Sinusoidal Steady
Sinusoidal Steady
State
State
Analysis
Analysis
Chapter 7
Artemio P. Magabo
Professor of Electrical Engineering
Professor of Electrical Engineering
Department
of Electrical and Electronics Engineering
University of the Philippines  Diliman
Department of Electrical and Electronics Engineering
The Sinusoidal Function
α
T
)
t
cos(
F
)
t
(
f
m
α
+
ω
=
where
F
m
= amplitude or peak value
ω
= angular frequency, rad/sec
α
= phase angle at t=0, rad
The sinusoid is described by the expression
ω
t, rad
F
m
F
m
π
2
π
Department of Electrical and Electronics Engineering
The sinusoid is generally plotted in terms of
ω
t,
expressed either in radians or degrees. Consider
the plot of the sinusoidal function f(t)=F
m
cos
ω
t.
F
m
F
m
ω
t, deg
180
360
ω
t, rad
π
2
π
T
When
ω
t=2
π
, t=T. Thus we get
ω
T=2
π
or
ω
=
.
The function may also be written as
T
2
π
t
T
2
cos
F
t
cos
F
)
t
(
f
m
m
π
=
ω
=
Department of Electrical and Electronics Engineering
The peak value of the voltage is V
m
=311 volts. The
angular frequency is
ω
=377 rad/sec. The frequency
is f=60 Hz. The period is T=16.67 msec.
Define:
The
frequency
of the sinusoid
T
1
f
=
sec
1
or cycles/sec or Hertz (Hz)
Then, the sinusoid may also be expressed as
ft
2
cos
F
t
cos
F
)
t
(
f
m
m
π
=
ω
=
Note:
The nominal voltage in the Philippines is a
sinusoid described by
V
)
t
377
cos(
311
)
t
(
v
α
+
=
Department of Electrical and Electronics Engineering
Leading and Lagging Sinusoids
Note:
We say either “f
1
(t)
leads
f
2
(t) by an angle
of 30
°
” or that “f
2
(t)
lags
f
1
(t) by an angle of 30
°
.”
)
30
t
cos(
F
)
t
(
f
m
2
°
+
ω
=
Consider the plot of the sinusoidal functions
and
)
60
t
cos(
F
)
t
(
f
m
1
°
+
ω
=
30
60
90
180
30
60
90
ω
t, deg
270
360
)
t
(
f
1
)
t
(
f
2
Department of Electrical and Electronics Engineering
Note:
The current is in phase with the voltage.
From Ohm’s law, we get
t
cos
RI
Ri
v
m
R
R
ω
=
=
Consider a resistor. Let the
current be described by
t
cos
I
i
m
R
ω
=
R
+

v
R
i
R
90
180
90
ω
t, deg
270
360
v
R
i
R
The Resistor
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The Inductor
Consider an inductor. Let the
current be described by
t
cos
I
i
m
L
ω
=
L
i
L
+

v
L
From v
L
=
, we get
t
sin
LI
v
m
L
ω
ω

=
dt
di
L
L
90
180
90
ω
t, deg
270
360
i
L
v
L
Note:
The current lags the voltage by 90
o
.
Department of Electrical and Electronics Engineering
The Capacitor
i
C
v
C
+

C
Consider a capacitor. Let the
current be described by
t
cos
I
i
m
C
ω
=
From v
C
=
, we get
t
sin
C
I
v
m
C
ω
ω
=
∫
dt
i
C
1
C
90
180
90
ω
t, deg
270
360
i
C
Note:
The current leads the voltage by 90
o
.
v
C
Department of Electrical and Electronics Engineering
Note:
We will show later that:
Summary:
In a resistor, i
R
and v
R
are in phase.
1.
In an inductor, i
L
lags v
L
by 90
o
. In a capacitor,
i
C
leads v
C
by 90
o
(
ELI the ICE man
).
2.
For an RL network, the current
lags
the
voltage by an angle between 0 and 90°.
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 Winter '10
 Co
 Electrical Engineering, Alternating Current, Electronics Engineering, Department of Electrical and Electronics Engineering

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