Application of Laplace to Circuits

7/30/2019 Application of Laplace to Circuits

1/45

1

APPLICATION OF THE LAPLACE TRANSFORM

TO CIRCUIT ANALYSISLEARNING GOALS

Laplace circuit solutionsShowing the usefulness of the Laplace transform

Circuit Element ModelsTransforming circuits into the Laplace domain

Analysis Techniques

All standard analysis techniques, KVL, KCL, node,loop analysis, Thevenins theorem are applicable

Transfer FunctionThe concept is revisited and given a formal meaning

Pole-Zero Plots/Bode PlotsEstablishing the connection between them

Steady State Response

AC analysis revisited


2/45

2

LAPLACE CIRCUIT SOLUTIONS

We compare a conventional approach to solve differential equations with atechnique using the Laplace transform

)()()( tdt

diLtRitvS +=:KVL

tCC

CC eKtit

dt

diLtRi ==+ )(0)()(

equationaryComplement

pC iii +=

LeLKeRK

tC

tC ==+

0)(

pp Kti =)(

casethisforsolutionParticular

pS RKv ==1

tL

R

CeKti

+=1)(

0000 =

=

teR

tit

L

R

Take Laplace transform of the equation

+=dt

diLsRIsVS L)()(

)()0()( ssIissIdt

di==

L

)()(1 sLsIsRIs

+=)(

1)(LsRs

sI+

=

LRs

K

s

K

sLRs

LsI

/)/(

/1)( 21

++=

+=

RsILRsK

RssIK

LRs

s

1|)()/(

1

|)(

/2

01

=+=

==

=

=

0;11)( >

=

teR

tit

L

R

)()()( tdt

diLtRitvS +=

LInitial conditionsare automatically

included

No need tosearch forparticularor comple-mentarysolutions

Only algebrais needed

P

ar

ticul

ar

Complementary


3/45

3

LEARNING BY DOING 0),( >ttvFind

KCLusingModel

dt

dvC

R

vv S

Sv

0=

+vv

dt

dvC S

Svvdt

dv

RC=+

L)()( sVsV

dt

dvRC S=+

L

)()0()( ssVvssVdt

dv==

L

ssVtuv SS1)()( ==

0)0(0,0)( =


4/45

4

CIRCUIT ELEMENT MODELS

The method used so far follows the steps:

1. Write the differential equation model2. Use Laplace transform to convert the model to an algebraic form

For a more efficient approach:1. Develop s-domain models for circuit elements

2. Draw the Laplace equivalent circuit keeping the interconnections and replacing

the elements by their s-domain models3. Analyze the Laplace equivalent circuit. All usual circuit tools are applicable and all

equations are algebraic.

Resistor

)()()()( sRIsVtitv ==

)()(

)()(

sIti

sVtv

SS

SS

sourcestIndependen

...

)()()()()()()()(

sBVsItBvtisAIsVtAitv

CDCD

CDCD

==

==

sourcesDependent


5/45

5

Capacitor: Model 1

svsI

CssV )0()(1)( +=)0()(1)(

0

vdxxiC

tvt

+=

Capacitor: Model 2

)0()(1)(0

vdxxiC

tvt

+=

Cs

)0()()( CvsCsVsI =

Source transformation

)0(1

)0(Cv

Cs

sv

Ieq ==

Impedance in serieswith voltage source

Impedance in parallelwith current source

s

sIdxxi

t)(

)(

0

=

L


6/45

6

Inductor Models

))0()(()()()( issILsVtdt

diLtv ==

)0()( issIdtdi =L

s

i

Ls

sVsI

)0()()( +=

s

i

Ls

sVsI

)0()()( +=)0()()( LisLsIsV =


7/45

7

)0()()0()()(

)0()()0()()(

22112

2211111

LisLsIMisMsIsV

MisMsIiLssILsV

+=

+=

Mutual Inductance

)()()(

)()()(

22

12

2111

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

+=

+=

Combine into a single source in the primary

Single source in the secondary


8/45

8

LEARNING BY DOING

Ai 1)0( =

Determine the model in the s-domain and the expression forthe voltage across the inductor

Inductor with

initial current

Equivalent circuit ins-domain

1

1

)()(1)( +== ssVsIsV

LawsOhm'

