1

CIRCUIT ANALYSIS USING LAPLACE TRANSFORM

2

METHODOLOGY

Examples of nonlinear circuits:logic circuits, digital circuits,or any circuits where theoutput is not linearlyproportional to the input.

Examples of linear circuits:amplifiers, lots of OPMcircuits, circuits made ofpassive components (RLCs).

If the circuit is a linear circuit

Laplace transform of the sourcesof excitation: s(t) S(s)

Laplace transform of the all theelements in the circuit

Find the output O(s) in theLaplace freq. domain

Obtain the time response O(t) bytaking the inverse Laplace

transform

Stop or approximatethe circuit into a linear

circuit and continue

NO

YES

3

THE s-DOMAIN CIRCUITS

Equation of circuit analysis: integrodifferential equations.

Convert to phasor circuits for AC steady state.

Convert to s-domain using Laplace transform.

KVL, KCL, Thevenin,etc.

4

KIRCHHOFF’S VOLTAGE LAW

Consider the KVL in time domain:

Apply the Laplace transform:

0)()()()( 4321 tvtvtvtv

0)()()()( 4321 sVsVsVsV

5

KIRCHHOFF’S CURRENT LAW

Consider the KCL in time domain:

Apply the Laplace transform:

0)()()()( 4321 tItItItI

0)()()()( 4321 titititi

6

OHM’S LAW

Consider the Ohm’s Law in time domain

Apply the Laplace transform

RsIsV RR )()(

Rtitv RR )()(

7

INDUCTOR

Inductor’s voltage– In the time domain:

– In the s-domain:

dt

diLtvL )(

)]0()([)( LLL issILsV

8

INDUCTOR

Inductor’s current– Rearrange VL(s) equation:

s

i

sL

sVsI L

L

)0()()(

9

CAPACITOR

Capacitor’s current– In the time domain:

– In the s-domain:

dt

dvCtic )(

)]0()([ ccc vssVC(s)I

10

CAPACITOR

Capacitor’s voltage– Rearranged IC(s) equation:

)(vs(s)IsC(s)V ccc 011

11

RLC VOLTAGE

The voltage across the RLC elements in the s-domain is the sum of a term proportional to its current I(s) and a term that depends on its initial condition.

)]0()([)( LLL issILsV

)(vs(s)IsC(s)V ccc 011

12

CIRCUIT ANALYSIS FOR ZERO INITIAL CONDITIONS (ICs = 0)

13

IMPEDANCE

If we set all initial conditions to zero, the impedance is defined as:

[all initial conditions=0]

)()()( sI

sVsZ

14

IMPEDANCE & ADMITANCE

The impedances in the s-domain are

The admittance is defined as:

sCsZ

sLsZ

RsZ

C

L

R

1)(

)(

)(

sCsYsL

sY

RsY

C

L

R

)(

1)(

1)(

15

Ex.

Find vc(t), t>0

H1

F5.0

3 )(tu

)(tvc

)(tvL

)(tvR

16

Obtain s-Domain Circuit

All ICs are zero since there is no source for t<0

ss

2

3s

1

)(sVc

)(sVL

)(sVR

)(sI

17

Convert to voltage sourced s-Domain Circuit

ss

23

s

3

)(sVc

)(sVL

)(sVR

)(sI

18

23

3)(

03

)(32

2

sssI

ssI

ss :KVL By

Find I(s)

19

Find Capacitor’s Voltage

The capacitor’s voltage:

Rewritten:

)23(

6)(

2)(

2

sss

sIs

sVc

)2)(1(

6

)23(

6)(

2

sssssssVc

20

Using PFE

Expanding Vc(s) using PFE:

Solved for K1, K2, and K3:

21)2)(1(

6)( 321

s

K

s

K

s

K

ssssVc

2

3

1

63

)2)(1(

6)(

ssssss

sVc

21

Find v(t)

Using look up table:

2

3

1

63

)2)(1(

6)(

ssssss

sVc

)(363)( 2 tueetv ttc

22

Ex. Find the Thevenin and Norton

equivalent circuit at the terminal of the inductor.

