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n 
to verify this.
f
LC
L
C
n
=
=
×
×
× ×
=
=
=
−
−
1
2
1
2
25 3 10
1 10
1
2
3
6
p
p
.
kHz
Critical resistance
3318 2
20
318 2
0 063
.
.
.
Ω
∴
=
=
Damping factor 
and Quality factor 
x
Q
==
=
1
2
7 95
x
.
Therefore, the values of x of interest to us are the odd integers. For x 
≥
3, we can approximate the 
gain and phase expressions as shown in the following.
V j
V j
j
x
x
j
x
V j
V j
x
x
x
R
S
R
S
(
)
(
)
(
)
(
)
(
)
(
)
w
w
x
x
w
w
x
x
=
−
+
=
−
+
≈
2
1
2
2
1
4
2
2 2
2 2
22
1
2
2
1
1
2
2
1
2
1
x
p
p
x
p
x
x
x
x
x
x
x
for
Phase
for
>>
=
−
−
−
>
≈ − +
=
−
−
(
tan
)
tan
−− +
>>
p
x
2
2
1
x
x
for
The following table shows gain and phase of output for x values from 1 to 11.
x
1
3
5
7
9
11
Gain
1
0.042
0.025
0.018
0.014
0.012
Phase(rad)
0
-
1.53
-
1.545
-
1.553
-
1.557
-
1.56
Now the first six terms of output may be constructed using the amplitude information.
v t
t
t
t
o
( )
=
×
+
×
−
+
×
1
2
10
0 014
6
10
1 53
0 005
10
10
3
3
3
sin
.
sin(
. )
.
sin(
p
p
p
−−
+
×
−
+
×
−
1 545
0 0026
14
10
1 553
0 0013
18
10
1 55
3
3
.
)
.
sin(
.
)
.
sin(
.
p
p
t
t
77
0 0011
22
10
1 56
3
)
.
sin(
. )
+
×
−
p
t
Obviously, the first three terms are the ones which are significant. All the others can easily be 
ignored (including the higher frequency components that we did not evaluate).
∴
≈
×
+
×
−
+
×
v t
t
t
o
( ) 1
2
10
0 014
6
10
1 53
0 005
10
10
3
3
3
sin
.
sin(
. )
.
sin(
p
p
p
tt
−
1 545
.
) V
This waveform has only about 1.5% of other frequency content. It is almost pure sinusoid at 1 kHz.

---

TheParallel
RLC
Circuit

12.43
Solution (b)
The 25.3 mH inductor has a reactance of 2
p
× 
10
3 
× 
25.3 
× 
10
-
3
=
159 
W
at 1 kHz and since its Q at 
that frequency is given as 50, it has a series resistance of 159/50 
=
3.18 
W
. The function generator 
that provides the square wave contributes 50 
W
. Thus, the total series resistance in the series RLC 
circuit is 73.18 
W
. The value of f
n
of the circuit remains unchanged at 1 kHz. But its 
x
factor becomes 
0.23 now. Also, the total voltage that develops across 73.18 
W
has to be multiplied by 20/73.18 
=
0.273 to calculate the voltage across the load resistance. Thus, the frequency-response expression
now is
V j
V j
j
x
x
j
x
R
S
(
)
(
)
.
(
)
w
w
x
x
=
−
+
0 547
1
2
2
We continue to use the same approximations for x >> 1.
The following table shows gain and phase of output for x values from 1 to 11.
x
1
3
5
7
9
11
Gain
0.273
0.042
0.025
0.018
0.014
0.011
Phase(rad)
0
-
1.418
-
1.479
-
1.505
-
1.52
-
1.53
Now the first six terms of output may be constructed using the amplitude information.
v t
t
t
o
( )
=
×
+
×
−
+
0 273
2
10
0 014
6
10
1 418
0 005
10
3
3
.
sin
.
sin(
.
)
.
sin(
p
p
p
××
−
+
×
−
+
×
10
1 479
0 0026
14
10
1 505
0 0015
18
10
3
3
3
t
t
t
.
)
.
sin(
.
)
.
sin(
p
p
−−
+
×
−
1 52
0 001
22
10
1 53
3
. )
.
sin(
. )
p
t
Higher frequency terms are neglected. The output 
waveforms for the two cases are plotted in Fig. 12.10-13.
The output in case (b) shows distortion clearly. 
This example illustrates the importance of high quality 
factor in filtering a square wave to a high Q sine wave. 
It also tells us that an otherwise high Q factor circuit 
may appear to be a low Q circuit if we forget about the 
output resistance of the signal generator that we use 
to test the circuit. The test signal should be passed on 
to the test circuit through a unity gain buffer amplifier 
with negligible output resistance.
12.11 
the pArALLeL 
RLC
 cIrcuIt
We have dealt with the series RLC circuit in great detail and have developed considerable insight into 
the time-domain behaviour of second-order circuits in general. Moreover, we have also studied the 
frequency-response of second-order circuits using series RLC circuit as an example. This will help us 
to draw parallels between the behaviour of series RLC circuit with another equally important, if not 
more important, circuit – the parallel RLC circuit.
Fig. 12.10-13 

