Zero-StateResponseof
RC
CircuitsforVariousInputs
11.13
Solution
The time constant of the circuit in Fig. 11.3-11 (a) is 100 ms
=
0.1 s and its step response is (1
-
e
-
10
t
) V
for
t
≥
0
+
.
Consider the circuit in (b). The Opamp is connected with negative feedback and we further assume
that the input voltage applied is of such magnitude that the Opamp does not enter voltage saturation at its
output. Moreover, we assume that the Opamp has sufficiently large slew rate capability such that it never
enters rate-limited operation. With these assumptions, we can analyse the Opamp using its ideal model.
The non-inverting terminal of Opamp is grounded and by
virtual short principle the inverting
input terminal is also
virtually grounded. Therefore, the current that flows through
R
1
is
u(
t)/
R
1
.
Since the current into the input terminals of an ideal Opamp is zero, this current flows into the
R
2
//
C
combination connected in the feedback path of the Opamp. The voltage developed across this parallel
combination is nothing but a scaled version of the step response of a Parallel
RC Circuit with step
current excitation. The scaling factor is 1/
R
1
. This step response is
R
2
(1
-
e
-
10
t
) since the time constant
involved is 0.1 s. Therefore, the voltage developed across the parallel combination in the feedback
path is (
R
2
/
R
1
) (1
-
e
-
10
t
) V with its positive polarity at the inverting input of operational amplifier.
Since the inverting input is at
virtual ground, the voltage of output terminal with respect to ground
(reference point) is the negative of this voltage.
∴
= −
−
(
)
= − −
(
)
≥
=
=
−
−
+
v t
R
R
e
e
t
R
R
k
t
t
o
V for
since
( )
2
1
10
10
1
2
1
1
0
10
Ω
Ω
Therefore, the two circuits in Fig. 11.3-11 have the same step response (and hence same dynamic
behaviour) except for a change in sign.
The voltage across
R in the circuit of Fig. 11.3-11 (a) is [1
-
(1
-
e
-
10
t
)]
=
e
-
10
t
V and therefore the
current drawn from the unit step voltage source by this circuit is 0.1
e
-
10
t
mA for
t
≥
0
+
. But the current
drawn by the second circuit is
u(
t)/
R
1
=
0.1 mA for
t
≥
0
+
. Thus, the second circuit presents a constant
input resistance level to the applied voltage source whereas the first circuit presents a time-varying
input resistance level to the source. If the voltage source is not an ideal one,
i.e., if it has a non-zero
internal resistance, the time constant of circuit in Fig. 11.3-11 (a) will change and hence the shape of
its step response will change. However, the shape of step response will not change in the case of circuit
Fig. 11.3-11 (b); but the final magnitude will change due to change in the ratio (
R
2
/
R
1
).
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