Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 13.10-4
(a) Find the transfer function 
V s
V s
o
s
( )
( )
in the Opamp circuit shown in 
Fig. 13.10-11 assuming ideal Opamp. (b) Show its pole-zero plot for 
k
= 
2.9 and 
k
= 
3.1 and find 
its zero-state response to 
v
S
(
t
) 
= 
0.01
d
(
t
) in both cases with 
R
= 
10k
W
and 
c
= 
1
m
F. (c) What is the 
maximum value of 
k
that can be used in the circuit without making it an unstable one?
v
S
(
t
)
v
o
(
t
)
–
–
–
+
+
+
2
R
(
k
– 1)
R
R
R
R
C
C
Fig. 13.10-11 
The Opamp circuit in Example: 13.10-4

---

13.38
Analysis of Dynamic Circuits by Laplace Transforms
Solution
(a) The Opamp circuit from its non-inverting input to its output is a simple non-inverting amplifier 
of gain 
k
and can be replaced with a dependent source as shown in the 
s-
domain equivalent circuit in 
Fig. 13.10-12.
V
S
(
s
)
V
1
(
s
)
V
2
(
s
)
V
(
s
)
kV
(
s
)
V
o
(
s
)
–
2
R
R
+
+
+
+
R
–
–
–
1
sC
1
sC
1
2
R
Fig. 13.10-12 
Transformed equivalent circuit of the circuit in Fig. 13.10-11
The node voltage transforms 
V
1
(
s
) and 
V
2
(
s
) are identified in the transformed equivalent circuit. 
Writing the node equations at these two nodes, we get
Second node:
sC V s
V s
V s
R
[ ( )
( )]
( )
2
1
2
2
0
−
+
=
Therefore, 
V s
V s
V s
sCR
V s
sCR
1
2
2
2
2
1
1
2
( )
( )
( )
( )
=
+
=
+




First node:
V s
V s
R
sCV s
sC V s
V s
V s
kV s
s
1
1
1
2
1
2
( )
( )
( )
[ ( )
( )]
( )
(
−
+
+
−
+
−
))
R
=
0
Substituting for 
V
1
(
s
) in terms of 
V
2
(
s
) and simplifying, we get
V s
V s
s
s
s
V s
V s
kV s
V
s
RC
k
RC
RC
o
s
2
1
2
3
1
2
2
( )
( )
( )
( )
( )
=
+
+
( )
=
−
( )
and
ss
k
RC
k
RC
RC
s
s
s
s
( )
=
+
+
( )
−
( )
2
3
1
2
This is the transfer function of the circuit.
(b) The denominator polynomial with 
R
= 
10 k
W
, 
C
= 
1 
m
F and 
k
= 
2.9 is 
s
s
2
10
10000
+
+
. 
Therefore, the poles are at 
s
= -
5 
±
j
99.88. The zero of the transfer function is at 
s
= 
0, 
i.e.,
at the 
origin in the 
s
-plane.
With 
k
= 
3.1, the denominator polynomial is 
s
s
2
10
10000
−
+
and the poles are at 
s
= 
5 
±
j
99.88. 
The zero of the transfer function is again at 
s
= 
0. The pole-zero plots are shown in Fig. 13.10-13.
Im(
s
)
Re(
s
)
x
(–5, 99.88)
x
(–5, –99.88)
k
= 2.9
–5
–100
5
100
Im(
s
)
Re(
s
)
x
(5, 99.88)
x
(5, –99.88)
k
= 3.1
–5
–100
5
100
Fig. 13.10-13 
Pole-zero plots for the Opamp circuit in Fig. 13.10-11 with 
k
= 
2.9 
and 
k
= 
3.1