)()1(1 sIs+=:KVL

Steady state for t


9/45

9

ANALYSIS TECHNIQUES

All the analysis techniques are applicable in the s-domain

LEARNING EXAMPLE Draw the s-domain equivalent and find the voltage in boths-domain and time domain

0)0(0,0)( =


10/45

10

LEARNING EXAMPLE Write the loop equations in the s-domain

=+ )0()0()0()( 1121 iLs

vs

vsVA

1Loop

))()(())()((1

)(1

)( 211212

1

1

11 sIsIsLsIsIsC

sIsC

sIR +++

=+ )()0()0(

)0( 222

11 sViLs

viL B

2Loop

)()())()((

1

))()(( 222122

121 sIRsLsIsIsCsIsIsL +++

Do not increasenumber of loops


11/45

11

LEARNING EXAMPLE Write the node equations in the s-domain

=+s

ivC

s

isIA

)0()0(

)0()( 2111

1VNode

)(1)(11 212

11

21

1 sVsCsL

sVsCsLsL

G +

+++

)(1

)(1

1

2

12

2

122 sVsL

sCsVsL

sCsCG

+

+++

=+s

ivCvCsIB

)0()0()0()( 21122

2VNode

Do not increase number

of nodes


12/45

12

LEARNING EXAMPLE

theorem.sNorton'andsThevenin'tion,transformasource

ion,superpositanalysis,loopanalysis,nodeusingFind )(tvo

Assume all initial conditions are zero

Node Analysis

01

)()(

12)(

4 11

=

+

+

s

sVsV

s

ssV

s

o

1V@KCL

01

)()(

2

)( 1=

+

s

sVsVsV oo

oKCL@V

s

2

0)()21()(2

124)()()1(

1

21

2

=++

+=+

sVsssV

s

ssVssVs

o

o

)1(2

s+

s2

2)1(

)3(8)(

s

ssVo

+

+=

Could haveused voltage

divider here

s

sV

12

)(1

+


13/45

13

Loop Analysis

s

sI4)(1 =

1Loop

ssIsI

ssIsIs

12)(2)(

1))()(( 2212 =++

2Loop

22)1()3(4)(

++=

sssI

22)1(

)3(8)(2)(

+

+==

s

ssIsVo


14/45

14

Source Superposition Applying current source

Current divider

'2I

s

s

ssVo

4

212

2)('

++

=

Applying voltage source

Voltage divider

sss

sVo12

12

2)("

++

=2

"'

)1(

)3(8)()()(

+

+=+=

s

ssVsVsV

ooo


15/45

15

Source Transformation

The resistance is redundant

Combine the sources and use currentdivider

+

++

=2124

21

2)(ss

ss

ssVo

2)1(

)3(8)(

+

+

=s

ssVo


16/45

16

Using Thevenins Theorem

Reduce this part

s

s

ss

ssVOC

124412)(

+=+=

Only independent sources

s

ss

sZ

Th

112+

=+=

s

s

s

ssVo124

12

2

)( 2+

++

=

Voltagedivider

2)1(

)3(8)(

+

+=

s

ssVo


17/45

17

Using Nortons Theorem

2

124/124)(

s

s

s

s

ssISC

+=+=

Reduce this part

sZTh =

2

124

21

2)(s

s

s

s

ssVo

+

++

=

2)1(

)3(8)(

+

+=

s

ssVo

Currentdivision


18/45

18

LEARNING EXAMPLE zerobetoconditionsinitialallAssumevoltagetheDetermine ).(tvo

. Three loops, three non-reference nodes

. One voltage source between non-referencenodes - supernode

. One current source. One loop current knownor supermesh

. If v_2 is known, v_o can be obtained with avoltage divider

Selecting the analysis technique:

Transforming the circuit to s-domain

ssVsV12)()( 12 =:constraintSupernode

01

)()(2

/2

)(

2

)( 211=

+++

s

sVsI

s

sVsV:supernodeKCL@

2

)()( 1

sVsI =:variablegControllin

)(11)( 20 sV

ssV

+=:dividerVoltage

ssVsV /12)()( 21 =:algebratheDoing

ssVsI /62/)()( 2 +=

( )

