AC steady state
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Transcript AC steady state
Lecture 16
AC Circuit Analysis (1)
Hung-yi Lee
Textbook
• Chapter 6.1
AC Steady State
y t y N t y F t
Second order circuits:
If the circuit is stable:
As t → ∞
y N t A1e 1t A 2 e 2t
y N t A1t A 2 e t
y N t e t a cos d t b sin d t
y N t 0
y t y F t
Steady State
In this lecture, we only care about the AC steady state
Source: xt A cost
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
vt
iR t
6 cos 4000t 20
R
vt 30 cos 4000t 20
iC t Cvt 25 30 4000 sin 4000t 20
t i t 6 cos4000t 20 3 sin 4000t 20
3 sin 4000t 20 3 cos 4000t 110
i t iR
C
6
3
6 3
cos 4000t 20
sin 4000t 20
2
2
2
2
6
3
6
3
6.7
sin
26
.
6
cos 26.6
2
2
6.7 cos 4000t 46.6
Example 6.4
1
i t iR t iC t vt Cvt
R
vt A cos 4000t B sin 4000t
i t 3 cos4000t
1
A cos 4000t B sin 4000t
5
25 4000- A sin 4000t B cos 4000t
3 cos 4000t
0.2 A 0.1B 3
0.1A 0.2 B 0
A 12
B6
vt 12 cos 4000t 6 sin 4000t 13.4cos 4000t - 26.6
Example 6.4
i t 3 cos4000t
1
i t iR t iC t vt Cvt
R
vt 13.4cos 4000t - 26.6
iR t 2.68cos 4000t - 26.6
1.34sin 4000t - 26.6
1.34cos4000t 63.4
iC t Cvt 25 13.4 4000sin 4000t - 26.6
AC Steady-State Analysis
Example 6.3
vt 30 cos 4000t 20
Example 6.4
i t 3 cos4000t
i t 3 cos4000t 110
i t 6.7 cos4000t 46.6
iR t 6 cos 4000t 20
C
t 2.68cos4000t - 26.6
t 1.34cos4000t 63.4
vt 13.4cos 4000t - 26.6
iR
iC
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
xt X m cost
Amplitude: Xm
Phase: ϕ
Phasor
• A sinusoidal function is a point on a x-y plane
xt X m cost
Polar form: X X m
Rectangular form:
X X m cos jX m sin
Exponential form: X X e j
m
Review
– Operation of Complex Number
A is a complex number
Review
– Operation of Complex Number
A is a complex number
rectangular polar:
A ar jai | A | A
ar | A | cos A , ai | A | sin A
| A | ar ai
2
ai
A tan
ar
2
1
Review
– Operation of Complex Number
A is a complex number
Complex conjugate: A ar jai | A | a
A ar jai | A | a
A A 2 ReA 2ar
A A ar ai | A |2
2
2
Review
– Operation of Complex Number
A ar jai
B br jbi
A B (ar br ) j (ai bi )
ReA B ReA ReB
ImA B ImA ImB
Addition and subtraction are difficult using the polar form.
Review
– Operation of Complex Number
A | A | A B | B | B
A B | A || B | ( A B )
A A
( A B )
B B
A ar jai
B br jbi
A B (ar jai ) (br jbi ) (ar br ai bi ) j (ar bi ai br )
ar bi ai br
B (br jbi )(ar jai ) br ar bi ai
j 2
2
2
2
A (ar jai )(ar jai )
ar ai
ar ai
Phasor
Sinusoid function:
xt X m cost
Phasor:
X X m
It is rotating.
At t=0, the phasor is at X m
Its projection on x-axis producing
the sinusoid function
2
f
Phasor - Summation
• KVL & KCL need summation
Textbook, P245 - 246
x1 t X 1 cos(t 1 ) X 1 X 11
x2 t X 2 cos(t 2 ) X 2 X 2 2
y t X 1 cos(t 1 ) X 2 cos(t 2 ) Y Y
Y X 1 X 2 X 11 X 2 2
X 1 X 2 1 2
Y X1 X 2 , X1 X 2
KCL and KVL for Phasors
KCL
input current
output current
ix1 t ix 2 t i y1 t i y 2 t
I x1 I x 2 I y1 I y 2
KVL
voltage rise
voltage drop
ix1 t ix 2 t i y1 t i y 2 t
I x1 I x 2 I y1 I y 2
Phasors also satisfy KCL and KVL.
Phasor - Multiplication
Phasor
Time domain
x(t )
Kx(t )
Multiply k
v(t ) Ri (t )
X
kX
Multiply k
V RI
Phasor - Differential
• We have to differentiate a sinusoidal wave due to
the i-v characteristics of capacitors and inductors.
x(t )
Differentiate
dx(t )
dt
j X
X
Multiplying jω
Phasor - Differential
• We have to differentiate a sinusoidal wave due to
the i-v characteristics of capacitors and inductors.
Time domain
Phasor
x(t ) X cos(t 1 )
X1
dx(t )
X sin(t 1 )
dt
X cos(t 1 90 )
X 1 90
Phasor - Differential
• We have to differentiate a sinusoidal wave due to
the i-v characteristics of capacitors and inductors.
