Analysis Using Laplace Function 3

8/12/2019 Analysis Using Laplace Function 3

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Circuit Analysis using

Laplace Transform


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Circuit Analysis using LT

All the linear networks can be represented by

linear constant coefficient differential

equations, and the use of LT provides aneasy method to obtain the solution.

The terminal characteristics of each element

can be described in s-domain by

transforming its time domain equations.


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(i) s-domain resistor model

The time domain terminal characteristic is

v(t) = i(t)R

s-domain equation will be

V(s) = I(s)R s-transformed resistor model

t-domain s-domain

v/s


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(ii) s-domain inductor model 1

The v-irelationship of inductor is given by

The s-domain equation: V(s) = sLI(s)-Li(0) s-transformed inductor model (voltage)

t-domain

s-domain

dt

tdiLtv

)()(

Using Theorem 4

Initial condition

v/s


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(ii) s-domain inductor model 2

Rearrange the s-domain equation:

s-transformed inductor model (current)

t-domain s-domain

Initial condition

( ) (0)( )

V s iI s

sL s

v/s


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(iii) s-domain capacitor model 1

The v-irelationship of capacitor is given by

The s-domain equation:

I(s) = sCV(s)-Cv(0) s-transformed capacitor model (current)

t-domain s-domain

dt

tdvCti

)()(

Using Theorem 4

Initial condition

v/s


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(iii) s-domain capacitor model 2

Rearrange the s-domain equation:

s-transformed capacitor model (voltage)

t-domain s-domain

( ) (0)( )

I s vV s

sC s

Initial condition

v/s


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Summary of RCL circuit


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Example 4.6.1: Draw the

s-domain transformed circuit for a

series RLC circuit.

For resistor

For inductor

For capacitor


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Transformed equation for voltage, V(s)

for series RLC

V(s)=I(s)R+sLI(s)-Li(0)+ I(s)/sC+v(0)/s

There may be situation when the initial conditionsare

zero, this, we called it simply as ZERO INITIALCONDITIONS.

Thus,

V(s)=I(s)R+sLI(s)+ I(s)/sC


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Applications of LT in Circuit

Problems

RC-circuit

Obtain the expression of current i(t)for the RC-circuit,

considering the capacitor as initially uncharged and

source voltage V dc.


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s-domain transformed circuit

s

ktkuL )

1( )

VI s Rs sC

( )

1

VI s

R sRC

1

1( ) [ ( )]t

RCVi t I s eR

L

Dc voltage

Initially uncharged, vC(0-)=0

as

dteeeL statat

1

0


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RL-circuit

Find the expression of current i(t)for t>0, if inductor

current at t=0-is IO. Voltage is dc.


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2 ( ) oII s

s R L

1 2( ) ( ) ( )( )

o oV s L I I V LI s I s I sR sL s s R L s R L

i(0-)=I0

1( )( )

V LI s

s s R L

Use partial fraction expansion


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1 0

0

( )s

s

V LA sI s V R

s R L

1( ) A B

I s

s s R L

1( ) ( ) s R Ls R L

V LB s R L I s V R

s

1

1 1( )

VI s

R s s R L


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1

1 2( ) [ ( ) ( )] 1

R Rt t

L Lo

RtL

o

Vi t I s I s e I e

R

V V I eR R

L

1 2

1 1( ) ( ) ( ) o

IVI s I s I s

R s s R L s R L


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Example1:

Determine the output voltage of the

network shown at t=0+


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Assume that right before the switch is flipped (that is att=0-), the circuit has achieved its steady state conditions.

So, the capacitor would be open and voltage acrosscapacitor would be vC(0-)=1V.

And, the inductor would be shorted and current acrossinductor would beiL(0-)=1A.

*We may denote t=0-as t


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When the switch is flipped at t=0, the topswitch closes and middle switch opens.

We can model the voltage source as 4u(t).

The circuit is as shown at the top.

We need to determine the output voltageafter the switch is flipped, that is at t=0+.

*We may denote t=0+as t>0as well

Li(0)=1 x 1


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The s-transformed circuit is as shown.

We will use mesh equations to solve.

Li(0)=1 x 1V(0)/s


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From Transform Pair 15


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Example 2:

Find the current i(t).

Assume zero initial conditions.


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s-transformed circuit. To find I(s),use current divider rule:


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Example 3:

Draw the s-domain circuit for the LC tuned circuitabove. Include the initial conditions. Then, find the

voltage v(t)using ILT assuming zero initial conditions

and C=0.5Fand L=1H.


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current


with initial conditions


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With zero initial conditions (shown above)

Substitute C=0.5Fand L=1H


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Pole-Zero Plot

)(

)()(

sB

sAsF

1

1 1 0 1 2

1

1 1 0 1 2

... ( )( ) ( )

... ( )( ) ( )

m m

m n m

n n

n n n

a s a s a s a s z s z s z

b s b s b s b s p s p s p


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Poles and zeros can be plotted in complex s-plane,

we called the plot pole-zero plot/diagram

Zeros:The roots of polynomialA(s), i.e. s =-z1,-z2,,

-zmofF(s)such that at these values of s, F(s)=0Poles:The roots of polynomial B(s), i.e. s= -p1, -p2,,-pnof F(s)such that at these values of s, F(s)=.


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Example 4.8.1

Plot the poles and zeros of function F(s)

)5(

)2()(

ss

ssF

Zero at s=-2

Poles at s=0 and -5

Complex s-plane, s=+j


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Example: Poles and zeros

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jsjss

jsjsssF

Let

Location of poles such that F(s) becomes infinity

sp1= 1 sp2= j2 sp3= j2

Location of zeros such that F(s) becomes 0

sz1= 0 sz2= 1 j sz3= 1 + j


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s plane plot j

sp1 = 1 sp2 = j2 sp3 = j2

sz1 = 0 sz2 = 1 j sz3 = 1 + j

Analysis Using Laplace Function 3

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