Lecture 16AC Circuit Analysis

(1)Hung-yi Lee

Textbook

• Chapter 6.1

AC Steady State tytyty FN

Second order circuits: ttN eety 21

21 AA t

N etty 21 AA

tbtaety ddt

N sincos

If the circuit is stable:

As t → ∞ 0tyN tyty F Steady State

In this lecture, we only care about the AC steady state

tAtx cosSource:

AC Steady State

• Why we care about AC steady state?• Fourier Series/Fourier Transform

AC Steady State

• Why we care about AC steady state?• Fourier Series/Fourier Transform• Most waveforms are the sum of sinusoidal waves with

different frequencies, amplitudes and phases• Compute the steady state of each sinusoidal wave• Obtaining the final steady state by superposition

Example 6.3

204000cos30 ttv 204000cos6 t

R

tvtiR

204000sin40003025CC ttvti 204000sin3 t 1104000cos3 t

tititi CR 204000sin3204000cos6 tt

204000sin

36

3204000cos

36

636

2222

22 tt

6.26cos6.26sin

7.6

6.464000cos7.6 t

Example 6.4 tti 4000cos3 tititi CR tvCtv

R

1

tBttv 4000sin4000cosA

t

tBt

tBt

4000cos3

4000cos4000sinA-400025

4000sin4000cosA5

1

31.02.0 BA02.01.0 BA 6B

12A

tttv 4000sin64000cos12 26.6-4000t13.4cos

Example 6.4 tti 4000cos3 tititi CR tvCtv

R

1

26.6-4000t13.4costv

26.6-4000tcos68.2tiR

26.6-4000t4000sin4.1325 tvCtiC

26.6-4000tsin34.1

63.44000tcos34.1

AC Steady-State Analysis

204000cos30 ttv

Example 6.3 Example 6.4

tti 4000cos3

204000cos6 ttiR

6.464000cos7.6 tti

1104000cos3 ttiC

26.6-4000t13.4costv

26.6-4000tcos68.2tiR

63.44000tcos34.1 tiC

AC Steady-State Analysis

• AC steady state voltage or current is the special solution of a differential equation.• AC steady state voltage or current in a circuit is a

sinusoid having the same frequency as the source.• This is a consequence of the nature of particular

solutions for sinusoidal forcing functions.• To know a steady state voltage or current, all we

need to know is its magnitude and its phase • Same form, same frequency

AC Steady-State Analysis

• For current or voltage at AC steady state, we only have to record amplitude and phase

ttx m cosX

Amplitude: Xm Phase: ϕ

Phasor

• A sinusoidal function is a point on a x-y plane

ttx m cosX

mXXPolar form:

Rectangular form: sincos mm jXXX

Exponential form: jmeXX

Review – Operation of Complex Number

A is a complex number

Review – Operation of Complex Number

A is a complex number

rectangular polar:

r

iA

ir

AiAr

Air

a

a

aaA

AaAa

AjaaA

1

22

tan

||

sin|| ,cos||

||

Review – Operation of Complex Number

Complex conjugate:

222 ||

2Re2

||

||

AaaAA

aAAA

AjaaA

AjaaA

ir

r

air

air

A is a complex number

Review – Operation of Complex Number

irir jbbBjaaA

)()( iirr bajbaBA

BABA

BABA

ImImIm

ReReRe

Addition and subtraction are difficult using the polar form.

Review – Operation of Complex Number

)(||||BA BABA BA BA ||B ||A

)( BAB

A

B

A

irir jbbBjaaA

)()()()( riiriirririr babajbabajbbjaaBA

2222))((

))((

ir

riir

ir

iirr

irir

irir

aa

babaj

aa

abab

jaajaa

jaajbb

A

B

Phasor

Sinusoid function:

ttx m cosX

Phasor: mXX

It is rotating.

Its projection on x-axis producing the sinusoid function

At t=0, the phasor is at mX

2

f

• KVL & KCL need summation

Phasor - Summation

2121 ,Y XXXX

111111 )cos( XXtXtx

222222 )cos( XXtXtx

)cos()cos( 2211 tXtXty

Textbook, P245 - 246

YY

21 XXY 2211 XX

2121 XX

KCL and KVL for Phasors

titititi yyxx 2121

KCL

KVL

2121 IIII yyxx

titititi yyxx 2121

2121 IIII yyxx

input current output current

voltage rise voltage drop

Phasors also satisfy KCL and KVL.

Phasor - Multiplication

)(tx )(K txMultiply k

X XkMultiply k

)()( tRitv IRV

Time domain Phasor

Phasor - Differential

• We have to differentiate a sinusoidal wave due to the i-v characteristics of capacitors and inductors.

)(txdt

tdx )(

Differentiate

X XjMultiplying jω

Phasor - Differential

• We have to differentiate a sinusoidal wave due to the i-v characteristics of capacitors and inductors.

)cos()( 1 tXtx

)sin()(

1 tXdt

tdx

)90cos( 1 tX

Time domain Phasor

1X

901 X

Phasor - Differential

• We have to differentiate a sinusoidal wave due to the i-v characteristics of capacitors and inductors.

