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AC STEADY-STATE ANALYSIS
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
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
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 "
BASIC TRIGONOMETRY
ESSENTIAL IDENTITIES
RADIANS AND DEGREES
sin( ) sin cos cos sin
2 radians 360 degrees
cos( ) cos cos sin sin
(rads)
sin( ) sin
cos( ) cos
SOME DERIVED IDENTITIES
sin( ) sin cos cos sin
cos( ) cos cos sin sin
1
1
sin cos sin( ) sin( )
2
2
1
1
cos cos cos( ) cos( )
2
2
APPLICATIO NS
cos t sin( t )
2
sin t cos( t )
2
cos t cos( t )
sin t sin( t )
180
(degrees)
ACCEPTED EE CONVENTION
sin( t ) sin( t 90)
2
LEARNING EXAMPLE
cos( t 45)
cos( t 45 360)
cos( t 45)
cos( t 45 180)
cos( t )
Leads by 45 degrees
Lags by 315
Leads by 225 or lags by 135
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 sec1
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
LEARNING EXTENSION
i1 (t ) 2 sin( 377 t 45)
i2 (t ) 0.5 cos(377 t 10)
i3 (t ) 0.25 sin( 377 t 60)
i1 leads i2 by_____?
i1 leads i3 by_____?
cos sin( 90)
0.5 cos(377t 10) 0.5 sin( 377t 10 90)
(377t 45) (377 100) 55
sin sin( 180)
i1 leads i2 by 55
0.25 sin( 377t 60) 0.25 sin( 377t 60 180)
(377t 45) (377t 120) 165
i1 leads i3 by 165
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
*/ 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 ( LA1 RA2 ) sin t ( LA2 RA1 ) cos t
the same frequency
VM cos t
LA1 RA2 0 algebraic problem
LA2 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
RVM
LVM
,
A
2
R 2 (L) 2
R 2 (L) 2
A
i (t )
A A12 A22 , tan
A2
A1
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)
y(t )
VM e j t Ae j (t ) Ae j e j t
If everybody knows the frequency of the sinusoid
then one can skip the term exp(jwt)
VM Ae j
Learning Example
R jL 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 ) jI M e ( j t )
dt
di
L (t ) Ri (t ) jLI M e ( j t ) RI M e ( j t )
dt
( jL R) I M e ( j t )
( j L R ) I M e j e j t
( jL R) I M e j e j t VM e j t
VM
R jL
I M e j
*/
jL R
R jL
I M e j
VM ( R jL)
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
x
y
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 jt 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 ) Re{YM cos( t )}
SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF
THE RESPONSE GIVEN THE INPUT PHASOR(S)
Learning Extensions
Learning 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 jt
I I M
j t
di
i
Ie
L (t ) Ri (t ) v
dt
L( jIe jt ) RIe jt Ve jt
In terms of phasors one has
jLI 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 1020 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
(V11 )(V2 2 ) V1V2(1 2 )
V11 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
jLI M e ( j t )
VM e ( j t ) L
Relationship between sinusoids
VM e j jLI M e j
V jLI
Learning 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 190 e j 90
V LI90
377
I
V
j L
i (t )
The voltage leads the current by 90 deg
The current lags the voltage by 90 deg
1220
( A)
L90
12
I
70( A)
3
377 20 10
V 1220 I
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 jCe j
I CV90
I jCV
