Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 12.8-2
A series RLC circuit has L 
=
10 mH, C 
=
 100 
m
F and R 
=
10 
W
. The capacitor is initially charged with 
-
10 V and the inductor has 1 A flowing in it at t 
=
0
-
. Find an expression for the voltage across the 
capacitor when the circuit is excited by 10 u(t). 
Solution
This is a problem that involves both zero-input response and zero-state response. The zero-input 
response component arises due to the initial voltage across the capacitor and initial current through 
inductor. The zero-state response component is nothing but step response scaled by 10.
Zero-Input Response Component
Characteristic equation is 
g
2
+
10
3
g
+
10
6
=
0
Natural frequencies of the circuit 
=
-
500 and 
-
500. This is an critically damped circuit.
\
v
C
(t) 
=
A
1
e
-
500 t
+
A
2
t e
-
500 t
and i(t) 
=
10
-
4
×
(
-
500 A
1
e
-
500 t
– 500 A
2
t e
-
500 t 
+
 A
2
e
-
500 t
)
Applying initial conditions of 
-
10 V and 1 A at 0
+
,
A
1 
=
-
10 and 
-
0.05 A
1
+
0.0001 A
2
=
1 
⇒
A
1 
=
-
10 and A
2
=
5000
\
v
C
(t) 
=
-
10 e
-
500 t
+
5000 t e
-
500 t
V for t 
≥
t 
=
0
+
is the zero-input response.
Zero-State Response Component
The initial conditions at t 
=
 0
-
are set to zero values for determining zero-state response. The final 
value of step response in a series RLC circuit is 1 V across the capacitor. Hence the final value of 
response here is 10 V across the capacitor in this example. The format of transient response is known 
from the values of natural frequencies. 
\
v
C
(t) 
=
10 
+
A
1
e
-
500 t
+
A
2
t e
-
500 t
is the format of zero-state 
response for 10 u(t) input.
The initial values of v
C
(t) and i(t) remain at zero at t 
=
0
+
as explained in the previous example. 
Applying these initial values on 
v
C
(t) 
=
10 
+
A
1
e
-
500 t
+
A
2
t e
-
500 t
and i(t) 
=
-
0.05A
1
e
-
500 t
–0.05A
2
t e
-
500 t
+
0.0001A
2
e
-
500 t
),
we get A
1 
=
-
10 and 
-
0.05 A
1
+
0.0001 A
2
=
0 
⇒
A
1 
=
-
10 and A
2
=
-
5000
\
v
C
(t) 
=
(10 – 10 e
-
500 t
– 5000 t e
-
500t
) V for t 
≥
0
+
is the zero-state response.
\
Total response of v
C
(t) 
=
zero-input response 
+
zero-state response
=
-
10 e
-
500 t
+
5000 t e
-
500 t
+
10 – 10 e
-
500 t
– 5000 t e
-
500 t
=
(10 
-
20 e
-
500t
) V for t 
≥
0
+

---

ExamplesonImpulseandStepResponseofSeries
RLC
Circuits

12.25
example: 12.8-3
Develop the differential equation governing v
C
(t) in 
the circuit in Fig. 12.8-1and find its step response 
for L 
=
16 mH, C 
=
10 
m
F and R 
=
100 
W
. Obtain 
the rise time, peak time, settling time and maximum 
percentage overshoot for this response.
Solution
Applying KCL at the junction between R, L and C 
along with the element equation of a capacitor yields i t
C
dv t
dt
v t
R
C
C
( )
( )
( )
.
=
+
Applying KVL in the first mesh yields L
di t
dt
v t
u t
C
( )
( )
( )
+
=
.
Substituting the first equation in the second one results in
LC
d v t
dt
L
R
dv t
dt
v t
u t
i e
d v t
dt
RC
dv t
C
C
C
C
C
2
2
2
2
1
( )
( )
( )
( )
. .,
( )
(
+
+
=
+
))
( )
( )
dt
LC
v t
LC
u t
C
+
=
1
1
The critical resistance of this circuiit is 
1
2
L
C
.
Substituting the parameter values,
d v t
dt
dv t
dt
v t
t
C
C
C
n
2
2
3
2
2
10
2500
2500
0
2500
( )
( )
( )
+
+
=
≥
=
+
for
rad/s
w
,, 
rad/s
2
10
0 2
1
2449 5
3
2
xw
x
w
x w
n
d
n
=
⇒ =
∴
=
−
=
.
.
The final value of step response is obtained by replacing the capacitor by open-circuit and the 
inductor by short-circuit. This shows that the final value is 1 V.
The initial conditions required for solving the differential equation are the values of capacitor 
voltage at t 
=
0
+
and of its first derivative at t 
=
0
+
.
v
v
dv t
dt
i t
C
i t
v t
R
C
i
i
d
C
C
C
C
C
( )
( )
( )
( )
( )
( )
( )
( )
0
0
0
0
0
0
+
−
+
−
=
=
=
=
−
=
=
∴
vv t
dt
C
( )
(
)
0
0
+
=
Fig. 12.8-1 
CircuitforExample:12.8-3
–
+
–
i
(
t
)
u
(
t
)
v
L
(
t
)
v
C
(
t
)
C
R
L
–
+
+

