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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 1045
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 1045
Supernode
I5
0 V2 1045
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
38 j
5
I o 3 8 j
1.46538.48
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 34.4166.8
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 n0t 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 1t
2 k 1 2k 1
Fourier Series
f t
Fourier Series
f t
Fourier Series
f t
cos n0t 90
Network
Network
Network
=
I1
i0
I2
Network
Capacitor = Open
Inductor = Short
……
Example
vs t
1 2 1
vs t
sin 2k 1t
2 k 1 2k 1
1 2
2
2
vs t sin t
sin 3t
sin 5t ...
2
3
5
Example
1 2
1
vs t
sin 2k 1t
2 k 1 2k 1
cos 2k 1t 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 1t
2 k 1 2k 1
v0 t 0
1 22k 1
cos 2k 1t tan
2 2
5
25 42k 1
2
Example
1 2
2
2
vs t sin t
sin 3t
sin 5t ...
2
3
5
1
0.64 sin t 0.21sin 3t 0.13 sin 5t ...
2
vo t 0.50 cos t 51.49
0.21 cos 3t 75.14 0.13 cos 5t 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 jL
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 jL 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 jL 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 cososc 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
josc
vout t a cos osct b sin osct Amplitude and phase
are determined by
vout t L cososct
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