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MAE140 Linear Circuits 132

s-Domain Circuit Analysis

Operate directly in the s-domain with capacitors, inductors and resistors Key feature – linearity – is preserved Ccts described by ODEs and their ICs

Order equals number of C plus number of L

Element-by-element and source transformation Nodal or mesh analysis for s-domain cct variables Solution via Inverse Laplace Transform

Why? Easier than ODEs

Easier to perform engineering design Frequency response ideas - filtering


Element Transformations

Voltage source Time 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 identically Constraints 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

R s Y R s Z R R 1 ) ( ) ( = =

sC s Y sC s Z C C = = ) ( 1 ) (

sL s Y sL s Z L L 1 ) ( ) ( = =

vC

iC

vL

iL + + +

svsI

sCsVCvssCVsI

vdiC

tvdttdvCti

CCCCCC

CtCC

CC

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

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

+=−−=

+== ∫ ττ


Element Transformations contd Resistor

Capacitor

Note the source transformation rules apply!

IR(s)

VR(s)

+

-

vR(t)

+

-

vC(t)

+

-

iR(t)

iC(t)

R R

C sC1

)0(CCvVC(s)

+

-

IC(s)

VC(s)

+

-

IC(s)

+ _ s

vC )0(

sC1

) ( ) ( s RI s V R R =

s v s I

sC s V Cv s sCV s I C C C C C C

) 0 ( ) ( 1 ) ( ) 0 ( ) ( ) ( + = - - =



Inductors s

i s V sL s I Li s sLI s V L

L L L L L ) 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 p 456

RC cct behavior Switch 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 +

-



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

Initial conditions s-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−

=+

=



Solve for i(t)

+ _

R

VAu(t) L i(t) + _

+ _

R

sL I(s) sVA

LiL(0)

)(sVL

+

-



Solve for i(t)

KVL around loop

Solve

Invert

+ _

R

VAu(t) L i(t) + _

+ _

R

sL I(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 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 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 resistance Admittance is like conductance


Circuit Analysis in s-Domain

Basic rules The 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) R C

A

B

v2(t) +

-

+ _

sL

V1(s) R

A

B

V2(s) +

- sC1

Ω +

+ + =

+ + = + =

1 1 1 1 ) (

2 RCs

R Ls RLCs sC

R

sL sC R sL s Z eq

)()()()()( 121

12 sV

RsLRLCsRsV

sZsZsV

eq

++=

=


Superposition in s-domain ccts

The s-domain response of a cct can be found as the sum of two responses 1.  The zero-input response caused by initial condition

sources with all external inputs turned off 2.  The zero-state response caused by the external sources

with initial condition sources set to zero Linearity and superposition

Another subdivision of responses 1.  Natural response – the general solution

Response representing the natural modes (poles) of the cct

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



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

+

-

V(s) R sL

+

- sI A

sC1

RCIA



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

+

-

V(s) R sL

+

- sI A

sC1

RCIA

R Ls RLCs RLs

sC R sL

s Z eq + +

= + +

= 2 1 1

1 ) (

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) R sL

+

- sI A

sC1

RCIA


Example

Formulate node voltage equations in s-domain

+ _

R1

v1(t) + -

R2 C2

C1 R3 vx(t) +

- µvx(t) v2(t)

+

-


Example

Formulate node voltage equations in s-domain

+ _

R1

v1(t) + -

R2 C2

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


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 Analysis Node 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

) ( ) ( 1 1 2 t u e

R R t v C R

t O

- -

=


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 stable if all of its poles are located in the open left half of the complex s-plane A key property of a system Stability: 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 stable With 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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