sinusoidal steady-state analysis phasor166.104.231.29/webpage_limdj/optoelectronics_lab/ac.pdf ·...
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AC Circuits
Sinusoidal Steady-State Analysis
Phasor
DC & AC
DC : Direct Current
AC : Alternating Current
Sinusoidal function
Sinusoidal function
Sine and Cosine Functions
Sine and Cosine Functions
Advancing in time
Delaying in time
Oscilloscope
Differential Equation with AC Source
:
cos ,
cos , (0) 1
KVL
dvRi v e t i C
dt
dvRC v t v
dt
Forced response(particular solution), steady state:
2 2
cos sin
cos sincos sin cos
sin cos cos sin cos
cos sin cos
1, 0
, 1
1,
1 1
v A t B t
d A t B tRC A t B t t
dt
RCA t RCB t A t B t t
RCB A t B RCA t
RCB A B RCA
B RCA RC RCA A
RCA B
RC RC
Forced response(particular solution), steady state:
2 2
2
1
2
1
2
cos sin
1cos sin
1 1
1cos( tan )
1
1cos( tan )
1
v A t B t
RCt t
RC RC
RCt RC
RC
t RCRC
Complete response
2 2
1
2
2
2 2
2
2 2 2
1( ) cos sin
1 1
1cos( tan )
1
1(0) 1,
1 1
1( ) cos sin
1 1 1
t
RC
t
RC
t
RC
RCv t Ke t t
RC RC
Ke t RCRC
RCv K K
RC RC
RC RCv t e t t
RC RC RC
Check Initial Condition
Check
2
2 2 2
2 2
2 2 2
2 2
2 2
0
1( ) cos sin
1 1 1
( ) 1sin cos
1 1 1
( ) 10
1 1
t
RC
t
RC
t
RC RCv t e t t
RC RC RC
RCdv t RCe t t
dt RC RC RC RC
RCdv t RC
dt RC RC RC
Phasors and Sinusoids
Sinusoidal source(a known fixed single frequency)
Steady state, forced response
Euler’s formula
Time domain, Frequency domain
cos sinje j
Euler Formula
2 3 4 5
2 4 3 5
12! 3! 4! 5!
12! 4! 3! 5!
cos sin
jj j j j
e j
j
j
Motivating Example: RC circuit
:
cos ,
cos , (0) 1
KVL
dvRi v e t i C
dt
dvRC v t v
dt
Motivating Example: RC circuit
Euler’s formula
Assume the solution:
cos sin
cos Re
cos
j t
j t
j t
e t j t
t e
dv dvRC v t RC v e
dt dt
( )j t j j tv Ve Ve e
Motivating Example: RC circuit
Assume the solution:
( )
( )( )
( ) ( )
tan
2
1 1 1
1 1
1 1
j t j j t
j tj t j t
j t j t j t j j t j j t j t
j j j
j
v Ve Ve e
dVeRC Ve e
dt
j RCVe Ve e j RCVe e Ve e e
j RCVe Ve j RC Ve
Ve ej RC RC
1
( ) 1
2
1Re cos( tan )
1
RC
j tVe t RCRC
Arithmetic using Phasors
KVL in Phasor
Impedances
Ohm’s Law for AC circuits
Impedance and admittance
( ) ( ) ( )
1 ( )( )
( ) ( )
V Z I
IY
Z V
Capacitor
( ) cos( ) V
( ) ( ) sin( ) cos( 90 ) A
C
C C
v t A t
di t C v t C A t C A t
dt
Inductor
( ) cos( ) A
( ) ( ) sin( ) cos( 90 ) V
L
L L
i t A t
dv t L i t L A t L A t
dt
Resistor
( ) cos( )
( )( ) cos( )
R
RR
v t A t
v t Ai t t
R R
Node analysis
KCL : ( ) ( ) ( )
48 75 ( ) ( ) ( )
50 80 80
( ) ( ) 50 5048 75 50 ( ) 1 ( )
80 80 80 80
C RI I I
V V V
j j
V V j jj V V
j j
Node analysis(MATLAB)
50 5048 75 1 ( )
80 80
j jV
j
>> V=48*exp(j*75*pi/180)/(j*50/(-j*80)+(j*50)/80+1)
V =
63.3158 +18.1122i
>> abs(V)
ans =
65.8555
>> angle(V)*180/pi
ans =
15.9638
Series Impedance
Parallel Impedance
1 21 2
1 2
1 2
1||
1 1eq
Z ZZ Z Z
Z Z
Z Z
Voltage and Current Division
In the frequency domain
( 80)80|| 40 40
80 80
|| 40 40 12 20
|| 50 (40 40) 17
(1.372 59.04 )(48 75 ) 65.9 15.96
C
CS S S
L C
jZ R j
j
Z R j jV V V V
Z Z R j j