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AC STEADY-STATE ANALYSIS
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
PHASOR DIAGRAMS
Representation of AC voltages and currents as complex vectors
BASIC AC ANALYSIS USING KIRCHHOFF LAWS
ANALYSIS TECHNIQUES
Extension of node, loop, Thevenin and other techniques
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
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
*/ R
If the independent sources are sinusoids di
*/ L
(t )   A1 sin  t  A2 cos t
of the same frequency then for any
dt
variable in the linear circuit the steady
state response will be sinusoidal and of ( LA1  RA2 ) sin  t  ( LA2  RA1 ) cos t 
the same frequency
 VM cos t
 LA1  RA2  0 algebraic problem
LA2  RA1  VM
To determine the steady state solution
we only need to determine the parameters A1  RVM , A2  LVM
2
2
2
2
R

(

L
)
R

(

L
)
B,
v (t )  Asin( t   )  iSS (t )  B sin( t   )
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   )
A1  A cos , A2   Asin 
A1 
A  A12  A22 , tan   
RVM
LVM
,
A

2
R 2  (L) 2
R 2  (L) 2
A
i (t ) 
VM
R 2  (L) 2
,    tan 1
VM
R  (L)
2
2
A2
A1
L
R
cos( t  tan 1
L
R
)
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)
VM  Ae j
Example
R  jL  R2  (L )2 e
I M e j 
v (t )  VM e j t
Assume i (t )  I M e ( j t  )
di
KVL : L (t )  Ri (t )  v (t )
dt
di
(t )  jI M e ( j t  )
dt
di
L (t )  Ri (t )  jLI M e ( j t  )  RI M e ( j t  )
dt
 ( jL  R) I M e ( j t  )
j
 ( jL  R) I M e e
jt
( jL  R) I M e j e j t  VM e j t
VM
R  jL
I M e j 
*/
jL  R
R  jL
I M e j 
VM ( R  jL)
R 2  (L) 2
IM 
VM
R 2  (L ) 2
VM
R 2  (L ) 2
 tan 1
e
L
R
 tan 1
L
R
,    tan 1
L
R
v (t )  VM cos t  Re{VM e j t }
 i (t )  Re{I M e ( j t  ) }  I M cos( t   )
C P
x  jy  re j
r  x 2  y 2 ,   tan 1
x  r cos  , y  r sin 
y
x
PHASORS
ESSENTIAL CONDITION
ALL INDEPENDENT SOURCES ARE SINUSOIDS OF THE SAME FREQUENCY
BECAUSE OF SOURCE SUPERPOSITION ONE CAN CONSIDER A SINGLE SOURCE
u(t )  U M cos( t   )
THE STEADY STATE RESPONSE OF ANY CIRCUIT VARIABLE WILL BE OF THE FORM
y(t )  YM cos( t   )
SHORTCUT 1
u(t )  U M e j ( t  )  y(t )  YM e
Re{U M e j ( t  ) }  Re{YM e
j (  t  )
j (  t  )
}
U M e j ( t  )  U M e j e jt u  U M e j  y  YM e j
SHORTCUT IN NOTATION
NEW IDEA:
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 )  YM cos( t   )
SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF
THE RESPONSE GIVEN THE INPUT PHASOR(S)
Example
It is essential to be able to move from
sinusoids to phasor representation
A cos(t   )  A  
A sin(t   )  A    90
V  V M 0
v  Ve jt
I  I M 
jt
di
i

Ie
L (t )  Ri (t )  v
dt
L( jIe jt )  RIe jt  Ve jt
In terms of phasors one has
jLI  RI  V
V
I
R  jL
The phasor can be obtained using
only complex algebra
We will develop a phasor representation
for the circuit that will eliminate the need
of writing the differential equation
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
Relationship between sinusoids
VM e j  jLI M e j
V  jLI
Example
The relationship between L  20mH , v (t )  12 cos(377t  20). Find i (t )
phasors is algebraic
For the geometric view
use the result
j  190  e j 90
V  LI90
  377
1220
( A)
V  1220 I 
L90
V
12
I
I

