AC Steady-State Analysis
Sinusoidal Forcing Functions,
Phasors, and Impedance
Kevin D. Donohue, University of Kentucky
1
The Sinusoidal Function
Terms for describing sinusoids:
x(t ) = X m sin(ωt + θ ) = X m sin(2πft + θ )
Maximum Value,
Amplitude, or
Magnitude
Phase
Frequency
in cycles/second or
Hertz (Hz)
Radian Frequency
in Radian/second
1
1
0.4π
0.8
sin( 2πt )
0.4
0.4
0.2
0.2
0
2π
sin 2πt +
5
-0.2
-0.4
-0.6
0
-0.2
-0.4
-0.6
-0.8
-0.8
-1
-1
-1.2566
-0.2566
0.7434
1.7434
2.7434
Radians
3.7434
4.7434
.2
0.6
Amplitude
Amplitude
0.6
0.8
-0.4
-0.2
Kevin D. Donohue, University of Kentucky
0
0.2
0.4
0.6
0.8
Seconds
1
1.2
1.4
1.6
2
Trigonometric Identities
π
cos(ω t ) = sin ωt +
2
(
o
− cos(ωt ) = cos ωt ± π ( or 180 )
π
sin(ωt ) = cos ω t −
2
(
)
− sin(ω t ) = sin ω t ± π (or 180 o )
)
sin(α ± β ) = sin(α ) cos( β ) ± cos(α ) sin( β )
Radian to degree conversion
cos(α ± β ) = cos(α ) cos( β ) m sin(α ) sin( β )
Degree to radian conversion
multiply by π/180
multiply by 180/π
X m sin (ω t ± θ ) = X m cos(θ ) sin(ω t ) ± X m sin(θ ) cos(ω t )
−1 − B
A cos(ω t ) + B sin(ω t ) = A + B cos ω t + tan
A
2
2
Kevin D. Donohue, University of Kentucky
3
Sinusoidal Forcing Functions
Determine the forced response for io(t) the circuit below
with vs(t) = 50cos(1250t):
io(t)
10 Ω
vs(t)
+
0.1 mF
vc(t)
40 Ω
-
Note:
20 mvs = 0.8m
Show:
dio
+ io
dt
io (t ) =
1
(sin(1250t ) + cos(1250t ) ) = 1 cos1250t − π
2
4
2
Kevin D. Donohue, University of Kentucky
4
Complex Numbers
Each point in the complex number plane can be
represented by in a Cartesian or polar format.
a + jb = r exp( jθ ) = r∠θ
IM
θ
r = a 2 + b2
a
r
RE
b
b
−
1
θ = tan
a
a = r cos(θ )
b = r sin(θ )
Kevin D. Donohue, University of Kentucky
5
Complex Arithmetic
Addition:
(a + jb ) + (c + jd ) = (a + c) + j (b + d )
Multiplication and Division:
(r∠θ )(v∠φ ) = rv∠(θ + φ )
r∠θ r
= ∠(θ − φ )
v∠φ v
Simple conversions:
j = 1∠90 o ,
1
= − j,
j
-1 = ∠180 o
Kevin D. Donohue, University of Kentucky
6
Euler’s Formula
Show: exp( jθ ) = cos(θ ) + j sin(θ )
A series expansion ….
jθ θ 2 jθ 3 θ 4 jθ 5
exp( jθ ) = 1 +
−
−
+
+
−L
1! 2!
3!
4!
5!
θ2 θ4 θ6
cos(θ ) = 1 −
+
−
+L
2! 4! 6!
θ θ3 θ5 θ7
sin(θ ) = −
+
−
+L
1! 3! 5! 7!
Kevin D. Donohue, University of Kentucky
7
Complex Forcing Function
Consider a sinusoidal forcing function given as a complex function:
X m exp( j (ωt + θ )) = X m cos(ωt + θ ) + jX m sin(ωt + θ )
Ø
Ø
Ø
Based on a signal and system’s concept (orthogonality), it can be shown that
for a linear system, the real part of the forcing function only affects the real
part of the response and the imaginary part of the forcing function only affect
the imaginary part of the response.
For a linear system excited by a sinusoidal function, the steady-state response
everywhere in the circuit will have the same frequency. Only the magnitude
and phase of the response will vary.
A useful factorization:
X m exp( j (ωt + θ )) = X m exp( j θ ) exp( jωt )
Kevin D. Donohue, University of Kentucky
8
Complex Forcing Function Example
Determine the forced response for io(t) the circuit below
with vs(t) = 50exp(j1250t):
io(t)
10 Ω
vs(t)
Note:
di
20 mvs = 0.8m o + io
dt
0.1 mF
40 Ω
Show:
π
exp j1250t −
4
2
1
π
Re[io (t )] =
cos1250t −
4
2
i o (t ) =
1
Kevin D. Donohue, University of Kentucky
Im[ io (t )] =
π
sin 1250t −
4
2
1
9
Phasors
Notation for sinusoidal functions in a circuit can be more
efficient if the exp(-jωt) is dropped and just the magnitude
and phase maintained via phasor notation:
Time Domain
x(t ) = A cos(ωt + θ )
x(t ) = A sin(ωt + θ )
Frequency Domain
X̂ = A∠θ
o
ˆ
X = A∠ θ − 90
Kevin D. Donohue, University of Kentucky
(
)
10
Impedance
The ω affects the resistive force that inductors and
capacitors have on the currents and voltages in the circuit.
This generalized resistance, which affects both amplitude
and phase of the sinusoid, will be called impedance.
Impedance is a complex function of ω.
ˆ
Given: V = Av exp( j (ωt + θ v ))
Iˆ = AI exp( j (ωt + θ I ))
using passive sign convention show:
For inductor relation Vˆ = ZˆIˆ :
Show
Zˆ = jωL
For capacitor relation Vˆ = ZˆIˆ :
Show
Kevin D. Donohue, University of Kentucky
Zˆ =
1
jωC
11
Finding Equivalent Impedance
Given a circuit to be analyzed for AC steady-state
behavior, all inductors and capacitors can be converted to
impedances and combined together as if they were
resistors.
2H
Find
Z eq ⇒
0.5F
0.1H
50Ω
Find
Z eq ⇒
5Ω
0.01F
10Ω
Show Zeq = .2039∠79.6°
Kevin D. Donohue, University of Kentucky
12
Impedance Circuit Example
Find the AC steady-state value for v1(t):
1µF
1H
2 sin(500t ) mA
+
v1
-
1kΩ
1µF
Kevin D. Donohue, University of Kentucky
13
Impedance Circuit Example
Find AC steady-state response for io(t):
3H
io
20 cos(10t ) V
60Ω
4H
25mF
80Ω
Kevin D. Donohue, University of Kentucky
14
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