s-Domain Circuit Analysis

MAE140 Linear Circuits132

s-Domain Circuit Analysis

Operate directly in the s-domain with capacitors,inductors and resistorsKey feature – linearity – is preservedCcts described by ODEs and their ICs

Order equals number of C plus number of L

Element-by-element and source transformationNodal or mesh analysis for s-domain cct variables

Solution via Inverse Laplace TransformWhy?

Easier than ODEsEasier to perform engineering designFrequency response ideas - filtering


Element Transformations

Voltage sourceTime domain

v(t) =vS(t)i(t) = depends on cct

Transform domain

V(s) =VS(s)=L(vS(t))I(s) = L(i(t)) depends on cct

Current source

I(s) =L(iS(t))V(s) = L(v(t)) depends on cct

+_

i(t)

vS

iS

v(t)


Element Transformations contd

Controlled sources

Short cct, open cct, OpAmp relations

Sources and active devices behave identicallyConstraints expressed between transformed variables

This all hinges on uniqueness of Laplace Transforms andlinearity

)()()()(

)()()()(

)()()()(

)()()()(

2121

2121

2121

2121

sgVsItgvti

srIsVtritv

sIsItiti

sVsVtvtv

=!=

=!=

=!=

=!=

""

µµ

)()()()(

0)(0)(

0)(0)(

sVsVtvtv

sIti

sVtv

PNPN

OCOC

SCSC

=!=

=!=

=!=



Resistors

Capacitors

Inductors

)()()()(

)()()()(

sGVsIsRIsV

tGvtitRitv

RRRR

RRRR

==

==

iR

vR

s

isV

sLsILissLIsV

idvL

tidt

tdiLtv

LLLLLL

t

LLLL

L

)0()(

1)()0()()(

)0()(1

)()(

)(

0

+=!=

+== " ##

RsYRsZ RR

1)()( ==

sCsYsC

sZ CC == )(1)(

sLsYsLsZ LL

1)()( ==

vC

iC

vL

iL+ + +

s

vsI

sCsVCvssCVsI

vdiC

tvdt

tdvCti

CCCCCC

C

t

CCC

C

)0()(

1)()0()()(

)0()(1

)()(

)(

0

+=!!=

+== " ##


Element Transformations contdResistor

Capacitor

Note the source transformation rules apply!

IR(s)

VR(s)

+

-

vR(t)

+

-

vC(t)

+

-

iR(t)

iC(t)

R R

CsC

1

)0(CCvVC(s)

+

-

IC(s)

VC(s)

+

-

IC(s)

+_s

vC )0(

sC

1

)()( sRIsV RR =

svsI

sCsVCvssCVsI C

CCCCC)0()(1)()0()()( +=−−=



Inductorss

isVsL

sILissLIsV LLLLLL

)0()(1)()0()()( +=−=

iL(t)+_

vL(t)

-

+sL

LiL(0)

IL(s)+

VL(s)

-

IL(s)

VL(s)+

-

sLs

iL )0(


Example 10-1 T&R p 456

RC cct behaviorSwitch in place since t=-∞, closed at t=0. Solve for vC(t).

Initial conditionss-domain solution using nodal analysis

t-domain solution via inverse Laplace transform

R

VA

t=0

C

R

sC

1)0(CCv

I2(s)

I1(s)

vC+

-

)(1

)()(

)()( 21 ssCV

sC

sVsI

R

sVsI C

CC ===

AC Vv =)0(

)()(1

)( tueVtv

RCs

VsV RC

t

AcA

C

!=

+

=



Solve for i(t)

KVL around loop

Solve

Invert

+_

R

VAu(t) Li(t) +_

+_

R

sLI(s)s

VA

LiL(0)

)(sVL

+

-

0)0()()( =++! LA

LisIsLRs

V

!

i(t) =VA

R"VA

Re"Rt

L + iL (0)e"Rt

L#

$ %

&

' ( u(t) Amps

!

I(s) =

VAL

s s+ RL( )

+iL (0)

s+ RL

=

VAR

s+

iL (0) "VA

R

# $ %

& ' (

s+ RL


Impedance and Admittance

Impedance is the s-domain proportionality factorrelating the transform of the voltage across a two-terminal element to the transform of the currentthrough the element with all initial conditions zero

Admittance is the s-domain proportionality factorrelating the transform of the current through atwo-terminal element to the transform of thevoltage across the element with initial conditionszero

Impedance is like resistanceAdmittance is like conductance


Circuit Analysis in s-Domain

Basic rulesThe equivalent impedance Zeq(s) of two impedances Z1(s)

and Z2(s) in series is

Same current flows

The equivalent admittance Yeq(s) of two admittances Y1(s)and Y2(s) in parallel is

Same voltage

)()()( 21 sZsZsZeq +=

I(s)

V(s) Z2

Z1+

-

( ) ( )sIsZsIsZsIsZsV eq )()()()()( 21 =+=

)()()( 21 sYsYsYeq +=

I(s)

V(s) Y1

+

-Y2)()()()()()()( 21 sVsYsVsYsVsYsI eq=+=



Find ZAB(s) and then find V2(s) by voltage division

+_

L

v1(t) RC

A

B

v2(t)+

-

+_

sL

V1(s) R

A

B

V2(s)+

-sC

1

Ω+

++=

++=+=

1111)(

2

RCsRLsRLCs

sCR

sLsC

RsLsZeq

)()()(

)()( 121

12 sV

RsLRLCs

RsV

sZ

sZsV

eq!"