---

Total Response of Circuits Using 
s 
-Domain Equivalent Circuit 
13.39
The zero-state response to 
v
S
(
t
) 
= 
0.01
d
(
t
) is nothing but the impulse response of the circuit scaled 
by 0.01.
With 
k
= 
2.9
The transfer function 
H
(
s
) 
= 
290
10
10000
2
s
s
s
+
+
. Inverse transform of 
H
(
s
) gives the impulse response 
of the circuit. We complete the squares in the denominator and identify the inverse transforms as 
shown below.
290
10
10000
290
5
5
5
99 88
290
5
5
99
2
2
2
2
s
s
s
s
s
s
s
+
+
=
+ −
+
+
=
+
+
+
(
)
(
)
.
(
)
(
)
..
.
.
(
)
.
88
5
99 88
99 88
5
99 88
2
2
2
−
+
+






s
∴
=
−
=
−
−
h t
e
t
t u t
e
t
t
( )
(cos
.
. sin
.
) ( )
.
cos(
290
99 88
0 05
99 88
290 36
9
5
5
99 88
2 86
0
.
.
) ( )
t
u t
+
This is a stable impulse response since 
lim ( )
t
h t
→∞
=
0
The zero-state response to 0.01
d
(
t
) 
= 
=
+
−
2 9
99 88
2 86
5
0
.
cos( .
.
) ( )
e
t
u t
t
.
With 
k
= 
3.1
The transfer function 
H
(
s
) 
= 
310
10
10000
2
s
s
s
−
+
. Inverse transform of 
H
(
s
) gives the impulse response 
of the circuit. We complete the squares in the denominator and identify the inverse transforms as 
shown below.
310
10
10000
310
5
5
5
99 88
310
5
5
99
2
2
2
2
s
s
s
s
s
s
s
−
+
=
− +
−
+
=
−
−
+
(
)
(
)
.
(
)
(
)
..
.
.
(
)
.
88
5
99 88
99 88
5
99 88
2
2
2
+
+
+






s
∴
=
+
=
h t
e
t
t u t
e
t
t
( )
(cos
.
. sin
.
) ( )
.
cos( .
310
99 88
0 05
99 88
310 39
99
5
5
888
2 86
0
t
u t
−
.
) ( )
This is an unstable impulse response since it is unbounded. The circuit is an unstable one as 
evidenced by its poles located in the right-half of 
s
-plane.
The zero-state response to 0.01
d
(
t
) 
= 
=
−
3 1
99 88
2 86
5
0
.
cos( .
.
) ( )
e
t
u t
t
.
(c) The transfer function 
V s
V s
s
s
s
o
s
k
RC
k
RC
RC
( )
( )
=
+
+
( )
−
( )
2
3
1
2
has poles on 
j
w
-axis when 
k
= 
3. The poles 
will lie on the right-half of 
s
-plane for all values of 
k
> 3. Therefore, 
k
< 3 is the constraint on 
k
value 
for stability in the circuit.
The circuit is marginally stable at 
k
= 
3. It can function as a sinusoidal oscillator with 
k
= 
3. But 
additional circuitry will be needed to stabilize its amplitude of oscillation.
Note that the circuit is a pure 
RC
circuit with one dependent source in it. A passive 
RC
circuit, 
i.e.,
a circuit containing only resistors and capacitors and no dependent sources, will have all its poles in 
the negative real axis. The dependent source is responsible for making the poles complex conjugate 

