Phasor Analysis
Phasors, Impedance, SPICE, and
Circuit Analysis
Kevin D. Donohue, University of Kentucky
1
Impedance
The conversion of resistive, inductive, and capacitive elements
to impedance for a sinusoidal excitation at frequency ω is
given by:
X L = ωL (Reactance)
1
(Reactance)
ωC
R = R (Resistance)
XC = −
In general impedance is a complex quantity with a resistive
component (real) and a reactive component (imaginary):
2
2
−1  X 
ˆ
Z = R + jX = R + X ∠ tan  
R
Kevin D. Donohue, University of Kentucky
2
Phasors
Sources can be converted to phasor notation as follows:
A cos(ωt + θ ) ⇔ A∠θ
A sin(ωt + θ ) ⇔ A∠θ − 90 o
This can be applied to all sources of the same frequency,
where ω is used in the impedance conversion of the circuit.
If sources of different frequencies exist, superposition must
applied to solve for a given voltage or current:
1. Select sources with a common ω and deactive all other sources.
2. Convert circuit elements to impedances.
3. Solve for desired voltage or current for selected ω.
4. Repeat steps 1 through 3 for new ω until all sources have been applied.
5. Add together all time-domain solutions solutions obtain in Step 3.
Kevin D. Donohue, University of Kentucky
3
Loop Analysis Example
Determine the steady-state response for vc(t) when vs(t) =
5cos(800πt) V
+ vc(t) 3 kΩ
vs(t)
114.86 nF
6 kΩ
Show:
π

Vˆc = 2.5000 - j1.4434 = 2.8868∠ − 30o ⇔ vc (t ) = 2.8868 cos 800πt −  V
6

Kevin D. Donohue, University of Kentucky
4
Nodal Analysis Example
Find the steady-state value of vo(t) in the
circuit below, if vs(t) = 20cos(4t):
10 Ω
vs
0.1 F
ix
1H
2 ix
+
vo
0.5 H
-
Show: v0(t) = 13.91cos(4t + 198.3º)
Kevin D. Donohue, University of Kentucky
5
Multiple Source Example
Find io if is = 3cos(10t) and vs = 6cos(20t + 60º)
io
5Ω
vs
0.01 F
0.5 H
is
3
1
Amps
Show io = 0.54cos(20t+123.4º)+2.7cos(10t-153.4º)
2
0
-1
-2
Kevin D. Donohue, University of Kentucky
0
0.2
0.4
0.6
S e conds
0.8
1
1.2
6
Equivalent Circuit Example
Find io steady-state using Norton’s Theorem, if
vs(t) = 2sin(10t):
10 Ω
vs
io
0.01 F
5Ω
.4 H
Show is(t)= .2sin(10t); Zth = 3-j = 3.2∠-18.4º;
io = 0.15cos(10t-153.4 º)
Kevin D. Donohue, University of Kentucky
7
Equivalent Circuit Example
Find vo steady-state using Thévenin’s Theorem, if
vs(t) = 20cos(4t):
10 Ω
vs
0.1 F
Iˆsc = 6.1∠33.7°
Vˆoc = 10.67∠ − 104°
ix
1H
2 ix
+
vo
0.5 H
-
Zˆ th = 1.75∠137.7 o
vo (t ) = 13.04 cos(4t − 161.56 o )
Kevin D. Donohue, University of Kentucky
8