#$%

&

++=

!!"

#

$$%

&=


Superposition in s-domain ccts

The s-domain response of a cct can be found as thesum of two responses1. The zero-input response caused by initial condition

sources with all external inputs turned off2. The zero-state response caused by the external sources

with initial condition sources set to zeroLinearity and superposition

Another subdivision of responses1. Natural response – the general solution

Response representing the natural modes (poles) of thecct

2. Forced response – the particular solutionResponse containing modes due to the input



The switch has been open for a long time and is closedat t=0.

Find the zero-state and zero-input components of V2(s)Find v(t) for IA=1mA, L=2H, R=1.5KΩ, C=1/6 µF

IA

v(t)

t=0

R CL

+

-

V(s)RsL

+

-s

I A

sC

1

RCIA

RLsRLCsRLs

sCRsL

sZeq ++=

++=

2111)(

LCs

RCs

sRIRCIsZsV

LCs

RCs

CI

s

IsZsV

AAeqzi

AA

eqzs

11)()(

11)()(

2

2

++

==

++

==


Example 10-6 contd

Substitute values

LCs

RCs

sRIRCIsZsV

LCs

RCs

CI

s

IsZsV

AAeqzi

AA

eqzs

11)()(

11)()(

2

2

++

==

++

==

!

Vzs(s) =6000

(s+1000)(s+ 3000)=

3

s+1000+

"3

s+ 3000

vzs(t) = 3e"1000t

" 3e"3000t[ ]u(t)

!

Vzi (s) =1.5s

(s+1000)(s+ 3000)=

"0.75

s+1000+

2.25

s+ 3000

vzi (t) = "0.75e"1000t + 2.25e"3000t[ ]u(t)

V(s)RsL

+

-s

I A

sC

1

RCIA


Example 10-11 T&R (not in 4th ed)

Formulate node voltage equations in s-domain

+_

R1

v1(t) +-

R2C2

C1 R3 vx(t)+

-µvx(t) v2(t)

+

-

+_

R1

V1(s) +-

R2

C2vC2(0)

R3 Vx(s)+

-µVx(s) V2(s)

+

-1

1

sC

C1vC1(0)2

1

sC

A B C D


Example 10-11 contd

Node A: Node D:

Node B:

Node C:

+_

R1

V1(s) +-

R2

C2vC2(0)

R3 Vx(s)+

-µVx(s) V2(s)

+

-1

1

sC

C1vC1(0)2

1

sC

A B C D

!

VB (s) "VA (s)

R1

+VB (s) "VD (s)

R2

+VB (s)

1sC1

+VB (s) "VC (s)

1sC2

"C1vC1(0) "C2vC2(0) = 0

)()( 1 sVsVA = )()()( sVsVsV CxD µµ ==

!

"sC2VB (s) + sC2 +G3[ ]VC (s) = "C2vC2(0)


Example 10-16 T&R (not in 4th ed)

Find vO(t) when vS(t) is a unit step u(t) and vC(0)=0

Convert to s-domain

+_R1

vS(t) +-

C R2 vO(t)+

A B C D

+_

R1

VS(s) +-

R2

VO(s)+

sC

1

CvC(0)

VA(s) VB(s) VC(s) VD(s)


Example 10-16 contd

Nodal AnalysisNode A:Node D:Node C:Node B:Node C KCL:Solve for VO(s)

Invert LT

+_

R1

VS(s) +-

R2

VO(s)+

sC

1

CvC(0)

VA(s) VB(s) VC(s) VD(s)

)()( sVsV SA =

)()( sVsV OD =

)0()()()( 11 CSB CvsVGsVsCG =!+

0)( =sVC

)0()()( 2 COB CvsVGssCV !=!!

CRsR

R

sCR

s

s

R

R

sV

CRs

s

R

RsV

sCG

G

CsG

sV SSO

12

1

12

1

12

1

1

21

1

11

1

)(1

)()(

+!

"=

##

$

%

&&

'

(

+!"=

##

$

%

&&

'

(

+!"=

###

$

%

&&&

'

(

+"=

)()( 121 tue

RRtv CR

t

O

−−=


Features of s-domain cct analysis

The response transform of a finite-dimensional,lumped-parameter linear cct with input being asum of exponentials is a rational function and itsinverse Laplace Transform is a sum of exponentials

The exponential modes are given by the poles of theresponse transform

Because the response is real, the poles are eitherreal or occur in complex conjugate pairs

The natural modes are the zeros of the cctdeterminant and lead to the natural response

The forced poles are the poles of the input transformand lead to the forced response


Features of s-domain cct analysis

A cct is stable if all of its poles are located in theopen left half of the complex s-planeA key property of a systemStability: the natural response dies away as t→∞

Bounded inputs yield bounded outputs

A cct composed of Rs, Cs and Ls will be at worstmarginally stableWith Rs in the right place it will be stable

Z(s) and Y(s) both have no poles in Re(s)>0

Impedances/admittances of RLC ccts are “PositiveReal” or energy dissipating

Documents

s-Domain Circuit Analysis