R+

v i v=Ri (Ohm’s Law).

V(s)=RI(sR

+

V I

An Inductor in the s DomainAn Inductor in the s Domain

+

v L i

Initial current of I0

div Ldt

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

+

VsL I

+ LI0

0( )( )IV sI s

sL s

+

V sL

I

I0s

A Capacitor in the s DomainA Capacitor in the s Domain

+

v Ci

Initially charged to V0 volts.dvi Cdt

0

0

( ) [ ( ) (0 )] ( )1( ) ( )

I s C sV s v sCV s CV

VV s I s

sC s

I

+V CV0

1/sC+ V0/s

1/sC+

V

I

The Natural Response of an RC CircuitThe Natural Response of an RC Circuit

Rt=0i

+vC +

V0

+

VRI1/sC

+V0/s

0

0

01

00

1 ( ) ( )

( )1

( ) ( )t tRC RC

VR

RC

VI s RI s

s sCCV

I sRCs s

Vi e u t v Ri V e u t

R

The Step Response of a Parallel CircuitThe Step Response of a Parallel Circuit

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The Step Response of a Parallel CircuitThe Step Response of a Parallel Circuit

2 1 1

2 1 1[ ]

dc

dc

dc

IC

RC LC

ILC

LRC LC

IV VsCVR sL s

Vs s

Is s s

5

2 8

5

*1 2 2

53

1 8

53 0

2

384 10( 64000 16 10 )

384 10( 32000 24000)( 32000 24000)

32000 24000 32000 24000

384 10 24 1016 10

384 10 20 10 126.87( 32000 24000)( 48000)

( ) [24 40

L

L

L

L

Is s s

Is s j s j

K K KI

s s j s j

K

Kj j

i t e

32000 0cos(24000 126.87 )] ( )t t u t mA

Transient Response of aTransient Response of a Parallel RLC Circuit Parallel RLC Circuit

Replacing the dc current source with a sinusoidal current source

2 2

1

2 1 1

2

2 2 2 1 1

2 2 2 1 1

cos A ( )

( ) ( )( ) ( )

( )( )[ ( ) ( )]

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

m

m

mg m g

Cg

RC LC

IC

RC LC

ILC

LRC LC

sIi I t I s

ss

V s I ss s

sV s

s s s

sV sI ssL s s s

5

2 8 2 8

* *1 1 2 2

53

1

24 , 40000 /

384 10( )( 16 10 )( 64000 16 10 )

( )40000 40000 32000 24000 32000 24000

384 10 ( 40000) 7.5 10 90( 80000)(32000 16000)(32000 64000)

m

L

L

I mA rad s

sI s

s s s

K K K KI s

s j s j s j s j

jKj j j

0

53 0

2

32000

384 10 ( 32000 24000) 12.5 10 90( 32000 16000)( 32000 64000)( 48000)

( ) (15sin40000 25 sin24000 ) ( )15sin40000

tL

Lss

jKj j j

i t t e t u t mAi t mA

Mesh AnalysisMesh Analysis

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Mesh Analysis (cont.)Mesh Analysis (cont.)

1 2

1 2

1

2

336 (42 8.4 ) 42

0 42 (90 10 )40( 9) 15 14 1

( 2)( 12) 2 12168 7 8.4 1.4

( 2)( 12) 2 12

s I Is

I s Is

Is s s s s s

Is s s s s s

2 121

2 122

1

2

(15 14 ) ( ) ,

(7 8.4 1.4 ) ( ) .336(90)( ) 15 ,42(48)15(42)( ) 7

90

t t

t t

i e e u t A

i e e u t A

i A

i A

Thevenin’s TheoremThevenin’s TheoremUse the Thevenin’s theorem to find vc(t)

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Thevenin’s Theorem (cont.)Thevenin’s Theorem (cont.)

