Sorry, the timer needs to reset even when no new peaks are detected, so
something like this:
peak_hold(HT, x) = loop
~ _
with {
loop(fb) = ba.sAndH(reset | (timer == 0), abs(x))
with {
reset = abs(x) >= fb;
timer = % (rint(HT * ma.SR)) ~ (+ (1) * (1 - reset));
};
};
Although, the hold time is correct when the timer cycles around, but it is
one sample behind when it is reset by a new peak.
For example, if the step function is 10 samples and the hold period is
longer, "reset" is 1 from n=0 to n=9, timer is 0 for those samples, but
then it starts at 1 at n=10.
I'll think of a better way.
Cheers,
Dario
On Sun, 21 Jun 2020 at 15:44, Dario Sanfilippo <sanfilippo.da...@gmail.com>
wrote:
> Hello, list.
>
> I noticed that the step response of ba.peakholder and the step function
> itself are the same when the step size and the hold time are the same.
>
> In other words, an input that is equal to the output is not detected as a
> new peak and it doesn't reset the countdown. Is that the desired behaviour?
>
> I would have thought that, if the input is 1 for 48 samples, and the hold
> time is 48 samples, we should see 1s from n = 0 to n = 95.
>
> This is how I'm doing it:
>
> peak_hold(HT, x) = loop
>
> ~ _
>
> with {
>
> loop(fb) = ba.sAndH(reset | timer, abs(x))
>
> with {
>
> reset = abs(x) >= fb;
>
> timer = fi.pole(1 - reset, 1 - reset) > HT * ma.SR;
>
> };
>
> };
>
>
> Best,
> Dario
>
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