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MAE140 Linear Circuits 168

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?

1. Easier than ODEs2. Easier to perform engineering design

3. Frequency response ideas - filtering


Element Transformations

Voltage sourceTime domain

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

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

Current sourceI(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 and linearity

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

2121

2121

2121

2121

sgVsItgvtisrIsVtritvsIsItitisVsVtvtv

=⇔=

=⇔=

=⇔=

=⇔=

ββ

µµ

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

sVsVtvtvsItisVtv

PNPN

OCOC

SCSC

=⇔=

=⇔=

=⇔=



Resistors

Capacitors

Inductors

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

RRRR

RRRR==

==

iR

vR

sisV

sLsILissLIsV

idvL

tidttdiLtv

LLLLLL

tLLL

LL

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

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

+=−=

+== ∫ ττ

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

+=−−=

+== ∫ ττ


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()()( +=−−=



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(


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−

=+

=


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


Impedance and Admittance

Impedance (Z) is the s-domain proportionality factor relating the transform of the voltage across a two-terminal element to the transform of the current through the element with all initial conditions zero

Admittance (Y) is the s-domain proportionality factor relating the transform of the current through a two-terminal element to the transform of the voltage across the element with initial conditions zero

Impedance is like resistanceAdmittance is like conductance

€

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

€

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


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=+=


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+

++=+

+=+=11

11)(2

RCsRLsRLCs

sCR

sLsC

RsLsZeq

)()()()()( 121

12 sV

RsLRLCsRsV

sZsZsV

eq!"

#$%

&

++=

!!"

#

$$%

&=


Example

Formulate node voltage equations in the 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


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)


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)


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

−−=


Superposition in s-domain ccts

The s-domain response of a cct can be found as the sum 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


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

The switch has been open for a long time and is closed at 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

RLsRLCsRLs

sCRsL

sZeq ++=

++=

2111)(

LCs

RCs

sRIRCIsZsV

LCs

RCs

CI

sIsZsV

AAeqzi

AA

eqzs

11)()(

11)()(

2

2

++==

++==


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?


Features of s-domain cct analysis

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

The exponential modes are given by the poles of the response transform

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

The natural modes are the zeros of the cct determinant and lead to the natural response

The forced poles are the poles of the input transform and lead to the forced response


Features of s-domain cct analysis

A cct is if all of its poles are located in the open 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 worst marginally 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 “Positive Real” or energy dissipating

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