Analog Filter Implementation Using Active Filters - Practical Analog and Digital Filter Design - page 98

Solution: A third-order equivalent lowpass filter is needed, which indicates
that a sixth-order bandpass function will result as shown below:

H a ( s ) =

3

( 7282 ⋅ s )

2 6 2 6 2 6

( s + 7 282 ⋅ s + 78 . 96 ⋅ 10 ) ⋅ ( s + 2 402 ⋅ s + 38 . 88 ⋅ 10 ) ⋅ ( s + 4 880 ⋅ s + 160 . ⋅ 10 )

By picking C = 0.01 µF and R A = 10 kΩ, the remaining values can be
determined by matching the three quadratic terms:

m R m K m R Bm GA m

0 15.92 kΩ 2.8410 18.41 kΩ 2.4512

1 22.68 kΩ 3.4550 24.55 kΩ 2.0919

2 11.17 kΩ 3.4550 24.55 kΩ 4.2481

The combined gain produced by the active filter stages at ω = ω o is
determined to be 21.783, which can be compensated by a voltage divider at the
output of the filter. If the output resistance is to be 10 kΩ, the voltage divider
resistor values are given below. (Note that in this case, if a voltage divider were
used, a gain of over 21 (over 40 dB) would be sacrificed. It would be better to
include this in the system gain.)

R = 218 k Ω , R = 10.5 k Ω

x y

The resulting bandpass filter is shown in Figure 4.9.

Figure 4.9 Butterworth bandpass active filter for Example 4.3.

4.5 BANDSTOP ACTIVE FILTERS USING OP-AMPS

Figure 4.10 shows a twin-tee bandstop active filter stage that can implement a
variety of second-order functions. The admittance labeled Y can represent a
conductance G , or a susceptance sC , or can have zero value (not be present at all).