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7.2.2 
steady-state solution to 
e 
j
v
 t
 and the 
j
v
 operator
Both the methods based on complex exponential function will be meaningful only if finding the 
steady-state response to complex exponential function is simpler and more elegant than finding the 
steady-state response to sinusoidal input straightaway. We show in the following paragraphs that this 
is true.
Consider the describing differential equation for the second mesh current in Fig. 7. 1-1 (b) with a unit 
amplitude complex exponential function driving the circuit. The equation is 
d i
dt
di
dt
i
e
j t
2
2
2
2
2
3
+
+ =
w
. 
This equation has to be true for all t and that will be possible only if both the sides of the equation 
have same waveshape in time. Therefore, the particular integral has to be chosen in such a way that, 
on substituting the trail solution in the differential equation, the left-hand side yields (.)e 
j
w
 t
. But this 
will imply that 
d i
dt
di
dt
2
2
2
2
,
and i
2
must have e 
j
w
t
in them. [In fact, there are special situations in circuits 
under which this is not strictly true. But we leave special cases to later chapters that deal extensively 
with time-domain analysis of circuits.] Therefore, we look for some function that has e 
j
w
t
in its zeroth-
order, first-order and second-order derivatives. The only function that comes to our mind is e 
j
w
t
itself. 
Therefore, the trial solution for particular integral is A e 
j
w
t
, where A can now be a complex number.
Substituting this solution in the differential equation, we get,

---

The Complex Exponential Forcing Function 
7.9
(
) [
]
(
)[
] [
]
j
Ae
j
Ae
Ae
e
j t
j t
j t
j t
w
w
w
w
w
w
2
3
+
+
=
Since this has to be true for all t, we cancel out e 
j
w
t
and get 
[(
)
(
)
]
.
j
j
A
w
w
2
3
1
1
+
+
=
The solution 
is completed by solving for A.
i
j
e
j
e
e
j t
j
j t
2
2
2 2
2
1
1
3
1
1
9
=
−
+
=
−
+








=
(
)
(
)
,
ta
w
w
w
w
f
w
w
f
where 
nn
−
−
1
2
3
1
w
w
We note that each differentiation in time has got replaced by a multiplication by j
w
in the equation 
determining the complex amplitude in the assumed particular solution. Once we appreciate that 
point, we can straightaway obtain the equation governing the complex amplitude A of the steady-state 
response to a complex exponential input by replacing every differentiation in the differential equation 
by a multiplication by j
w
and solve for A easily.
Let us generalise this. We consider a linear circuit with one sinusoidal source at angular frequency 
of 
w
driving it. If there are more sources, we employ superposition principle and solve many single-
source circuits. Hence, the basic problem is to solve the circuit for a single source. The most general 
differential equation governing a chosen response variable for a linear circuit is
d y
dt
a
d
y
dt
a
dy
dt
a y
b
d x
dt
b
d
x
dt
n
n
n
n
n
m
m
m
m
+
+ +
+
=
+
−
−
−
−
−
−
1
1
1
1
0
1
1
1
m
m
++ +
+
b
dx
dt
b x
1
0
(7.2-4)
where y is the chosen response variable, x is the input function and a’s and b’s are real positive 
constants decided by circuit parameters. The order of differential equation, n, is equal to the number 
of independent energy storage elements in the circuit.
If x 
=
e 
j
w
t
, then y 
=
A e 
j
w
 t
, where A is a complex amplitude decided by the equation
A j
a
j
a j
a
b
j
b
j
b
n
n
n
m
m
m
[(
)
(
)
(
)
]
(
)
(
)
(
w
w
w
w
w
+
+ +
+
=
+
+ +
−
−
−
−
1
1
1
0
1
1
1
m
jj
b
w
)
.
+
0
Therefore, the complex amplitude of steady-state response to a unit amplitude complex exponential 
input comes out as a ratio of rational polynomials in the variable j
w
. The desired steady-state response 
is obtained as 
y
b j
j
a j
e
k
k
k
m
n
i
i
i
n
j t
=
+
=
=
−
∑
∑
(
)
(
)
(
)
.
w
w
w
w
0
0
1
Hence, solving for steady-state response to complex exponential function is much simpler and 
more elegant than solving for steady-state response to cosine or sine input functions. Therefore, 
trying to obtain sinusoidal steady-state response indirectly from steady-state response to complex 
exponential input function is worthwhile.
The method based on Eqn. 7.2-2 and superposition principle – we called it the first method – is the 
Electronics and Communication Engineer’s favourite. Electrical Power Engineers usually prefer the 
second method that is based on Eqn. 7.2-3 and superposition principle. Both lead to the same result 
of course. But, the analysis of sinusoidal steady-state response from the so-called phasor concept 
evolves from the second method. Hence, we take up the second method for detailed discussion.
In this method, we obtain sinusoidal steady-state response for cos
w
 t by obtaining the real part of 

---

7.10
The Sinusoidal Steady-State Response
steady-state response to e 
j
w
 t
input. The sinusoidal steady-state response for sin
w
 t input is similarly 
the imaginary part of steady-state response to e 
j
w
 t
input. A separate determination of response for 
cosine and sine input is unnecessary. sin
w
 t is 90
°
phase delayed version
of cos
w
 t and therefore it must 
be possible to obtain the steady-state response for sin
w
 t input by delaying the steady-state response 
for cos
w
 t input by 90
°
. We will verify this in later sections.