Chapter 6: AC Circuits

Chapter 6: Outline

Phasors and the AC Steady State

AC Circuits

l A stable, linear circuit operating in the steady state with sinusoidal excitation (i.e., sinusoidal steady state).

l Complete response = forced response + natural response.

l In the steady state, natural response à 0.

)cos( 11 θω +tA )cos( 22 θω +tA

Same frequency, different amplitudes and phases

Sinusoids

l Three parameters are needed to determine a sinusoid.l x(t)=Xmcos(ωt+φ)=Re[Xmej(ωt+φ)].l Xm: amplitude, ω=2πf=2π/Τ: angular frequency, φ:

phase angle (radian).

Phasors

l The three parameters can be represented by a rotating phasor in a two-dimensional plane.

l At a given time (e.g., t=0 ), the nonrotating phasor is represented by .

l The frequency information is not included.φ∠= mXX

AC Forced Response

l The forced response of any branch variable (current or voltage) is at the same frequency as the excitation frequency ω for a linear circuit.

l In other words, any branch variable has the general form y(t)=Ymcos(ωt+φy).

l Circuit analysis becomes manipulation of complex numbers.

Complex Numbers in the Complex Plane

ai

ar

Aa

Aa

φ

φ

sin

cos

=

=

1−≡j

Rotating phasor

( )

0 tan180

0 tan

10

1

2/122

<

−±=

>

=

+=

−

−

rr

ia

rr

ia

ir

aaa

aaa

aaA

φ

φ

Complex Addition and Subtraction

[ ] [ ] [ ][ ] [ ] [ ]BABA

BABAImImImReReRe

±=±±=±

Complex Conjugate

222*

*

AaaAA

ajaAA

ajaAA

ir

ira

ira

=+=⋅

⋅−=−∠≡

⋅+=∠=

φ

φ

Complex Multiplication

[ ] [ ]]Im[]Im[],Re[]Re[ real, is if

ImRe)()(

BkkBBkkBk

ABABbabajbabaBA riiriirr

==+=−⋅+−=⋅

Complex Division (Rationalization)

2222*

*

ir

riir

ir

iirr

aaabab

jaa

ababAABA

AB

+−

+++

==

Complex Number in Exponential Form

l Euler’s formula: l Complex number in exponential form:

ααα sincos jje ±=±

ajeAA

φ=

0901∠=

−∠=

+∠=⋅

j

BA

BA

BABA

ba

ba

φφ

φφ

Phasor Representation

l A sinusoid can be represented by a phasor:

==+=+ tjeXtjejXetjXetX ωωφφωφω ReReRe)cos(

l The sum of two sinusoids at the same frequency can be represented by another phasor. The new phasor is simply the sum of the two original phasors.

+=+=+ tjeXXtjeXtjeXtjeXtjeX ωωωωω )21(Re21Re2Re1Re

Phasor Representation

l The steady-state response of any branch variable in a stable circuit with a sinusoidal excitation will be another sinusoid at the same frequency (forced response in Chapter 5)

l Kirchhoff’s laws hold in phasor form (only additions are involved).

Phasor with Differentiation

[ ]XjX

Xejdt

dXedt

ationdifferenti

tjtj

tjXed

ω

ω φωφω

φω

→

=

=

++

+

ReReRe

Example 6.3: Parallel Network with an AC Voltage Source

VjCIdt

tdvCti

VI

vi

Vttv

Cc

C

RR

ω⋅=⇒=

=⇒=

∠=⇒+=

)()(

55

2030)204000cos(30)( 00

))(6.464000cos(71.6)(

)(6.4671.60 Atti

AIII CR

+=

∠=+=

Example 6.4: Parallel Network with an AC Current Source

VjCIV

I

Itti

CR ω⋅==

=⇒=

,5

34000cos3)(

))(6.264000cos(4.13)(

1.02.03

)1.02.0(3

0 Vttv

jV

VjIII CR

−=

⋅+=

⋅+=+==

Impedance and Admittance

Phasor Representation

l Under ac steady-state, both the voltage and the current of a branch are sinusoids at the same frequency.

