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S-Domain Analysis

s-Domain Circuit Analysis

Time domain

(t domain)

Complex frequency domain (s domain)

Algebraictechniques

Responsetransform

Kirchhoff’s Laws in s-Domain

Kirchhoff’s current law (KCL)

Kirchhoff’s voltage law (KVL)

) (

1 t i

) (

4 t i

) (

2 t i

) (

3 t i

0 )

( )

( )

( )

1 t + i t − i t + i t =

i I1( s ) + I2( s ) − I3( s ) + I4( s ) = 0

0 )

( )

( )

− v t v t v t − V1( s ) + V2( s ) + V3( s ) = 0

− +v2(t) +v4(t) −

−

+

) (

1 t v

−

+

) (

3 t v

−

+

) (

5 t v

Signal Sources in s Domain

Voltage Source:

_

+ )

depends )

(

) ( )

(

= =

s I

s V s

+

_ )

depends

)

(

) ( )

depends )

(

) ( )

(

=

=

s V

s V s

Time and s-Domain Element Models

Impedance and Voltage Source for Initial Conditions

t

v L( ) = L( )

Inductor:

_ +

_

+ )

(s

V L

Ls

) 0 (

) ( )

(

L

L L

Li

s LsI s

Inductor:

_

+ )

) (

1 )

(

0

C

t C C

v

d

i C

s

I Cs

s V

C

C C

0

) (

1 )

Capacitor:

Impedance and Voltage Source for Initial

Conditions

R s

I

s V s

) ( )

(

0 ) 0 (

th

wi )

(

) ( )

s V s

Z

0 0

th wi

1 )

(

) ( )

Cs s

I

s V s

C

C C

• Impedance Z(s)

with all initial conditions set to zero

ansform current tr

transform voltage

) (s =

Z

• Impedance of the three passive elements

Time and s-Domain Element Models

Admittance and Current Source for Initial Conditions

1 )

R

s

I R = RResistor:

_

+ )

) (

i

d

v L

s

V Ls

s I

L

L L

0

) (

1 )

_

+ )

) ( )

(

C

C C

Cv

s CsV s

Capacitor:

_

+ )

Admittance and Current Source for Initial

Conditions

R s

V

s I s

Y

R

R R

1 )

(

) ( )

0 ) 0 (

th wi

1 )

(

) ( )

V

s I s

Y

0 0

th

wi )

(

) ( )

s V

s I s

C

C C

• Admittance Y(s)

with all initial conditions set to zero

) (

1 transform

voltage

ansform current tr

)

(

s Z

s

• Admittance of the three passive elements

Example: Solve for Current Waveform i(t)

(s

V L

) 0 (

L

Li

− +V R (s)

0 )

( )

A

By KVL:

) ( )

Resistor: Inductor: V L(s) = LsI(s) − Li L( 0 )

0 )

0 ( )

( )

+

s V

L R s

i L

R s

R V

s

R V

L R s

i L

R s

s

L

V s

I

L A

A

L A

+

+ +

−

=

+

+ +

=

) 0 (

) 0 ( )

(

) (

) ( )

0 ( )

R

V R

V t

R L

t L

R A

Series Equivalence and Voltage Division

) ( ) ( )

( ) ( )

1 s Z s I s Z s I s

) ( )) ( )

( (

) ( )

( )

(

2 1

2 1

s I s Z s

Z

s V s

V s

)

( )

(

)

( ) (

)

( )

(

2 2

1 1

s

V s Z

s

Z s

V

s

V s Z

s

Z s

( ) ( )

2 s Z s I s Z s I s

) ( )

( )

−

+

) (

)

(s I

) (

1 s I

) (

2 s I

)

(s I

2

1 Z Z

Parallel Equivalence and Current Division

) ( ) ( )

1 s Y s V s

) ( )) ( )

( (

) ( )

( )

(

2 1

2 1

s V s Y s

Y

s I s

I s

)

( )

(

)

( ) (

)

( )

(

2 2

1 1

s

I s Y

s

Y s

I

s

I s Y

s

Y s

2 s Y s V s

) ( )

( )

)

(s I

2

1 Y Y

)

(s I

) (

2 s I

) (

1 s I

2 s V

2 s V

(

2 1

=

RCs

R Ls

RLCs

RCs

R Ls

s Z

Ls s

R

RCs Cs

R s

Z

s Y

EQ EQ

1 1

) (

1 )

(

1 1

+

= +

2 t v

2 s V

2 s V

A

B

) (

) (

)

( )

(

1 2

1

1 2

s

V R Ls

RCLs

R

s

V Z

s

Z s

V

EQ EQ

+ +

2 s V

A

General Techniques for s-Domain Circuit

Analysis

• Node Voltage Analysis (in s-domain)

– Use Kirchhoff’s Current Law (KCL)– Get equations of node voltages

– Use current sources for initial conditions– Voltage source current source

• Mesh Current Analysis (in s-domain)

– Use Kirchhoff’s Voltage Law (KVL)– Get equations of currents in the mesh– Use voltage sources for initial conditions– Current source voltage source

(Works only for “Planar” circuits)

Formulating Node-Voltage Equations

Step 0: Transform the circuit into the s domain using current sources to represent capacitor and inductor initial conditions