0)1/()(

)/62/)((2/12)()1)(2/1(

2

22

=++

++

ssV

ssVssVs

)54(

)3)(1(12)(

22++

++=

sss

sssV

)54(

)3(12)(

2++

+=

sss

ssVo


19/45

19

Continued ... theoremsThevenin'usingCompute )(sVo

-keep dependent source and controlling

variable in the same sub-circuit

-Make sub-circuit to be reduced as simpleas possible

-Try to leave a simple voltage divider afterreduction to Thevenin equivalent

sVOC /12

2

/12'

0'2/2

/12

2

/12

sVI

Is

sVsV

OC

OCOC

=

=

+

ssVOC12)( =

0'0'2)/2/()'2(' ==+ IIsII

0)/2/("2""2 = sIIIISC

+

s/12

sI /6"=

s

sISC

)3(6 +=

3

2

)(

)(

+==

ssI

sVZ

SC

OCTH

s

ss

sVo 12

3

21

1)(

+++

=


20/45

20

Continued Computing the inverse Laplace transform

Analysis in the s-domain has established that the Laplace transform of the

output voltage is

)54(

)3(12)(

2++

+=

sss

ssVo

)12)(12(542 jsjsss +++=++

)12()12()12)(12(

)3(12)(

*11

js

K

js

K

s

K

jsjss

ssV oo

+++

++=

+++

+=

)()cos(||2)()(

11

*11 tuKteK

js

K

js

K t+

+++

+

536|)( 0== =soo ssVK

==

=

+

+=

+=+=

57.16179.343.19879.3)902(43.1535

45212

)2)(12(

)11(12

12|)()12(

1

jj

jjs

so

VjsK

)(57.161cos(59.75

36)(

2 tutetv to

++=

1)2(2++= s

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

12

)3(12)(

2

2

2

1

2++

+++

++=

++

+=

s

C

s

sC

s

C

ss

ssV oo

)(]sincos[)()(

)(21222221 tutCtCes

C

s

sCt

+

+++

++

+ 5/36|)( 0== =soo ssVC

])2([)1)2(()3(12 212

CsCssCs o +++++=+ 5/12610/362122 22 ==== CCCs o

5/360: 11 =+= CCCo2softscoefficienEquating

)(sin512)cos1(

536)( 22 tutetetv tto

=

One can also use

quadratic factors...


21/45

21

LEARNING EXTENSION equationsnodeusingFind )(tio

supernode

oVSo VV +

Assume zero initial conditionsImplicit circuit transformation to s-domain

SV

KCL at supernode

02

)()(2))()(( =+++

sV

Ls

sV

ssVsVCs ooSo

2

)()(,

12)(

sVsI

ssV ooS ==

1615

41

61

15.0

61)(

22

+ +

=

++

=

s

s

ss

ssIo

algebratheDoing

++

+

+

=

++

+

=

4

15

4

1

4

15

4

1

4

15

4

1

4

15

4

1

61)(

*

1 1

js

K

js

K

jsjs

ssIo

)()cos(||2)()(

11

*11 tuKteK

js

K

js

K t+

+++

+

4

152

4

15

4

161

|)(415

41

4

15

4

11

j

j

sIjsKjs

o

+

=

+=

+=

=

= 72.15653.6

9097.0

72.6633.61K

=

72.1564

15cos06.13)( 4 teti

t

o


22/45

22

LEARNING EXTENSION equationsloopusingFind )(tvo

supermesh

12

2II

s=

sourcetodueconstraint

0212

21

2211 =+++ Is

sIIIs

supermeshonKVL

Solve for I2

)73.3)(27.0(216

)14(216)(

22++

+=++

+=sss

ssss

ssI

73.327.0)( 2102

++

++=

s

K

s

K

s

KsI

2|)( 020==

=sssIK

48.2)73.327.0)(27.0(

2)27.0(16|)()27.0( 27.021 =

+

+=+=

=ssIsK

47.4)27.073.3)(73.3(

2)73.3(16

|)()73.3( 73.322=

+

+=+=

=ssIsK

( ))(2)(

)(47.448.22)(

2

73.327.02

titv

tueeti

o

tt

=

+=

Determine inverse transform


23/45

23

TRANSIENT CIRCUIT ANALYSIS USING LAPLACE TRANSFORM

For the study of transients, especially transients due to switching, it is importantto determine initial conditions. For this determination, one relies on the properties:

