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