lecture 17 ac circuit analysis (2) hung-yi lee. textbook ac circuit analysis as resistive circuits...

Lecture 17AC Circuit Analysis

(2)Hung-yi Lee

Textbook

• AC Circuit Analysis as Resistive Circuits• Chapter 6.3, Chapter 6.5 (out of the scope)

• Fourier Series for Circuit Analysis• Resonance • Chapter 6.4 (out of the scope)

• Oscillator• Example 9.7 and 6.10

Systematic Analysis for AC Steady State

Example – Node Analysis

1V 2V

1I2I 3I

4I

0IIII 4321

3I1 j

V

3

0I

1

2

4I

12

3

VV ?I4

Example – Node Analysis

1V 2V

Supernode

1I2I

3I 4I

5I6I

0IIIIII 654321

43 II

4510V

V

2

1

Example – Node Analysis

1V 2V

Supernode

1I2I

3I 4I

5I6I

0IIII 6521

3I1

j

V

3

45100I

2

2

4510V

V

2

1

j

V

6

0I

2

5

12

0I

2

6

V

Thevenin and Norton Theorem

for AC Steady State

Thevenin & Norton Theorem• DC circuit

TwoTerminalNetwork

Thevenin Theorem

Norton Theorem

tRsci

t

oc

R

vsci

Thevenin & Norton Theorem• DC circuit• Find the Thevenin parameters

TwoTerminalNetwork

ocv

TwoTerminalNetwork

sciTwo

TerminalNetwork

tv

ti

t

tt i

vR

Suppress Sources

TZ

Thevenin & Norton Theorem• AC steady state

TwoTerminalNetwork

Thevenin Theorem

Norton Theorem

TZocV

scI

t

oc

sc

VI

Z

Thevenin & Norton Theorem• AC steady state• Find the Thevenin parameters

TwoTerminalNetwork

TwoTerminalNetwork

TwoTerminalNetwork

t

t

t

V

IZ

Suppress Sources

ocV scI

tI

tV

• Obtain Io by Norton Theorem

Example - Norton Theorem

TZ

scI

Example - Norton Theorem

• Obtain Io by Norton Theorem• Find Zt

Suppress Sources

5Z tTwo-terminal Network

Two-terminal Network

Example - Norton Theorem

• Obtain Io by Norton Theorem• Find scI

scI

jsc 83I

1I

2I

• Obtain Io by Norton Theorem

Example - Norton Theorem

5j83

jjo 15205

583I

48.38465.1

Superpositionfor AC Steady State

AC Superposition – Example 6.17

Find vc

However, what is the value of ω?

Cj1

ZC

LZL j

?2?,5

AC Superposition – Example 6.17

Superposition Principle

1-Cv 2-Cv

2-C1-CC vvv

AC Superposition – Example 6.17

C2C1C vvv

The same element has different impedances.

C1v C2v

AC Superposition – Example 6.17

V )2.1585cos(7.551 ttvC

j

j

1050

1050

j

j

5

50

jjjjj

V C

205

505

50

601

2.1587.55

V )8.1662cos(4.342 ttvC

j

jjj

jj

17

200

17

200

258

258

j3

jjC

17200

50

503I 2

8.1664.3417

200IV 22 jCC

j3

Fourier Series for Circuit Analysis

Beyond Sinusoids

1. Fourier Series: periodic function is a linear combination of sinusoids

tvs

2. Superposition: find the steady state of individual sinusoids, and then sum them together

Fourier Series

• Periodic Function: f(t) = f(t+nT)• Period: T• Frequency: f0 = 1/T• Circular Frequency: ω0 = 2πf0 = 2π/T

Fourier Series:

You will learn how to find a0, an and bn in other courses.

90cos 0 tn

Fourier Series tf

1

12sin12

12

2

1

k

tkk

tf

Fourier Series tf

Fourier Series tf

Fourier Series tf

Network Network

90cos 0 tn

Network

……

=

Network

Capacitor = OpenInductor = Short

0i 1I 2I

Example

tvs

1

12sin12

12

2

1

ks tk

ktv

...5sin5

23sin

3

2sin

2

2

1 ttttvs

1

12sin12

12

2

1

ks tk

ktv

Example

00 tv

9012cos tk

12 k

5

212 kj

9012

12V

ks

5

122tan12cos

12425

2 1

22

ktk

k

1

12sin12

12

2

1

ks tk

ktv

Example

00 tv

Example

...5sin5

23sin

3

2sin

2

2

1 ttttvs

...5sin13.03sin21.0sin64.02

1 ttt

...96.805cos13.014.753cos21.0

49.51cos50.0

tt

ttvo

Example

Example

Application:Resonance

Communication

How to change audio into different frequency?

