Page 192 - Analog and Digital Filter Design

189
Bandpass Filters

From before, Q = /;16m' 5.388756
=
8111-
FF' = QUZ + @ZT = 1.039375
hl
=
The frequencies are fRI = - 9.57306 and fR: = Vlfb = 10.34178.
m7
Consider the pole with-f,, = 9.57306 and GR = 2Q' = 58.07738.
Let c = 1pF

Note this is somewhat higher than the nominal lOOldz maximum, so the value
of C must be increased to allow the value of R3 to be reduced. However, you
also have to take into account the suggested minimum resistance of 1 kQ and
R3
the fact that Rl = 7 and 4Q' = 116.15476. Thus, reducing the value of R3 to
4Q-
below lOOkR will also reduce the value of R1 below 1 WL. If C is increased
in value to 1.5pF (multiplying by a factor of IS), R3 is reduced in value to
119.453 kQ and Rl then becomes 1.028 kQ. This is a reasonable compromise.

Now the gain, Go, at the bandpass center frequency,.fo, needs to be found. The
gain is given by:

GR =2Q2 = 58.07738, at fRI = 9.57306radIs.

The center frequency is given by fo =,/m and is 9.95rads. Substituting
these values into the equation for the center frequency gain gives, Go = 53.61636.

The gain at resonance, GR was previously found to be GR = 58.07738. The gain
needed at resonance in order to give a gain of 1 at the passband center frequency
is given by:

Attenuation is required, so a potential divider formed from two resistors Rl and
R2 is needed. The value of R1 has to be changed from the initial value calcu-
lated earlier, and resistor R2 has to be introduced.