feeding current output to current tick

[David Lowenfels]

I've been looking at this, and don't see how it works...
by definition the feedback should be delayed... ie it should come out 
only after you have made an output.. hence Y(n-1) <-> y * z^-1
So what is the problem?
Plus, "delay free" loops are impossible to implement in the digital domain.

>The other EE guy Tony and I were having discussion about
>the way to feed the current output to current tick, when using delay.
>It can be solved using nextOut method. just FYI.
>
>the probelm is, when designing the feedback comb filters,
>
>for(i=0; i< max ; i++){
>current_out = delay->tick( last_out + current_in);
>last_out = current_out;
>}
>
>the code above adds an extra delay on lastoutput until it is
>fed to the delay tick method. (so it's like having one delay
>on the feedback path.)
>
>for(i=0l i<max; i++){
>current_out = delay->tick(next_but_current_out + current_in);
>next_but_current_out = delay->nextOut();
>}
>
>it will solve the problem.


-- 
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   David Lowenfels       it has as its ultimate goal the betterment
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