使用加速框架 vDSP 的 iPhone FFT

Question

我在使用 vDSP 实现 FFT 时遇到困难。我理解这个理论，但我正在寻找一个具体的代码示例。

我的 wav 文件数据如下：

问题 1. 如何将音频数据放入 FFT 中？

问题 2. 如何从 FFT 中获取输出数据？

问题 3. 最终目标是检查低频声音。我该怎么做？

-(OSStatus)open:(CFURLRef)inputURL{
OSStatus result = -1;

result = AudioFileOpenURL (inputURL, kAudioFileReadPermission, 0, &mAudioFile);
if (result == noErr) {
    //get  format info
    UInt32 size = sizeof(mASBD);

    result = AudioFileGetProperty(mAudioFile, kAudioFilePropertyDataFormat, &size, &mASBD);

    UInt32 dataSize = sizeof packetCount;
    result = AudioFileGetProperty(mAudioFile, kAudioFilePropertyAudioDataPacketCount, &dataSize, &packetCount);
    NSLog([NSString stringWithFormat:@"File Opened, packet Count: %d", packetCount]);

    UInt32 packetsRead = packetCount;
    UInt32 numBytesRead = -1;
    if (packetCount > 0) { 
        //allocate  buffer
        audioData = (SInt16*)malloc( 2 *packetCount);
        //read the packets
        result = AudioFileReadPackets (mAudioFile, false, &numBytesRead, NULL, 0, &packetsRead,  audioData); 
        NSLog([NSString stringWithFormat:@"Read %d  bytes,  %d packets", numBytesRead, packetsRead]);
    }
}
return result;
}

FFT代码如下：

log2n = N;
n = 1 << log2n;
stride = 1;
nOver2 = n / 2;

printf("1D real FFT of length log2 ( %d ) = %d\n\n", n, log2n);

/* Allocate memory for the input operands and check its availability,
 * use the vector version to get 16-byte alignment. */

A.realp = (float *) malloc(nOver2 * sizeof(float));
A.imagp = (float *) malloc(nOver2 * sizeof(float));
originalReal = (float *) malloc(n * sizeof(float));
obtainedReal = (float *) malloc(n * sizeof(float));

if (originalReal == NULL || A.realp == NULL || A.imagp == NULL) {
printf("\nmalloc failed to allocate memory for  the real FFT"
"section of the sample.\n");
exit(0);
}

/* Generate an input signal in the real domain. */
for (i = 0; i < n; i++)

    originalReal[i] = (float) (i + 1);

/* Look at the real signal as an interleaved complex vector  by
 * casting it.  Then call the transformation function vDSP_ctoz to
 * get a split complex vector, which for a real signal, divides into
 * an even-odd configuration. */

vDSP_ctoz((COMPLEX *) originalReal, 2, &A, 1, nOver2);

/* Set up the required memory for the FFT routines and check  its
 * availability. */

setupReal = vDSP_create_fftsetup(log2n, FFT_RADIX2);

if (setupReal == NULL) {

printf("\nFFT_Setup failed to allocate enough memory  for"
"the real FFT.\n");

exit(0);
}

/* Carry out a Forward and Inverse FFT transform. */
vDSP_fft_zrip(setupReal, &A, stride, log2n, FFT_FORWARD);
vDSP_fft_zrip(setupReal, &A, stride, log2n, FFT_INVERSE);

/* Verify correctness of the results, but first scale it by  2n. */
scale = (float) 1.0 / (2 * n);
vDSP_vsmul(A.realp, 1, &scale, A.realp, 1, nOver2);
vDSP_vsmul(A.imagp, 1, &scale, A.imagp, 1, nOver2);

/* The output signal is now in a split real form.  Use the  function
 * vDSP_ztoc to get a split real vector. */
vDSP_ztoc(&A, 1, (COMPLEX *) obtainedReal, 2, nOver2);

/* Check for accuracy by looking at the inverse transform  results. */
Compare(originalReal, obtainedReal, n);

谢谢

Answer 1

1. 您将音频样本数据放入输入的实部，并将虚部归零。

2. 如果您只对频域中每个 bin 的大小感兴趣，那么您可以为每个输出 bin 计算 sqrt(re*re + im*im)。如果您只对相对幅度感兴趣，那么您可以删除 sqrt 并仅计算平方幅度，(re*re + im*im)。

3. 您将查看与您感兴趣的一个或多个频率相对应的一个或多个箱的大小（参见 (2)）。如果您的采样率为 Fs，并且 FFT 大小为 N，则输出 bin i 的相应频率由 f = i * Fs / N 给出。相反，如果您对特定频率 f 感兴趣，则感兴趣的区间 i 由以下公式给出：i = N * f / Fs。

