Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

11.3.5 
Zero-state response of parallel 
RC
 circuit for sinusoidal Input
The zero-state response for sinusoidal input for any linear time-invariant circuit can be obtained in 
three ways.
1. Let the input be sin
w
t u(t). Obtain the zero-state response of the circuit for a complex exponential 
input function e
st
, where s is a complex number. Substitute s 
= 
j
w
in the solution and accept the 
imaginary part of the solution as the zero-state response for sin
w
t u(t) .
2. Express sin (
w
t) as [(e
j
w
t
-
 e
-
 
j
w
t
)/2j] by using Euler’s formula, get the zero-state responses for the 
two exponential functions separately and use superposition principle.
3. Use phasor method to obtain the sinusoidal steady-state response and add a transient response 
term such that the total response satisfies initial conditions. Initial conditions will be zero valued 
since we are dealing with zero-state response.
The first two methods were already illustrated in the context of sinusoidal response of RL 
circuits in Chapter 10. We use the third method here to obtain the zero-state response for the voltage 

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11.14


First-Order
RC
Circuits
appearing across a Parallel RC Circuit excited by a sinusoidal current source with source function of 
sin
w
t u(t) A.
The circuit in time-domain and phasor-domain are shown in of Fig. 11.3-12 (a) and (b), respectively.
(b)
+
–
1
R
V
0
(
j
ω
)
C
j
ω
1
2
0 A
(a)
+
–
sin 
ω
t
u
(
t
)
i
S
(
t
)
R
CV
0
(
t
)
Fig. 11.3-12 
Parallel
RC
circuitwithsinusoidalexcitationanditsphasorequivalent
V j
R
j C
R
j RC
R
jk
k
RC
R
o
(
)
/ /
;
w
w
wt
=
( )
×
∠ =
+
×
∠ =
+
×
∠
=
=
=
+
1
1
2
0
1
1
2
0
1
1
2
0
1
w
w
kk
k
2
1
1
2
0
∠ − ×
∠
=
−
f
f
;
tan
Going back to time-domain, we get the steady-state component of voltage across the Parallel RC 
Circuit as
R
k
t
k
k
1
2
1
+
−
=
−
sin(
tan
)
w
wt
with 
.
Now we add a transient component of known form Ae
-
 
t
 
/
 
t
and evaluate A such that the total 
solution is zero at t 
=
0
+
. We get,
A
R
k
k
=
+
−
1
2
1
sin(tan
)
Since tan
lies in first quadrant ,
1
−
−
=
+