of Microprocessors

486 MUSICAL ApPLICATIONS OF MICROPROCESSORS
Note that the expression for finding K or L depends on the ratio of
cutoff frequency to sample frequency. This should have been expected, since
the same string of sample values can represent entirely different signal frequencies
if the sample rates are different. Thus, it is customary in digital
filter work to always specify frequency as a fraction of the sample rate rather
than in hertz. Amplitude response plots, therefore, have the frequency axis
calibrated from zero (or some lower limit if a log scale) to 0.5.
There is still one undesirable effect in the digital filter. It has a substantial
amount of passband gain. In fact, as K is adjusted for lower cutoff
frequencies, the gain increases in inverse proportion to K. This is of no
immediate concern with the floating-point arithmetic in BASIC but later,
when the filter arithmetic is converted to integers for increased speed, it can
become a real headache. The amount of dc gain is easily determined by
noting that for a constant input of 1.0 the output will rise until the amount
removed from the accumulator each sample period via leakage is equal to the
amount added via the input. Thus, the dc gain is 11K or 1/(l-L). The best
way to counteract the gain is to multiply the input samples by the inverse, K,
before adding. The final filter statement therefore is:
1000 LET A=A*L+K*I
Note that two multiplications and one addition are required for each
sample processed. By rearranging constants and allowing large numbers in
the accumulator, one of the multiplications can be eliminated:
1000 LET O=K*A
1001 LET A=A-O+I
0
-2
-4
-6
-8
-10
-12
-14
-16
_18~
-20
I
0005 0.01 0.02 0.05
FREQUENCY F/Fs
0.1 0.2
DIGITAL
FILTER
I
! II'
0.5
Fig. 14-3. Measured response of digital low-pass filter