AM

Frequency at f

Frequency close to f

FMFrequency at f

Frequency close to f

Communication

How to design a circuit that can only receive the signal of a specific frequency?

Series RLCCj

LjRZ

1)(

CLjR

1

22 1

)(

CLRZ

RC

L

1

tan 1

ZIV

imI I

22 1||

I

CLR

V

Z

V mmm

vm VV

Series RLC2

2 1)(

CLRZ

RC

L

1

tan 1

LC

1

LC

10

||I

ZVm

m

mIFix Vm Change ω

imI I

vm VV

Resonance

imI I

vm VV

mI

Antenna

LC

1

If the frequency of the input signal is close to ω0

Large current

Otherwise

Like open circuit

Series RLC - Bandwidth

22 1||

I

CLR

V

Z

V mmm

mI

RmV

RmV

2

1

RC

LR 21

22

LC

10

12B L/R

Quality

BQ 0

Using quality factor Q to define the selectivity

L

RB

R

LQ 0

Quality

• For radio, cell phone, etc., the quality should be• 1. As high as possible?• 2. As low as possible?• 3. None of the above?

Application:Oscillator

Oscillator

• Oscillator (Example 9.7 and 6.10)• An oscillator is an electric circuit that generate a

sinusoidal output with dc supply voltage• DC to AC

Remote Controller,Cell phone

Oscillator - Example 6.10

?V

V

in

xFirst Find

Oscillator - Example 6.10

221 2 IR

LjII

2

2IL

Cj

RC

LRVin

2

1

1 1 IR

Lj

R

VI

21 ILjRV

2IRVx 121 VIIC

jVin

222 ILjRIR

Lj

C

j

2

2ILjR

RC

L

C

j

Oscillator - Example 6.10

2

2IL

Cj

RC

LRVin

R

LCj

CR

L

V

V

x

in

2

12

2IRVx

If we want vin and vx in phase

0R

2

L

C

LCosc

2

xin VV

Oscillator - Example 6.10

CR

L

KV

V

out

in

21

1 If we want vin = vout

CR

LK

21

LCosc

2 (vin and vx in phase)

CR

L

R

LCj

CR

L

V

V

x

in

221

2

1

xout VV K

Oscillator - Example 6.10

vin = vout

CR

LK

21

LCosc

2

SetCR

LK

21

tVtv oscmin cosInput:

tvtv inout Use output as input

CR

LK

21

LCosc

2

Oscillator - Example 6.10

Generate sinusoids without input!Will the oscillation attenuate with time?

Yes. R dissipate the energy No. Who supply the power? Amplifier

CR

LK

21

LCosc

2

Oscillator - Example 6.10

TV remote controller Battery of controller

CR

LK

21

LCosc

2

Oscillator - Example 9.7

02

11

2

2

out

outout vLCdt

dvK

L

R

RCdt

vd

011

KL

R

RC

CR

LK

21

LCosc

2

CR

LK

21Set Undamped

Oscillator - Example 9.7

02

11

2

2

out

outout vLCdt

dvK

L

R

RCdt

vd

CR

LK

21

LCosc

2

LCj

2 tbtatv oscoscout sincos

oscj ttv oscout cosL

Amplitude and phase are determined by initial condition

Homework

• 6.46• 6.52• 6.44

Homework – Mesh Analysis 1

𝐼𝑓𝑖𝑛𝑑𝐼

Homework – Mesh Analysis 2

𝑓𝑖𝑛𝑑𝑉

𝑉

Homework – Thevenin 1

• Find the Thevenin equivalent of the following network

Homework – Thevenin 2

• Find the Thevenin equivalent of the following network

Homework – Superposition 1• (out of the scope) Calculate vo

Homework – Superposition 2• (out of the scope) Calculate vo

Thank you!

Answer

• 6.46: v2=8cos(5t+53.1 。 )• 6.52:• 6.44

8.219.26V,05V 21 VV

RCosc

1

6

1

Answer – Mesh Analysis 1

Answer – Mesh Analysis 2

𝑉=¿

Answer – Thevenin 1

• Find the Thevenin equivalent of the following network

Answer – Thevenin 2

• Find the Thevenin equivalent of the following network

Answer – Superposition 1

• Using superposition

1.125sin33.279.302cos498.21 tt

Answer – Superposition 2

• Using superposition

Acknowledgement

• 感謝 陳俞兆 (b02)• 在上課時指出投影片中的錯誤

• 感謝 趙祐毅 (b02)• 在上課時指出投影片中的錯誤

• 感謝 林楷恩 (b02)• 修正作業的答案