=+= tjeVvtmVtv ωφω Re)cos()(

=+= tjeIitmIti ωφω Re)cos()(

Resistors

l Current and voltage are collinear (in phase).

=×== tjeIRtjeIRtjeVv ωωω ReReRe

imRIvmVIRV φφ ∠=∠==

ivmRImV φφ == ,

Inductors

l Current lags voltage by 90 degrees.

==×=== tjeVtjeILj

dt

tjdeILtjeIdtdL

dtdiLv ωωω

ωω ReReReRe

ILjV ω=

o90 , +== ivmLImV φφω

Capacitors

l Current leads voltage by 90 degrees.

==×=== tjeItjeVCj

dt

tjdeVCtjeVdtdC

dtdvCi ωωω

ωω ReReReRe

ICjI

CjV

ωω−== 1

o90 , −== ivCmI

mV φφω

Phasor Relations (Resistor)

Phasor Relations (Inductor)

Phasor Relation (Capacitor)

Impedance

l In general, we can define a quantity Z and Ohm’s law for ac circuits as IZV =

o90/1/1

o90

−∠==

∠==

=

CCjcZ

LLjLZ

RRZ

ωω

ωω

Time Domain vs. Frequency Domain

Admittance

l Similarly, another quantity admittance Y can be defined.

ZY /1≡

VYI =

Equivalent Impedance and Admittance

)21( NVVVV +++= L

•Series equivalent impedance:

•Parallel equivalent impedance:

NZZZserZ +++= L21

NYYYparY +++= L21

)21( NIIII +++= L

Load Network

IVeqZ /≡

Impedance and Admittance

l Impedance and admittance are complex functions of frequency.

[ ] [ ] )()(ImRe)( ωωω jXRZjZjZZ +=+==

Resistance (Ω) Reactance (Ω)

l Inductors and capacitors are reactive elements, inductive reactance is positive and capacitive reactance is negative.

Impedance and Admittance

Conductance(Siemens)

Susceptance(Siemens)

l Inductors and capacitors are reactive elements, inductive reactance is positive and capacitive reactance is negative.

[ ] [ ] )()(ImRe)( ωωω jBGYjYjYY +=+==

Impedance Triangle

Example 6.6: Impedance Analysis of a Parallel RC Circuit.

0

0

0

6.4671.6

6.26224.015.02.010

151

6.2647.410||5

∠===

∠=+=−

+=

−∠Ω=−=

AYVZV

I

Sjj

Y

jZ

Example 6.7: Frequency Dependence of a Parallel RC Network.

2

2

2

2

2

)(1]Im[)(

)(1]Re[)(

)(1)(

CRCR

ZX

CRR

ZR

CRCRjR

ZZZZ

jZCR

CR

ωω

ω

ωω

ωω

ω

+−==

+==

+−

=+

=

Example 6.8: AC Ladder Calculationsl AC ladder networks can be analyzed by series-parallel

reduction (by replacing resistance with impedance).

Cjω1

Ljω

)(50000cos10)( mAtti =

Example 6.8: (Cont.)

0

0

0

9.364010)24(

24

7.794.8910)24(

101.53806448

−∠=+−

−=

∠=+−

=

∠=+==

VVjj

jV

VVjj

jV

VjIZV

C

L

AC Circuit Analysis

AC Circuit Analysis

l Sources at the same frequency:– Phasor transform method: the time domain sinusoids

are transformed to the frequency domain and represented by phasors.

Time domain à Frequency domain

AC Circuit Analysis

l Sources at the same frequency:– With the transformation, all resistive circuit analysis

techniques are applicable. Resistance is replaced by impedance and conductance is replaced by admittance.

