7.6
The Sinusoidal Steady-State Response
The total solution
=
v
Ae
t
t
t
C
for
=
+
+
−
>
−
−
1
1
0
2
1
w
w
w
cos(
tan
)
.
This solution must approach
zero value as
t
→
0 from right side since the initial value of voltage across capacitor is zero. Therefore,
0
1
1
0
2
1
2
1
=
+
+
−
⇒ = −
+
−
−
−
−
Ae
V
A
V
m
m
w
w
w
w
cos( tan
)
cos( tan
)
∴
=
+
−
−
+
−
−
−
−
v
V
t
V
e
m
m
t
C
V
É
2
1
2
1
1
1
cos(
tan
)
cos( tan
)
w
w
w
w
Once again we see that the total response contains a transient term that vanishes as
t
→
∞
and a
steady-state term that persists.
The steady-state term is a sinusoidal waveform of same frequency as
that of input sinusoid. But it has a phase difference with respect to input sinusoid. It is a phase lag.
Both the amplitude and phase of steady-state response component depend on the angular frequency
w
of input sinusoid. The steady-state response is given by particular integral and the transient response
is contributed by complementary function.
We can proceed the same way in the case of Fig. 7.1-1 (b) too. We solve only for the particular
integral of differential equation this time since we know that the transient response will vanish in
this circuit. We try out the solution
i
2
=
a sin
w
t
+
b cos
w
t in
d i
dt
di
dt
i
V
t
2
2
2
2
2
3
+
+ =
m
cos
w
to find the
sinusoidal steady-state solution for second mesh current in the circuit when the circuit is driven by
V
m
cos
w
t V. Substituting the trail solution in the differential equation and collecting terms, we get,
(
) cos
(
)sin
cos
−
+
+
+ −
−
+
=
w
w
w
w
w
w
w
2
2
3
3
b
a b
t
a
b a
t V
t
m
Equating
coefficients of sin
w
t and cos
w
t on both sides of the equation, we get
(
)
(
)
−
+
+ =
−
−
+
=
w
w
w
w
2
2
3
3
0
b
a b
V
a
b a
m
and
Solving for
a and
b we get
a
V
b
V
=
−
+
=
−
−
+
3
1
9
1
1
9
2 2
2
2
2 2
2
w
w
w
w
w
w
m
m
and
(
)
(
)
(
)
Therefore, the sinusoidal
steady-state response for i
2
in Fig. 7.1-1 is
i t
V
t
V
t
V
2
2 2
2
2
2 2
2
3
1
9
1
1
9
1
( )
(
)
sin
(
)
(
)
cos
(
=
−
+
+
−
−
+
=
−
w
w
w
w
w
w
w
w
w
m
m
m
22 2
2
1
2
9
3
1
)
cos
tan
+
−
−
−
w
w
w
w
t
We observe once again that the response is a pure sinusoidal waveform at same frequency as that
of input sinusoid. But the response has a phase lag with respect to the input. Both the amplitude and
phase of response depend on the angular frequency
w
of the input sinusoid.
The steady-state response of a circuit variable in a linear dynamic circuit under sinusoidal
excitation is a sinusoidal waveform of same frequency as that of input. The response will,
in general, have a phase difference with respect to input.
The amplitude and phase of
response under steady-state condition will depend on the amplitude of input and the
angular frequency of input sinusoid.
The Complex Exponential Forcing Function
7.7
7.2
the complex exponentIal ForcInG FunctIon
The method outlined in the previous section to determine the sinusoidal steady-state response of a
dynamic circuit can be summarized as follows:
1. Use mesh or nodal analysis to obtain integro-differential equations of the circuit.
2. Differentiate equations
again to eliminate integrals, if needed.
3. Choose one of the mesh currents (or node voltages) as the describing variable for the circuit.
Eliminate all the other variables and obtain an
n
th
order linear constant-coefficient differential
equation describing the chosen circuit variable.
4. Assume solution in the form (
a sin
w
t
+
b cos
w
t). Substitute the assumed solution in the differential
equation. Equate coefficients of cosine and sine on both sides of the equation. Solve for
a and
b
using the resulting equations.
5. Express the solution in the form of a single sinusoid with phase shift.
6. Find the other mesh currents (or node voltages) which were eliminated earlier by using the
elimination equations in the reverse. Once all the mesh currents (node voltages) are available, any
element variable can be obtained from them.
There is nothing wrong with this method – except that it is going to be very tedious if the circuit
contains more than two energy storage elements.
Hence, we look for another simpler and more elegant method to obtain sinusoidal steady-state.
Euler’s identity, which relates a complex exponential time-function to trigonometric time-functions,
is the key to this new method.
Euler’s
Identity
e
j
j
q
q
q
=
+
cos
sin .
(7.2-1)
By letting
q
=
w
t and using Euler’s Identity, we can express
e
j
w
t
as
e
t
j
t
j t
w
w
w
=
+
cos
sin
and by
letting
q
=
-
w
t and using Euler’s Identity, we can express
e
-
j
w
t
as
e
t
j
t
j t
−
=
−
w
w
w
cos
sin
. Therefore,
cos
sin
w
w
w
w
w
w
t
e
e
t
e
e
j
j t
j t
j t
j t
=
+
=
−
−
−
2
2
and
(7.2-2)
e
j
w
t
is the
complex exponential function of unit amplitude. Equation. 7.2-2 expresses unit amplitude
cosine and sine functions of time with an angular frequency of
w
in terms of two complex exponential
functions of time with
w
and
-
w
in the indices of exponential functions. We can also express sine and
cosine functions in terms of complex exponential function in another way too as in Eqn. 7.2-3.
cos
Re[
]
sin
Im[
]
w
w
w
w
t
e
t
e
j t
j t
=
=
and
(7.2-3)
Do'stlaringiz bilan baham: 
