Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

12.5 
ImpuLSe reSponSe oF SerIeS 
RLC
 cIrcuIt
Impulse response of a circuit is the zero-state response with unit impulse input. The relevant circuit is 
shown in Fig. 12.5-1 (a).
The capacitor can not absorb the impulse voltage. It can not absorb even a finite voltage change 
without infinite current flow, let alone impulse voltage. Resistor can not absorb impulse voltage, for in 
that case the circuit current will become infinite and the inductor won’t permit that. Hence, the entire 
impulse voltage appears across the inductor. This results in 1 Wb-T of flux linkage getting dumped 
into the inductor instantaneously, resulting in a sudden change in its current from zero to 1/L A. 
Therefore i(t) at t 
=
0
+
becomes 1/L A and v
C
(t) at t 
=
0
+
remains at 0. And the input source becomes 
short-circuit for all t 
≥
t 
=
0
+
. Thus, the circuit effectively becomes a source-free circuit shown in 
(b) Fig. 12.5-1 (b), with initial energy of 1/2L J in inductor and 0 J in the capacitor. Thereafter the 
response is similar to zero-input response.

---

12.20


SeriesandParallel
RLC
Circuits
–
–
–
–
v
R
(
t
)
v
L
(
t
)
v
C
(
t
)
i
(
t
)
C
(a)
(
t
)
R
L
+
+
+
+
δ
–
–
–
v
R
(
t
)
v
L
(
t
)
v
C
(
t
)
i
(
t
)
C
(b)
R
L
+
+
+
i
(
t
) 0
+
=
L
A
1
Fig. 12.5-1 
Series
RLC
circuitwithunitimpulsevoltageinput
v t
e
t
t
i t
C
dv t
C
n
t
n
C
n
( )
sin
( )
( )
=
−








−
≥
=
−
+
w
x
x w
xw
1
1
0
2
2
V for 
ddt
L
e
t
t
n
t
n
=
−
−
+
−








≥
−
−
1
1
1
1
1
0
2
2
1
2
x
x w
x
x
xw
cos
tan
amps for 
++
(12.5-1)
12.6 
Step reSponSe oF SerIeS 
RLC
 cIrcuIt
Step response of any linear electrical circuit is the zero-state response with unit step input. This can 
be found out by integrating the impulse response of the same circuit. Only the under-damped case is 
considered here.
Impulse Response of a under-damped Series RLC Circuit is
vv t
e
t
t
C
n
t
n
n
n
( )
sin
=
−








−
≥
=
−




−
+
w
x
x w
w
x
xw
1
1
0
1
2
2
2
V for 




−
≥
∴
−
+ −
−
− −
+
e
e
j
t
n
n
t
t
(
)
(
)
xw
x
xw
x
1
1
2
2
2
0
V for 
Step Responsee of the same circuit is
C
v t
j
e
n
( )
(
=
−








−
w
x
xw
2
1
2
nn
n
t
t
t
e
dt
+ −
−
− −
−
(
)
+
∫
1
1
0
2
2
x
xw
x
)
(
)
Carrying out the required integration and simplifying the result, we get,
v t
e
t
t
C
t
n
n
( )
cos
tan
= −
−
−
−
−








≥
−
−
+
1
1
1
1
0
2
2
1
2
xw
x
x w
x
x
V for 
(12.6-1)
The step response waveforms of a series RLC circuit with L 
=
1 H and C 
=
1 F are shown in 
Fig. 12.6-1 for various values of resistance resulting in damping factors of 0.5, 0.2, 0.05 and 0. Solid 
curve shows v
C
(t) and dotted curve shows i(t) in all cases.

---

StandardTime-DomainSpecificationsforSecond-OrderCircuits

12.21
1
0.5
5
10
(a)
15
20
Time (s)
V, A
1.5
1
0.5
–0.5
5
10
(c)
15
20
Time (s)
V, A
1.5
1
0.5
–0.5
5
10
(b)
15
20
Time (s)
V, A
1.5
2
1
0.5
–0.5
1
5
10
(d)
15
20
Time (s)
V, A
Fig. 12.6-1 

Stepresponseofseries
RLC
circuit(
L
=
1H,C
=
1F):(a)
x

=
0.5,
(b)
x

=
0.2(c)
x

=
0.05(d)
x

=
0
12.7 
 StAndArd tIme-domAIn SpecIFIcAtIonS For 
Second-order cIrcuItS
Certain measures of speed and nature of response of a second-order system, defined with respect to 
step response, are in wide use in various areas of electrical and electronic engineering. These measures 
are as follows:
1. Delay time, t
d
– is the time required by step response to cross 50% of final value for the first 
time.
2. Rise time, t
r
– is the time required for step response to reach 100% of final value for the first time 
starting from 0% in the case of under-damped response. In the case of over-damped response and 
critically damped response, it is the time required for step response to reach 90% of final value 
from 10% of final value.
3. Peak time, t
p
– is the time required for step response to reach its first peak.
4. Settling time, t
s
– is the time required for the step response to reach a 
±
5% band around the final 
value and remain within that band thereafter.
5. Maximum (percentage) overshoot, M
p
– is the difference between the maximum peak value 
attained by the step response and the final value of step response expressed as a fraction or 
percentage of the final value of step response.