---

13.40
Analysis of Dynamic Circuits by Laplace Transforms
in such a circuit. Complex conjugate poles are often necessary in filter circuits to tailor the filter 
frequency response function suitably to meet filtering specifications.
This circuit is used as a band-pass filter in practice. The value of 
k
will be decided by the bandwidth 
required in the band-pass filter and will be < 3 at any rate.
example: 13.10-5
(a) Obtain the transfer function of the filter 
circuit shown in Fig. 13.10-14 and identify 
the type of filter. (b) Determine the zero-state 
response for 
v
S
(
t
) 
= 
0.1
d
(
t
). 
R
= 
10k
W
, 
C
= 
1
m
F 
and 
k
= 
1
Solution
The transformed equivalent circuit is shown 
in Fig. 13.10-15. Initial condition sources are 
not required since the circuit is needed for 
determining transfer function and for evaluating zero-state response. The Opamp is assumed to be 
ideal. There is only one node that has a free node voltage variable. This node and the assigned node 
voltage transform are also indicated in the equivalent circuit.
1
0
sC
1
sC
V
(
s
)
R
V
(
s
)
R
1
C
+
+
V
S
(
s
)
V
(
s
)
–
–
–
+
+
R
R
kR
V
(
s
)
sCR
v
o
(
s
) = –
–
Fig. 13.10-15 
The transformed equivalent circuit of circuit in Fig. 13.10-14 
Virtual short across the Opamp input terminals and zero input current drawn by the Opamp 
inverting pin makes the current in the feedback capacitor equal to 
V
(
s
)/
R. 
This results in the output 
voltage transform becoming equal to 
-
V
(
s
)/
sCR.
Now, we write the node equation at the node-1 marked in Fig. 13.10-15.
V s
R
sCV s
V s
R
V s
V s
sRC
kR
V s
R
s
( )
( )
( )
( ) (
( )
)
( )
+
+
+
− −
=
Algebraic simplification leads to 
V s
V s
skRC
k sRC
k
sRC
s
( )
( )
(
)
(
)(
)
=
+
+
+
2
2
1
1
The output transform 
V
o
(
s
) 
= -
V
(
s
)/
sCR.
Therefore,
V s
V s
k
k sRC
k
sRC
k
s
s
o
s
k
RC
k
R
( )
( )
(
)
(
)(
)
(
)
=
−
+
+
+
=
− ×
( )
+
+
2
1 1
2
2
1
1
2
1
1
2
C
C
k
RC
( )
+
( )
1 1
2
Fig. 13.10-14 
The Opamp-rc filter circuit 
in Example: 13.10-5 
C
C
+
+
v
S
(
t
)
–
–
+
R
R
kR
v
o
(
t
)
–

---

Total Response of Circuits Using 
s 
-Domain Equivalent Circuit 
13.41
The denominator has its roots in the left-half of 
s
-plane for all positive values of 
k
since a 
second-order polynomial with positive coefficients will have both roots in left-half of 
s
-plane. 
Therefore, the impulse response is stable and will be absolutely integrable. Therefore, the impulse 
response will have a Fourier transform. If Fourier transform of a time-function exists, then, the Laplace 
transform of the same function evaluated on 
j
w
-axis is its Fourier transform. Fourier transform of 
impulse response is the frequency response function. 
Therefore, the sinusoidal steady-state frequency 
response function of a stable circuit is given by its Laplace transform
evaluated with s 
= 
j
w
.
Therefore, 
V j
V j
k
j
o
s
k
RC
k
RC
k
RC
(
)
(
)
(
)
(
)
w
w
w
w
= −
( )
−
+ +
( )
+
( )
1
1
2
2
1
1
1
1
2
2
. This frequency response function has 
a magnitude of unity at 
w
= 
0 and 0 at 
w
= ∞
. Therefore, it is a low-pass filter.
Substituting the numerical values, we get
V s
V s
s
s
o
s
( )
( )
=
−
+
+
10
300
10
4
2
4
The impulse response of the circuit is obtained by inverting the transfer function. The roots of the 
denominator are at 
s
= -
261.8 and 
s
= -
38.2.
V s
V s
s
s
A
s
B
s
o
s
( )
( )
.
.
=
−
+
+
=
+
+
+
10
300
10
261 8
38 2
4
2
4
A
and 
B 
can be evaluated as –44.72 and 44.72, respectively.
Therefore, 
h t
e
e
u t
t
t
( )
. (
) ( )
.
.
=
−
−
−
44 72
38 2
261 8
Therefore, zero-state response for 
v
S
(
t
) 
= 
0.1
d
(
t
) is 
4 47
38 2
261 8
. (
) ( )
.
.
e
e
u t
t
t
-
-
-
V