AC STEADY-STATE ANALYSIS
(Part I)
LEARNING GOALS
SINUSOIDS
Review basic facts about sinusoidal signals
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
Behavior of circuits with sinusoidal independent sources
and modeling of sinusoids in terms of complex exponentials
PHASORS
Representation of complex exponentials as vectors. It facilitates
steady-state analysis of circuits.
IMPEDANCE AND ADMITANCE
Generalization of the familiar concepts of resistance and
conductance to describe AC steady state circuit operation
SINUSOIDS
x(t )  X M sin  t
Adimensional plot
As function of time
X M  amplitude or maximum value
  angular frequency (rads/sec)
 t  argument (radians)
T
2

" leads by  "
 Period  x (t )  x (t  T ), t
1 

 frequency in Hertz (cycle/sec )
T 2
  2 f
f 
" lags by  "
LEARNING EXAMPLE
v1 (t )  12 sin(1000t  60), v2 (t )  6 cos(1000t  30)
FIND FREQUENCY AND PHASE ANGLE BETWEEN VOLTAGES
Frequency in radians per second is the factor of the time variable
f ( Hz) 
  1000 sec1

 159.2 Hz
2
To find phase angle we must express both sinusoids using the same
trigonometric function; either sine or cosine with positive amplitude
take care of minus sign with cos( )   cos(  180)
 6 cos(1000t  30)  6 cos(1000t  30  180)
Change sine into cosine with cos( )  sin(  90)
6 cos(1000t  210)  6 sin(1000t  210  90)
We like to have the phase shifts less than 180 in absolute value
6 sin(1000t  300)  6 sin(1000t  60)
v1 (t )  12 sin(1000t  60)
(1000t  60)  (1000t  60)  120
v2 (t )  6 sin(1000t  60)
(1000t  60)  (1000t  60)  120
v1 leads v2 by 120
v2 lags v1 by 120
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
Learning Example
KVL : L
di
(t )  Ri (t )  v (t )
dt
In steady state i (t )  A cos( t   ), or
i (t )  A1 cos t  A2 sin  t
If the independent sources are sinusoids
of the same frequency, then for any
variable in the linear circuit the steady
state response will be sinusoidal and of
the same frequency.
*/ R
di
*/ L
(t )   A1 sin  t  A2 cos t
dt
( LA1  RA2 ) sin  t  ( LA2  RA1 ) cos t 
 VM cos t
To determine the steady state solution
 LA1  RA2  0 algebraic problem
LA2  RA1  VM
we only need to determine the parameters
A1 
v (t )  Asin( t   )  iSS (t )  B sin( t   )
B,
RVM
LVM
,
A

2
R 2  (L) 2
R 2  (L) 2
Determining the steady state solution can
be accomplished with only algebraic tools!
FURTHER ANALYSIS OF THE SOLUTION
The solution is i (t )  A1 cos t  A2 sin  t
The applied voltage is v (t )  VM cos t
For comparison purposes one can write i (t )  A cos( t   )
A
2
2
A1  A cos , A2   Asin  A  A1  A2 , tan    2
A1
A1 
RVM
LVM
,
A