Step 1: Select a reference node Identify a node voltage at each

of the non-reference nodes and a current with every element

in the circuit

Step 2: Write KCL connection constraints in terms of the

element currents at the non-reference nodes

Step 3: Use the element admittances and the fundamental

property of node voltages to express the element currents in terms of the node voltages

Step 4: Substitute the device constraints from Step 3 into the KCL connection constraints from Step 2 and arrange the

resulting equations in a standard form

Example: Formulating Node-Voltage Equations

inductor initial conditions

Step 1: Identify N-1=2 node voltages and a current with each element

Step 2: Apply KCL at nodes A and B:

0 ) ( )

( )

0 ( )

0 (

: B Node

0 ) ( )

( )

0 ( )

( : A Node

3 1

2 1

S

=

− +

I s

i Cv

s I s

I s

i s I

L C

( ) ( )

(

1 where

) ( )

( ) ( )

(

) ( )

(

1 )

( )

( ) ( )

(

3 2 1

s CsV s

V s Y s

I

R G

s GV s

V s Y s

I

s V s

V Ls

s V s

V s Y s

I

B B

C

A A

R

B A

B A

Formulating Node-Voltage Equations (Cont’d)

Step 2: Apply KCL at nodes A and B:

0 ) ( )

( )

0 ( )

0 (

: B Node

0 ) ( )

( )

0 ( )

( : A Node

3 1

2 1

S

=

− +

I s

i Cv

s I s

I s

i s I

L C

L

Step 3: Express element equations in terms of node voltages

) ( )

( ) ( )

(

1 where

) ( )

( ) ( )

(

) ( )

(

1 )

( )

( ) ( )

(

3 2 1

s CsV s

V s Y s

I

R G

s GV s

V s Y s

I

s V s

V Ls

s V s

V s Y s

I

B B

C

A A

R

B A

B A

Step 4: Substitute eqns in Step 3 into eqns in Step 2 and collect

common terms to yield node-voltage eqns.

s

i Cv

s V

Cs Ls

s

V Ls

s

i s I s

V Ls

s

V Ls G

L C

B A

L S

B A

) 0 ( )

0 ( )

(

1 )

(

1

: B Node

) 0 ( )

( )

(

1 )

(

1

: A Node

Solving s-Domain Circuit Equations

• Circuit Determinant:

Ls

G Cs GLCs

Ls Ls

Cs Ls

G

Ls Cs

Ls

Ls Ls

G s

+ +

=

− +

1 (

1 1

1

1 )

(

Depends on circuit element parameters: L, C, G=1/R,

not on driving force and initial conditions

• Solve for node A using Cramer’s rule:

G Cs GLCs

Cv LCsi

G Cs GLCs

s I LCs

s

Ls Cs

Cv s

i

Ls s

i s I

s

s s

V

C L

S

C L

L S

A A

+ +

+

− + +

− +

2

) 0 ( )

0 ( )

( ) 1 (

) (

1 )

0 ( )

0 (

1 )

0 ( )

(

) (

) ( )

(

Zero State when initial condition sources are turned off

Zero input when input sources are turned off

Solving s-Domain Circuit Eqns (Cont’d)

• Solve for node B using Cramer’s rule:

G Cs GLCs

Cv GLs

GLi G

Cs GLCs

s I

s

Cv s

i Ls

s i

s I Ls

G

s

s s

V

C L

S

C L

L S

B B

+ +

+ +

+ +

) 0 ( )

1 (

) 0 ( )

(

) (

) 0 ( )

0 ( 1

) 0 ( )

( 1

) (

) ( )

(

Zero State Zero input

Network Functions

• Driving-point function relates the voltage and

current at a given pair of terminals called a port

Transform Signal

Input

Transform Response

state -

Zero function

) (

1 )

(

) ( )

(

s Y s

I

s V s

• Transfer function relates an input and response at

different ports in the circuit

) (

) ( Function

Transfer Voltage

) (

1

2

s V

s V s

Circuit

in the zero-state

) ( Function

Transfer Current

) (

1

2

s I

s I s

) (

) ( Admittance

Transfer )

(

1

2

s V

s I s

) (

) ( Impedance

Transfer )

(

1

2

s I

s V s

Calculating Network Functions

2 s V

2 s I

) ( )

( )

) ( )

(

) ( )

(

) ( )

(

2 1

2 1

2

s Z s

Z

s Z s

V

s V s

(

) ( )

2 1

2

s Z s

Z

s Z s

( )

) ( )

(

) ( )

(

) ( )

(

2 1

2 1

2

s Y s

Y

s Y s

I

s I s

(

) ( )

2 1

2

s Y s

Y

s Y s

Impulse Response and Step Response

T(s) Circuit

(s T s X s

• Input-output relationship in s-domain

)(1

)()

• When input signal is an impulse

– Impulse response equals network function

– H(s) = impulse response transform

– h(t) = impulse response waveform

s

s H s

s

T s

G( ) = ( ) = ( )

• When input signal is a step

– G(s) = step response transform

– g(t) = step response waveform

)()

(t u t

)()

(=) means equal almost everywhere,

excludes those points at which g(t)

has a discontinuity

dt

t

dg t

h d

h s

g( ) t ( ) , ( )( ) ( )

= ∫ τ τ