  70( A)
jL
377  20 103
i (t ) 
The voltage leads the current by 90 deg
The current lags the voltage by 90 deg
12
cos(377t  70)
3
377  20 10
CAPACITORS
I M e ( j t  )  C
d
(VM e ( j t  ) )
dt
Relationship between sinusoids
I M e j  jCe j
I  CV90
I  jCV
C  100 F , v (t )  100 cos(314t  15). Find i (t )
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
  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)
L  0.05 H , I  4  30( A), f  60 Hz
Find the voltage across the inductor
  2 f  120
V  jLI
V  120  0.05 190  4  30
V  2460
v (t )  24 cos(120  60)
Now an example with capacitors
C  150 F , I  3.6  145, f  60 Hz
Find the voltage across the capacitor
  2 f  120
I  jCV  V 
V
I
jC
3.6  145
120 150 106 190
200
V
  235

v (t ) 
200

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
X
 z  tan 1
R
(INPUT) IMPEDANCE
V V  V
Z   M v  M ( v   i ) | Z |  z
I I M  i I M
(DRIVING POINT IMPEDANCE)
The units of impedance are OHMS
Impedance is NOT a phasor but a complex
number that can be written in polar or
Cartesian form. In general its value depends
on the frequency
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
KVL AND KCL HOLD FOR PHASOR REPRESENTATIONS
 v2 (t ) 


v1 ( t )
v3 ( t )


i0 (t )
i1 (t )
i2 ( t )
i3 (t )
KVL: v1(t )  v2 (t )  v3 (t )  0
KCL :  i0 (t )  i1 (t )  i2 (t )  i3 (t )  0
vi (t )  VMie j ( t i ) , i  1,2,3
ik (t )  I Mk e j ( t k ) , k  0,1,2,3
In a similar way, one shows ...
KVL : (VM 1e j1  VM 2e j 2  VM 3e j 3 )e jt  0
VM11  VM 2 2  VM 33  0
V1  V2  V3  0 Phasors!
 V2 


V1
V3


 I0  I1  I 2  I3  0
I0
I1
I2
I3
The components will be represented by their impedances and the relationships
will be entirely algebraic!!
SPECIAL APPLICATION:
IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED
FOR RESISTORS
I
 V1 
Z1
I
 V2 
I
Zs  Z1  Z2
Z2
Z1
Z2 V

V


1
1
k
Zp
Zk
Z s   k Zk
LEARNING EXAMPLE

I
Zp 
Z1Z 2
Z1  Z 2
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 L  j120  20  103  , ZC 
Z s  Z R  Z L  ZC  25  j 45.51
I
V
5030
5030

( A) 
( A)
Z s 25  j 45.51
51.93  61.22
I  0.9691.22( A)  i (t )  0.96 cos(120 t  91.22)( A)
Parallel Combinatio n of Admittances
(COMPLEX) ADMITTANCE
Y p  Yk
1
 G  jB (Siemens)
Z
G  conductanc e
B  Suceptanc e
Y
k
YR  0.1S
1
1
R  jX
R  jX


 2
Z R  jX
R  jX R  X 2
R
R2  X 2
X
B 2
R  X2
G
L
C
V  RI
V  jLI
1
V
I
jC
ZR
Z  jL
Z
1
jC
1
 j1( S )
 j1
Y p  0.1  j1( S )
Series Combinatio n of Admittanc es
1
1

Ys k Yk
0.1S
Element Phasor Eq. Impedance
R
YC 
Admittance
1
Y  G
R
1
Y
j L
Y  j C
 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 
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
( P  R) Z12  1.384  j1.077
Z2  2  j 2 ( R  P ) Z2  2.82845
Y2  0.354  45
( P  R)Y2  0.250  j 0.250
1
4  j2
ZT  2  (1.384  j1077)  3.383  j1.077
Y1 
 2
4  j 2 (4)  (2) 2
1
2  j2
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
PHASOR DIAGRAMS
Display all relevant phasors on a common reference frame
Very useful to visualize phase relationships among variables.
Especially if some variable, like the frequency, can change
SKETCH THE PHASOR DIAGRAM FOR THE CIRCUIT
Any one variable can be chosen as reference.
For this case select the voltage V
KCL : I S 
V
V

