2012/11/13

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Sinusoidal Steady-State Analysis•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:–The natural response (due to initial

conditions) is ignored.–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.

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Nodal Analysis•Variables = Node Voltages.•Applying KCL to each node

gives each independentequation.

•If supernodes included,–Applying KCL to each

supernode gives 1 equation.–Applying KVL at each

supernode gives 1 moreequation.

Supernode

Example 1

5.2F1.02H.504H1

1

/40204cos20

jjj

Cj

Lj

sradt

C

L

Z

Z

Find ix.

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Example 1 (Cont’d)

(2)015115.2

242

2,nodeatKCLApplying(1)205.2)5.11(

45.21020

1,nodeatKCLApplying

211

221

21

2111

VVV

I

VVVI

VV

VVVV

j

jj

jjjj

x

x

)4.1084cos(59.7

4.10859.75.2

3.19891.1343.1897.18

020

15115.25.11

asformmatrixinputbecan(2)and(1)

1

2

1

2

1

tij

jj

x

xV

I

VV

VV

Example 2

•Applying KCL for thesupernode gives 1equations.

•Applying KVL at thesupernode gives 1equations.

•2 variables solved by 2equations.

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Example 2 (Cont’d)

48.7078.254510

18.8741.314510But

)21(436or1263

3

supernode,at theKCLApplying:Sol

21

2

21

21

221

VVV

VV

VV

VVV

jjjj

Mesh Analysis

Supermesh

SIii 12

Excluded

•Variables = Mesh Currents.•Applying KVL to each

mesh gives eachindependent equation.

•If supermeshes included,–Applying KVL to each

supermesh gives 1 equation.–Applying KCL at each

supernode gives 1 moreequation.

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Example 1

Find Io.

78.14412.622.3512.6

3050

442288

53,meshFor

(2)09020)2()2()224(

2,meshFor

(1)010)2(

)2108(1,meshtoKVLApplying

:Sol

2

2

2

1

3

3

12

32

1

III

II

I

III

II

I

o

jj

jjjj

jjjj

jj

jj

Example 2 Find Vo.

•Applying KVL for mesh1 & 2 gives 2 equations.

•Applying KVL for thesupermesh gives 1equations.

•Applying KCL at node Agives 1 equations.

•4 variables solved by 4equations

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Example 2 (Cont’d)

(3)0I5I)56(I8I)48(supermesh,For

(2)3I2,meshFor

(1)10I8I2I)28(

0I8I)2(I)28(101,meshFor

:Sol

2413

2

321

321

jjj

jj

jj

)II(2Vequations.4using

bysolvedbecanvariables4

(4)4IIgivesAnodeatKCLApplying

21

34

jo

Superposition Theorem•Since ac circuits are linear, the superposition

theorem applies to ac circuits as it applies to dccircuits.

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

frequency.–Total response = summation of individual

responses in the time domain.–Total response summation of individual

responses in the phasor domain.

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Example 1

Find Io.

= +

"0

'00 III

Example 1 (Cont’d)

78.14412.6529.35

)176.1353.2()647.2353.2(

176.1647.2

1050

442288

(3)53,meshFor

(2)0)2()2()44(2,meshFor

)1(0210)88(1,meshtoKVLApplying

,getTo

353.2353.225.425.4

202420

25.225.0)108(||)2(:Sol

"0

'00

2"0

2

1

3

312

231

"0

'0

j

j

j

jj

jjjj

jjj

jjj

jj

jjj

jjj

III

II

II

I

III

III

I

ZI

Z

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Example 2

sourcecurrent5sin2thetodueissourcevoltage2cos10thetodueis

sourcevoltagedcV-5thetodueiswhere

3

2

1

3210

tvtv

v

vvvv

Example 2 (Cont’d)

circuitopen1

circuit-short0,0Since

Cj

Lj

V1)5(41

1division,By voltage

1

v

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Example 2 (Cont’d)