附加说明：您需要应用合适的窗口函数（例如Hann aka Hanning) 到您的 FFT 输入数据，然后再计算 FFT 本身。

Answer 2

你可以查看Apple的文档并妥善保管数据打包。

这是我的例子：

//  main.cpp
//  FFTTest
//
//  Created by Harry-Chris Stamatopoulos on 11/23/12.
//  

/* 
 This is an example of a hilbert transformer using 
 Apple's VDSP fft/ifft & other VDSP calls.
 Output signal has a PI/2 phase shift.
 COMPLEX_SPLIT vector "B" was used to cross-check
 real and imaginary parts coherence with the original vector "A"
 that is obtained straight from the fft.
 Tested and working. 
 Cheers!
*/

#include <iostream>
#include <Accelerate/Accelerate.h>
#define PI 3.14159265
#define DEBUG_PRINT 1

int main(int argc, const char * argv[])
{
    float fs = 44100;           //sample rate
    float f0 = 440;             //sine frequency
    uint32_t i = 0;

    uint32_t L = 1024;

    /* vector allocations*/
    float *input = new float [L];
    float *output = new float[L];
    float *mag = new float[L/2];
    float *phase = new float[L/2];

    for (i = 0 ; i < L; i++)
    {
        input[i] = cos(2*PI*f0*i/fs);
    }

    uint32_t log2n = log2f((float)L);
    uint32_t n = 1 << log2n;
    //printf("FFT LENGTH = %lu\n", n);

    FFTSetup fftSetup;
    COMPLEX_SPLIT A;
    COMPLEX_SPLIT B;
    A.realp = (float*) malloc(sizeof(float) * L/2);
    A.imagp = (float*) malloc(sizeof(float) * L/2);

    B.realp = (float*) malloc(sizeof(float) * L/2);
    B.imagp = (float*) malloc(sizeof(float) * L/2);

    fftSetup = vDSP_create_fftsetup(log2n, FFT_RADIX2);

    /* Carry out a Forward and Inverse FFT transform. */
    vDSP_ctoz((COMPLEX *) input, 2, &A, 1, L/2);
    vDSP_fft_zrip(fftSetup, &A, 1, log2n, FFT_FORWARD);

    mag[0] = sqrtf(A.realp[0]*A.realp[0]);

    //get phase
    vDSP_zvphas (&A, 1, phase, 1, L/2);
    phase[0] = 0;

    //get magnitude;
    for(i = 1; i < L/2; i++){
        mag[i] = sqrtf(A.realp[i]*A.realp[i] + A.imagp[i] * A.imagp[i]);
    }

    //after done with possible phase and mag processing re-pack the vectors in VDSP format
    B.realp[0] = mag[0];
    B.imagp[0] = mag[L/2 - 1];;

    //unwrap, process & re-wrap phase
    for(i = 1; i < L/2; i++){
        phase[i] -= 2*PI*i * fs/L;
        phase[i] -= PI / 2 ;
        phase[i] += 2*PI*i * fs/L;
    }

    //construct real & imaginary part of the output packed vector (input to ifft)
    for(i = 1; i < L/2; i++){
        B.realp[i] = mag[i] * cosf(phase[i]);
        B.imagp[i] = mag[i] * sinf(phase[i]);
    }

#if DEBUG_PRINT
    for (i = 0 ; i < L/2; i++)
    {
       printf("A REAL = %f \t A IMAG = %f \n", A.realp[i], A.imagp[i]);
       printf("B REAL = %f \t B IMAG = %f \n", B.realp[i], B.imagp[i]);
    }
#endif
    //ifft
    vDSP_fft_zrip(fftSetup, &B, 1, log2n, FFT_INVERSE);

    //scale factor
    float scale = (float) 1.0 / (2*L);

    //scale values
    vDSP_vsmul(B.realp, 1, &scale, B.realp, 1, L/2);
    vDSP_vsmul(B.imagp, 1, &scale, B.imagp, 1, L/2);

    //unpack B to real interleaved output
    vDSP_ztoc(&B, 1, (COMPLEX *) output, 2, L/2);

    // print output signal values to console
    printf("Shifted signal x = \n");
    for (i = 0 ; i < L/2; i++)
        printf("%f\n", output[i]);

    //release resources
    free(input);
    free(output);
    free(A.realp);
    free(A.imagp);
    free(B.imagp);
    free(B.realp);
    free(mag);
    free(phase);
}

Answer 3

您需要注意的一件事是计算的 FFT 的直流分量。我将我的结果与 fftw 库 FFT 进行了比较，并且使用 vDSP 库计算的变换的虚部在索引 0 处始终具有不同的值（这意味着 0 频率，因此 DC）。
我应用的另一项措施是将实部和虚部除以 2。我猜这是由于函数中使用的算法所致。而且，这两个问题都发生在FFT过程中，而不是IFFT过程中。

我使用了vDSP_fft_zrip。