Lecture 17
AC 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
I1  I 2  I 3  I 4  0
I2 
I1  3
I3 
V2  V1
4
0  V1
3j
I4  ?
V1 I 4
I1 I
2
V2
I3
Example – Node Analysis
I1  I 2  I 3  I 4  I 5  I 6  0
V1
 V2  1045
Supernode
I3  I 4
V1
I1
I2
V2
I6
I3 I 4
I5
Example – Node Analysis
I1  I 2  I 5  I 6  0
I1  3
I2 
V1
 V2  1045
Supernode
I5 

0  V2  1045

3j
I6 
V1
I1
I2
V2
I6
I3 I 4
I5
0  V2
6j
0  V2
12
Thevenin and Norton Theorem
for AC Steady State
Thevenin & Norton Theorem
• DC circuit
Thevenin
Theorem
Two
Terminal
Network
voc
i sc 
Rt
i sc
Rt
Norton
Theorem
Thevenin & Norton Theorem
• DC circuit
• Find the Thevenin parameters
Suppress
Sources it
Two
Terminal
Network

voc

Two
Terminal
Network
isc
Two
Terminal
Network
vt
Rt 
it

vt

Thevenin & Norton Theorem
• AC steady state
ZT
Voc
Two
Terminal
Network
I sc 
Thevenin
Theorem
Voc
Zt
I sc
ZT
Norton
Theorem
Thevenin & Norton Theorem
• AC steady state
• Find the Thevenin parameters
Suppress
Sources
Two
Terminal
Network

Voc

Two
Terminal
Network
I sc
Two
Terminal
Network
Zt 
Vt
It
It

Vt

Example - Norton Theorem
• Obtain Io by Norton Theorem
I sc
ZT
Example - Norton Theorem
• Obtain Io by Norton Theorem
• Find Zt
Suppress
Sources
Two-terminal Network
Zt  5
Example - Norton Theorem
• Obtain Io by Norton Theorem
I sc  3  8 j
• Find I sc
I2
I1
Two-terminal Network
I sc
Example - Norton Theorem
• Obtain Io by Norton Theorem
38 j
5

I o  3  8 j 
 1.46538.48
5  20  15 j 
5
Superposition
for AC Steady State
AC Superposition – Example 6.17
Z L  j L
Find vc
ZC 
1
j C
However, what is the value of ω?   5 ?,   2 ?
AC Superposition – Example 6.17
Superposition Principle
v C-1
v C-2
v C  v C-1  v C-2
AC Superposition – Example 6.17
v C  v C1  v C2
v C1
v C2
The same element has different impedances.
AC Superposition – Example 6.17
3j
 3 j
50 10 j   50 j

50  10 j
5 j
 50 j
5 j
 55.7  158.2
V C 1  60
 50 j
 20 j
5 j
vC 1 t   55.7 cos(5t  158.2 ) V
8 j  25 j  200 200


j
8 j  25 j  17 j 17
50
200
50 
j
17
200
 IC 2 
j  34.4166.8
17
IC 2  3 j
V C 2
vC 2 t   34.4 cos(2t  166.8 ) V
Fourier Series
for Circuit Analysis
Beyond Sinusoids
vs t 
1. Fourier Series: periodic function is a linear combination
of sinusoids
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:

cos n0t  90
You will learn how to find a0, an and bn in other courses.

Fourier Series
f t 
1 2  1
f t    
sin 2k  1t
2  k 1 2k  1
Fourier Series
f t 
Fourier Series
f t 
Fourier Series
f t 

cos n0t  90
Network

Network
Network
=
I1
i0
I2
Network
Capacitor = Open
Inductor = Short
……
Example
vs t 
1 2  1
vs t    
sin 2k  1t
2  k 1 2k  1
1 2
2
2
vs t    sin t 
sin 3t 
sin 5t  ...
2 
3
5
Example

1 2
1
vs t    
sin 2k  1t
2  k 1 2k  1

 cos 2k  1t  90
v0 t   0
5
2
1
Vs 
  90
 2k  1
  2k  1
j 2k  1  2


Example

1 2
1
vs t    
sin 2k  1t
2  k 1 2k  1
v0 t   0

1 22k  1 
cos 2k  1t  tan

2 2
5


25  42k  1 
2
Example
1 2
2
2
vs t    sin t 
sin 3t 
sin 5t  ...
2 
3
5
1
  0.64 sin t  0.21sin 3t  0.13 sin 5t  ...
2

vo t   0.50 cos t  51.49






 0.21 cos 3t  75.14  0.13 cos 5t  80.96  ...


