[4.1]
4. SINOSOIDAL STEADY - STATE ANALYSIS
4.1 Sinusoids
A graphical approach may be used to relate or compare
sinusoids as an alternative to using the trigonometric
identities. (See Fig. 4.1).
- Note that
cos(ωt – 90o) = sin(ωt)
sin(ωt – 180o) = – sin(ωt)
also
A cos(ωt) + B sin (ωt) = C cos(ωt − θ)
Where
B
C = A 2 + B 2 , θ = tan −1
(See Fig. 4.2)
A
Example 4.1: Add 3cos(ωt) and – 4sin(ωt).
Solution: 3cos(ωt) – 4sin(ωt) = 5cos(ωt+53.1o)
(See Fig. 4.3)
Example 4.2: Find the amplitude, phase, period, and
frequency of the sinusoid v(t) = 12 cos(50t + 10o)
Solution:
The amplitude is V m = 12 V.
The phase is φ = 10◦.
The angular frequency is ω = 50 rad/s.
The period T =2π/ω =2π/50 = 0.1257 s.
The frequency is f =1/T = 7.958 Hz.
Figure 4.1
Figure 4.2
Figure 4.3
4.2 Phasors
A phasor is a complex number that represents the amplitude and phase of a
sinusoid. The idea of phasor representation is based on Euler’s identity. In general,
e±jφ = cos φ ± j sin φ
(4.1)
we may write
cos φ = Re(ejφ)
sin φ = Im(ejφ)
Given a sinusoid v(t) = V m cos(ωt + φ), we use Eq. (4.2a) to express v(t) as
Dr. Ahmed Abdul-Kadhim
(4.2a)
(4.2b)
[4.2]
v(t) = V m cos(ωt + φ) = Re(V m e
or
j (ωt+φ)
)
v(t) = Re(V m ejφ ejωt )
Thus,
v(t) = Re(Vejωt )
(4.3)
where
(4.4)
V = V m ej φ =
V is thus the phasor representation of the sinusoid v(t). The plot of the sinor Vejωt
on the complex plane is shown in Fig. 4.4. As time increases, the sinor rotates on a
circle of radius V m at an angular velocity ω in the counterclockwise direction.
Figure 4.4
Since a phasor V has magnitude and phase (“direction”), it behaves as a vector. For
example, phasors V =
are graphically represented in Fig.
4.5. Such a graphical representation of phasors is known as a phasor diagram.
From Eqs. (4.3) and (4.4),
v(t) = Re(Vejωt ) = V m cos (ω t+ φ),
so that
dv
= −ωV m sin(ωt+φ)=ωV m cos(ωt+φ + 90◦)
dt
= Re(ωV m ejωt ejφ e j 90 ) = Re(jωVejωt )
o
This shows that the derivative v(t) is
transformed to the phasor (frequency)
domain as jωV
Figure 4.5
Dr. Ahmed Abdul-Kadhim
[4.3]
Similarly, the integral of v(t) is transformed to the phasor domain as V/jω
Example 4.3:
Using the phasor approach, determine the current i(t) in a circuit described by the
integrodifferential equation
Solution:
We transform each term in the equation from time domain to phasor domain.
But ω = 2, so
Converting this to the time domain,
Keep in mind that this is only the steady-state solution, and it does not require
knowing the initial values.
4.3 Phasor Relationships for Circuit Elements
- The Resistor R:
Let i(t) = I m cos(ωt + φ), the voltage is given by Ohm’s law as;
V(t) = i(t)R = RI m cos(ωt + φ)
The phasor form of this voltage is
Dr. Ahmed Abdul-Kadhim
[4.4]
But the phasor representation of the current is
Hence, V = RI
This means that the voltage and current are in phase, as illustrated in the phasor
diagram in Fig. 4.6.
- The Inductor L:
For the inductor L, assume the current through it is
i(t) =I m cos(ωt + φ). The voltage across the
inductor is
Recall that −sinA = cos(A + 90◦). We can write the
voltage as
v(t) = ωLI m cos(ωt + φ+ 90◦)
Figure 4.6
which transforms to the phasor
But
and
. Thus V = jωLI
The current lags the voltage by 90o. Fig. 4.7 shows the phasor diagram.
- The Capacitor C:
For the capacitor C, assume the voltage
across it is v(t)=V m cos(ωt + φ). The
current through the capacitor is
Therefore,
The current leads the voltage by 90◦. Fig.
4.8 gives the phasor diagram.
Figure 4.7
Table 4.1 summarizes the time-domain and phasordomain representations of the circuit elements.
Figure 4.8
Dr. Ahmed Abdul-Kadhim