Question 7.4: Derive a mathematical expression for the output voltage in t...

The circuit shown in Fig. 7.22 is similar to the inverting amplifier circuit shown in Fig. 7.17, except that in this active filter a capacitor is used instead of resistance R_{2} in the feedback path. To follow the steps used in the analysis of an inverting amplifier in Subsection 7.5.1, start with Kirchhoff’s current law equation at terminal 2,

i_{R} =-i_{C},          (7.66)

where the two currents can be expressed as

i_{R}=\frac{e_{i} }{R},          (7.67)

i_{C} =C\frac{de_{o} }{dt}.          (7.68)

Substituting the expressions for currents into Eq. (7.66) gives

\frac{e_{i} }{R}=-C\frac{de_{o} }{dt},          (7.69)

and hence the output voltage is

e_{o} =-\frac{1}{RC } \int_{0}^{t}{e_{i}dt } +e_{o} \left(0\right) .         (7.70)

Thus the output voltage is proportional to the integral of the input voltage with a gain of -\left({1}/{RC}\right) . To change the gain to a positive number, so that the output signal is positive when an integral of the input signal is positive, the output of this circuit can be connected to an inverting amplifier with R_{1}= R_{2}, which results in a voltage gain of –1. A complete circuit diagram is shown in Fig. 7.23. With the initial condition set to zero, the output voltage from this circuit is

e_{o} =-\frac{1}{RC } \int_{0}^{t}{e_{i}dt } .          (7.71)

A simulation block diagram for this filter acting as a noninverting integrator is shown in Fig. 7.24.