AC steady state

Report
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: xt   A cost   
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
vt 
iR t  
 6 cos 4000t  20
R


vt   30 cos 4000t  20



iC t   Cvt   25  30  4000  sin 4000t  20





t   i t   6 cos4000t  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   vt   Cvt 
R
vt   A cos 4000t  B sin 4000t
i t   3 cos4000t 
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
B6

vt   12 cos 4000t  6 sin 4000t  13.4cos 4000t - 26.6

Example 6.4
i t   3 cos4000t 
1
i t   iR t   iC t   vt   Cvt 
R

vt   13.4cos 4000t - 26.6


iR t   2.68cos 4000t - 26.6


 1.34sin 4000t - 26.6 
 1.34cos4000t  63.4 
iC t   Cvt   25 13.4  4000sin 4000t - 26.6



AC Steady-State Analysis
Example 6.3

vt   30 cos 4000t  20
Example 6.4

i t   3 cos4000t 


i t   3 cos4000t  110 
i t   6.7 cos4000t  46.6 
iR t   6 cos 4000t  20

C



t   2.68cos4000t - 26.6 
t   1.34cos4000t  63.4 
vt   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
xt   X m cost   
Amplitude: Xm
Phase: ϕ
Phasor
• A sinusoidal function is a point on a x-y plane
xt   X m cost   
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 ReA  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 )
ReA  B   ReA  ReB 
ImA  B   ImA  ImB 
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:
xt   X m cost   
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 11
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 11  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 )
X1
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
X1
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
dvt 
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  jL  0, short circuit
Z L  jL  , 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 ReZ   R, ac resistances
ImZ   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 ReZ   R, ac resistances
ImZ   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 
3020
V
I

Z
4.47  26.6
30

 20  (26.6 )  6.7146.6 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/jC
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  jL  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
 853.1 kΩ
Example 6.8
V  8053.1 V
Zeq = 4.8kΩ
+ j6.4k Ω
Zeq  8053.1 V
V  I  Z eq

 
 10m0  8k53.1
 8053.1 V

Example 6.8
VL 
V  8053.1 V
j10
V  3.2  j88  89.479.7  V
(4  j 2)  j10
4  j2
VC 
V  32  j 24  40  36.9 V
(4  j 2)  j10
Example 6.8
V  8053.1 V
v (t )  80 cos(50000t  53.1 )V
VL  89.479.7 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
t0
   
i 0  i 0  0 A
i t 
   
i 0  i 0  0 A
Complete Response
i t   1.03 cos5t  59 
i t 

F
t 0
I
Z L  jL  j10
t  0 Reach steady state
120
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 cos5t  59 

F
t 0
I
Natural Response:
iN t   Ke

t

 Ke 3t
  L / R  1/ 3

vt   vN t   vF t   Ke 3t  1.03 cos 5t  59

K  0.53
Complete Response
 5
t0


1
1
ZC 

  j4
jC j 5  50m

4j
48


90

V
120 

8
.
485


45
4   4 j 
5.66  45


t  0 vt   8.485 cos5t  45 
V
Complete Response
v0   6V

 5
t0



vt   8.485 cos 5t  45

Initial Condition:
 

v0   v0   6V

v 0   8.485 cos  45  6V


V
Complete Response
v0   6V

t 0


t  0 Reach steady state

4j
48


90

V
120 

10
.
74


26
.
6
2   4 j 
4.47  63.4


t  0 vF t   10.74 cos5t  26.6


Forced
Response
V
Complete Response
v0   6V

t 0


Natural Response:
vN t   Ke

t
RC
 Ke 10t
RC  2  50mF  0.1


vt   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
vt   
10 t


3
.
6
e

10
.
74
cos
5
t

26
.
6



t0
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

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