2
R 2  (L) 2
R 2  (L) 2
A
i (t ) 
VM
1 L
,


tan
R
R 2  (L) 2
VM
1 L
cos(

t

tan
)
2
2
R
R  (L)
For L  0 the current ALWAYS lags the voltage.
If R  0 (pure inductor), the current lags the voltage by 90 .
SOLVING A SIMPLE ONE LOOP CIRCUIT CAN BE VERY LABORIOUS
IF ONE USES SINUSOIDAL EXCITATIONS.
TO MAKE ANALYSIS SIMPLER ONE RELATES SINUSOIDAL SIGNALS
TO COMPLEX NUMBERS. THE ANALYSIS OF STEADY STATE WILL BE
CONVERTED TO SOLVING SYSTEMS OF ALGEBRAIC EQUATIONS ...
… WITH COMPLEX VARIABLES.
ESSENTIAL IDENTITY : e j  cos  j sin  (Euler identity)
v (t )  VM cos t  y (t )  A cos( t   )
v (t )  VM sin  t  y (t )  A sin( t   ) * / j (and add)
VM e j t  Ae j (t  )  Ae j e j t
y(t )
If everybody knows the frequency of the sinusoid
then one can skip the term exp(jwt).
j
VM  Ae
PHASORS
ESSENTIAL CONDITION
ALL INDEPENDENT SOURCES ARE SINUSOIDS OF THE SAME FREQUENCY
SHORTCUT IN NOTATION
INSTEAD OF WRITING u  U M e j WE WRITE u  U M 
... AND WE ACCEPT ANGLES IN DEGREES
U M  IS THE PHASOR REPRESENTA TION FOR U M cos( t   )
u(t )  U M cos( t   )  U  U M   Y  YM   y(t )  Re{YM cos( t   )}
SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF
THE RESPONSE GIVEN THE INPUT PHASOR(S).
Learning Extensions
It is essential to be able to move from
sinusoids to phasor representation
A cos(t   )  A  
A sin(t   )  A    90
v (t )  12 cos(377t  425)  12  425
y(t )  18 sin( 2513t  4.2)  18  85.8
Given f  400 Hz
V1  1020  v1 (t )  10 cos(800 t  20)
V2  12  60  v2 (t )  12 cos(800 t  60)
Phasors can be combined using the
rules of complex algebra.
(V11 )(V2 2 )  V1V2(1   2 )
V11 V1
 (1   2 )
V2 2 V2
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS
RESISTORS v (t )  Ri (t )
VM e ( j t  )  RI M e ( j t  )
VM e j  RI M e j
V  RI
Phasor representation for a resistor
Phasors are complex numbers. The resistor
model has a geometric interpretation.
The voltage and current
phasors are colineal.
In terms of the sinusoidal signals, this
geometric representation implies that
the two sinusoids are “in phase.”
INDUCTORS
d
( I M e ( j t  ) )
dt
 jLI M e ( j t  )
VM e ( j t  )  L
VM e j  jLI M e j
V  jLI
The relationship between
phasors is algebraic.
For the geometric view,
use the result.
j  190  e j 90
V  LI90
The voltage leads the current by 90 deg.
The current lags the voltage by 90 deg.
Relationship between sinusoids
INDUCTORS
Learning Example
L  20mH , v (t )  12 cos(377t  20). Find i (t )
  377
1220
( A)
L90
12
I
  70( A)
3
377  20 10
V  1220 I 
I
V
j L
i (t )  1.59 cos(377t  70)
CAPACITORS
I M e ( j t  )  C
d
(VM e ( j t  ) )
dt
I M e j  jCe j
I  CV90
I  jCV
The relationship between
phasors is algebraic.
In a capacitor, the
current leads the
voltage by 90 deg.
The voltage lags
the current by 90 deg.
Relationship between sinusoids
CAPACITORS
Learning Example
C  100  F , v(t )  100 cos(314t  15).
Find i (t ).
  314
V  10015
I  jCV
I  C 190 10015
I  314 100 106 100105( A)
i (t )  3.14 cos(314t  105)( A)
LEARNING EXTENSIONS
L  0.05 H , I  4  30( A), f  60 Hz
Now an example with capacitors
Find the voltage across the inductor
C  150 F , I  3.6  145, f  60 Hz
  2 f  120
V  jLI
Find the voltage across the inductor
V  120  0.05 190  4  30
V  75.460
v(t )  75.4 cos(377t  60)
  2 f  120
I  jCV  V 
I
jC
3.6  145
V
120 150 106 190
V  63.66  235
v(t )  63.66 cos(120 t  235)
IMPEDANCE AND ADMITTANCE
For each of the passive components the relationship between the voltage phasor
and the current phasor is algebraic. We now generalize for an arbitrary 2-terminal
element.
Z ( )  R( )  jX ( )
R( )  Resistive component
X ( )  Reactive component
| Z | R 2  X 2
(INPUT) IMPEDANCE
Z
V VM  v VM


( v   i ) | Z |  z
I I M  i I M
(DRIVING POINT IMPEDANCE)
The units of impedance are OHMS.
 z  tan 1
X
R
IMPEDANCE AND ADMITTANCE
Element Phasor Eq. Impedance
R
L
C
V  RI
V  jLI
1
V
I
j C
ZR
Z  jL
1
Z
j C
SPECIAL APPLICATION:
IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED
FOR RESISTORS
I
 V1 
 V2 
Z1
Z2
I
Zs  Z1  Z2
Z s   k Zk
I