---

12.22


SeriesandParallel
RLC
Circuits
t
d
t
r
t
p
M
p
t
s
1.6
Volts
1.05
0.95
Time
1.4
1.2
1
0.8
0.6
0.4
0.5
0.2
Fig. 12.7-1 
Time-domainspecificationsforasecondordercircuit
These measures are illustrated on a typical step response plot of a series RLC circuit (capacitor 
voltage is plotted) in the Fig. 12.7-1.
It is possible to derive expressions for these measures starting from Eqn. 12.6-1. The algebra 
involved is skipped here and the final expressions are given in Eqn. 12.7-1.
t
t
t
M
e
r
n
p
n
s
n
p
=
−
−
−
=
−
=
=
−
−
−
p
x
x
x w
p
x w
xw
p x
x
tan
,
,
,
1
2
2
2
1
1
1
1
3
2
(12.7-1)
9
8
7
6
5
4
3
2
1
0.2
(a)
0.4 0.6 0.8
τ
P
τ
r
ξ
t
p
, 
t
r
s
M
P
1
0.8
0.6
0.4
0.2
0.2
(b)
0.4 0.6
0.8
ξ
Fig. 12.7-2 

Risetime,peaktimeandmaximumovershootina
Second-ordercircuitvs.dampingfactor
The relationship between various time-domain specifications and the damping factor is highly non-
linear. Refer Fig. 12.7-2 that shows the variation of rise time (the lower curve), peak time (the upper 
curve) and maximum overshoot in step response of capacitor voltage in a series RLC circuit with an 
undamped natural frequency of 1 rad/s. 
x
≈
0.7 is a critical value. When 
x
increases above 0.7, the 
response becomes sluggish. When 
x
decreases below 0.7, the maximum overshoot increases rapidly 
without much change in the rise time. Thus, 
x
=
0.7 is a value of damping factor that results in a step 
response which has optimal speed without incurring too heavy a penalty in the form of excessive 
overshoot.

---

ExamplesonImpulseandStepResponseofSeries
RLC
Circuits

12.23
12.8 
exAmpLeS on ImpuLSe And Step reSponSe oF SerIeS 
RLC
 cIrcuItS
The circuit variables and polarity conventions for specifying initial conditions in the following 
examples are the same as in the series RLC circuit in Fig. 12.1-1. 
example: 12.8-1
A series RLC circuit has L 
=
10 mH, C 
=
 1000 
m
F and R 
=
10 
W
. The capacitor is initially charged 
with 10 V. 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. 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
5
=
0
Natural frequencies of the circuit 
=
-
112.7 and 
-
887.3. This is an over-damped circuit.
\
v
C
(t) 
=
A
1
e
-
112.7 t
+
A
2
e
-
887.3 t
and i(t) 
=
10
-
3
×
(
-
112.7 A
1
e
-
112.7 t
– 887.3 A
2
e
-
887.3 t
)
Applying initial conditions of 10 V and 0 A at 0
+
,
A
1 
+
A
2
=
10 and 
-
0.1127 A
1
– 0.8873 A
2
=
0 
⇒
A
1 
=
11.455 and A
2
=
-
1.455
\
v
C
(t) 
=
11.455 e
-
112.7 t
– 1.455 e
-
887.3 t
V for t 
≥
0
+
is the zero-input response.
Zero-State Response Component
The initial conditions at t 
=
 0
-
are set to zero 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. Therefore, v
C
(t) 
=
10 
+
A
1
e
-
112.7 t
+
A
2
e
-
887.3 t
is the format of zero-state 
response for 10 u(t) input.
The applied voltage suddenly becomes 10 V at t 
=
0
+
. Since the inductor current can not change 
instantaneously unless impulse voltage comes across it and capacitor voltage can not change 
instantaneously unless impulse current flows through it, the capacitor voltage and inductor current 
remain at zero values at t 
= 
0
+
. The 10 V step appears across the inductor first, thereby changing the 
initial value of derivative of current. But, we need initial values only for capacitor voltage and inductor 
current to solve for A
1
and A
2
in the zero-state response above. Applying these initial values on 
v
C
(t) 
=
10 
+
A
1
e
-
112.7 t
+
A
2
e
-
887.3 t 
and i(t) 
=
-
0.1127 A
1
e
-
112.7 t
– 0.8873 A
2
e
-
887.3 t
),
we get A
1 
+
A
2
=
-
10 and 
-
0.1127 A
1
– 0.8873 A
2
=
0 
⇒
A
1 
=
-
11.455 and A
2
=
1.455
\
v
C
(t) 
=
10 
-
11.455 e
-
112.7 t
+
1.455 e
-
887.3 t
V for t 
≥
0
+
is the zero-state response.
\
Total response of v
C
(t) 
=
zero-input response 
+
zero-state response
=
11.455 e
-
112.7 t
– 1.455 e
-
887.3 t
+
10 
-
11.455 e
-
112.7 t
+
1.455 e
-
887.3 t
=
10 V for t 
≥
0
+

---

12.24


SeriesandParallel
RLC
Circuits
We didn’t have to go through all this to see that v
C
(t) 
=
10 ! If the expected final state of a circuit is 
same as the given initial state, then, there is no transient response in the circuit. 
Note the method employed in solving this problem. We could have obtained the zero-input response 
by direct substitution in Eqn. 12.1-5. But such direct substitutions in equations should be avoided 
as far as possible and problems should be worked out employing basic principles governing the
circuit.