4

4

(480/ )(0.002 ) 48020 0.002 10

0.002 (20) 80( 7500)6020 0.002 10

Th

Th

s sV

s ss sZ

s s

4

4 5

2 6 2

2

5000 5000

480/( 10 )[80( 7500) /( 10 )] [(2 10 ) / ]

6 610000 25 10 ( 5000)

30000 65000( 5000)

( ) ( 30000 6 ) ( )

C

C

C

t tc

sIs s s

s sIs s s

Iss

i t te e u t A

5 5

2 2

5 5000

1 2 10 6 12 10( 5000) ( 5000)

( ) 12 10 ( )

c C

tc

sV IsC s s s

v t te u t V

MUTUAL INDUCTANCE EXAMPLEMUTUAL INDUCTANCE EXAMPLE

1 260(0 ) 5 , (0 ) 012

i A i

i2(t)=?

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MUTUAL INDUCTANCE EXAMPLEMUTUAL INDUCTANCE EXAMPLE

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

2

32

2.5 1.25 1.25( 1)( 3) 1 3

( ) 1.25( ) ( )t t

Is s s s

i t e e u t A

1 2

1 2

(3 2 ) 2 102 (12 8 ) 10

s I sIsI s I

THE TRANSFER FUNCTIONTHE TRANSFER FUNCTION

The 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.

( )( )( )

Y sH sX s

Y(s) is the Laplace transform of the output,X(s) is the Laplace transform of the input.

THE TRANSFER FUNCTION THE TRANSFER FUNCTION (cont.)(cont.)

1

2

2 2

( ) 1( )( ) 1/

1( ) 1( )( ) 1

g

g

I sH sV s R sL sC

sCs LC RCsV sH sV s s LC RCs

EXAMPLEEXAMPLE

Find the transfer function V0/Vg and determine the poles and zeros of H(s).

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EXAMPLEEXAMPLE

0 0 06

0 2 6

02 6

1

2

1

01000 250 0.05 10

1000( 5000)6000 25 10

1000( 5000)( )6000 25 10

3000 4000, 3000 4000

5000

g

g

g

V V V V ss

sV V

s sV sH sV s s

p jp jz

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

0 2 6 2

*31 1 2

2

4 0 41 2 3

4 3000 00

4

1000( 5000) 50( ) ( ) ( )( 6000 25 10 )

3000 4000 3000 4000

5 5 10 79.7 , 10, 4 10

[10 5 10 cos(4000 79.7 )

10 4 10 ] ( )

g

t

sV s H s V ss s s

KK K Ks j s j ss

K K K

v e t

t u t V

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

4 3000 010 5 10 cos(4000 79.7 )te t

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

4(10 4 10 ) ( )t u t

Time Invariant SystemsTime Invariant Systems

If the input delayed by a seconds, then

1

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

as

as

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

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 ResponseImpulse Response

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

1

( ) ( ) ( ) 1( ) ( )

( ) ( ) ( )

x t t X sY s H s

y t L H s h t

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 INTEGRALCONVOLUTION INTEGRAL

( ) ( )x t t

y(t)

a b t

x(t)

a b t

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

( ) ( )y t h t

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 ( )h t

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)

( ) ( )x t N ( ) ( )x h t

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

( ) ( )x t d

N ( ) ( )x h t d

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

N ( ) ( )x h t d

X(t)

( ) ( ) ( ) ( ) ( )y t x t z h z dz x t z h z dz

( ) ( )* ( )y t x t h t

( ) ( ) ( )y t x h t d

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

( ) ( )* ( ) ( ) ( ) ( ) ( )y t x t h t x z h t z dz x t z h z dz

( ) ( )* ( ) ( ) ( ) ( ) ( )y t x t h t x z h t z dz x t z h z dz

Convolution and Realizable SystemsConvolution 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

0( ) ( )* ( ) ( ) ( ) ( ) ( )ty t x t h t x z h t z dz x t z h z dz

EXAMPLEEXAMPLE

( ) 2 ( )th t e u th(t)

x(t)

ty(t)

0

0

( ) ( ) ( 1)

( ) ( )* ( ) ( ) ( )

