[chuck-users] analog to digital conversion simulation

eduard aylon eduard.aylon at gmail.com
Fri Oct 12 14:05:21 EDT 2007


Sorry... I think I get were the problem is. I am using the same lpf  
before and after conversion. Using a separate lpf gives better  
results. One of the problems is that I was passing lpf.last to  
pulse.next, but pulse is also connected to lpf in the patch.


On Oct 12, 2007, at 6:41 PM, eduard wrote:

> Hi list,
>
> I'm trying to simulate how adc conversion works using chuck. However,
> I'm not succeeding as I'm getting too many artifacts in the output
> sound. Don't know whether i'm getting this wrong from a conceptual
> point of view or I'm using/setting chuck's UGens wrongly. The patch I
> use is:
>
> <<<"usage:", "sample_and_hold.ck:output_gain:sr" >>>;
> adc => OnePole lpf => blackhole;
> Impulse pulse => lpf => Gain g =>  dac;
> .8 => g.gain;
> if( me.args() ) Std.atof( me.arg(0) ) => g.gain;
> 1::second/1::samp => float chuck_sr => float new_sr;
> if( me.args() > 1 ) Std.atof( me.arg(1) ) => new_sr;
> chuck_sr/new_sr => float next_samp;
> 1. - new_sr/chuck_sr => lpf.pole;
> <<<lpf.pole()>>>;
> <<< next_samp>>>;
> while(true)
> {
>      now => time T;
>      next_samp::samp +=> T;
>      lpf.last() => pulse.next;
>      while( now < T )
>      {
>          1::samp => now;
>      }
> }
>
>
>
> Then the user should be able to give a certain SR and experience how
> the input sound would sound when sampled at SR.
> I'd like to be able to set lpf cutoff frequency at SR/2 and what I
> thougt of is something like:
>
> 1. - new_sr/chuck_sr => lpf.pole;
>
> thus,
> 	if user_sr= 44100, then pole=0 and cutoff freq=chuck_sr/2 (should be
> 22050 if running at 44.1).
> 	if user_sr=22050, then pole=0.5 and cutoff freq=chuck_sr/4 (should
> be 11025 if running at 44.1)	
> 	etc...
>
> Am I doing this right? On the other hand, a Onepole filter will not
> suffice to get rid off all frequencies above the cutoff. Is there a
> way of having a kind of brickwall filter in chuck?
>
> Thanks,
>
> eduard
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