Phasor

1X

901 X

Multiply ω Rotate 90 。

Differentiate on time domain = phasor multiplying jω

Equivalent to multiply jω

Phasor - Differential

• Capacitor

dt

tdvti C)(

Time domain Phasor

VCI j

i leads v by 90 。

Phasor - Differential

• Inductor

dt

tditv L)(

Time domain Phasor

ILV j

v leads i by 90 。

For C, i leads vbut v leads i for L

Capacitor & Inductor

i-v characteristicsTi

me

dom

ain

Phas

or

Phasors satisfy Ohm's law for resistor, capacitor and inductor.

i-v characteristics

RZR

Resistor Capacitor Inductor

LjZ L Cj

CjZ

11

C

Impedance

Admittance is the reciprocal of impedance.

circuitshort ,01

Z

circuitopen ,

,frequency)-(hightWhen

C

Cj

LjZL

circuitopen ,1

Z

circuitshort ,0

0(DC),When

C

Cj

LjZL

Equivalent impedance and admittance

Series equivalent impedance

Nser ZZZZ 21

Parallel equivalent impedance

Npar ZZZZ

1111

21

Impedance

RZR

jXRZ

reactances ,Im

sresistance ac ,Re where

XZ

RZ

Resistor Capacitor Inductor

LjZ L Cj

CjZ

11

C

Inductors and capacitors are called reactive elements. Inductive reactance is positive, and capacitive

reactance is negative.

After series and parallel, the equivalent impedance is

33

R

XZ

XRZ

ZZ

1

22

tan

Impedance Triangle

jXRZ

reactances ,Im

sresistance ac ,Re where

XZ

RZ

After series and parallel, the equivalent impedance is

AC Circuit Analysis

• 1. Representing sinusoidal function as phasors • 2. Evaluating element impedances at the source

frequency• Impedance is frequency dependent

• 3. All resistive-circuit analysis techniques can be used for phasors and impedances• Such as node analysis, mesh analysis, proportionality

principle, superposition principle, Thevenin theorem, Norton theorem

• 4. Converting the phasors back to sinusoidal function.

Example 6.6 μF 25 5

)204000cos(30

CR

tv

5RZ

1025μ4000

1jj

Cj

CjZC

11

105

105)10(||)5(||

j

jjZZZ CR

6.2647.4

2030

24 j 6.2647.4

Z

VI

A6.4671.6 )6.26(2047.4

30

Example 6.7

• Impedance is frequency dependent

Find equivalent network

||CR

CRCReq ZZ

ZZZZZ

/1R

R/

Cj

Cj

R1

R

Cj Zeq should be Zeq(ω) or

Zeq(jω)

Cj

Cj

CjCjj

R1

R1

R1

R

R1

RZeq

22

R1

R

C

CjR

22

2 R1

R

R1 C

Cj

C

R

Example 6.7

• Impedance is frequency dependent

Find equivalent network

2

2

2 R1

R

R1Z

C

Cj

C

Rjeq

If ω → 0

RZ jeq

For DC, C is equivalent to open circuit

If ω → ∞

0Z jeq C becomes short

Example 6.8

kjmkjjZL 1020050L

mA)50000cos(10 tti

Find v(t), vL(t) and vC(t)

kjnk

jjj

Z 10250

1

C

1

C

1C

Example 6.8

kΩ10||510||40 jjZeq

kΩ 4.68.4 j

eqZ

kΩ 1.538

Zeq = 4.8kΩ + j6.4k Ω

Example 6.8

eqZIV

Zeq = 4.8kΩ + j6.4k Ω

1.538010 km

V1.5380

V1.5380Z eq

V1.5380V

Example 6.8 V1.5380V

V7.794.89882.310)24(

10

jVjj

jVL

V9.3640243210)24(

24

jV

jj

jVC

Example 6.8V1.5380V

V7.794.89 LV

V9.3640 CV

V)1.5350000cos(80)( tv t

V)7.7950000cos(4.89)( tv tL

V)9.3650000cos(40)( tv tC

Complete Response ti

0t

Aii 000

Aii 000

Complete Response ti

0t I

10ZL jLj 0t Reach steady state

5903.1106

012I

j

0t 595cos03.1F tti Forced Response

595cos03.1F tti

Aii 000

Complete Response ti

0t I

Natural Response:

t

eti

KN 3/1/ RLte 3K

595cos03.1K 3N tetvtvtv t

F

53.0K

595cos03.1F tti

Aii 000

Complete Response

0t 5

4505

11ZC j

mjCj

45485.8

4566.5

9048012

44

4V

j

j

V

0t 455cos485.8 ttv

Complete Response

0t 5

V

455cos485.8 ttv

Initial Condition:

Vv 645cos485.80

Vvv 600

Vv 60

Complete Response

0t

V

0t Reach steady state

6.2674.10

4.6347.4

9048012

42

4V

j

j

0t 6.265cos74.10F ttvForced

Response

Vv 60

Complete Response

0t

V

Natural Response:

RC

t

etv

KN 1.0502RC mFte 10K

Vv 60

6.265cos74.10K 10N tetvtvtv t

F

6.3K

Complete Response

Summarizing the results:

06.265cos74.106.3

0455cos485.810 tte

tttv

t

Three Terminal Network

Homework

• 6.20• 6.22• 6.24• 6.26• 6.36 (b)

Thank you!

Answer

• 6.20: 10, 0.002• 6.22: Z=10Ω, v=40cos500t, i1=5.66cos(500t-45 。 ),

i2=4cos(500t+90 。 )• 6.24:Z=7.07<-45, i=1.41cos(2000t+45 。 ),

vc=14.1cos(2000t-45 。 ), v1=10cos(2000t+90 。 )• 6.26:Z=18 Ω, v=36cos2000t, iL=2cos(2000t-90 。 ),

i2=2.83cos(2000t+45 。 )• 6.36 (b): L=12mH