---

12.26


SeriesandParallel
RLC
Circuits
The step response for v
C
(t) will have the following format with A
1 
and A
2
decided by initial 
conditions.
v
C
(t) 
=
1
+
e
-
500 t
(A
1 
cos 2449.5 t 
+
 A
2
sin 2449.5 t ). Then,
v
C
‘
(t) 
=
-
500 e
-
500 t
(A
1 
cos 2449.5 t 
+
 A
2
sin 2449.5 t ) 
+
 e
-
500 t
(
-
2449.5 A
1
sin 2449.5 t 
+
2449.5 A
2
cos 2449.5 t)
Substituting initial conditions,
A
1
=
-
1 and 
-
500 A
1
+
2449.5 A
2
=
0 
⇒
 A
1
=
-
1 and A
2
=
-
0.2041
\
Step response v
C
(t) 
=
1
-
e
-
500 t
(cos 2449.5 t 
+
0.2041 sin 2449.5 t ) 
=
[1 – 1.0206 e
-
500 t
cos (2449.5 t 
-
0.2013 rad)] V for t 
≥
0
+
Rise time, 
rad
t
r
n
=
−
−
−
=
−
×
=
−
p
x
x
x w
p
tan
.
.
.
1
2
2
1
1
1 37
0 978 2500
0 725 m
ms
Peak time, 
ms
Settling time,
t
t
p
n
s
=
−
=
×
=
p
x w
p
1
0 978 2500
1 29
2
.
.
==
=
×
=
=
×
=
−
−
3
3
0 2 2500
6
100
1
2
xw
p x
x
n
p
M
e
.
ms
Percentage overshoot, 
1100
52 6
0 6425
×
=
−
e
.
. %
12.9 
Frequency reSponSe oF SerIeS 
RLC
 cIrcuIt
Consider a series RLC circuit excited by a voltage source v
S
(t) 
=
1 sin
w
 t V for t 
≥
0
+
with a set of 
specified initial conditions. We continue to use v
C
(t), the voltage across capacitor, as the describing 
variable for the circuit.
The total response will contain natural response terms of exponentially damped sinusoidal nature 
and forced response term. We expect the natural response terms to vanish when time increases without 
limit. Only the forced response term acts in the long run, and of course, it is termed steady-state 
response. The sinusoidal steady-state response of series RLC circuit in this section. 
12.9.1 
Sinusoidal Forced-response from differential equation
The differential equation governing v
C
(t), the capacitor voltage, in a series RLC circuit excited by a 
voltage source v
S
(t) was derived earlier in this chapter.
d v t
dt
dv t
dt
v t
v t
LC
R
L
C
n
C
n
C
n
S
n
2
2
2
2
2
1
2
( )
( )
( )
( )
,
+
+
=
=
=
xw
w
w
w
x
where
C
C
(12.9-1)

---

FrequencyResponseofSeries
RLC
Circuit

12.27
The steady-state component of response when v
S
(t) 
=
1 sin
w
t u(t) is obtained by ‘method of 
undetermined coefficients’. We assume a trial solution of the form v
C
(t) 
=
A sin
w
 t 
+
B cos
w
 t and 
determine the values of A and B by substituting the assumed solution in the differential equation and 
equating coefficients of sin
w
 t and cos
w
 t on both sides of the resulting equation.
Trial Solution,
Substituting in the
v t
A
t B
t
C
( )
sin
cos
=
+
w
w
differential equation and collecting terms,
(
)
w
w
n
2
2
−
A
A
B
t
B
A
t
t
n
n
n
n
−


+
−
+


=
2
2
2
2
2
xw
w
w
w
xw
w w
w
sin
(
)
cos
sin
The only way this equation can remain true independent of t is by the coefficients of sin
w
t (and 
cos
w
t ) becoming equal on both sides of the equation. Therefore,
(
)
(
)
w
w
xw
w
w
w
xw
n
n
n
n
n
A
B
B
A
2
2
2
2
2
2
2
0
−
−
=
−
+
=
.
Solving these two equations simultaneously we get,
A
B
n
n
n
n
n
n
n
=
−
−
+
=
−
−
+
w w
w
w
w
x w w
xw w w
w
w
x
2
2
2
2
2 2
2
2
2
2
2
2 2
2
4
2
4
(
)
(
)
,
(
)
(
)
w
w w
n
2
2
Substituting these in the assumed solution and simplifying the solution to a ‘single sinusoid with 
phase’ form we get,
v t
t
C
n
n
n
C
n
n
( )
(
)
sin(
)
tan
=
−
+
+
= −
−
w
w
w
x w w
w f
f
xw w
w
2
2
2 2
2
2
2
1
4
2
C
where
22
2
−
w
(12.9-2)
Therefore, the magnitude part of the frequency-response function for v
C
(t) is 
w
w
w
x w w
n
n
n
2
2
2 2
2
2
2
4
(
)
−
+
and the phase part of frequency-response function is 
-
-
-
tan
1
2
2
2
xw w
w
w
n
n
.