1. Voltage across capacitors cannot change discontinuously2. Current through inductors cannot change discontinuously

LEARNING EXAMPLE 0),( >ttvoDetermine

AiVv LC 1)0(,1)0( ==

itshortcircuareinductors

circuitopenarecapacitorscaseDCFor

Assume steady state for t0


24/45

24

Circuit for t>0Laplace

Use mesh analysis

14

)1( 21 +=+s

sIIs

11

)21( 21 =+++

sI

sssI

Solve for I2s

)1( + s

232

12)(

22

++

=

ss

ssI

s

sI

s

sVo1)(

2)( 2 +=

232

72)(

2++

+=

ss

ssVo

Now determine the inverse transform

rootsconjugatecomplex


25/45

25

LEARNING EXTENSION 0),(1 >ttiDetermine

Initial current through inductor

Aii LL 1)0()0(=+=

12

6 s2s1)(1 sI

ss

ssI

1

182

2)(1

+

=Current

divider

)()(

9

)(9

11 tueti

s

ssI t=

+

=


26/45

26

LEARNING EXTENSION 0),( >ttvoDetermine

Determine initial current through inductor

)0(LiUse sourcesuperposition

Ai V 212=

Ai V3

24 =

AiL

3

4)0( =

s2

+

V38

+

)(sVo

divider)(voltage +

+=

3812

242)(

sssVo

=+

+=

)2(3

)368()(

ss

ssVo

2

21

++

s

K

s

K

6|)( 01 == =so ssVK

3

10|)()2( 22 =+= =so sVsK

)(3

86)(

2 tuetv t

o

=


27/45

27

TRANSFER FUNCTION

)(sX )(sYSystem with allinitial conditionsset to zero

)(

)()(

sX

sYsH =

xadtdxa

dtxda

dtxda

ybdt

dyb

dt

ydb

dt

ydb

om

m

mm

m

m

on

n

nn

n

n

++++=

++++

11

1

1

11

1

1

...

...

equationaldifferentiaissystemtheformodeltheIf

)(sYs

dt

yd kk

k

=

L

zeroareconditionsinitialallIf

)()(...)(

)()(...)(

01

01

sXassXasXsa

sYbssYbsYsb

mm

nn

+++=

+++

)(...

...

)(01

01

sXasasa

bsbsb

sY mm

nn

+++

+++

=

01

01

...

...)(

asasa

bsbsbsH

mm

nn

+++

+++=

1)()()( == sXttx

functionimpulsetheFor

The inverse transform of H(s) is alsocalled the impulse response of the system

If the impulse response is known then onecan determine the response of the systemto ANY other input

H(s) can also be interpreted as the Laplace

transform of the output when the input isan impulse and all initial conditions are zero


28/45

28

LEARNING EXAMPLE responseimpulsehasnetworkA u(t)eth t=)(

)(10)()(2 tuetvtv tio

=inputthefor,response,theDetermine

In the Laplace domain, Y(s)=H(s)X(s)

)()()( sVsHsV io =

1

1)()()(

+==

ssHtueth t

2

10)()(10)(

2

+==

ssVtuetv i

ti

)2)(1(

10

)( ++= sssVo

21

21

+++= s

K

s

K

10|)()1( 11 =+= =so sVsK

10|)()2( 22 =+= =so sVs

)(10)(2

tueetvtt

o

=


29/45

29

Impulse response of first and second order systems

First order system

t

Keths

KsH

=

+

= )(1

)(

Normalized second order system

2

00

2

20

2

)(

++

=

ss

sH

12

002,1 = s:poles

tt

eKeKth

)1(

2

)1(

1

200

200

)(

++=

>

networkOverdamped:1:1Case

networkdUnderdampe:1:2Case


30/45

30

LEARNING EXAMPLE

)(

)()(

sV

sVsH

i

o=functiontransfertheDetermine

Transform the circuit to the Laplacedomain. All initial conditions set to zero

Mesh analysis

212)( IIsVi =

21

110 I

sCsI

+++=

)(sVi

)(1)( 2 sI

sCsVo =

Css

C

sVo /1)2/1(

)2/1(

)( 2 ++=

25.025.02,1s == :poles8FCa)

25.02,1 == s:poles16FCb)

073.0,427.02,1 == s:poles32FCc)

