MAE140 Linear Circuits165

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

MAE140 Linear Circuits166

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)

MAE140 Linear Circuits167

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

sgVsItgvtisrIsVtritvsIsItitisVsVtvtv

=!=

=!=

=!=

=!=

""

µµ

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

sVsVtvtvsItisVtv

PNPN

OCOC

SCSC

=!=

=!=

=!=

MAE140 Linear Circuits168

Element Transformations contd

Resistors

Capacitors

Inductors

)()()()()()()()(sGVsIsRIsVtGvtitRitv

RRRR

RRRR==

==

iR

vR

sisV

sLsILissLIsV

idvL

tidttdiLtv

LLLLLL

tLLL

LL

)0()(1)()0()()(

)0()(1)()()(0

+=!=

+== " ##

RsYRsZ RR

1)()( ==

sCsYsC

sZ CC == )(1)(

sLsYsLsZ LL

1)()( ==

vC

iC

vL

iL+ + +

svsI

sCsVCvssCVsI

vdiC

tvdttdvCti

CCCCCC

CtCC

CC

)0()(1)()0()()(

)0()(1)()()(0

+=!!=

+== " ##

MAE140 Linear Circuits169

Element Transformations contdResistor

Capacitor

Note the source transformation rules apply!

IR(s)

VR(s)

+

-

vR(t)

+

-

vC(t)

+

-

iR(t)

iC(t)

R R

CsC1

)0(CCvVC(s)

+

-

IC(s)

VC(s)

+

-

IC(s)

+_s

vC )0(

sC1

)()( sRIsV RR =

svsI

sCsVCvssCVsI C

CCCCC)0()(1)()0()()( +=−−=

MAE140 Linear Circuits170

Element Transformations contd

Inductorss

isVsL

sILissLIsV LLLLLL

)0()(1)()0()()( +=−=

iL(t)+_

vL(t)

-

+sL

LiL(0)

IL(s)+

VL(s)

-

IL(s)

VL(s)+

-

sL siL )0(

MAE140 Linear Circuits171

Example 10-1, T&R, 5th ed, p 456

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

R

VA

t=0

C

R

sC1

)0(CCv

I2(s)

I1(s)

vC+

-

MAE140 Linear Circuits172

Example 10-1 T&R, 5th ed, 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

sC1

)0(CCv

I2(s)

I1(s)

vC+

-

)(1)()()()( 21 ssCV

sC

sVsIRsVsI C

CC ===

AC Vv =)0(

)()(1)( tueVtv

RCs

VsV RCt

AcA

C!

=+

=

MAE140 Linear Circuits173

Example 10-2 T&R, 5th ed, p 457

Solve for i(t)

+_

R

VAu(t) Li(t) +_

+_

R

sLI(s)sVA

LiL(0)

)(sVL

+

-

MAE140 Linear Circuits174

Example 10-2 T&R, 5th ed, p 457

Solve for i(t)

KVL around loop

Solve

Invert

+_

R

VAu(t) Li(t) +_

+_

R

sLI(s)sVA

LiL(0)

)(sVL

+

-

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

!

i(t) =VAR"VARe"RtL + iL (0)e

"Rt L#

$ %

&

' ( u(t) Amps

!

I(s) =VA

Ls s+ RL( )

+iL (0)s+ RL

=VA

Rs

+iL (0) "

VAR

# $ % &

' (

s+ RL

MAE140 Linear Circuits175

Impedance and Admittance

Impedance (Z) is the s-domain proportionalityfactor relating the transform of the voltage acrossa two-terminal element to the transform of thecurrent through the element with all initialconditions zero

Admittance (Y) is the s-domain proportionalityfactor relating the transform of the current througha two-terminal element to the transform of thevoltage across the element with initial conditionszero

Impedance is like resistanceAdmittance is like conductance

!

V (s) = Z(s)I(s)

!

I(s) = Y (s)V (s)

MAE140 Linear Circuits176

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=+=

MAE140 Linear Circuits177

Example 10-3 T&R, 5th ed, p 461

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)+

-sC1

Ω+

++=

++=+=

1111)(

2

RCsRLsRLCs

sCR

sLsC

RsLsZeq

)()()()()( 121

12 sV

RsLRLCsRsV

sZsZsV

eq!"

#$%

&

++=

!!"

#

$$%

&=

MAE140 Linear Circuits178

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 cct2. Forced response – the particular solution

Response containing modes due to the input

MAE140 Linear Circuits179

Example 10-6, T&R, 5th ed, p 466

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

Find the zero-state and zero-input components of V(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

+

-sI A

sC1

RCIA

MAE140 Linear Circuits180

Example 10-6, T&R, 5th ed, p 466

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

+

-sI A

sC1

RCIA

RLsRLCsRLs

sCRsL

sZeq ++=

++=

2111)(

LCs

RCs

sRIRCIsZsV

LCs

RCs

CI

sIsZsV

AAeqzi

AA

eqzs

11)()(

11)()(

2

2

++==

++==

MAE140 Linear Circuits181

Example 10-6 contd

Substitute values

LCs

RCs

sRIRCIsZsV

LCs

RCs

CI

sIsZsV

AAeqzi

AA

eqzs

11)()(

11)()(

2

2

++==

++==

!

Vzs(s) =6000

(s+1000)(s+ 3000)=

3s+1000

+"3

s+ 3000vzs(t) = 3e"1000t " 3e"3000t[ ]u(t)

!

Vzi (s) =1.5s

(s+1000)(s+ 3000)=

"0.75s+1000

+2.25

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

V(s)RsL

+

-sI A

sC1

RCIA

What are the natural and forced responses?

MAE140 Linear Circuits182

Example

Formulate node voltage equations in the s-domain

+_

R1

v1(t) +-

R2C2

C1 R3 vx(t)+

-µvx(t) v2(t)

+

-

MAE140 Linear Circuits183

Example

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

1sC

C1vC1(0)2

1sC

A B C D

MAE140 Linear Circuits184

Example contd

Node A: Node D:

Node B:

Node C:

+_

R1

V1(s) +-

R2

C2vC2(0)

R3 Vx(s)+

-µVx(s) V2(s)

+

-1

1sC

C1vC1(0)2

1sC

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)

MAE140 Linear Circuits185

Example

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

MAE140 Linear Circuits186

Example

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)+

sC1

CvC(0)

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

MAE140 Linear Circuits187

Example

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

Invert LT

+_

R1

VS(s) +-

R2

VO(s)+

sC1

CvC(0)

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

)()( sVsV SA =

)()( sVsV OD =

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

0)( =sVC

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

!

VO(s) = "

sG1CG2

G1 + sC

#

$

% % %

&

'

( ( ( VS (s) = "

R2R1)

ss+ 1

R1C

#

$

% %

&

'

( ( VS (s)

= "R2R1)

ss+ 1

R1C

#

$

% %

&

'

( ( 1s

="R2R1

)1

s+ 1R1C

)()( 112 tue

RRtv CR

t

O

−−=

MAE140 Linear Circuits188

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

MAE140 Linear Circuits189

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