Z1
I

Z2 V
V


Zp 
1
1
k
Zp
Zk
Z1Z 2
Z1  Z 2
LEARNING EXAMPLE
f  60 Hz, v (t )  50 cos( t  30)
Compute equivalent impedance and current
  120 , V  5030, Z R  25
ZR  R
Z L  jL
ZC 
1
jC
1
j120  50  106
Z L  j 7.54, ZC   j53.05
Z s  Z R  Z L  ZC  25  j 45.51
V
5030
5030
I

( A) 
( A)
Z s 25  j 45.51
51.93  61.22
Z L  j120  20  103  , ZC 
I  0.9691.22( A)  i (t )  0.96 cos(120 t  91.22)( A)
LEARNING ASSESSMENT
FIND i (t )
  377
Z R  20
V  120(60  90)
Z L  j377  40 103  j15.08
ZC 
Zeq  ZC || ( Z R  Z L )
Zeq  30.5616 + j 4.9714  30.9639.239
I
V
120  30

 3.876  39.239( A)
Zeq 30.9639.239
j
  j53.05
6
377  50 10
(COMPLEX) ADMITTANCE
1
 G  jB (Siemens)
Z
G  conductanc e
Y
B  Suceptanc e
Element
R
L
C
Impedance
ZR
Z  j L
Z
1
j C
Admittance
1
Y  G
R
1
Y
j L
Y  j C
Parallel Combinatio n of Admittanc es
Y p  Yk
k
YR  0.1S
1
 j1( S )
 j1
Y p  0.1  j1( S )
YC 
Series Combinatio n of Admittanc es
1
1

Ys k Yk
0.1S
 j 0.1S
1
1
1


Ys 0.1  j 0.1
 10  j10
(0.1)( j 0.1) 0.1  j 0.1

0 . 1  j 0 .1 0 .1  j 0 . 1
1
10  j10
Ys 

10  j10
200
Ys  0.05  j 0.05 S
Ys 
LEARNING EXAMPLE
VS  6045(V )
FIND Y p , I
Y p  YR  YL
 0.5  j 0.25
OR
Zp 
2  j4
2  j4
 
Yp 
2  j4
 0.5  j 0.25( S )
j8
I  Y pV  (0.5  j 0.25)  6045( A)
I  0.559  26.565  6045( A)
I  33.5418.435( A)
LEARNING EXAMPLE
SERIES-PARALLEL REDUCTIONS
1
2  j4
 2
2  j 4 (2)  (4) 2
1
4  j2
Y34 

4  j2
20
Y2 
Z3  4  j 2
Y4   j 0.25  j 0.5  j 0.25
Z 4  1 / Y4   j 4
1 ( j 2)
1 j2
1
Z1 
1  j 0.5
Z1 
1  j 0.5
Z1 
1  (0.5) 2
Z1  0.8  j 0.4()
Z4 
j 4  ( j 2) 8

j4  j2
j2
Y2  0.1  j 0.2( S )
Y34  0.2  j 0.1
Z2  2  j 6  j 2  2  j 4
Z34  4  j 2
Z 234 
Y234  0.3  j 0.1( S )
Z 234 
1
1
0.3  j 0.1


Y234 0.3  j 0.1
0.1
Z 2 Z 34
 3  j1
Z 2  Z 34
Zeq  Z1  Z234  3.8  j 0.6  3.8478.973
LEARNING EXTENSION
FIND THE IMPEDANCE ZT
Z1  4  j 6  j 4
Z1  4  j 2
Y12  Y1  Y2
( R  P ) Z1  4.47226.565
Y1  0.224  26.565
( P  R)Y1  0.200  j 0.100
Y12  Y1  Y2  0.45  j 0.35
( R  P )Y12  0.570  37.875
Z12  1.75437.875
Z2  2  j 2 ( R  P ) Z2  2.82845 ( P  R) Z12  1.384  j1.077
Y2  0.354  45
( P  R)Y2  0.250  j 0.250
1
4  j2
Y1 
 2
4  j 2 (4)  (2) 2
1
2  j2
ZT  2  (1.384  j1077)  3.383  j1.077
Y2 
 2
2  j 2 (2)  (2) 2
1
1
0.45  j 0.35
Z12 


Y12 0.45  j 0.35
0.325
1
Z12 
Y12