Learning Example
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 10015
I C 190 10015
I jCV
I 314 100 106 100105( A)
i (t ) 3.14 cos(314t 105)( A)
LEARNING EXTENSIONS
L 0.05 H , I 4 30( A), f 60 Hz
Find the voltage across the inductor
2 f 120
V jLI
V 120 0.05 190 4 30
V 2460
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 inductor
2 f 120
I jCV V
V
I
jC
3.6 145
120 150 106 190
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
z tan 1
(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
X
R
Element Phasor Eq. Impedance
R
L
C
V RI
V jLI
1
V
I
j C
ZR
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 j1 VM 2e j 2 VM 3e j 3 )e jt 0
VM11 VM 2 2 VM 33 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
Z2 V
Z1
Z s k Zk
LEARNING EXAMPLE
f 60 Hz, v (t ) 50 cos( t 30)
I
V
1
1
k
Zp
Zk
Zp
Z1Z 2
Z1 Z 2
Compute equivalent impedance and current
120 , V 5030, Z R 25
ZR R
Z L jL
ZC
1
jC
1
j120 50 106
Z L j 7.54, ZC j53.05
Z L j120 20 103 , ZC
Z s Z R Z L ZC 25 j 45.51
I
V
5030
5030
( A)
( A)
Z s 25 j 45.51
51.93 61.22
I 0.9691.22( A) i (t ) 0.96 cos(120 t 91.22)( A)
LEARNING EXTENSION
FIND i (t )
377
Z R 20
Z L j377 40 103 j15.08
V 120(60 90)
ZC
Zeq ZC || ( Z R Z L )
I
j
j53.05
6
377 50 10
Zeq 30.5616 + j 4.9714 30.9639.239
V
120 30
3.876 39.924( A)
Z eq 30.9639.239
Parallel Combinatio n of Admittanc es
(COMPLEX) ADMITTANCE
Y p Yk
1
G jB (Siemens)
Z
G conductanc e
Y
k
YR 0.1S
YC
B Suceptanc e
1
1
R jX
R jX
2
Z R jX
R jX R X 2
L
C
V jLI
1
V
I
jC
Series Combinatio n of Admittanc es
0.1S
Element Phasor Eq. Impedance
V RI
Y p 0.1 j1( S )
1
1
Ys k Yk
R
R2 X 2
X
B 2
R X2
G
R
1
j1( S )
j1
ZR
Z jL
Z
1
jC
Admittance
1
Y G
R
1
Y
jL
Y jC
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 6045(V )
LEARNING EXTENSION
FIND Y p , I
Y p YR YL
0.5 j 0.25
Y p 0.5 j 0.5 j1 0.25 0.75 j 0.5( S )
Y p 0.901433.69( S )
2 j4
2 j4
Yp
0.5 j 0.25( S )
2 j4
j8
I Y pV 0.901433.69 1020
I Y pV (0.5 j 0.25) 6045( A)
I 9.01453.79( A)
Zp
I 0.559 26.565 6045( A)
I 33.5418.435( 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 Z34
3 j1
Z 2 Z 34
Zeq Z1 Z234 3.8 j 0.6 3.8478.973
LEARNING EXTENSION FIND THE IMPEDANCE ZT
Z1 4 j 6 j 4
Z1 4 j 2
Y12 Y1 Y2
( R P ) Z1 4.47226.565
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 1.75437.875
( P R) Z12 1.384 j1.077
Z2 2 j 2 ( R P ) Z2 2.82845
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
LEARNING EXAMPLE
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
jCV
R jL
(capacitiv e)
| I L || IC |
| I L || IC |
IC jCV
IL
V
jl
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 jLI
It is convenient to select
1
the current as reference
VC
I
jC
VS VR VL VC
1. DRAW ALL THE PHASORS
| VL || VC |
DIAGRAM WITH REFERENCE VS 12 290
VL 18135(V )
Read values from
diagram!
I 345( A)
VR 1245(V )
(Pythagora s)
VC 6 45
FIND THE FREQUENCY AT WHICH v (t ) AND i (t )
LEARNING BY DOING
C
v (t )
ARE IN PHASE
i.e., the phasors for i (t ), v (t ) are co - lineal
1
V
I jLI RI
jC
L
PHASOR DIAGRAM
R
jLI
1
I
j C
V
RI
V and I are co - lineal iff jL
2
Notice that I was
chosen as reference
1
I jLI RI
jC
I
1
1
0 2
jC
LC
1
9
4
10
3
.