 jCV
R jL
  (capacitiv e)
| I L || IC |
| I L || IC |
IC  jCV
IL 
V
jl
INDUCTIVE CASE
CAPACITIVE CASE
  (inductive )
LEARNING EXAMPLE
DO THE PHASOR DIAGRAM FOR THE CIRCUIT
  377( s 1 )
2. PUT KNOWN NUMERICAL VALUES
| VL  VC || VR |
VR  RI
VL  jLI
It is convenient to select
1
the current as reference
VC 
I
j C
VS  VR  VL  VC
1. DRAW ALL THE PHASORS
| VL || VC |
DIAGRAM WITH REFERENCE VS  12 290
VL  18135(V )
Read values from
diagram!
 I  345( A)
VR  1245(V )
(Pythagoras)
VC  6  45
BASIC ANALYSIS USING KIRCHHOFF’S LAWS
PROBLEM SOLVING STRATEGY
For relatively simple circuits use
Ohm' s law for AC analysis; i.e., V  IZ
The rules for combining Z and Y
KCL AND KVL
Current and voltage divider
For more complex circuits use
Node analysis
Loop analysis
Superposit ion
Thevenin' s and Norton' s theorems
MATLAB
PSPICE
ANALYSIS TECHNIQUES
PURPOSE: TO REVIEW ALL CIRCUIT ANALYSIS TOOLS DEVELOPED FOR
RESISTIVE CIRCUITS; I.E., NODE AND LOOP ANALYSIS, SOURCE SUPERPOSITION,
SOURCE TRANSFORMATION, THEVENIN’S AND NORTON’S THEOREMS.
COMPUTE I0
V2  60
V
 20  V2  2  0
1  j1
1  j1
 1
1 
6
V2 
1

2

1  j1
1  j1
1  j1
V2
1. NODE ANALYSIS
V1
V
V
 20  2  2  0
1  j1
1 1  j1
V1  V2  60
I0 
V2
( A)
1
(1  j1)  (1  j1)(1  j1)  (1  j1) 2(1  j1)  6

(1  j1)(1  j1)
1  j1
V2
4
 8  j2
1 j
V2 
(4  j )(1  j )
2
3
5
I 0    j ( A)
2
2
I0  2.92  30.96
Circuit with voltage source
set to zero (SHORT CIRCUITED)
SOURCE SUPERPOSITION
I
I L2
1
L
=
V
1
L
+
Circuit with current
source set to zero(OPEN)
Due to the linearity of the models we must have
I L  I L1  I L2
VL  VL1  VL2
Principle of Source Superposition
The approach will be useful if solving the two circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient
VL2
3. SOURCE SUPERPOSITION
I 0'  10( A)
Z '  (1  j ) || (1  j ) 
(1  j )(1  j )
1
(1  j )  (1  j )
COULD USE SOURCE TRANSFORMATION
TO COMPUTE I"0
V1"
1 j
1 j
I 0" 
6
2

j
(
1

j
)

3

j
I 0" 
6 ( A)
1 j
6 6
"
1 j
I

 j ( A)
0
2 j
4 4
5 3 
I 0  I 0'  I 0"    j ( A)
2 2 
Z"
Z "  1 || (1  j )
Z"
Z"
"
 "
60(V ) I 0  "
60( A)
Z 1 j
Z 1 j
THEVENIN’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
ZTH


RTH
vTH

i
a
vO
b
_

i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
PART B
PART A
Phasor
Thevenin Equivalent Circuit
for PART A
vTH
Thevenin Equivalent Source
RTH
Thevenin Equivalent Resistance
Impedance
5. THEVENIN ANALYSIS
Voltage Divider
VOC 
10  6 j
1 j
(8  2 j ) 
2
(1  j )  (1  j )
ZTH  (1  j ) || (1  j )  1
8 2j
I0 
53j
( A)
2
EXAMPLE
Find the current i(t) in steady state
The sources have different frequencies!
For phasor analysis MUST use source superpositio
Frequency domain
SOURCE 2: FREQUENCY 20r/s
Principle of superposition
LEARNING BY DESIGN
USING PASSIVE COMPONENTS TO CREATE GAINS LARGER THAN ONE
PRODUCE A GAIN=10
AT 1KhZ WHEN R=100
 2 LC  1 
 L  1.59mH
 C  15.9 F