boris

I am trying to make an audio rate counter. If I connect [sig~ 1] to a [+~] and the outlet of the [+~] to a [delay~ 1 1] and then back into the right inlet of the [+~], I would expect the output to be 1 higher every time.

But, it doesn't work. Why not? Is this because of signal vector things?

with-delay-and-feedback-doesnt-work

You can't have a loop in DSP chain. Try with tap{in,out}~, but keep in mind it will (at best) count vectors, not samples. Look up +=~, and of course gen~ is natural environment for this kind of processing.

it works fine with a send~ and receive~ object. just place the output of the delay to the send~ and the receive~ back into the delay~

That's because a send~ receive~ pair automatically introduces a sample's delay. It will still work obviously but it's just something to bare in mind if you want sample precision.

Ah no kidding? Well that's useful to know in the future. Thanks!

Just to clarify, in this case (because there's a loop), send~/receive~ introduce one "signal vector" of delay.

I might be oversimplifying here but if your goal is to have an audio rate counter checking out the [count~] object might be useful.

[+~] with [delay~] and feedback doesn't work