S domain analysis

8/18/2019 S domain analysis

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MAE140 Linear Circuits132

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) =V S (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

s gV s I t gvt i

srI sV t rit v

s I s I t it i

sV sV t vt v

=!=

=!=

=!=

=!=

" "

µ µ

)()()()(

0)(0)(

0)(0)(

sV sV t vt v

s I t i

sV t v

P N P N

OC OC

SC SC

=!=

=!=

=!=


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Resistors

Capacitors

Inductors

)()()()(

)()()()(

sGV s I s RI sV

t Gvt it Rit v

R R R R

R R R R

==

==

i R

v R

s

i sV

sL s I Li s sLI sV

id v L

t idt

t di Lt v

L L L L L L

t

L L L

L

L

)0()(

1)()0()()(

)0()(

1

)(

)(

)(0

+=!=

+==

" # #

R sY R s Z R R

1)()( ==

sC sY sC

s Z C C == )(

1)(

sL sY sL s Z L L

1)()( ==

vC

iC

v L

i L+ + +

s

v s I

sC sV Cv s sCV s I

vd iC

t vdt

t dvC t i

C C C C C C

C

t

C C

C

C

)0()(

1)()0()()(

)0()(

1

)(

)(

)(0

+=!!=

+==

" # #


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Resistor

Capacitor

Note the source transformation rules apply!

I R(s)

V R(s)

+

-

v R(t)

+

-

vC (t)

+

-

i R(t)

iC (t)

R R

C sC

1

)0(C CvV C (s)

+

-

I C (s)

V C (s)

+

-

I C (s)

+ _ s

vC )0(

sC

1

)()( s RI sV R R

=

s

v

s I sC sV Cv s sCV s I C

C C C C C

)0(

)(

1

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

=


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Inductors si sV sL s I Li s sLI sV L L L L L L )0()(1)()0()()( +=!=

i L(t)

+ _

v L(t)

-

+

sL

Li L(0)

I L(s)

+

V L(s)

-

I L(s)

V L(s)

+

-

sL s

i L )0(


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Example 10-1 T&R 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

V A

t=0

C

R

sC

1)0(C Cv

I 2(s)

I 1(s)

vC +

-

)(1

)()(

)()( 21 s sCV

sC

sV s I

R

sV s I C

C C ===

AC V v =)0(

)()(1

)( t ueV t v

RC s

V sV RC

t

Ac A

C

!

=

+

=


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Solve for i(t)

KVL around loop

Solve

Invert

+ _

R

V Au(t) Li(t) + _

+ _

R

sL I(s) s

V A

Li L(0)

)( sV L

+

-

0)0()()( =++! L A

Li s I sL R s

V

i(t ) =V A

R" V A

Re" Rt

L+ i L (0)e

" Rt L

#

$ %

&

' ( u(t ) Amps

I (s) =

V A L

s s + R

L( )

+

i L (0)

s+ R L

=

V A R

s+

i L (0) "V A

R

# $ %

& ' (

s+ R L


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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 resistance

Admittance is like conductance


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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 s Z s Z s Z eq +=

I(s)

V(s) Z2

Z1+

-( ) ( ) s I s Z s I s Z s I s Z sV eq )()()()()( 21 =+=

)()()( 21 sY sY sY eq +=

I(s)

V(s) Y1

+

-

Y2)()()()()()()( 21 sV sY sV sY sV sY s I 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

V 1(s) R

A

B

V 2(s)

+

-

sC

1

#+

++=

+

+=+=

11

11)(

2

RCs

R Ls RLCs

sC R

sL sC

R sL s Z eq

)()()(

)()( 121

12 sV

R sL RLCs

R sV

s Z

s Z sV

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 the

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 V2(s)

Find v(t) for IA=1mA, L=2H, R=1.5K#, C=1/6 µF

I A

v(t)

t = 0

R C L

+

-

V(s)

R sL

+

- s

I A

sC

1

RCI A

R Ls RLCs

RLs

sC R sL

s Z eq++

=

++

=211

1)(

LC s

RC s

s RI RCI s Z sV

LC s

RC s

C

I

s I s Z sV

A Aeq zi

A

Aeq zs

11)()(

11)()(

2

2

++

==

++

==


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Example 10-6 contd

Substitute values

LC s

RC s

s RI RCI s Z sV

LC s

RC s

C

I

s

I s Z sV

A Aeq zi

A

Aeq zs

11)()(

11)()(

2

2

++

==

++

==

V zs(s) =6000

(s+1000)(s+ 3000)=

3

s+1000+

"3

s+ 3000

v zs(t ) = 3e"1000t

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

V zi (s) =1.5s

(s+1000)(s+ 3000)=

"0.75s+1000

+2.25

s+ 3000

v zi (t ) = "0.75e"1000t

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

V(s)

R sL

+

- s

I A

sC

1

RCI A


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

Formulate node voltage equations in s-domain

+ _

R1

v1(t) +

-

R2C 2

C 1 R3 v x(t)

+

-

µ v x(t) v2(t)

+

-

+ _

R1

V 1(s) +

-

R2

C 2vC2(0)

R3 V x(s)

+

-

µ V x(s) V 2(s)

+

-1

1

sC

C 1vC1(0)

2

1

sC

A B C D


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Example 10-11 contd

Node A: Node D:

Node B:

Node C:

+ _

R1

V 1(s) +

-

R2

C 2vC2(0)

R3 V x(s)

+

-

µ V x(s) V 2(s)

+

-1

1

sC

C 1vC1(0)

2

1

sC

A B C D

V B (s)"V A (s)

R1

+

V B (s)"V D(s)

R2

+

V B (s)

1sC 1

+

V B (s) "V C (s)

1sC 2

"C 1vC 1(0)"C 2vC 2(0) = 0

)()( 1 sV sV A = )()()( sV sV sV C x D µ µ ==

"sC 2V B (s)+ sC 2 +G3[ ]V C (s) = "C 2vC 2(0)


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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

+ _

R1vS (t) +

-C

R2 vO(t)+

A B C D

+ _

R1

V S (s) + -

R2

V O(s)

+

sC

1

CvC (0)

V A(s) V B(s) V C (s) V D(s)


18/20


Example 10-16 contd

Nodal AnalysisNode A:

Node D:

Node C:

Node B:

Node C KCL:Solve for VO(s)

Invert LT

+ _

R1

V S (s) + -

R2

V O

(s)+

sC

1

CvC (0)

V A(s) V B(s) V C (s) V D(s)

)()( sV sV S A =

)()( sV sV O D =

)0()()()( 11 C S B Cv sV G sV sC G =!+

0)( = sV C

)0()()( 2 C O B Cv sV G s sCV !

=

!!

C R s R

R

sC R

s

s

R

R

sV

C R s

s

R

R sV

sC G

G

C sG

sV S S O

12

1

12

1

12

1

1

21

111

1

)(1

)()(

+

!"=##

$

%

&&

'

(

+

!"=

##

$

%

&&

'

(

+

!"=

###

$

%

&&&

'

(

+

"=

)()( 1

2

1 t ue R

Rt v

C R

t

O

!!

=


19/20


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

S domain analysis

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