Example
Example
Application:
Resonance
Communication
How to change audio into different frequency?
AM
Frequency at f
Frequency close to f
FM
Frequency at f
Frequency close to f
Communication
How to design a circuit that can only receive
the signal of a specific frequency?
Series RLC
Z ( )  R  jL 
1 

j  L 


C


1
 R
j C
1 

2
Z ( )  R   L 

C 

    tan 1
I
V
Z
1
C
R
L 
I  I m  i
Vm
Im 

| Z   |
2
V  Vm  v
Vm
1 

2
R   L 

C 

2
Series RLC
1 

Z ( )  R 2   L 


C



1
LC
2
1
L 
C
    tan 1
R
0 
I  I m  i
V  Vm  v
Im
Vm
Im 
| Z   |
Fix Vm
Change ω
1
LC
Resonance
Im
Antenna

1
LC
If the frequency of the input
signal is close to ω0
Large current
I  I m  i
V  Vm  v
Otherwise
Like open circuit
0 
Series RLC - Bandwidth
Vm
Im 

| Z   |
Im
Vm
R
1 Vm
2 R
1
LC
Vm
1 

R   L 

C 

2
2
2
1 

R   L 
  2R
C 

2
B  2  1  R / L
Quality
Using quality factor Q to
define the selectivity
Q
0
B
R
B
L
0 L
Q
R
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
Vx
?
First Find
V in
Oscillator - Example 6.10
Vx  R I 2
V1  R  jL I 2
j L 

I1   1 
I 2
R 
R 
V1
j L 

I1  I 2   2 
I 2
R 



 j
Vin  
 I1  I 2  V1
 C 
j L 
  j 

2


 I 2  R  jL I 2
R 
 C 
  j2 L



 R  j L  I 2
 C RC


L   2

Vin   R 
 L   I 2
  j
RC   C


Oscillator - Example 6.10

L   2

Vin   R 
 L   I 2
  j
RC   C


Vx  R I 2
 2



L


Vin
L

C


 1  2  j
Vx
RC
R






If we want vin and vx in phase
   
 Vin   Vx
2
 L
C
 0  osc 
R
2
LC
Oscillator - Example 6.10
osc
2

(vin and vx in phase)
LC
Vout  KVx
Vin
Vout
 2



L


Vin
L
L

C


 1 2  j
 1 2
Vx
RC
R
RC






1
L 
 1  2 
K R C
If we want vin = vout
L
K  1 2
RC
Oscillator - Example 6.10
osc
2

LC
L
K  1 2
RC
osc
2

LC
L
K  1 2
RC
L
Set K  1 2
RC
vin = vout
Input: vin t   Vm cososc t   
vout t   vin t 
Use output as input
Oscillator - Example 6.10
osc
2

LC
L
K  1 2
RC
Generate sinusoids without input!
 Will the oscillation attenuate with time?
 Yes.
R dissipate the energy
 No.
Who supply the power? Amplifier
Oscillator - Example 6.10
TV remote controller
osc
2

LC
L
K  1 2
RC
Battery of
controller
Oscillator - Example 9.7
osc
2

LC
L
K  1 2
RC
d 2vout  1 R
 dvout 2

 K  1

vout  0
2
dt
 RC L
 dt LC
L
Set K  1 2
RC
1 R
 K  1  0
RC L
Undamped
Oscillator - Example 9.7
osc
2

LC
L
K  1 2
RC
d 2vout  1 R
 dvout 2

 K  1

vout  0
2
dt
 RC L
 dt LC
2
j
LC
  josc
vout t   a cos osct  b sin osct Amplitude and phase
are determined by
vout t   L cososct   
initial condition
Homework
• 6.46
• 6.52
• 6.44
Homework – Mesh Analysis 1
𝑓𝑖𝑛𝑑 𝐼
𝐼
Homework – Mesh Analysis 2
𝑓𝑖𝑛𝑑 𝑉
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。)


V

5
V

0
,
V

26
.
9
V

21
.
8
• 6.52: 1
2
• 6.44
1 1
osc 
6 RC
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  2.498 cos 2t  30.79  2.33 sin 5t  12.1

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