feeding current output to current tick

[Jeffrey Traer Bernstein]

ps Hiroko...

use ++i in loops instead of i++, it's pre-increment so it's faster

David Lowenfels wrote:

> 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.
>
>
>