带有 Accelerate 框架 vDSP 的 iPhone FFT

Question

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

我有来自 wav 文件的数据，如下所示：

问题1.如何将音频数据放入FFT？

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 的幅度感兴趣，那么您可以为每个输出 bin 计算 sqrt(re*re + im*im)。如果您只对 relative 幅度感兴趣，那么您可以放弃 sqrt 并计算平方幅度，(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 bin 的幅度（参见 (2)）。如果您的采样率为 Fs，并且您的 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，那么感兴趣的 bin，i，由i = N * f / Fs 给出。

附加说明：在计算 FFT 本身之前，您需要将合适的 window function（例如 Hann aka Hanning）应用于您的 FFT 输入数据。

Answer 2

您可以查看 Apple 的文档并注意数据打包。

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出的 FFT 的直流分量。我将我的结果与 fftw 库 FFT 进行了比较，使用 vDSP 库计算的变换的虚部始终在索引 0 处具有不同的值（这意味着 0 频率，因此是 DC）。 我应用的另一项措施是将实部和虚部都除以 2。我猜这是由于函数中使用的算法。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。