Peretmuan 12 Laplace in Circuits

8/2/2019 Peretmuan 12 Laplace in Circuits

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THE LAPLACE TRANSFORM

IN CIRCUIT ANALYSIS


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A Resistor in the s Domain

R

+

vi

v=Ri (Ohms Law).

V(s)=RI(s)

R

+

VI


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An Inductor in the s Domain

+

v L i

Initial current of I0di

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

+

V

sL I

+ LI0

0( )( ) IV sI ssL s

+

V sL

I

I0s


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A Capacitor in the s Domain

+

v C

i

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 CV01/sC

+V0/s

1/sC+

V

I


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The Natural Response of an RC Circuit

Rt=0

i

+

vC+

V0

+

VR

I1/sC

+V0/s

0

0

0

1

0

0

1 ( ) ( )

( ) 1

( ) ( )t t

RC RC

VR

RC

V I s RI ss sC

CV

I s RCs s

Vi e u t v Ri V e u t

R


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The Step Response of a Parallel Circuit

iL (t) = ?


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The Step Response of a Parallel Circuit

2 1 1

2 1 1[ ]

dc

dc

dc

I

C

RC LC

I

LCL

RC LC

IV VsCV

R sL s

Vs s

Is s s


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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 1024 10

16 10

384 1020 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


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Transient Response of a

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

L

RC LC

sIi I t I s

s

s

V s I ss s

sV s

s s s

sV sI s

sL s s s


10/56

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

jK

j j j

05

3 02

32000

384 10 ( 32000 24000)12.5 10 90

( 32000 16000)( 32000 64000)( 48000)

( ) (15sin40000 25 sin24000 ) ( )

15sin40000

tL

Lss

jK

j j j

i t t e t u t mA

i t mA


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Mesh Analysis


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Mesh Analysis (cont.)

1 2

1 2

336(42 8.4 ) 42

0 42 (90 10 )

s I Is

I s I


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1

2

40( 9) 15 14 1

( 2)( 12) 2 12

168 7 8.4 1.4

( 2)( 12) 2 12

sI

s s s s s s

Is s s s s s


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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


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Thevenins Theorem

Use the Thevenins theorem to find vc(t)


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Thevenins Theorem (cont.)

4

4

(480 / )(0.002 ) 480

20 0.002 10

0.002 (20) 80( 7500)6020 0.002 10

Th

Th

s sV

s s

s sZs s

IC= ?

4


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4

4 5

2 6 2

2

5000 5000

480 / ( 10 )

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

6 6

10000 25 10 ( 5000)

30000 6

5000( 5000)

( ) ( 30000 6 ) ( )

C

C

C

t tc

sI

s s s

s sIs s s

I ss

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 I

sC s s s

v t te u t V


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MUTUAL INDUCTANCE EXAMPLE

1 260(0 ) 5 , (0 ) 012

i A i

i2(t)=?


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MUTUAL INDUCTANCE EXAMPLE

Using the T-equivalent of theinductors, and s-domainequivalent gives the following circuit


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2

3

2

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 10

2 (12 8 ) 10

s I sI

sI s I

2

1 1 2

10 2 2 (12 8 ) 103 2

sII sI s I

s


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THE TRANSFER FUNCTION

The transfer function is defined as the ratio of the Laplace

transform 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.


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THE TRANSFER FUNCTION (cont.)

1

2

22

( ) 1( )

( ) 1 /

1

( ) 1( )

( ) 1

g

g

I sH s

V s R sL sC

sC

s LC RCs

V sH s

V s s LC RCs


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EXAMPLE

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


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EXAMPLE

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 s

s

s

V Vs s

V sH s

V s s

p j

p j

z


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

0 2 6 2

*

31 1 22

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 s

s s s

KK K Ks j s j ss

K K K

v e t

t u t V


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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


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Time 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.


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Impulse Response

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

1( ) ( ) ( ) 1

( ) ( )

( ) ( ) ( )

x t t X s

Y 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.


