Question 14.12: Design a lowpass active filter with a dc gain of 4 and a cor......

From Eq. (14.61), we find

$ω_{c} = \frac{1}{R_{f}C_{f}}$                 (14.61)

$ω_{c} = 2πf_{c} = 2π (500) = \frac{1}{R_{f}C_{f}}$              (14.12.1)

The dc gain is

$H(0) = – \frac{R_{f}}{R_{i}} = – 4$            (14.12.2)

We have two equations and three unknowns. If we select $C_{f} = 0.2  μF ,$ then

$R_{f} = \frac{1}{2π(500) 0.2  ×  10^{-6}} = 1.59  kΩ$

and

$R_{i} = \frac{R_{f}}{4} = 397.5  Ω$

We use a  1.6-kΩ resistor for $R_{f}$  and a 400-Ω  resistor for $R_{i}$  Figure 14.42 shows the filter.