ECE 3336
Introduction to Circuits & Electronics
Note Set #10
Phasors Analysis
Fall 2012,
TUE&TH 4:00-5:30 pm
Dr. Wanda Wosik
1
Review of Phasor Analysis and Circuit Solutions
A phasor is a transformation of a sinusoidal voltage or current.
Using phasor analysis, we can solve for the steady-state solution for
circuits that have sinusoidal sources.
It means that the frequency will be the same so we only need to find the
amplitude and phase.
All previous techniques used for DC conditions will be applicable to
phasors:
•
Ohms law
•
Kirchhoff laws
•
Node Voltage Method
•
Mesh Currents
•
Thevenin and Norton equivalents.
2
Table of Phasor Transforms
The phasor transforms can be summarized to be used in circuits analyzes. We
will use “static” phasors so the frequency will be remembered for V and I.
Do not mix the domains! Notation important! Impedance for passive elements!
Component
Value
Transform
(time domain – NO j’s)
(phasor domain – NO t’s)
Voltages
vX (t ) = Vm cos(wt + fv )
Vx ( jw) = VmÐfv )
Currents
iX (t ) = I m cos(wt + fi )
Ix ( jw) = ImÐfi )
Resistors
RX
Z RX = RX
Inductors
LX
Z LX = jw LX
Capacitors
CX
ZCX = 1
jw C X
3
Graphical Correlation Between
Time Dependent Signals and Their Phasors

Rotation of the phasor (voltage vector)
V with the angular frequency 
In general the vector’s
length is r (it is the
amplitude) so
V=a+jb in the
rectangular form:
V=rcos(t+)+jrsin(t+)
At t=0
t=0
Or in the polar form:
V=rejr/ 
Corresponds to the time
dependent voltage changes
4
Impedances Represented by Complex Numbers
Current lagging
voltage by 90°
Current leading
voltage by 90°
5
Inductance
Current lagging
voltage by 90°
Capacitance
Current leading
voltage by 90°
For resistance R both vectors VR(jt) and IR (jt) are the same and there is
no phase shift!
6
Solve Circuits in the Phasor Domain
Use Known Methods
Replace all elements defined in the time
domain v(t), i(t), R, C, and L by the
corresponding elements in the frequency
domain V(j), I(j), R, ZC, and ZL.
7
Equivalent Circuits; Thevenin and Norton Equivalent
All elements here are
in the Phasor Domain
8
Use Equivalent Admittances for Parallel Connections
and Impedances for the Series Connections.
9
Previous Example Solution: find i(t)
The circuit here has a
sinusoidal source. What is the
steady state value for the current
i(t)?
R
vS
vS (t ) = Vm cos(wt + f ).
Solution: We use the phasor analysis
technique.
The first step is to transform the
problem into the phasor domain.
Note time does not appear in this
diagram.
i(t)
+
L
-
Phasor Domain diagram.
R
Vsm(w)
+
Im(w)
-
jwL
10
Previous Example Solution
Phasors Used in the Transformed Circuit
Still looking for the steady state
value for the current i(t)?
R
vS
vS (t ) = Vm cos(wt + f ).
i(t)
+
L
-
Phasor Domain diagram.
We replace the phasors with
their complex numbers,
R
Vsm = VmÐf , and
I m = I mÐq ,
where Im and  are the values we
want, specifically, the magnitude
and phase of the current.
I mÐq
Vsm
+
-
VmÐf
Im
jwL
11
Previous Example Solution
Phasors Bring Back Ohm’s Law
R
I mÐq
Vsm
+
Im
VmÐf
-
jwL
Two impedances are in series. We can combine them in the same way
we would combine resistances. We can then write the complex version of
Ohm’s Law,
Vsm
VmÐf
= Im =
= I mÐq .
Z
R + jw L
(
)
Magnitudes
and phases of
both sides
have to be
equal.
where Im and  are the unknowns.
12
Finding Phasor Magnitude and Phase
To find the steady state current i(t)
we inverse transform the current
phasor. We find the amplitude (Im)
and the phase of i(t).
VmÐf
( R + jw L)
R
I mÐq
Vsm
= I mÐq
+
Im
VmÐf
-
jwL
reminder
Z1 | Z1 |
=
Ð(q1 - q 2 )
Z2 | Z2 |
Magnitude
Vm
R 2 + w 2 L2
Phase
= Im.
VmÐf
( jw L + R)
= I mÐq
æ
ö æ
æ
öö
Vm
-1 w L
I = I mÐq = çç
÷÷Ðçf - tan ç
÷.
÷
è R øø
è R2 + w 2 L2 ø è
13
Previous Example Solution
Inverse Transformation of Phasor Gives i(t)
The complete expression for the
steady state value of the current
i(t) will now be calculated as iss(t).
R
vS
vS (t ) = Vm cos(wt + f ).
æ
Vm
I = I mÐq = ç
2
2 2
è R +w L
+
i(t)
L
-
ö æ
-1 æ w L ö ö
÷ Ð ç f - tan ç
÷ ÷.
è R øø
ø è
The inverse phasor transform give us the solution in time domain
æ
Vm
iSS (t ) = ç
2
2 2
R
+
w
L
è
ö æ
-1 æ w L ö ö
÷ cos ç w t + f - tan ç
÷ ÷.