2V

V010

4j

5j

51

F1.0

4H2rad/s2,0102cos10

jCj

jLjt

V)79.302cos(498.279.302.498

049.2439.310

)010()4||5(41

1

2

2

tv

j

jjV

Example 2 (Cont’d)

3V

10j

2jA902

21

F1.0

10H2rad/s5

9025sin2

jCj

jLj

t

V)805cos(33.28033.2

)2(4.88.1

101

)902()4||2(110

10division,currentBy

3

13

1

tv

jj

jjj

j

IV

I

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Source Transformation

Example 1

V28519.5

)105(1013425.15.2

10division,By voltage

105)25.15.2(448

)43(54)43(||5

49045

9020

jjj

jjjjj

jj

j

x

ss

s

V

IV

I

sI

sV

Find Vx.

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Thevenin & Norton Equivalent Circuits

N

ThNTh I

VZZ

Example 1

V31.22095.37

)75120(124

1268

8

64.248.6)12||4()6||8(

Th

Th

jj

j

jjj

V

Z

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Example 2

Example 2 (Cont’d)

3)6(2

)6(2)4234(givesKVLApplying

25.03

,simplicityfor3Set

sTh

0s

000

j

jjj

s

s

s

IV

Z

IV

IIII

I

V905555

)34(5)42(10

0)34(5.0)42(givesloopthetoKVLApplying

105.015gives1nodeatKVLApplying

Th

Th00

000

j

jj

jj

V

VII

III

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Example 3

NN

N0 )1520(

division,currentBy

IZ

ZI

j

Example 3 (Cont’d)

A48.38465.115205

5

83give)3(and),2(),1(

(3)3givesnodeatKCLApplying

(2)0)218()410()213(givessupermeshfor theKVLApplying

)1(0)410()28()218(40gives1meshforKVLApplying

.gettoanalysismeshApply)2(

5,easilyfoundbecan(1)

N0

3N

23

132

321

N

NN

II

II

II

III

III

I

ZZ

j

j

a

jjj

jjjj

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OP AMP AC Circuits: Example 1•Ideal op amps with negative feedback assumed.

–Zero input current & zero differential input voltage.

V)04591000cos(0291)(

04.59029.153

6give)2(and)1(

(2)10

010

02,nodeatKCLApplying

)1(-)45(62010

0510

31,nodeatKCLApplying

1

1

1

1111

.t.tvj

jj

jj

o

o

o

o

o

o

V

VV

VV

VV

VVVVV

V1000cos3 tvs

Example 2

.shiftphaseandgainloop-closetheFind

rad/s200F1F2

k10

2

1

21

CC

RR

130.6:shiftPhase434.0:gainloopClosed

6.130434.0)21)(41(

4

)1)(1(

1

1||

:Sol

2211

21

11

22

jjj

CRjCRjRCj

CjR

CjR

i

f

s

o

Z

Z

VV

G

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Applications: Capacitance Multiplier

i

ooi

i

i

ii

oioi

i

CjCj

CjCj

VVVV

VIV

Z

VVVV

I

1

1)(

)(1

CRR

C

Cj

RR

i

i

o

1

2eq

eq

1

2

1where

1

But

Z

VV

Applications: Oscillators

v1

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Oscillation Conditions• Barkhausen criteria must be meet for oscillators.

1. Overall phase shift (input-to-output-to-input) = AH = 0

• The oscillation frequency can be determined.

2. Overall gain of the oscillator = |AH| 1

• Loses must be compensated by an amplifying device.

• The minimum gain requirement can be determined.

H

A++

H

Avi vo

vf

Example

gfg

f

o

o

o

RRR

R

RCCR

CRjRCRC

CCCRRR

CjRCjRCjR

21131

givestrequiremengainThe

for31

101

givestrequiremenphaseThe

)1(3

,andIf

1||11||

02

0222

0

2222

2121

2211

222

VV

VV

VV