iphone - 峰值滤波器有咔哒声和爆裂声

Question

OSStatus MusicPlayerCallback ( 
                       void *                     inRefCon, 
                       AudioUnitRenderActionFlags *   ioActionFlags, 
                       const AudioTimeStamp *         inTimeStamp, 
                       UInt32                     inBusNumber, 
                       UInt32                     inNumberFrames, 
                       AudioBufferList *            ioData) { 


MusicPlaybackState *musicPlaybackState = (MusicPlaybackState*) inRefCon; 

double sampleinp; 

double A,omega,sn,cs,alpha,Bandwidth; 

double dbGain; 

double   a0,a1,a2,b0,b1,b2; 

dbGain = 1.0; 

A=pow(10.0,dbGain/40.0); 

Bandwidth = 2.0; 

omega=2 * M_PI * 800.0/44100.0; 

sn=sin(omega); 

cs=cos(omega); 

alpha=sn*sinh(((M_LN2/2)*Bandwidth*omega)/sn); 


//Peak Filter Biquad 

b0 =1.0 + alpha * A; 

b1 = (-2.0 * cs); 

b2 = 1.0 - alpha * A; 

a0 = 1.0 + (alpha /A); 

a1 = -2.0 * cs; 

a2 = 1.0 - (alpha /A); 



double b0a0, b1a0, b2a0, a1a0, a2a0; 

double static x1; 

double static x2; 

double static y1; 

double static y2; 


b0a0=b0/a0;   

b1a0=b1/a0; 

b2a0=b2/a0; 

a1a0=a1/a0; 

a2a0=a2/a0; 


for (int i = 0 ; i < ioData->mNumberBuffers; i++){ 


  AudioBuffer buffer = ioData->mBuffers[i]; 
  AudioSampleType *outSample = buffer.mData; 

  for (int j = 0; j < inNumberFrames*2; j++){ 

     sampleinp = *musicPlaybackState->samplePtr++; 

     outSample[j] =  b0a0 * sampleinp + 
     b1a0 * x1 + 
     b2a0 * x2 - 
     a1a0 * y1 - 
     a2a0 * y2; 


     x2=x1; 
     x1=sampleinp; 

     y2=y1; 
     y1=outSample[j]; 

  }} 





return noErr; 
} 

有点击/弹出问题。有人请帮助......我不知道我做错了什么。这是在 Xcode 中使用 Objective-C 中的 C。我尝试制作 Coeff Global 和 Static 但没有成功。使用的音频文件是 .caf 我试过 .wav 但还是不行....

谢谢，对于普遍的求救声感到抱歉。我是这个网站的新手。我想在我的应用程序中添加一个峰值过滤器，但每次我使用 slider 或将增益保持在 1 时，我都会听到爆裂声和咔哒声。似乎一切都在那里，并且就保持以前的样本等而言正常工作。当改变频率或带宽时，我也会得到某种类型的相位。几个月来我一直在研究 dsp，我很困惑，我认为这是 Objective-C 的问题和一个小的用户错误。将样本更改为 SInt32 时它似乎消失了，但在更改频率时左声道消失了。

Dsp.h

typedef struct { 

  void* audioData; 

   UInt32 audioDataByteCount; 

   SInt16 *samplePtr; 

} MusicPlaybackState; 

Answer 1

根据 hotpaw2 的回答，这是您的过滤器响应的图表:

from pylab import *
import scipy.signal as signal

def biquad_peak(omega, gain_db, bandwidth):
    sn = sin(omega)
    cs = cos(omega)
    alpha = sn * sinh(log(2) / 2 * bandwidth * omega / sn)
    gain_sqrt = 10.0 ** (gain_db / 40.0)

    # feed-forward coefficients
    b0 = 1.0 + alpha * gain_sqrt
    b1 = -2.0 * cs
    b2 = 1.0 - alpha * gain_sqrt
    # feedback coefficients
    a0 = 1.0 + (alpha / gain_sqrt)
    a1 = -2.0 * cs
    a2 = 1.0 - (alpha / gain_sqrt)
    # normalize by a0
    B = array([b0, b1, b2]) / a0
    A = array([a0, a1, a2]) / a0
    return B, A

omega = 2 * pi * 800.0 / 44100.0
gain_db = 1.0
bandwidth = 2.0

B, A = biquad_peak(omega, gain_db, bandwidth)
w, H = signal.freqz(B, A)
f = w / pi * 22050.0
plot(f, abs(H), 'r')
gain = 10.0 ** (gain_db / 20.0)
print  "Gain:", gain
plot(f, gain*ones(len(f)), 'b--'); grid()

峰值增益设置为 1.1220184543(即 1 dB)。您可以看到滤波器如何使大部分可听范围的增益大于 1。

编辑 2:如果这是用于可调 EQ，则由用户设置避免失真的增益。另外，我怀疑您描述的极端问题是由典型音轨在窄带上 1 dB 的轻微增益引起的。我认为这是因为您的音频具有交错的立体声数据。这些 channel 中的每一个都需要单独过滤。我已经尝试修改您的嵌套循环来完成此任务:

a0 = 1.0 + alpha / A; 
a1 = -2.0 * cs / a0;     
a2 = (1.0 - alpha / A) / a0; 
b0 = (1.0 + alpha * A) / a0; 
b1 = -2.0 * cs / a0; 
b2 = (1.0 - alpha * A) / a0;

double static x11, x12, x21, x22;
double static y11, y12, y21, y22;
double x0, y0;

for (int i = 0; i < ioData->mNumberBuffers; i++) {  

    AudioBuffer buffer = ioData->mBuffers[i]; 
    AudioSampleType *outSample = buffer.mData; 

    for (int j = 0; j < inNumberFrames*2; j++) { 

        /* x0 is in the range of SInt16: -32768 to 32767 */

        x0 = *musicPlaybackState->samplePtr++;

        y0 = b0 * x0 + 
             b1 * x11 + 
             b2 * x12 - 
             a1 * y11 - 
             a2 * y12; 

        outSample[j] = fmax(fmin(y0, 32767.0), -32768.0); 

        x12 = x11; 
        x11 = x0;
        y12 = y11;
        y11 = y0
        j++;          

        x0 = *musicPlaybackState->samplePtr++;

        y0 =  b0 * x0 + 
              b1 * x21 + 
              b2 * x22 - 
              a1 * y21 - 
              a2 * y22; 

        outSample[j] = fmax(fmin(y0, 32767.0), -32768.0); 

        x22 = x21; 
        x21 = x0;          
        y22 = y21; 
        y21 = y0; 
    }
}