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

x t u t u t

y t x t h t x t z h z dz

u t z u t z e u z dz

1

Graphical Method of ConvolutionGraphical Method of Convolution

0( ) 2 2(1 ) 0 1t z ty t e dz e t

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( ) 2 2( 1) 1t z t

ty t e dz e e t

EXAMPLEEXAMPLE

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

?

h(t)=u(t)-2u(t-1)+u(t-2),

0( ) 1 0 1ty t dz t t

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

1 10 10 1

( ) 1 1 2 , 1 2t t t ttt

y t dz dz z z 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

12 1( ) 1 1 0 2t t

t ty t dz dz t

Convolution and the Laplace TransformConvolution and the Laplace Transform

1 2 1 2( )* ( ) ( ) ( )L f t f t L f f t d

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-

1 2 1 20 0( )* ( ) [ ( ) ( ) ]stL f t f t e f f t dt d

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

1 2 1 20 0

( )1 20 0

1 20 0

1 20

2 10

1 2

( )* ( ) ( )[ ( ) ]

( )[ ( ) ]

( ) [ ( ) ]

( ) [ ( )]

( ) ( )

( ) ( )

st

s x

s sx

s

s

L f t f t f e f t dt d

f e f x dx d

f e e f x dx d

f e F s d

F s f e d

F s F s

STEADY-STATE SINUSOIDAL RESPONSESTEADY-STATE SINUSOIDAL RESPONSE

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

2 2 2 2

2 2

2 2

( ) cos cos sin sin( cos ) ( sin )( )

cos sin ) =

cos sin )( ) ( )

x t A t A tA s A

X ss s

A(ssA(sY s H s

s

The partial fraction expansion of Y(s) is*

1 1( ) terms generated by the poles of H(s)K KY s

s j s j

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.

1( ) ( cos sin )

( ) ( cos sin )2

( ) (cos sin ) 1 ( )2 2

s j

j

H s A sK

s j

H j A jj

H j A j H j Ae

( )

[ ( )]1

( ) ( )

( )2

( ) ( ) cos[ ( )]

j

j

ss

H j H j eA

K H j e

y t A H j t

EXAMPLEEXAMPLE

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

2 6

6 6

0

0 00

0

1000( 5000)( )6000 25 10

1000(5000 5000)( 5000)25 10 5000(6000) 25 10

1 1 2 456 6

120 2 cos(5000 30 45 )6

=20 2 cos(5000 15 )

ss

sH s

s sjH j

j

jj

v t

t V

THE IMPULSE FUNCTION THE IMPULSE FUNCTION IN CIRCUIT ANALYSISIN 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

0 0

1 2

1 1 1

1 2

1 2

( ) ( ) ( )e

V Vs R

sC sC RC

e

IR s

C CC

C C

/0( ) ( )et RCVi t e u t

R

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.

/000

Area= et RCe

Vq e dt V C

R

Thus, as R approaches zero, the current approaches an impulsestrength V0Ce. 0 ( )ei V C t

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

1 2(0 ) 10 , (0 ) 0i A i

10000

0

0

50

[( ) 30]0

2 15 3 1040( 7.5) 12( 7.5)

( 5) 560 1012

5( ) 12 ( ) (60 10 ) ( )

s

t

VVs s

s sVs s s

Vs s

v t t e u t V

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

100

5

( ) 30 4 25 25 5

( ) (4 2 ) ( )

s

t

Is s s

i t e u t A

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 L1changes 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

( )ddtv

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

1 2( ) (0 ) 5 (0 )30(0 ) 65

L L i i

i A

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

Impulsive SourcesImpulsive Sources

When 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

000

1 ( ) (0 )t Vi V x dx i A

L L

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

2 20 01 1

2 2V V

w L JL L

( )0 ( )RL tV

i e u tL

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

EXAMPLEEXAMPLE

100

5

0

50

50 3025 5

12 45

( ) (12 4) ( )60 60(15 2 ) 32

5( ) 32 ( ) (60 60) ( )

s

t

t

Is

s si t e u t A

V s Is s

v t t e u t V

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