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

    boris
    Jan 24 2014 | 12:20 am
    Hello,
    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?
    Thanks in advance!
    Boris
    • ak
      Jan 24 2014 | 4:06 pm
      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.
    • tonydouglas
      Jan 25 2014 | 1:04 am
      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~
      like this
    • Cphas
      Jan 25 2014 | 6:42 pm
      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.
    • tonydouglas
      Jan 26 2014 | 5:44 am
      Ah no kidding? Well that's useful to know in the future. Thanks!
    • Emmanuel Jourdan
      Jan 26 2014 | 1:10 pm
      Just to clarify, in this case (because there's a loop), send~/receive~ introduce one "signal vector" of delay.
    • Luigi Castelli
      Jan 27 2014 | 7:29 am
      I might be oversimplifying here but if your goal is to have an audio rate counter checking out the [count~] object might be useful.
      - Luigi