162
10
(rad / s)
3
6
10 10
f
5.033 103 Hz
2
LEARNING EXTENSION
Draw a phasor diagram illustrating all voltages and currents
j4
4 90
I
445
2 j4
4.472 63.435
I1 3.57818.435( A)
I1
I2
Current
divider
1
20
I
445
2 j4
4.472 63.435
I 2 1.789108.435 Simpler than I 2 I I1
V 2 I1 7.15618.435(V )
DRAW PHASORS. ALL ARE
KNOWN. NO NEED TO SELECT
A REFERENCE
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
Source transformation
Thevenin' s and Norton' s theorems
MATLAB
PSPICE
LEARNING EXAMPLE
COMPUTE ALL THE VOLTAGES AND CURRENTS
Compute I1
Use current divider for I2 , I3
Ohm' s law for V1 , V2
V1 690 I 2
Zeq 4 ( j 6 || 8 j 4)
V1 16.2678.42(V )
24 j 48 32 j8 24 j 48
Z eq 4
8 j2
8 j2
V2 7.2815(V )
56 j56
79.19645
9.60430.964()
8 j2
8.24614.036
V
2460
I1 S
2.49829.036( A)
Zeq 9.60430.964
Z eq
j6
690
I1
2.49829.036( A)
8 j2
8.24614.036
8 j4
8.944 26.565
I2
I1
2.49829.036( A)
8 j2
8.24614.036
I3
I1 2.529.06
I 2 2.71 11.58
V2 4 90 I3
I3 1.82105
LEARNING EXTENSION
IF VO 845, COMPUTE VS
THE PLAN...
COMPUTE I3
COMPUTE V1
COMPUTE I2 , I1
COMPUTE VS
VO
( A) 445( A)
2
V1 (2 j 2) I3 8 45 445
I3
V1 11.3140(V )
I2
V1 11.3140
5.657 90( A)
j2
290
I1 I 2 I3 5.657 90 445
I1 j5.657 (2.828 j 2.828)( A)
I1 2.828 j 2.829( A)
VS 2 I1 V1 2(2.828 j 2.829) 11.3140
VS 16.97 j5.658(V )
VS 17.888 18.439
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 60
V
20 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
20 2 2 0
1 j1
1 1 j1
V1 V2 60
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
NEXT: LOOP ANALYSIS
2. LOOP ANALYSIS
ONE COULD ALSO USE THE SUPERMESH
TECHNIQUE
SOURCE IS NOT SHARED AND Io IS
DEFINED BY ONE LOOP CURRENT
LOOP 1 : I1 20
I2
I 0 I3
LOOP 2 : (1 j )( I1 I 2 ) 60 (1 j )( I 2 I3 ) 0
LOOP 3 : (1 j )( I 2 I3 ) I3 0
CONSTRAINT : I1 I 2 20
SUPERMESH : (1 j ) I1 60 ( I 2 I 3 ) 0
MUST FIND I3
2 I 2 (1 j ) I3 6 (1 j )(2)
(1 j ) I 2 (2 j ) I3 0
(1 j)
/* (1 j )
/* (2)
2(2 j ) I3 (1 j )(8 2 j )
5 3
10 6 j
I
j ( A)
I3
0
2 2
4
2
MESH 3 : ( I 3 I 2 ) (1 j ) I 3 0
I 0 I 2 I3
NEXT: SOURCE SUPERPOSITION
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' 10( A)
Z ' (1 j ) || (1 j )
(1 j )(1 j )
1
(1 j ) (1 j )
COULD USE SOURCE TRANSFORMATION
TO COMPUTE I"0
V1"
Z"
Z"
"
"
60(V ) I 0 "
60( A)
Z 1 j
Z 1 j
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 )
NEXT: SOURCE TRANSFORMATION
Source transformation is a good tool to reduce complexity in a circuit ...
WHEN IT CAN BE APPLIED!!