D t i th t f f ti th t f d i d th


31/45

31

LEARNING EXAMPLE Determine the transfer function, the type of damping and the

unit step response

Transform the circuit to the Laplacedomain. All initial conditions set to zero

0=+

V

01

)(

1

)(1=+

s

sVsV o

01

)()(

1

)(

1

)(

1

)()( 01111=

+++

sVsVsV

s

sVsVsV S

= )()(1 ssVsV o

16

1

2

132

1

)(

)(

2++

=

ss

sV

sV

S

o

25.002

=o

o2 1=

ssVS1)( =responsestepUnit

2

4

1

)32/1()(

+

=

ss

sVo 21211

)25.0(25.0 ++

++=

s

K

s

K

s

Ko

5.0|)( 0== =soo ssVK

125.0|)()25.0( 25.02

12 =+= =so sVsK[ ]

5.0)(

25.0

2

11 ==

=s

o

ds

sVsdK

( )( ) )(5.0125.05.0)( 25.0 tuettv to +=


32/45

32

LEARNING EXTENSION

84

10)(2

++

+=

ss

ssH

Determine the pole-zero plot, the type of damping and theunit step response

10-z =:zero

22084 2,12

jsss ==++

:poles

x

x

2

2O

10

842

++ ss

2o

o22

2=

)84(

101)()( 2

++

+

==sss

s

ssHsY

2222

)(*221

js

K

js

K

s

KsY

++

+

+

+=

)22)(22(842 jsjsss +++=++

8

10|)( 01 == =sssYK

)4)(22(

28|)()22( 222

jjsVjsK jso

+

+=+=

+=

)()cos(||2)()(

11

*11 tuKteK

js

K

js

K t+

+++

+

=

= 21173.0

90413583.2

1425.82K

+= )2112cos(46.1

8

10)( ttvo


33/45

33

Second order networks: variation of poles with damping ratio

Normalized second order system

200

2

20

2)(

++=

sssH

12

002,1 = s:poles

networkdUnderdampe:1:2Case


34/45

34

LEARNING EXAMPLE The Tacoma Narrows Bridge Revisited

Previously the eventwas modeled as a

resonance problem.More detailed studiesshow that a modelwith a wind-dependentdamping ratio provides

a better explanation

022

2

2

=++

oodt

d

dt

d

(mph)speedwind=

=

U

U00013.00046.0

Torsional ResonanceModel

45mincollapsetotime

12twist

42mphspeedwindfailureatConditions

=

=

=

Problem: Develop a circuit that models this event

2 integrator

adder


35/45

35

model02

2

2

2

=++

oodt

d

dt

d

)2(02 2...

2

...

oooo +==++

iv

( )

)(1

)(

01

)()(

sVsCR

sV

Cs

sV

R

sV

i

i

=

=+

integrator

1v

2v

adder

+=

=++

2

2

1

1

2

2

1

1 0

VR

RV

R

RV

R

V

R

V

R

V

ff

f

dt

d

2

2

dt

d

Simulation usingdependent sources

579.1)00013001156.0( =

U

Using numerical values

Simulation

buildingblocks


36/45

36

Simulation results

Wind speed=20mph

initial torsion=1 degree

Wind speed=35mphinitial torsion =1degree

POLE ZERO PLOT/BODE PLOT CONNECTION


37/45

37

LCs

L

Rs

LCsV

sVsG

in

o

1

1

)(

)()(

2+

+

==

POLE-ZERO PLOT/BODE PLOT CONNECTION

Bode plots display magnitude and phase information of jssG =|)(

They show a cross section of G(s)

52)(2

2

++=

ss

ssG

Cross sectionshown by Bode

If the poles get closer toimaginary axis the peaksand valleys are morepronounced


38/45

38

Cross section

Front view

Due to symmetryshow only positivefrequencies

Amplitude Bode plot

Uses log scales

STEADY STATE RESPONSE


39/45

39

STEADY STATE RESPONSE

)()()( sXsHsY = Response when all initial conditions are zero

Laplace uses positive time functions. Even for sinusoids the response containstransitory terms

EXAMPLE ))(][cos)(()(,1

1)(

22tuttx

s

ssX

ssH =

+=

+=

jsK

jsK

sK

jsjssssY

+

++

+=

++=

*221

1))()(1()(

)()cos(||2)( 22 tuKtKKetyt

++=

transient Steady state response

If interested in the steady state responseonly, then dont determine residuesassociated with transient terms

termstransient++

+

=

++

=

oo

x

o

M

o

M

js

K

js

K

js

X

js

XsHsY x

*

2

1)()(

)(2

1|)()( oMjsox jHXsYjsK o == =

termstransient++= )cos(||2)( 2KtKty ox

))(cos(|)(|)( oooMss HtHXty +=

( )

++

=+=

o

M

o

MtjtjMM

js

X

js

XsXee

XtutX

2

1)(

2)(cos

For the general case))(cos(|)(||)(

)()cos()(

++=

+=

oooMss

oM

jHtjHXty

tutXtxIf

LEARNING EXAMPLE Determine the steady state response


40/45

40

LEARNING EXAMPLE Determine the steady state response

Transform the circuit to the Laplace domain.

