Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

case-2 
`
1
 and 
`
2
 real, negative and equal
This case occurs when R
L
C
=
2
and then 
a
a
1
2
2
=
= −
R
L. The limiting form of expression for 
v
C
(t) given in Eqn. 12.1-4 has to be found in this case. We had found out such a limiting form in 
Section 11.3 in Chapter 11 in the context of zero-state response of a RC circuit excited by an exponential 
signal with a time constant equal to the time constant of the circuit itself. We had employed the series 
expansion of exponential function for that purpose. Here we try a different approach. We denote 
a
2
as 
a
1
+
D
a
and determine the limit of v
C
(t) as 
D
a
→
0.
The expression for v
C
(t) in a general case is repeated in Eqn. 12.1-5.
v t
V
e
e
I
C
e
e
t
C
o
t
t
o
t
t
( )
=
−
−
−
−
−
≥
+
a
a
a
a
a
a
a
a
a
a
2
1
2
1
2
1
1
2
1
2
0
V for 
(12.1-5)
This equation is first recast in the following equivalent form.
Fig. 12.1-2 

Zero-inputresponse
ofseries
RLC
circuit
inExample:12.1-1
3
Volts
Amps
Time (s)
2
1
1
2
3
–1
–2
–3
–4
–5
v
R
(
t
)
v
C
(
t
)
i
(
t
)
v
L
(
t
)

---

12.6


SeriesandParallel
RLC
Circuits
v t
V
e
e
e
e
I
C
e
e
C
o
t
t
t
t
o
t
t
( )
=
−
+
−
−
−
−
−
a
a
a
a
a
a
a
a
a
a
a
a
a
a
2
1
1
1
2
1
2
1
1
1
1
2
1
2
V
V for t
V e
e
e
I
C
e
e
o
t
t
t
o
t
t
≥
=
+
−
−





−
−
−
+
0
1
1
2
1
2
1
2
1
2
a
a
a
a
a
a
a
a
a
a
11
1
1
2
0
1
1
2
1






≥
=
−
−
−





+
−
+
V for t
V e
e
e
I
C
e
e
o
t
t
t
o
t
a
a
a
a
a
a a
aa
a a
2
1
2
0
t
t
−






≥
+
V for 
Now, let 
a
1
=
a
and 
a
2
=
a
+
D
a
. Then,
v t
V e
e
e
I
C
e
e
C
o
t
t
t
o
t
t
( )
=
−
−
−








+
−
−


+
(
)
+
(
)
a
a
a
a
a
a
a
a
a
a
1
∆
∆
∆
∆






≥
−
=
−

≈− ×
+
+
(
)
V for 
for sm
t
e
e
e
e
e
t
t
t
t
t
t
0
1
a
a
a
a
a
a
a
∆
∆
∆
aall 
V for 
where
∆
∆
a
a
a
a
a
lim
( )
→
+
=
+
−




≥
0
0
v t
V e
I
C
V t e
t
C
o
t
o
o
t
a
= −
−
R
L
2
1
sec
Therefore, the circuit solution in this case is
v t
V e
I
C
R
L
V t e
t
C
o
R
L t
o
o
R
L t
( )
=
+
+




≥
−
( )
−
( )
+
2
2
2
0
V for 
. The other circuit variables may be readily 
obtained now. The next example illustrates this case.
example: 12.1-2
A series RLC circuit has R 
=
2 
W
, L 
=
1 H and C
=
1 F. The capacitor is initially charged to 2 V and the 
initial current in the inductor is 2 A at t 
=
0
-
. Find the zero-input response of capacitor voltage and 
circuit current.
Solution
The differential equation governing the capacitor voltage v
C
(t) is
d v
dt
dv
dt
v
t
C
C
C
2
2
2
0
0
+
+
=
≥
+
for
. 
The characteristic equation is 
g
g
2
2
1 0
+
+ =
and its roots 
a
1,2 
=
-
1 s
-
1
and –1 s
-
1
. The trial solution 
to be attempted is of the form v
C
(t) 
=
A
1
e
-
t
+
A
2
t e
-
t
V. Applying initial conditions at t 
=
0
+
, we get 
two equations in A
1 
and A
2
. 
v
A
dv
dt
I
C
A e
A te
A e
C
C
o
t
t
t
( )
(
)
(
0
2
2
1
0
1
2
2
0
+
−
−
−
=
=
=
=
= −
−
+


+
V;
V
s
++
= −
+
)
A
A
1
2

---

TheSeries
RLC
Circuit–Zero-InputResponse

12.7
∴ =
=
∴
=
+
≥
=
−
−
−
+
A
A
v t
e
te
t
i t
C
t
t
1
2
1
2
4
2
4
0
V and
V
V for
and
s
( )
( )
C
C
dv
dt
e
te
e
e
te
t
v t
L
di
dt
C
t
t
t
t
t
L
= −
−
+
=
−
≥
=
=
−
−
−
−
−
2
4
4
2
4
0
A for
( )
11
2
4
4
6
4
0
2
di
dt
e
te
e
e
te
t
v t
Ri
i
t
t
t
t
t
R
= −
+
−
= −
+
≥
= =
−
−
−
−
−
+
V for
( )
==
−
≥
−
−
+
4
8
0
e
te
t
t
t
V for
These waveforms are shown in Fig. 12.1-3.
case-3 
`
1
and 
`
2
complex and conjugates 
with negative real parts
Now we come up against the most interesting case 
of all. This occurs when R
L
C
<
2
. The quantity 
R
L
LC
2
2
4
1
−




