Sinusoidal Steady-State Analysis

Instructor: Chia-Ming TsaiElectronics Engineering

National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.

Contents• Introduction

• Nodal Analysis

• Mesh Analysis

• Superposition Theorem

• Source Transformation

• Thevenin and Norton Equivalent Circuits

• OP-amp AC Circuits

• Applications

Introduction• Steps to Analyze ac Circuits:

– Transform the circuit to the phasor (frequency) domain.

– Solve the problem using circuit techniques (nodal/mesh analysis, superposition, etc.).

– Transform the resulting phasor to the time domain.

Nodal Analysis• Variables = Node Voltages

• Applying KCL to each node gives each independent equation

• If supernodes included,– Applying KCL to each supern

ode gives 1 equation.– Applying KVL at each supern

ode gives 1 more equation.

Supernode

Example 1

5.2 F 1.0

2H .50

4 H 1

1

/ 4

0204cos20

j

j

j

Cj

Lj

srad

t

C

L

Z

Z

Find ix.

Example 1 (Cont’d)

(2) 015115.2

242

2, nodeat KCL Applying

(1) 205.2)5.11(

45.210

20

1, nodeat KCL Applying

211

221

21

2111

VVV

I

VVVI

VV

VVVV

j

jj

jj

jj

x

x

)4.1084cos(59.7

4.10859.75.2

3.19891.13

43.1897.18

0

20

1511

5.25.11

as formmatrix in put becan (2) and (1)

1

2

1

2

1

ti

j

jj

x

x

VI

V

V

V

V

Example 2

• Applying KCL for the supernode gives 1 equations.

• Applying KVL at the supernode gives 1 equations.

• 2 variables solved by 2 equations.

Example 2 (Cont’d)

48.7078.25

4510

18.8741.31

4510But

)21(436or

12633

supernode, at the KCL Applying

:Sol

21

2

21

21

221

VV

V

VV

VV

VVV

jj

jj

Mesh Analysis

Supermesh

SIii 12

Excluded

• Variables = Mesh Currents

• Applying KVL to each mesh gives each independent equation

• If supermeshes included,– Applying KVL to each sup

ermesh gives 1 equation.– Applying KCL at each supe

rnode gives 1 more equation.

Example 1

Find Io.

78.14412.6

22.3512.6

30

50

442

288

5 3,mesh For

(2) 09020)2(

)2()224(

2,mesh For

(1) 010)2(

)2108(

1,mesh toKVL Applying

:Sol

2

2

2

1

3

3

12

32

1

II

I

I

I

I

I

II

II

I

o

j

j

jj

jj

j

jjj

jj

jj

Example 2Find Vo.

• Applying KVL for mesh 1 & 2 gives 2 equations.

• Applying KVL for the supermesh gives 1 equations.

• Applying KCL at node A gives 1 equations.

• 4 variables solved by 4 equations

Example 2 (Cont’d)

(3) 05)56(

8)48(

(2) 32,mesh For

(1) 1082)28(

08)2()28(10

1,mesh For

:Sol

24

13

2

321

321

II

II

I

III

III

jj

j

jj

jj

)(2

equations. 4 using

by solved becan variables4

(4) 4

givesA nodeat KCL Applying

21

34

IIV

II

jo

Superposition Theorem• Since ac circuits are linear, the superposition th

eorem applies to ac circuits as it applies to dc circuits.

• The theorem becomes important if the circuit has sources operating at different frequencies.– Different frequency-domain circuit for each frequen

cy– Total response = summation of individual responses

in the time domain– Total response summation of individual response

s in the phasor domain

Example 1

Find Io.