“ideal sources” are not good models for real behavior of sources
A real battery does not produce infinite current when short-circuited
ZV
+
-
RV
VS
a
ZI
a
RI
b
IS
THE MODELS ARE EQUIVALENT S WHEN
RV RI R
b
VS RI S
Z I Z VZ
S IZ SV
Improved model
Improved model
for voltage source for current source
Source Transformationcan be used to determine the Thevenin or Norton Equivalent...
BUT THERE MAY BE MORE EFFICIENT TECHNIQUES
4. SOURCE TRANSFORMATION
IS
8 2j
1 j
Z (1 j ) || (1 j ) 1
V ' 8 2 j
Now a voltage to current transformation
NEXT: THEVENIN
I0
I S 4 j (4 j )(1 j ) 5 3 j
2 1 j (1 j )(1 j )
2
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
53j
( A)
2
NEXT: NORTON
NORTON’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
i
a
vO
b
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
Phasors
ZN
iN
RN
i
a
LINEAR CIRCUIT
vO
_
b
PART B
PART A
Norton Equivalent Circuit
for PART A
iN
Thevenin Equivalent Source
RNZ N Thevenin Equivalent Resistance
Impedance
6. NORTON ANALYSIS
ZTH (1 j ) || (1 j ) 1
I0
Possible techniques: loops, source
transformation, superposition
BY SUPERPOSTI ON
I SC 20
60 8 2 j
( A)
1 j
1 j
I SC 4 j (4 j )(1 j ) 5 3 j
2
1 j (1 j )(1 j )
2
LEARNING EXAMPLE FIND V0 USING NODES, LOOPS, THEVENIN, NORTON
WHY SKIP SUPERPOSITION AND TRANSFORMATION?
Supernode constraint : V1 V3 120
KCL @ Supernode
40
V3 V0 V3 V2 V1 V2 V3
0
1
1
j
j
KCL@V2
V2 V1
V V3
2I x 2
0
j
1
NODES
KCL@ V0
V0 V0 V3
40 0 V3 2V0 4
1
1
V1 V3 12
Controllin g variable
V V0
Ix 3
1
V1 2V0 16
V3 V0 V0 4
j (V2 2V0 16) 2(V0 4) (V2 2V0 4) 0
Notice choice of ground
j (V2 2V0 16) (V2 2V0 4) (V0 4) j (2V0 4) 4
8 4j
Adding : V0
1 2 j
LOOP ANALYSIS
MESH CURRENTS DETERMINED BY SOURCES
I 2 40
I3 2 I x
I3 2( I 4 4)
MESH 1 :
jI1 120 1( I1 I 3 ) 0
MESH 4 :
1( I 4 I 2 ) 1 I 4 j ( I 4 I 3 ) 0
CONTROLLING VARIABLE : I x I 4 I 2
VARIABLE OF INTEREST : V0 1 I 4 (V )
MESH CURRENTS ARE ACCEPTABLE
I 4 4 I 4 j ( I 4 2( I 4 4)) 0
( 2 j ) I 4 ( 4 8 j ) I 4
V0
84j
1 2 j
48 j j
2 j
j
Alternative procedure to compute Thevenin
impedance:
1. Set to zero all INDEPENDENT sources
2. Apply an external probe
THEVENIN
ZTH
I "x
FOR OPEN CIRCUIT VOLTAGE
Vtest
I "x
KVL
Vtest I "x jI "x ZTH 1 j ()
I x' 40
ZTH 1 j
2I x
80
VOC 4 8 j (V )
V0
1
(4 8 j )(V )
2 j
Supernode constraint
V1 V3 12
V1 V3 120
NORTON
KCL@ Supernode
V3 V3 V3 V2 V1 V2
40 0 / j
1
j
1
j
V2 V3 V2 V1
0 / ( j )
1
j
V
Controllin g Variable : I x''' 3
1
KCL@ V2 : 2 I X'''
I SC
2 jV3 j (V2 V3 ) (V2 V3 12) 0
(1 j )V2 (1 3 j )V3 12
(1 j )V3 jV3 jV2 (V3 12) V2 4 j
I x'''
V3
( A)
1
I SC I x''' 4
USE NODES
(1 j )V2 2 jV3 12 4 j
4j
48j
(1 j )V3 4 j V3
I SC
1 j
1 j
I SC
(4 8 j ) j
8 4j
(1 j ) j
1 j
Now we can draw the Norton
Equivalent circuit ...