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CONVOLUTION INTEGRAL

( ) ( )x t t

( ) ( )y t h t

y(t)

a b t

x(t)

a bt

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

Nx(t) y(t)

Nx(t) y(t)(1)

t t


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( )t ( )h t

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

N

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

constant x()

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


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

possible values of

( ) ( )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)


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( ) ( ) ( ) ( ) ( )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

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

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

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

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


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


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Convolution and Realizable Systems

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

For a physically realizable system, the response of the systemcannot 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


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EXAMPLE

http://www.youtube.com/watch?v=MEDjw6VcDTY
http://240p%20%20%20128%20kbit%20convolution%20examples%20_%20convolution%20integ.mp4/http://www.youtube.com/watch?v=MEDjw6VcDTYhttp://www.youtube.com/watch?v=MEDjw6VcDTYhttp://240p%20%20%20128%20kbit%20convolution%20examples%20_%20convolution%20integ.mp4/


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EXAMPLE

( ) 2 ( )th t e u 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

h(t)

x(t)

t

y(t)

1


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Graphical Method of Convolution


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0( ) 2 2(1 ) 0 1

t z ty t e dz e t

1( ) 2 2( 1) 1

t z t

ty t e dz e e t

Since h(z) does not exist prior to t=0 and vi(t-z) does not existfor z>t, product of these functions has nonzero values only inthe interval of 0


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EXAMPLE

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


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


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0( ) 1 0 1

ty t dz t t

1 1

0 10 1( ) 1 1 2 , 1 2

t t t t

tty t dz dz z z t t

When t


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1

2 1( ) 1 1 0 2

t t

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


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Convolution and the Laplace Transform

1 2 1 2( )* ( ) ( ) ( )L f t f t L f f t d 1 2 1 20 0( )* ( ) [ ( ) ( ) ]stL f t f t e f f t dt 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 existprior to t=0-, the lower limit can be changed to 0-


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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

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


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STEADY-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 t

A s AX ss s

A(s

s

A(sY s H s

s


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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 terms

determine the steady-state response.


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1

( ) ( cos sin )

( ) ( cos sin )

2

( ) (cos sin ) 1

( )2 2

s j

j

H s A sK

s j

H j A j

j

H j A j

H j Ae

( )

[ ( )]1

( ) ( )

( )2

( ) ( ) cos[ ( )]

j

j

ss

H j H j e

AK H j e

y t A H j t


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EXAMPLE

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 1 245

6 6

120 2cos(5000 30 45 )

6

=20 2 cos(5000 15 )

ss

sH s

s s

jH jj

j

j

v t

t V


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THE IMPULSE FUNCTION

IN CIRCUIT ANALYSIS

The capacitor is charged to an initial voltageV0 at the time the switch is closed. Find theexpression 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

I R s

C CC

C C

/0( ) ( )et RCV

i t e u tR

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.


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

/000Area=

et RC

e

V

q e dt V C R

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

strength V0Ce. 0 ( )ei V C t


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Series Inductor Circuit

Find v0. Note that opening the switch forcesan instantaneous change in the current L2.

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

10000

0

0

5

0

[( ) 30] 02 15 3 10

40( 7.5) 12( 7.5)

( 5) 5

60 1012

5

( ) 12 ( ) (60 10 ) ( )

s

t

VVs s

s sV

s s s

Vs s

v t t e u t V


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

100

5

( ) 30 4 2

5 25 5

( ) (4 2 ) ( )

s

t

I

s 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?


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

After switching

( )d dtv

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

1 2( ) (0 ) 5 (0 )

30(0 ) 6

5

L L i i

i A

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


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Impulsive 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 AL L

Thus, in an infinitesimal

moment, the impulsive voltagesource has stored

2 20 01 1

2 2

V Vw L J

L L

( )0 ( )

RL tVi e u t

L

Current in the circuit decays to

zero in accordance with thenatural response of the circuit


56/56

EXAMPLE

100

5

0

50

50 30

25 5

12 4

5

( ) (12 4) ( )

60 60(15 2 ) 325

( ) 32 ( ) (60 60) ( )

s

t

t

Is

s s

i 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

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Peretmuan 12 Laplace in Circuits