Phasor
X1
Equivalent to multiply jω
Rotate 90。
Multiply ω
X 1 90
Differentiate on time domain
= phasor multiplying jω
Phasor - Differential
i leads v by 90。
• Capacitor
Time domain
Phasor
dvt
i (t ) C
dt
I j C V
Phasor - Differential
v leads i by 90。
• Inductor
Time domain
Phasor
di t
v(t ) L
dt
V j L I
Capacitor & Inductor
For C, i leads v
but v leads i for L
Phasor
Time domain
i-v characteristics
Phasors satisfy Ohm's law for resistor, capacitor and inductor.
i-v characteristics
Impedance
Resistor
Capacitor
ZR R
Z L j L
Inductor
1
1
ZC
j
j C
C
Admittance is the reciprocal of impedance.
When 0(DC),
When (hight - frequency),
Z L jL 0, short circuit
Z L jL , open circuit
1
ZC
, open circuit
j C
1
ZC
0, short circuit
j C
Equivalent impedance and
admittance
Series equivalent impedance
Z ser Z1 Z 2 Z N
Parallel equivalent impedance
1
1
1
1
Z par Z1 Z 2
ZN
Impedance
After series and parallel,
the equivalent impedance is Z R jX
where ReZ R, ac resistances
ImZ X , reactances
Resistor
Capacitor
ZR R
Z L j L
Inductor
1
1
ZC
j
j C
C
Inductors and capacitors are called reactive elements.
Inductive reactance is positive, and capacitive
reactance is negative.
Impedance Triangle
After series and parallel,
the equivalent impedance is Z R jX
where ReZ R, ac resistances
ImZ X , reactances
Z Z
Z R2 X 2
Z tan
1
X
R
33
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
v 30 cos(4000t 20 )
R 5 C 25 μF
Z R 5
1
1
ZC
j
j C
C
1
j
j10
4000 25μ
Z Z R || Z C (5) || ( j10)
5 j10
5 j10
4 j 2 4.47 26.6
3020
V
I
Z
4.47 26.6
30
20 (26.6 ) 6.7146.6 A
4.47
Example 6.7
• Impedance is frequency dependent
Z R ZC
Find equivalent
Z eq Z R || Z C
network
Z R ZC
R
R/jC
Zeq should be Zeq(ω) or
Zeq(jω)
R 1 / j C 1 j R C
R
R
1 j R C
Zeq j
1 j R C 1 j R C 1 j R C
2
2
R
R
C
R j R C
j
2
2
2
1 RC
1 RC
1 RC
Example 6.7
• Impedance is frequency dependent
R
R 2 C
Zeq j
j
2
2
1 RC
1 RC
If ω → 0
Find equivalent
network
Zeq j R
For DC, C is equivalent to
open circuit
Zeq j 0
C becomes short
If ω → ∞
Example 6.8
i t 10 cos(50000t )mA
Find v(t), vL(t) and
vC(t)
Z L jL j 50k 200m j10k
1
1
1
ZC
j
j
j10k
j C
C
50k 2n
Example 6.8
Z eq
Z eq 40 || j10 5 || j10 kΩ
Zeq = 4.8kΩ
+ j6.4k Ω
4.8 j 6.4 kΩ
853.1 kΩ
Example 6.8
V 8053.1 V
Zeq = 4.8kΩ
+ j6.4k Ω
Zeq 8053.1 V
V I Z eq
10m0 8k53.1
8053.1 V
Example 6.8
VL
V 8053.1 V
j10
V 3.2 j88 89.479.7 V
(4 j 2) j10
4 j2
VC
V 32 j 24 40 36.9 V
(4 j 2) j10
Example 6.8
V 8053.1 V
v (t ) 80 cos(50000t 53.1 )V
VL 89.479.7 V
vL (t ) 89.4 cos(50000t 79.7 )V
VC 40 36.9 V
vC (t ) 40 cos(50000t 36.9 )V
i 0 i 0 0 A
Complete Response
t0
i 0 i 0 0 A
i t
i 0 i 0 0 A
Complete Response
i t 1.03 cos5t 59
i t
F
t 0
I
Z L jL j10
t 0 Reach steady state
120
I
1.03 59
6 10 j
t 0
iF t 1.03 cos 5t 59
Forced
Response
i 0 i 0 0 A
i t
Complete Response
i t 1.03 cos5t 59
F
t 0
I
Natural Response:
iN t Ke
t
Ke 3t
L / R 1/ 3
vt vN t vF t Ke 3t 1.03 cos 5t 59
K 0.53
Complete Response
5
t0
1
1
ZC
j4
jC j 5 50m
4j
48
90
V
120
8
.
485
45
4 4 j
5.66 45
t 0 vt 8.485 cos5t 45
V
Complete Response
v0 6V
5
t0
vt 8.485 cos 5t 45
Initial Condition:
v0 v0 6V
v 0 8.485 cos 45 6V
V
Complete Response
v0 6V
t 0
t 0 Reach steady state
4j
48
90
V
120
10
.
74
26
.
6
2 4 j
4.47 63.4
t 0 vF t 10.74 cos5t 26.6
Forced
Response
V
Complete Response
v0 6V
t 0
Natural Response:
vN t Ke
t
RC
Ke 10t
RC 2 50mF 0.1
vt vN t vF t Ke 10t 10.74 cos 5t 26.6
K 3.6
V
Complete Response
Summarizing the results:
8.485 cos 5t 45
vt
10 t
3
.
6
e
10
.
74
cos
5
t
26
.
6
t0
t 0
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