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MAE140 Linear Circuits165

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 transformationNodal or mesh analysis for s-domain cct variables

Solution via Inverse Laplace Transform

Why?

Easier than ODEsEasier to perform engineering design

Frequency response ideas - filtering

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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)

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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 andlinearity

)()()()(

)()()()(

)()()()(

)()()()(

2121

2121

2121

2121

sgVsItgvti

srIsVtritv

sIsItiti

sVsVtvtv

=!=

=!=

=!=

=!=

""

)()()()(

0)(0)(

0)(0)(

sVsVtvtv

sIti

sVtv

PNPN

OCOC

SCSC

=!=

=!=

=!=

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Resistors

Capacitors

Inductors

)()()()(

)()()()(

sGVsIsRIsV

tGvtitRitv

RRRR

RRRR

==

==

iR

vR

s

isV

sLsILissLIsV

idvL

tidt

tdiLtv

LLLLLL

t

LLL

L

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

CC

C

C

)0()(

1)()0()()(

)0()(

1

)(

)(

)(0

+=!!=

+==

"##

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Resistor

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

)()( sRIsVRR

=

s

v

sIsCsVCvssCVsIC

CCCCC

)0(

)(

1

)()0()()( +=

=

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Inductors sisVsLsILissLIsV LLLLLL )0()(1)()0()()(+==

iL(t)

+_

vL(t)

-

+

sL

LiL(0)

IL(s)

+

VL(s)

-

IL(s)

VL(s)

+

-

sL s

iL )0(

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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

sC

1)0(CCv

I2(s)

I1(s)

vC+

-

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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 conditions

s-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

!

=

+

=

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Solve for i(t)

+_

R

VAu(t) Li(t) +_

+_

R

sLI(s)s

VA

LiL(0)

)(sVL

+

-

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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

"RtL

#

$%

&

'(u(t) Amps

I(s) =

VAL

s s+ R

L( )

+

iL (0)

s+ RL

=

VAR

s+

iL(0)"VA

R

#$%

&'(

s+ RL

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Impedance and Admittance

Impedance (Z) is thes

-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 resistance

Admittance is like conductance

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

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

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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=+=

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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

+

++=

+

+=+=

11

11)(

2

RCs

RLsRLCs

sCR

sLsC

RsLsZeq

)()()(

)()( 1211

2 sVRsLRLCs

RsV

sZ

sZsV

eq!"

#$%

&

++

=

!!"

#

$$%

&=

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Superposition in s-domain ccts

The s-domain response of a cct can be found as thesum of two responses

1. The zero-input response caused by initial conditionsources, with all external inputs turned off

2. The zero-state response caused by the external sources,

with initial condition sources set to zeroLinearity and superposition

Another subdivision of responses

1. 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

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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=1.5K, C=1/6 F

IA

v(t)

t=0

R CL

+

-

V(s)

RsL

+

-s

IA

sC

1

RCIA

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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=1.5K, C=1/6 F

IA

v(t)

t=0

R CL

+

-

V(s)

RsL

+

-s

IA

sC

1

RCIA

RLsRLCs

RLs

sCRsL

sZeq++

=

++

=211

1)(

LCs

RCs

sRIRCIsZsV

LCs

RCs

C

I

sIsZsV

AAeqzi

A

Aeqzs

11)()(

11)()(

2

2

++

==

++

==

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Example 10-6 contd

Substitute values

LCs

RCs

sRIRCIsZsV

LCs

RCs

C

I

s

IsZsV

AAeqzi

A

Aeqzs

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)=

"0.75s+1000

+2.25

s+ 3000

vzi (t) = "0.75e"1000t

+ 2.25e"3000t[ ]u(t)

V(s)

RsL

+

-s

IA

sC

1

RCIA

What are the natural and forced responses?

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Example

Formulate node voltage equations in the s-domain

+_

R1

v1(t)+-

R2C2

C1 R3 vx(t)

+

-

vx(t) v2(t)

+

-

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Example

Formulate node voltage equations ins

-domain

+_

R1

v1(t)+-

R2C2

C1 R3 vx(t)

+

-

vx(t) v2(t)

+

-

+_

R1

V1(s)+-

R2

C2vC2(0)

R3 Vx(s)

+

-

Vx(s) V2(s)

+

-1

1

sC

C1vC1(0)

2

1

sC

A B C D

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Example contd

Node A: Node D:

Node B:

Node C:

+_

R1

V1(s) +-

R2

C2vC2(0)

R3 Vx(s)

+

-

Vx(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 ==

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

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Example

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

Convert to s-domain

+_

R1vS(t) +

-C

R2 vO(t)+

A B C D

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Example

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

Convert to s-domain

+_

R1vS(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)

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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)+

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!

=

!!

VO(s) = "

sG1C

G2

G1 +

sC

#

$

%%%

&

'

(((

VS(s) = "

R2

R1

)s

s+ 1R

1C

#

$

%%

&

'

((VS(s)

= "R

2

R1

)s

s+ 1R

1C

#

$

%%

&

'

((1

s=

"R2

R1

)1

s+ 1R

1C

)()( 1

1

2 tueR

Rtv

CR

t

O

=

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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 the

response transformBecause 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 responseThe forced poles are the poles of the input transform

and lead to the forced response

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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-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 worstmarginally 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 PositiveReal or energy dissipating

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