1 H

0 .5 F

3 u (t )

23

Obtain s-domain circuit

s

2 /s

3 1 /s

24

Find ZTH

2 /s

3

sZTH

23

25

Find VTH or Voc

2 /s

3 1 /s+V T H

-

ssVTH

313

26

Draw The Thevenin Circuit

Using ZTH and VTH:

2 /s 3+- 3 /s

27

Obtain The Norton Circuit The norton current is:

2 /s

33 /(3s + 2)

23

323

3

ss

sZ

VI

TH

THN

28

Ex.

Find v0(t) for t>0.

29

s-Domain Circuit Elements

Laplace transform all circuit’s elements

ssC

s

F

ssLH

tu

3131

1

1

)(

30

s-Domain Circuit

31

Apply Mesh-Current Analysis

21

3

5

31

1I

sI

s

Loop 1

Loop 2

22

1

21

353

1

35

30

IssI

Is

sIs

32

Substitute I1 into eqn loop 1

sssI

Isss

Is

Isss

188

3

1883

335

3

1

5

31

1

232

223

222

33

Find V0(s)

22

2

20

)2()4(

2

2

3

188

3

)(

s

ss

sIsV

34

Obtain v0(t)

tetv t 2sin2

3)( 4

0

22 )2()4(

2

2

3)(

ssVo

35

Ex. The input, is(t) for the circuit below is

shown as in Fig.(b). Find i0(t)

1

0 2 t(s)

is(t)

(b)

)(tis

1H1

)(tio

(a)

36

s-Domain Circuit

)(sIs1s

)(sIo

37

Using current divider:

)1()(1

)(

sI

s

ssIo

38

1

0 2 t(sec)

is(t)

Derive Input signal, Is

0t

is1(t)

0 2t

is2(t)

39

Obtain Is(t) and Is(s)

Expression for is(t):

Laplace transform of is(t):

)2()()( tututis

)2(1111

)( 22 sss e

sse

ssI

40

Substitute eqn. (2) into (1):

11

1

)1(

)1()(

2

2

0

s

e

s

ss

essI

s

s

41

)2()()( )2( tuetueti tto

Inverse Laplace transform

42

CIRCUIT ANALYSIS FOR NON-ZERO INITIAL CONDITION (ICs ≠ 0)

43

TIME DOMAIN TO s-DOMAIN CIRCUITS

s replaced t in the unknown currents and voltages.

Independent source functions are replaced by their s-domain transform pair.

The initial condition serves as a second element, the initial condition generator.

44

THE ELEMENTS LAW OF s-DOMAIN

)0(1

)(1

)(

)0()()(

)()(

CCC

LLL

RR

vsC

sIsC

sV

LissLIsV

sRIsV

45

THE ELEMENTS LAW OF s-DOMAIN

)0()()(

)0()()(

)()(

CC

LLL

RR

CvssCVsI

si

sLsVsI

RsVsI

46

TRANSFORM OF CIRCUITS- RESISTOR

In the time domain:

In the s-domain:

i (t ) + v (t )-

R v (t )= i (t )R

I (s ) + V (s )-

R

V (s )= I (s )R

47

TRANSFORM OF CIRCUITS- INDUCTOR

In the time domain:

48

TRANSFORM OF CIRCUITS- INDUCTOR

Inductor’s voltage: Inductor’s current:

49

TRANSFORM OF CIRCUITS- CAPACITOR

In the time domain:

50

TRANSFORM OF CIRCUITS- INDUCTOR

Capacitor’s voltage: Capacitor’s current:

51

Ex.

Find v0(t) if the initial voltage is given as v0(0-)=5 V

52

s-Domain Circuit

53

Apply nodal analysis method

5.21

1)2(

10

1

5.210101

1

10

5.021010

0

10

)1(10

0

ssV

sVV

s

V

VVV

o

oo

s

oos

54

Cont’d

)2)(1(

3525

251

10)2(

0

ss

sV

ssVo

55

Using PFE

Rewrite V0(s) using PFE:

Solved for K1 and K2:

21)2)(1(

3525 21

s

K

s

K

ss

sVo

15;10 21 KK

56

Obtain V0(s) and v0(t)

Calculate V0(s):

Obtain V0(t) using look up table:

2

15

1

10)(

sssVo

)()1510()( 2 tueetv tto