One pole filter code - does this look right?

[seancostello]

Hi all:

I'm working on a simple one pole filter, where the frequency is variable from a pot or other code. This is what I have right now:

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equ		lpcoef	reg10	; frequency coefficient
equ		lpstate	reg11	; filter state

rdax	lpstate, -1.0	; subtract lpstate from current value in accum
mulx	lpcoef		; multiply by lpcoef
rdax	lpstate, 1.0	; add lpstate back to (input-lpstate)*lpcoef
wrax	lpstate, 1.0	; write filter output back to lpstate, 
				; while keeping value in the accumulator

Does this look about right? Any suggestions to make this more efficient? 4 instructions isn't that bad, but I like to shave cycles wherever possible.

Also, the coefficient is computed (in C code) as

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1.f-exp(-2.f*PI*cutoff/samplingRate);

Any hints for calculating this on the FV-1? I don't need a precise cutoff, as this would vary with clock speed. Just wondering if there are any fixed point tricks I can keep in mind. I figure the EXP instruction is the goto one for this calculation, but any hints of taking a knob input, and transforming it into a usable range for a tone control or variable LPF, are most welcome.

Thanks,

Sean Costello

[frank]

Hmmm, since you want to calculate the cut-off coefficient on the fly I think that is about the best that can be done. If the coefficient was fixed then RDFX would knock the instruction count down.

Generally what I do to use a pot as a coefficient is to determine the extrema values I want as coefficients, read the pot and maybe adjust the curve of it if linear is not what I want (i.e. square the pot value to make it more exponential) then use something like SOF to scale and offset the result to range between the min and max value. Avoids any exponential or power calculations in the code.