Page 353 - Analog and Digital Filter Design

350 Analog and Digital Filter Design

G,
G,, = VO,~. - G,.
SO, G2 = 0.14204 x 20.433 - GI

G2 = 1.902092
G2)
and G3 = 0.2701 75 x 20.433 - (G1 i-
G3 = 2.617992

This process continues for the remaining conductance values:

G4 = 3.07766
G5 = 3.236025

G6 = 3.07766
G7 = 2.617992
G8 = 1.902092
G9= 1.

Since the signal generated will be symmetrical, the conductance values are also
symmetrical. There is an odd resistor value, R5 in this case (corresponding
to conductance G5), that will have the lowest resistance. This value can then
be denormalized to have a suitable resistance and other resistor values can be
calculated in relation to it. In the examples R5 is 22.1 kQ, an E96 value. To find
the resistor values, the ratio of conductance values, G5/G?z, must be multiplied
by the smallest resistor value used:

Rn = R5 x G5/Gn
This means that R1= R9 = 3.236025 x 22.1 kQ/Gl= 71.516 kQ, the
nearest E96 value is 71.5 kQ. In other words, since GI=l, Rn = Rl/Gfi
R2 = R8 = 71.516 k0/1.902092 = 37.5986 kQ, the nearest E96 value is
37.4 kQ.
R3 = R7 = 71.516 kQl2.617992 = 27.317 k0, the nearest E96 value is
27.4 kQ.
R4 = R6 = 71.516 kQ/3.07766= 23.237 kQ, the nearest E96 value is
23.2 kQ.

Synthesizer Filtering

If the sine wave generator has a fixed frequency, perhaps deriving its clock from
a crystal oscillator, the lowpass filter on its output can be a simple active filter
using op-amps. If precision resistors are used in the summing circuit, the har-