ProportionalityThévenin-NortonNode AnalysisMesh Analysis…

AC Circuit Analysis

l Sources at the same frequency:– After analysis, the resultant phasors are converted back

to the time-domain sinusoids.

Frequency domain à Time domain

AC Circuit Analysis

l Sources at different frequencies:– Due to the linearity, the proportionality method is still

applicable.– The phasor analysis is performed at each individual

frequency

Example 6.9: AC Network with a Controlled Source

)(1000cos20 Vtv =

2/2636

3132)4/(

812)01(01Let

1

21

11

1

02

jIVZjVVIV

jIIIjjVI

jVjI

x

−==+=+−=

+−=+=+−=−=

+=+=∠=

Ati

AIIII

AZVI

)3.1051000cos(4.11

3.1054.11)/(

9010/

01

011

0

+=

∠==

∠==

Use proportionality

Example 6.10: Phase-Sift Oscillator

l Oscillator: Generator a sinusoidal output without an independent input source with initial stored energy.

l Design goal: At one particular frequency, vout=vin.

vx=vin/K

Example 6.10: (Cont.)

RC

LjCR

LVV

I

x

in /2

1

,1let andality proportion Use

2

2

−+

+=

=

ωω

+===

=

=

21,when

/2

/2 :requiresn Oscillatio

CRL

KVV

LC

CL

x

inOSC

OSC

ωω

ω

ωω

Superposition

l An ac source network is any two-terminal network that consists entirely of linear elements and sources. If there are more than one independent source, all of them must be at the same frequency so that the phasor method can be applied.

Frequency Domain Thévenin Parameters

l Frequency-domain Thevenin parameters: – the open-circuit voltage phasor:– the short-circuit current phasor:– Thévenin impedance:

ocV

ocV

scI

scIocVtZ /=

Example 6.11: Application of an AC Norton Network.

Thévenin parameters:

0

0

0

8.81495.0

13.89.91040280

280

7.73202028040

−∠==

−∠=+

=

∠Ω=−=

AZ

VI

Vj

V

jjZ

t

OCSC

OC

t

V Maximize

Example 6.11: (Cont.)

08.814.35

048.00

)048.0()014.0(

minimum is if maximum is

/ and 11

−∠=

+=

−++=

=+=

VV

SjY

BjGY

YV

YIVZZ

Y

eq

eq

eqSCt

eq

Yeq

AC Mesh Analysis

l By using phasors, impedance and admittance, node analysis and mesh analysis are still applicable assuming all independent sources are at the same frequency.

l AC mesh analysis:

[ ][ ] [ ]sVIZ = or [ ][ ] [ ]sVIZZ~~ =−

with controlled sources

Example 6.12: Systematic AC Mesh Analysis

Find i1

Two sources at the same frequency

Example 6.12: (Cont.)

)3.10410cos(9.3)(

3.1049.3

3615)10(2615681048

:equationmesh Single

01

01

1

+=

∠=

+=−−+=Ω−=−+=

=

tti

AI

VjjjVjjjZ

VIZ

S

S

AC Node Analysis

[ ][ ] [ ]sIVY =

[ ][ ] [ ]sIVYY~~

=−

or

with controlled sources

Example 6.13: Systematic AC Node Analysis

Find V1 and Z1

1 2

Constraint equation: I=(V-V1)/j4

Example 6.13: (Cont.)

[ ]

[ ]

Ω+=∠Ω==

−∠=

−∠=

−=

++−−−

−

+

−+

−+

=

+−−−

=

39.192.88.803.9

1.3115.1

3.224.10

20180

3310226

05.005.0

69

24/2

332256

201

011

0

01

2

1

2

1

jI

VZ

AI

VV

jj

VV

jjj

VV

jj

jj

IjVI

I

jj

Y

s

Chapter 6: Problem Set

l 7, 17, 24, 32, 36, 41, 44, 47, 51, 53, 57, 59