Outputwaveforms
inExample:12.10-2
1
–1
0.2 0.4 0.6 0.8 1
Time
(ms)
v
o
(
t
)
(b)
(V)
(a)

---

12.44


SeriesandParallel
RLC
Circuits
Parallel RLC circuit finds application in almost all communications equipment (starting from radio 
receiver), sinusoidal oscillators, low-power and high-power filters and electrical power systems. In 
fact, one can even state that analog communications will be impossible without using parallel RLC 
circuit 
12.11.1 
Zero-Input response and Zero-State response of parallel 
RLC
 circuit 
Figure 12.11-1 shows a parallel RLC circuit excited 
by a current source i
S
(t). There are four other circuit 
variables apart from i
S
(t) – they are the three current 
variables and one common voltage variable.
We choose i
L
(t) as the variable for deriving the 
differential equation. However, the variable that is 
commonly used as output variable in practice is v(t). 
All the three possible output voltage variables are used 
in practical applications in the case of series RLC circuit. But in the case of parallel RLC circuit, it is 
v(t) that is almost used invariably. However, v(t) can be obtained easily once we solve for i
L
(t).
v t
L
di t
dt
i t
L
R
L
R
( )
( )
(
( )
=
∴
=
by element equation of inducatnce)
ddi t
dt
i t
LC
d i t
dt
L
C
L
( )
( )
( )
(
and 
By element equations of R 
=
2
2
aand C)
Now we apply KCL at the positive node of current source and make use of expressions for i
C
(t) 
and i
R
(t) in terms of i
L
(t).
i t
i t
i t
i t
t
LC
d i t
dt
L
R
di t
dt
i
C
R
L
S
L
L
( )
( )
( )
( )
( )
( )
+
+
=
≥
∴
+
+
+
for
0
2
2
L
L
S
L
L
L
S
t
i t
t
d i t
dt
RC
di t
dt
LC
i t
LC
i
( )
( )
( )
( )
( )
(
=
≥
∴
+
+
=
+
for 
0
1
1
1
2
2
tt
t
LC
RC
n
n
n
) for 
and 
are identified as 
and 2
≥
=
=
⇒
+
0
1
1
x
w
w
xw
xx
=
=
=
1
2
1
2
L
C
cr
cr
L
C
R
R
R
R
Critical Resistance 
Ω
We see that if we write the differential equations in the standard format using 
x
and 
w
n
, the 
differential equations for v
C
(t) in series RLC circuit and i
L
(t) in parallel RLC Circuit are identical 
except that v
C
(t) gets replaced by i
L
(t) and v
S
(t) gets replaced by i
S
(t). The only point we have to 
remember is that the damping factor is R
L C
/ 2
in the series RLC circuit, whereas it is 
1
2
L
C
R  
for parallel RLC Circuit. Thus, increasing resistance will increase damping in series RLC circuit, 
whereas it will decrease damping in parallel RLC Circuit. The initial conditions needed for solving the 
differential equation remain the same – inductor current I
o
at t 
=
0
-
and capacitor voltage V
o
at t 
=
0
-
.
The parallel RLC circuit is over-damped for 
x
> 1, critically damped for 
x
=
1 and under-damped 
for 
x
< 1.
Fig. 12.11-1 
Theparallel
RLC
circuit
–
R
C
L
+
i
S
(
t
)
i
R
(
t
)
i
L
(
t
)
i
C
(
t
)
v
(
t
)

---

TheParallel
RLC
Circuit

12.45
Zero-input response and zero-state response of parallel RLC circuit is illustrated by a series of 
examples that follow. The polarity conventions for specifying initial conditions is as per Fig. 12.11-1.