è R øø
ø è
14
Another Example
What is the steady state value for the voltage vX(t)?
L1=10[H]
vS(t)
+
-
R2=2.2[kW]
iX
R1=
1[kW]
C1=
50[mF]
iS=
30 iX
+
C2=
10[mF]
vX(t)
-
vS(t) = 30 cos(50[rad/s] t + 38º)[V]
15
Phasor Transformation of all
Components
ZL1=
500j[W]
ZR2=
2.2[kW]
+
Ix,m
+
-
Vs,m(w)=
30Ð38º[V]
ZR1=
1[kW]
ZC1=
-400j[W]
Is,m=
30 Ix,m
Vx,m(w)
ZC2=
-2j[kW]
-
Phasor Domain Version
All components have been transformed to the phasor domain, including
the current, iX, that the dependent source depends on.
16
ZL1=
500j[W]
Numerical Solution: Use NVM
ZR2=
2.2[kW]
+
+
Va,m
Vs,m(w)=
30Ð38º[V]
ZR1=
1[kW]
-
+
Ix,m
Vx,m(w)
Is,m=
30 Ix,m
ZC1=
-400j[W]
ZC2=
-2j[kW]
-
-
Phasor Domain Version
There are only two essential nodes, so we will write the node-voltage
equations,
Va,m - 30Ð38°[V ]
500 j[W]
+
Va,m
( 2,200 - 2,000 j )[W]
æ V
ö
a,m
÷÷.
I x,m = - çç
è -400 j[W] ø
- 30I x,m +
Va,m
+
Va,m
-400 j[W] 1000[W]
= 0, and
17
Solution in the Phasor Domain
ZL1=
500j[W]
ZR2=
2.2[kW]
+
+
Va,m
Vs,m(w)=
30Ð38º[V]
ZR1=
1[kW]
-
+
Ix,m
ZC1=
-400j[W]
Is,m=
30 Ix,m
-
Vx,m(w)
ZC2=
-2j[kW]
-
Phasor Domain Version
Now, we can substitute Ix,m back into this equation, and we get
Va ,m - 30Ð38°[V ]
500 j[W]
Va ,m
Va ,m
æ Va ,m ö
+
+ 30 ç
+
= 0,
÷+
(2, 200 - 2,000 j )[W] è -400 j[W] ø -400 j[W] 1000[W]
Va ,m
18
Solve Equations
We can solve. We collect terms on each side, and get
æ 1
æ 30 ö
1
1
1 ö 30Ð38°
Va ,m çç
+
-ç
+
,
÷÷ =
÷+
500 j
è 500 j (2, 200 - 2, 000 j ) è 400 j ø -400 j 1000 ø
which can be simplified to
1
æ
ö 30Ð38°
Va ,m ç -0.002 j +
+ (0.075 j ) + 0.0025 j + 0.001÷ =
.
2,973Ð - 42.27°
è
ø 500Ð90°
(
(
)
)
Va,m -0.002 j + 0.000336Ð42.27° + 0.075 j + 0.0025 j + 0.001 = 0.06Ð - 52°.
(
(
)
)
Va,m -0.002 j + 0.000249 + 0.000226 j + 0.075 j + 0.0025 j + 0.001 = 0.06Ð - 52°, or
(
)
Va,m 0.001249 + 0.07573 j = 0.06Ð - 52°.
19
Simplify the Circuit (Still in the Phasor Domain)
Now, we need to solve for Va,m. We get
Va ,m =
0.06Ð - 52°
0.06Ð - 52°
=
= 0.79Ð - 141°[V].
(0.001249 + 0.07573 j ) 0.07574Ð89°
ZL1=
500j[W]
ZR2=
2.2[kW]
+
+
-
Vs,m(w)=
30Ð38º[V]
Va,m
ZR1=
1[kW]
+
Ix,m
ZC1=
-400j[W]
Is,m=
30 Ix,m
Vx,m(w)
ZC2=
-2j[kW]
-
-
Phasor Domain Version
Next, we note that we can get Vx,m from Va,m by using the complex version of the
voltage divider rule, since ZR2 and ZC2 are in series.
20
Find the Complete Phasor
Using the complex version of the voltage divider rule, we have
Vx ,m
-2000 j
2000Ð - 90°
= Va ,m
= (0.79Ð - 141° )
2973Ð - 42.27°
(2200 - 2000 j )
Vx ,m = 0.53Ð - 188.7° = 0.53Ð171.3°[V].
Note: there is no information about the frequency in this expression. But we remember!
ZL1=
500j[W]
ZR2=
2.2[kW]
+
+
-
Vs,m(w)=
30Ð38º[V]
Va,m
ZR1=
1[kW]
ZC1=
-400j[W]
-
Phasor Domain Version
+
Ix,m
Is,m=
30 Ix,m
Vx,m(w)
ZC2=
-2j[kW]
21
Inverse Transformation Gives v(t)
The final step is to inverse transform. We need to remember that the
frequency was 50[rad/s], and we can write,
vx (t) = 0.53cos(50[rad / s]t +171.3°)[V ]
L1=
10[H]
R2=
2.2[kW]
+
iX
vS(t)
+
R1=
1[kW]
vX(t)
C1 =
50[mF]
iS=
30 iX
C2 =
10[mF]
vS(t) = 30 cos(50[rad/s] t + 38º)[V]
22