IIR 梳状滤波器帮助
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);
}
}
}
好吧，我得到了很多信息，但我在实现它时遇到了困难。有人可以帮忙吗，这可能是我很容易忘记的事情。它只是正常播放，有一点增强，但没有延迟。
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.
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评论（1）
您对 x0[] 变量的处理看起来不正确 - 按照您的方式，左通道和右通道将混合在一起。您分配给左声道的
x0[j]，然后用右通道数据覆盖
x0[j]。因此延迟信号x0[jD]将始终对应于右通道，延迟的左通道数据丢失。
您没有说您的采样率是多少，但对于典型的音频应用程序，
三个样本的延迟可能不会产生太大的听觉效果。在 44.1 ksamp/秒时，
使用 3 个样本延迟时，滤波器响应的波峰和波谷将位于
14,700 Hz 的倍数。您将得到的只是音频中的单个峰值
范围，在频谱的一部分，几乎没有任何功率（假设
信号是语音或音乐）。
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 tox0[j]for the left channel, thenoverwrite
x0[j]with the right channel data. So the delayed signalx0[j-D]willalways 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).