is negative under this condition 
and the roots of characteristic equation become 
a
1 2
2
2
2
1
4
,
= −
±
−
R
L
j
LC
R
L
.
We define three new symbols – mainly for convenience at present. But they will turn out to be 
important parameters in circuit studies soon. They are defined as in Eqn. 12.1-6.
Let 
, 
and 
x
w
w
x w
=
=
=
−
= −
R
L
C
LC
LC
R
L
n
d
n
2
1
1
4
1
2
2
2
(12.1-6)
The roots of characteristic equation can be expressed in terms of these new symbols as 
a
xw
x w
a
xw
x w
1
2
2
2
1
1
= −
+
−
= −
−
−
n
n
n
n
j
j
and 
.
v t
V
e
e
I
C
e
e
t
C
o
t
t
o
t
t
( )
,
=
−
−
−
−
−
≥
+
a
a
a
a
a
a
a
a
a
a
a
2
1
2
1
2
1
1 2
1
2
1
2
0
V for 
== −
±
−
= −
±
⇒
−
= −
xw
x w
xw
w
a
a
w
n
n
n
d
d
j
j
j
1
2
2
2
1
Simplifying
a
a
a
x w
xw
w
a
xw
w
2
2
1
2
1
1
2
1
e
j
j
e
t
n
n
d
j
t
n
d
−
=
−
−
−
−
− −
−
+
(
)
(
(
)
xxw
w
xw
w
n
d
j
t
j
e
n
d
+


−
−
)
(
)
Now we use Eqn. 12.1-4 for the zero-input response for v
C
(t) and substitute these expressions for 
a
1
and 
a
2 
and employ Euler’s formula for algebraic simplification of the resulting expression.
Fig. 12.1-3 

Zero-inputresponse
waveformsofaseries
RLC

circuitinExample:12.1-2
3
4
Volts
Amps
Time (s)
2
1
1
2
3
–1
–2
–3
–4
–5
–6
v
R
(
t
)
v
C
(
t
)
i
(
t
)
v
L
(
t
)

---

12.8


SeriesandParallel
RLC
Circuits
=
−
−
−
−
(
)
− −
+
(
)




=
−
−
( )
−
(
)
e
j
j
e
j
e
n
d
d
t
n
n
d
j
t
n
d
j
t
xw
w
w
x w
xw
w
xw
w
2
1
2
ee
j
e
e
j
e
e
n
d
d
d
d
t
n
n
j
t
j
t
d
j
t
j
t
−
( )
−
(
)
( )
−
(
)
−
−
−
(
)
−
+
(
xw
w
w
w
w
x w
xw
w
2
1
2
))




=
−
−
−
(
)
−
(
)


−
e
j
j
t
j
t
n
t
n
n
d
d
d
xw
x w
xw
w
w
w
2
1
2
2
2
sin
cos
(by Euler’s formula)
=
−
+








=
−
−
e
t
t
n
t
d
d
d
xw
x
x
w
w
w
1
1
2
sin
cos
(
∵
xx w
a
a
a
a
a
a
a
a
2
2
1
2
1
1
2
1
2
n
t
t
t
t
e
e
e
e
)
Similarly simplifying
,
−
−
−
−
= −
ee
t
v t
V e
t
t
n
n
t
n
d
C
o
t
d
d
−
−
−
∴
=
−
+






xw
xw
x w
w
x
x
w
w
1
1
2
2
sin
( )
sin
cos


+
−
−
I
C
e
t
o
t
n
d
n
xw
x w
w
1
2
sin
=
LC
n
w
1
Substituting 
in the second tterm ,
v t
V e
t
t
I
e
C
o
t
d
d
o
L
C
t
n
n
( )
sin
cos
=
−
+








+
−
−
xw
xw
x
x
w
w
1
2
11
1
2
−
x
w
sin
d
t
V
Therefore, the zero-input response for capacitor voltage in this case with 0 
≤
x
< 1 is given by the 
following expression in terms of 
x
and 
w
n
.
v t
e
V
I
t V
t
C
t
o
o
L
C
n
o
n
n
( )
sin
cos
=
+
−








−
+
−



−
xw
x
x
x w
x w
1
1
1
2
2
2





≥
=
=
+
V for 
where 
t
LC
R
L
C
n
0
1
2
w
x
,
(12.1-7)
That, indeed, looks slightly complicated. We will take it up through special cases soon. But before 
that, here is how it looks for a series RLC circuit with L
=
1 H, C 
=
1 F, R 
=
0.5 
W
, V
o
=
2 V and I
o
=
2 A. 
The value of 
x
is 0.25 and 
w
n
is 1.
The zero-input response has an exponentially damped sinusoidal shape. All the circuit 
variables oscillate with the amplitude of oscillation decreasing with time. An exponential function 
governs the decrease in amplitude with time. The numbers 
x
and 
w
n
decide the decay rate of this 
exponential function. They also decide the time-interval between successive zero-crossings of circuit
variables.

---

TheSeriesLCCircuit–ASpecialCase

12.9
3
Volts
Amps
2.5
2
1.5
1
0.5
–0.5
2
4
6
8
10
Time (s)
The exponential envelope
12
–1
–1.5
–2
–2.5
–3
v
R
(
t
)
v
C
(
t
)
i
(
t
)
v
L
(
t
)
Fig. 12.1-4 

ZIRUnderdampedzero-inputresponseofaseries
RLC
circuitwith
R

= 
0.5
W
,
L

=
1H,
C

=
1F,
V
o

=
2Vand
I
o

=
2A