Assume all initial conditions are zero

0

1222: 111 =

+

++

s

VVVV i

1

KCL@V

1

12

1V

s

Vo+

=:dividerVoltage

443)()(443)( 2

2

2

2

++=

++=

ss

s

sHsVss

s

sV io

10,2 == Mo X

=++

= 45354.04)2(4)2(3

)2()2(

2

2

jj

jjH

))(cos(|)(||)(

)()cos()(

++=

+=

oooMss

oM

jHtjHXty

tutXtxIf

Vttys )452cos(54.3)( +=

LEARNING EXTENSION 0),( >ttvossDetermine


41/45

41

LEARNING EXTENSION )(oss

))(cos(|)(||)(

)()cos()(

++=

+=

oooMss

oM

jHtjHXty

tutXtxIf

12,2 == Mo X

Transform circuit to Laplace domain.Assume all initial conditions are zero

s

s

1

)(sVi

Thevenin

)(1

1

)(11

1

)( sVssV

s

ssV iiOC +=

+=

1

1

1

1||1,1||)(

2

+

++=

++=+=

s

ss

ss

sssZTh

)()(2

2)( sV

sZsV OC

Th

o+

=

)(33

2)(2

sVss

sV io++

=

)(sH

=

+=

++=

46.9908.6

2

61

2

364

2)2(

jjjH

)(

1

1

112

2)(

2sV

ss

ss

sV io

+

++++

=

)46.992cos(08.6

212)( = ttvoss

LEARNING BY APPLICATION De-emphasize bass


42/45

42

G C O De-emphasize bass

Enhances bass to original level

)1(

)1)(1()( 21

p

zzvR

s

ssKsG

+

++=

filterrecordingRIAA

s

s

s

p

z

z

318

3180

75

2

1

=

=

=

1|)(| 10002 == jssKG

]500[/1346,3

]50[/46.313

],12.2[/33.13

Hzsr

Hzsr

kHzskr

=

=

=

p

z1

z1

:pole

:zeros

Pole-zero map for )(sGvR

The playback filter is the reciprocal

)(

1

)( sGsG rpvp

=

Pole/zero of playback filter cancelspole/zero of recording filter

LEARNING BY DESIGN Filtering noise in a data transmission line


43/45

43

Data bits at 1000bps Noise source is 100kHz

SOLUTION: Insert a second order low-passfilter in the path. Should not affect datasignal and should attenuate noise

Proposed filter

1V

0=V

0)/1(

3

1

2

1

1

1

1

1=

+++

R

VV

R

V

sC

V

R

VV odata

0)/1( 22

1=+

sC

V

R

V o

2132131211

2

21321

3

1111

1

)(

)(

CCRRCRCRCRss

CCRRR

R

sV

sV

data

o

+

+++

=

22

2 ooss ++

Below 100k

Well below 100kabove 1k

Filter designcriteria 1>

1

2

21

3212

3,

1

C

C

CCRRRR o ====

Design equations

Selected pole location or filter


44/45

44

Selected pole location or filter

nFCnFC

kR

33.1,75.0

.2,000,25,40

21

0

==

===

determineandequationsdesignUse

Select

Circuit simulating the filter

noise

5102 =

The filter eliminates noise

but smooths data pulse

25kbpsdata transferrate

Filtered signalis useless

REDESIGN!

N l l ti 1


45/45

45

New pole-zero selection 1=

Simulation for 25kbps

1

2

210

2

31

000,40

1

000,150

C

C

CC

==

==

equationsDesign

APPLI CATI ONAPPLI CATI ONAPPLI CATI ONAPPLI CATI ON

LAPLACELAPLACELAPLACELAPLACE

Application of Laplace to Circuits

Documents

Transcript of Application of Laplace to Circuits