bab 16 sistem linier

8/20/2019 Bab 16 Sistem Linier

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Chapter 16Chapter 16

Applications of theApplications of theLaplace TransformLaplace Transform

Copyright © The McGraw-Hill Companies, Inc. Permission require !or reprouction or isplay.


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#pplication o! the $aplace Trans!orm#pplication o! the $aplace Trans!orm

Chapter 1%Chapter 1%

1%.1 Circuit &lement Moels

1%." Circuit #nalysis 1%.' Trans!er (unctions

1%.) *tate +ariales


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1%.1 Circuit &lement Moels 11%.1 Circuit &lement Moels 1

*teps in #pplying the $aplace Trans!orm/

1.Trans!orm the circuit !rom the time omainto the s-omain

".*ol0e the circuit using noal analysis, meshanalysis, source trans!ormation,superposition, or any circuit analysistechnique with which we are !amiliar

'.Tae the in0erse trans!orm o! the solutionan thus otain the solution in the timeomain.


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1%.1 Circuit &lement Moels "1%.1 Circuit &lement Moels "

#ssume 2ero initial conition !or the inuctor an capacitor,

3esistor / +s43Is

Inuctor/ +s4s$Is

Capacitor/ +s 4 Is5sC

The impeance in the s-omainis e!ine as 6s 4 +s5Is

The amittance in the s-omainis e!ine as 7s 4 Is5+s

Time-domain and s-domain representations of

passive elements under zero initial conditions.


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1%.1 Circuit &lement Moels '1%.1 Circuit &lement Moels '

&9ample 1/(in v 0(t) in the circuit shown elow, assuming

2ero initial conitions.


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1%.1 Circuit &lement Moels )1%.1 Circuit &lement Moels )

*olution/Trans!orm the circuit !rom the time omain tothe s-omain, we ha0e

s

L

t u

3sC1 F

31

ss H1

s

1 )(

=⇒

=⇒

⇒


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*olution/#pply mesh analysis, on sol0ing !or V 0(s)

Taing the in0ersetrans!orm gi0e

220

)2()4(

2

2

3)(V

++

= s

s

1%.1 Circuit &lement Moels 81%.1 Circuit &lement Moels 8

0 V, )2sin(2

3)( 40 ≥=

− t t et v t


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1%.1 Circuit &lement Moels %1%.1 Circuit &lement Moels %

&9ample "/


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1%.1 Circuit &lement Moels :1%.1 Circuit &lement Moels :

&9ample '/(in v 0(t) in the circuit shown elow. #ssume

v 0(0)=5V .

V )()1510()( v:Ans2

0 t ueet t t −− +=*Refer to in-class illustration, textbook


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1%.1 Circuit &lement Moels ;1%.1 Circuit &lement Moels ;

&9ample )/The switch shown elow has een in position b!or a long time. It is mo0e to position a at t=0..

where0, t,I)IV()( v:Ans 0/

00 RC Re Rt t =>+−= − τ τ

*Refer to in-class illustration, textbook


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1%." Circuit #nalysis 11%." Circuit #nalysis 1

@ Circuit analysis is relati0ely easy to o in the s-omain.

@Ay trans!orming a complicate set o! mathematicalrelationships in the time omain into the s-omain

where we con0ert operators eri0ati0es anintegrals into simple multipliers o! s an 15s.

@This allow us to use algera to set up an sol0ethe circuit equations.

@In this case, all the circuit theorems anrelationships e0elope !or c circuits are per!ectly0ali in the s-omain.


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1%." Circuit #nalysis "1%." Circuit #nalysis "

&9ample 8/Consier the circuit elow.(in the 0alue o! the 0oltageacross the capacitor

assuming that the 0alue o!v s(t)=10u(t) + an assume

that at t4>, -1# !lowsthrough the inuctor an B8is across the capacitor.


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*olution/Trans!orm the circuit !rom time-omain a intos-omain using $aplace Trans!orm. nrearranging the terms, we ha0e

Ay taing the in0erse trans!orm, we get

2

30

1

35V1

+−

+=

s s

1%." Circuit #nalysis '1%." Circuit #nalysis '

V )()3035()(v 21 t ueet t t −−−=


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1%." Circuit #nalysis )1%." Circuit #nalysis )

&9ample %/The initial energy in the circuit elow is 2ero at t=0. #ssumethat v s=5u(t) +. a (in +>s using the the0enin theorem.

#pply the initial- an !inal-0alue theorem to !in v 0(0) an

v 0(∞). c tain v 0(t).

Ans: (a) V 0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c)

(3.333+0.6667e-0.3t)u(t) V.



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@ The trans!er !unction Hs is the ratio o! theoutput response 7s to the input response Ds,assuming all the initial conitions are 2ero.

, h(t) is the impulse response function.

@ (our types o! gain/

1. Hs 4 0oltage gain 4 +>s5+is

". Hs 4 Current gain 4 I>s5Iis

'. Hs 4 Impeance 4 +s5Is

). Hs 4 #mittance 4 Is5+s

1%.' Trans!er (unctions 11%.' Trans!er (unctions 1

)(

)()(

s X

sY s H =


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&9ample :/

The output o! a linear system is y(t)=10e-t cs4t when theinput is !(t)=e-t u(t . (in the trans!er !unction o! the system

an its impulse response.

*olution/

Trans!orm yt an 9t into s-omain an applyHs47s5Ds, we get

#pply in0erse trans!orm !or Hs, we get

16)1(

44010

16)1(

)1(10

)(

)()(

22

2

++

−=

++

+==

s s

s

s X

sY s H

1%.' Trans!er (unction "1%.' Trans!er (unction "

)()4sin(40)(10)( t ut et t h t −−= δ


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&9ample ;/The trans!er !unction o! a linear system is

(in the output yt ue to the input e-'tut an

its impulse response.

6

2

)( += s

s

s H

1%.' Trans!er (unction '1%.' Trans!er (unction '

)(12e-(t)2 0; t,42 :Ans -6t63 t uee t t δ ≥+− −−



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1%.) *tate +ariales 11%.) *tate +ariales 1

@ It is a physical property that characteri2es thestate o! a system, regarless o! how the systemgot to that state.

@ *teps to apply the *tate +ariale Metho to

Circuit #nalysis1. *elect the inuctor current i an capacitor 0oltage 0 as

the state 0ariales, maing sure they are consistentwith the passi0e sign con0ention.

". #pply EC$ an E+$ to the circuit an otain circuit0ariales in terms o! state 0ariales. This shoul lea toa set o! !irst orer i!!erential equations necessary ansu!!icient to etermine all the state 0ariales.

'. tain the output equations an put the !inal result in a

state-space representation.


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1%.) *tate +ariales "1%.) *tate +ariales "

&9ample =/

tain the state 0ariale moel !or the circuitshown elow. $et 3141Ω, 3"4" Ω,C4>.8( an

$4>."H an otain the trans!er !unction.

[ ]3012

20)H ; 0 ,

0

1

1

11

Ans2201

2

1

++=

=

+

−

−

=

s s(s

i

v RvvC R

i

v

L

R

L

C C R

i

v s


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