= +

"0

'00 III

Example 1 (Cont’d)

78.14412.6529.35

)176.1353.2()647.2353.2(

176.1647.2

10

50

442

288

(3) 5 3,mesh For

(2) 0)2()2()44(

2,mesh For

)1( 0210)88(

1,mesh toKVL Applying

,get To

353.2353.2

25.425.4

20

24

20

25.225.0)108(||)2(

:Sol

"0

'00

2"0

2

1

3

312

231

"0

'0

j

j

j

j

j

jj

jj

jjj

jjj

j

j

j

j

j

jjj

III

II

I

I

I

III

III

I

ZI

Z

Example 2

sourcecurrent 5sin2 the todue is

source voltage2cos10 the todue is

source voltagedc V-5 the todue is

where

3

2

1

3210

tv

tv

v

vvvv

Example 2 (Cont’d)

circuitopen1

circuit-short 0

,0 Since

Cj

Lj

V 1)5(41

1

division, By voltage

1

v

Example 2 (Cont’d)

2V

V 010

4j

5j

51

F 1.0

4 H 2

rad/s 2 ,0102cos10

jCj

jjωω

t

V )79.302cos(498.2

79.302.498

049.2439.3

10

)010()4||5(41

1

2

2

tv

j

jjV

Example 2 (Cont’d)

3V

10j

2jA 902

21

F 1.0

10 H 2

rad/s 5 ,9025sin2

jCj

jLj

t

V )805cos(33.2

8033.2

)2(4.88.1

101

)902()4||2(110

10

division,current By

3

13

1

tv

jj

j

jj

j

IV

I

Source Transformation

Example 1

V 28519.5

)105(1013425.15.2

10

division, By voltage

105)25.15.2(4

48

)43(54)43(||5

49045

9020

jjj

jjj

j

jjj

j

x

ss

s

V

IV

I

sI

sV

Find Vx.

Thevenin & Norton Equivalent Circuits

N

ThNTh I

VZZ

Example 1

V 31.22095.37

)75120(124

12

68

8

64.248.6

)12||4()6||8(

Th

Th

j

j

j

j

jj

V

Z

Example 2

Example 2 (Cont’d)

3

)6(2

)6(2)4234(

gives KVL Applying

25.03

,simplicityfor 3Set

sTh

0s

000

j

jjj

s

s

s

I

VZ

IV

IIII

I

V 9055

55

)34(5)42(10

0)34(5.0)42(

gives loop the toKVL Applying

105.015

gives 1 nodeat KVL Applying

Th

Th00

000

j

jj

jj

V

VII

III

Example 3

NN

N0

)1520(

division,current By

IZ

ZI

j

Example 3 (Cont’d)

A 48.38465.1 15205

5

83

give )3( and ),2( ),1(

(3) 3

gives nodeat KCL Applying

(2) 0)218()410()213(

givessupermesh for the KVL Applying

)1( 0)410()28()218(40

gives 1mesh for KVL Applying

.get toanalysismesh Apply )2(

5 ,easily found becan (1)

N0

3N

23

132

321

N

NN

II

II

II

III

III

I

ZZ

j

j

a

jjj

jjjj

OP AMP AC Circuits: Example 1• Ideal op amps assumed

– Zero input current & zero differential input voltage

V )04591000cos(0291)(

04.59029.1 53

6

give )2( and )1(

(2)

10

0

10

0

2, nodeat KCL Applying

)1( -)45(6

2010

0

510

3

1, nodeat KCL Applying

1

1

1

1111

.t.tv

j

j

j

j

j

o

o

o

o

o

o

V

VV

VV

VV

VVVVV

V 1000cos3 tvs

Example 2

.shift phase andgain

loop-close theFind

rad/s 200

F 1

F 2

k 10

2

1

21

C

C

RR

130.6 :shift Phase

434.0:gain loopOpen

6.130434.0 )21)(41(

4

)1)(1(

1

1||

:Sol

2211

21

11

22

jj

j

CRjCRj

RCj

CjR

CjR

i

f

s

o

Z

Z

V

VG

Applications: Capacitance Multiplier

i

ooi

i

i

ii

oioi

i

CjCj

CjCj

V

VVV

V

I

VZ

VVVV

I

1

1

)(

)(1

CR

RC

Cj

R

R

i

i

o

1

2eq

eq

1

2

1 where

1

But

Z

V

V

Applications: Oscillators• Barkhausen criteria must be meet for oscillators

– (1) Overall gain 1– (2) Overall phase shift = 0

gfg

f

o

o

o

RRR

R

RCCR

CRjRC

RC

CCCRRR

CjRCjR

CjR

2113

1:)1(

for 3

1

101:)2(

)1(3

, and If

1||1

1||

02

0222

0

2222

2121

2211

222

V

V

V

V

V

V