Page 197 - Analog and Digital Filter Design

1 94 Analog and Digital Filter Design

6J
J = ~ = 0.18571
QBP
n = m2 + J' + 4 = 4.0404056
Q=d 8nz2 = 13.0556778
n -t 2/12' - 16m2

H7 = Qnz + 4- = 1.09723.

The frequencies are fR1= f. = 9.068286 and fRZ = W& = 10.917444.
w

I will use fRI = 9.068286 to find the filter stage gain, Go, given that the gain at
resonance will be GR = 2@ = 340.9014456.

340.9014456
= 129.9005655.
l+Q-
fR fo

Let capacitor, C = IpF.
A

This value is too high, so let C = 2.2 pF.
R1= 104.153kQ.
R1
R2 = R3 = - = 7.977kQ.
Q
R1
R4 = - 39.687kQ.
=
GRR
Let the value of R' = 10 kQ.

Cauer and Inverse Chebyshev Active Filters

Designing bandpass filters with a Cauer or an Inverse Chebyshev response is
slightly more difficult because each filter stage must provide both poles and
zeroes close to the filter center frequency. Moreover, the pole and zero pairing
must also be considered. A filter may have a number of poles and zeroes
and, in principle, any zero could be associated with any pole. In practice the
pole-zero pairing affects performance. Pole and zero pairing are illustrated in
Figure 6.14.