NORTON’S EQUIVALENT CIRCUIT
ZTH
I SC
V0 (1) I 0 (V )
1 j 8 4 j
(V ) Current Divider
2 j 1 j
EQUIVALENCE OF SOLUTIONS
12 4 j
(8 4 j )(1 j )
3 j
(1 2 j )(1 j )
Using Norton’s method
V0
Using Thevenin’s
48j j
V0
j
2 j
Using Node and Loop methods V0
84j
1 2 j
LEARNING EXTENSION
COMPUTE V0
USE THEVENIN
USE NODAL ANALYSIS
V1
4j
VOC
V1 1230 V1 V1 V1 V0
0 / 2 j
2
1 j2
j
3 4 j 4 j (2 6 j )
2 j2 2 6 j
40
3
1 || j 2
j2
1230
1230
2 (1 || j 2)
2(1 2 j ) 2 j
24120 12120
26j
1 3 j
ZTH 2 || 1 || j 2
VOC
V0 V1 V0
0 V1 (1 j )V0
j
1
ZTH
j (V1 1230) 2 jV1 V1 2(V1 V0 ) 0
2V0 (1 2 2 j j )(1 j )V0 j1230
(2 (1 3 j )(1 j ))V0 190 1230
12120 12120
V0
2.1275(V )
44j
5.6645
j1
VOC
1
+
-
V0
V0
ZTH
1
VOC
1 j
LEARNING EXTENSION COMPUTE V0 USING MESH ANALYSIS
V1
USING NODES
V1 240
V
290 1 0
2
22j
2
V0
V1
22j
USING SOURCE SUPERPOSITION
CONSTRAINT
I1 I 2 290 I1 I 2 2 j
SUPERMESH
240 2 I1 2 jI 2 2 I 2 0
2( I 2 2 j ) (2 2 j ) I 2 24 (4 2 j ) I 2 24 4 j
V0 2 I 2
24 4 j
24.339.46
10.8636.03
2 j
2.24 26.57
V0V
2
240
222j
V0I 2
2
290
42j
V0 V0V V0I
LEARNING EXTENSION
COMPUTE V0
2V
V0"
V0 V0' V0"
1. USING SUPERPOSITION
2 || (2 2 j )
V2
V1
V1
2 || (2 2 j )
(120)
j 2 (2 || 2 2 j )
V0'
V0'
2
V1
22j
(2 j ) || (2 2 j )
240
2 (2 j || (2 2 j )
V0"
2
V2
22j
2. USE SOURCE TRANSFORMATION
2j
Z
I eq
I1
2 V0
I eq 120 6 90 12 6 j
I1
V0 2I1
j 2
120 2
I1 2
j 2 6 90
Z 2 || j 2
Z
I eq
Z 22j
V0
USE NORTON’S THEOREM
j2
I1
I SC
ZTH 2 || j 2
ZTH
I1
2
ZTH
I SC
ZTH 2 2 j
V0 2I1
120
6 90
I SC
V0
USING MATLAB
MATLAB recognizes complex numbers
in rectangular representation. It does
NOT recognize Phasors
Unless previously re-defined, MATLAB
recognizes “i” or “j” as imaginary units
» z2=3+4j
z2 =
3.0000 + 4.0000i
» z1=4+6i
Phasors Rectangula r z 1045
» a=45; % angle in degrees
» ar=a*pi/180, %convert degrees to radians
ar =
0.7854
» m=10; %define magnitude
» x=m*cos(ar); %real part
x =
7.0711
» y=m*sin(ar); %imaginary part
y =
7.0711
» z=x+i*y
z =
7.0711 + 7.0711i
z = 7.0711 + 7.0711i;
» mp=abs(z); %compute magnitude
z1 =
mp =
10
4.0000 + 6.0000i
» arr=angle(z); %compute angle in RADIANS
