Help with IIR Comb Filter - iphone

Question

Reverb.m
#define D 1000
OSStatus MusicPlayerCallback(
void* inRefCon,
AudioUnitRenderActionFlags * ioActionFlags,
const AudioTimeStamp * inTimeStamp,
UInt32 inBusNumber,
UInt32 inNumberFrames
AudioBufferList * ioData){
MusicPlaybackState *musicPlaybackState = (MusicPlaybackState*) inRefCon;
//Sample Rate 44.1
float a0,a1;
double y0, sampleinp;
//Delay Gain
a0 = 1;
a1 = 0.5;
for (int i = 0; i< ioData->mNumberBuffers; i++){
AudioBuffer buffer = ioData->mBuffers[i];
SIn16 *outSampleBuffer = buffer.mData;
for (int j = 0; j < inNumberFrames*2; j++) {
//Delay Left Channel
sampleinp = *musicPlaybackState->samplePtr++;
/* IIR equation of Comb Filter
y[n] = (a*x[n])+ (b*x[n-D])
*/
y0 = (a0*sampleinp) + (a1*sampleinp-D);
outSample[j] = fmax(fmin(y0, 32767.0), -32768.0);
j++;
//Delay Right Channel
sampleinp = *musicPlaybackState->samplePtr++;
y0 = (a0*sampleinp) + (a1*sampleinp-D);
outSample[j] = fmax(fmin(y0, 32767.0), -32768.0);
}
}
}
Ok, I got a lot of info but I'm having trouble implementing it. Can someone help, it's probably something really easy i'm forgeting. It's just playing back as normal with a little boost but no delays.

Answer 1

Your treatment of the x0[] variables doesn't look right -- the way you have it, the left and right channels will be intermingled. You assign to x0[j] for the left channel, then
overwrite x0[j] with the right channel data. So the delayed signal x0[j-D] will
always correspond to the right channel, with the delayed left channel data being lost.
You didn't say what your sample rate is, but for a typical audio application, a
three-sample delay might not have much of an audible effect. At 44.1 ksamp/sec,
with a 3-sample delay the peaks and troughs of the filter response will be at
multiples of 14,700 Hz. All you'll get is a single peak in the audio frequency
range, in a part of the spectrum where there's hardly any power (assuming the
signal is speech or music).