In its output MATLAB always uses “i” for arr =
0.7854
the imaginary unit
» adeg=arr*180/pi; %convert to degres
adeg =
45
x=real(z)
x=
7.0711
y=imag(z)
y=
7.70711
LEARNING EXAMPLE
COMPUTE ALL NODE VOLTAGES
V1 1230
V1 (1 j 0.5 j )V2 jV3 0.5 jV5 0
jV2 ( j 0.5 1)V3 0.5V5 0
0.5V1 (0.5 1 j )V4 V5 0
0.5 jV2 0.5V3 V4 (1 0.5 j 0.5 j )V5 245
V1 1230
0
0
0 0
1
V1 1230
1 1 j 0.5
V
j1
0 j 0.5
0
2
j1 1.5 j1
0 0.5
0
V3 0
0.5
0
0
1.5 j1
1 V4 0
2
45
0
j
0
.
5
0
.
5
1
1
.
5
j
0
.
5
V
5
V2 V1 V2 V3 V2 V5
0
1
j1
j2
V3 V2 V3 V5 V3
0
j1
2
1
V4 V1 V4 V5 V4
0
2
1
j1
V V4 V5 V2 V5 V3 V5
245 5
0
1
j2
2
j1
YV I R
V Y 1I R
%example7p17
%define the RHS vector.
ir=zeros(5,1); %initialize and define non zero values
ir(1)=12*cos(30*pi/180)+j*12*sin(30*pi/180);
ir(5)=2*cos(pi/4)+j*2*sin(pi/4), %echo the vector
%now define the matrix
y=[1,0,0,0,0; %first row
-1,1+0.5j,-j,0,0.5j; %second row
0,-j,1.5+j,0,-0.5; %third row
-0.5,0,0,1.5+j,-1; %fourth row
0,0.5i,-0.5,-1,1.5+0.5i] %last row and do echo
v=y\ir %solve equations and echo the answer
ir =
10.3923 + 6.0000i
0
0
Echo of RHS
0
1.4142 + 1.4142i
y =
Columns 1 through 4
1.0000
0
0
-1.0000
1.0000 + 0.5000i
0 - 1.0000i
0
0 - 1.0000i
1.5000 + 1.0000i
-0.5000
0
0
0
0 + 0.5000i -0.5000
Column 5
0
Echo of Matrix
0 + 0.5000i
-0.5000
-1.0000
1.5000 + 0.5000i
Echo of Answer
v =
10.3923
7.0766
1.4038
3.7661
3.4151
+
+
+
-
6.0000i
2.1580i
2.5561i
2.9621i
3.6771i
0
0
0
1.5000 + 1.0000i
-1.0000
AC PSPICE ANALYSIS
Circuit ready to be simulated
Select and place components
L
R
Ground set, meters specified
VAC
C
Wire and set correct attributes
****
AC ANALYSIS
TEMPERATURE =
27.000 DEG C
************************************************************************
******
Results in output file
FREQ
VM($N_0003) VP($N_0003)
6.000E+01
2.651E+00
-3.854E+01
**** 05/20/01 09:03:41 *********** Evaluation PSpice (Nov 1999)
**************
* C:\ECEWork\IrwinPPT\ACSteadyStateAnalysis\Sec7p9Demo.sch
****
AC ANALYSIS
TEMPERATURE =
27.000 DEG C
************************************************************************
******
FREQ
6.000E+01
IM(V_PRINT2)IP(V_PRINT2)
2.998E-03
5.146E+01