FFT iOS 值加倍？

Question

您好，我是运行 iOS 平台模拟器中的 fft 函数

喜欢

-(Matrix*) spectralDensityWithVec:(Matrix*) vector {

    NSLog(@"Vector:%@",vector);
    int sampleSize = [vector rows];

    double *r  = [vector array]; // arc will handle the memory

    float * f = (float*) malloc(sampleSize * sizeof(float));

    for (int i = 0;i< sampleSize;i++) {
        f[i] = r[i];
    }

    // Working with 128 samples which is 2^7
    // 7
    vDSP_Length log2n = log2([self nextPowerOf2WithNumber:sampleSize]);

    FFTSetup fftSetup = vDSP_create_fftsetup(log2n,FFT_RADIX2);

    int nOver2 = sampleSize/2;

    COMPLEX_SPLIT A;

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

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

    vDSP_fft_zrip(fftSetup, &A,1 ,log2n, FFT_FORWARD);

    Matrix* PSD = [Matrix matrixOfRows:sampleSize/2 Columns:1];

    Matrix* fftResult = [Matrix matrixOfRows:sampleSize Columns:1];

    // obtain the imaginary / real parts
    for (int i = 0;i < sampleSize/2;i++) {
        float realp = A.realp[i];
        float imagp = A.imagp[i];

        NSLog(@"real:%g img:%g",realp,imagp);

        float val = sqrtf((realp * realp) + (imagp * imagp)) * 2.0f /(float)sampleSize ;

        [PSD setValue:val Row:i Column:0];
        [fftResult setValue:realp Row:i Column:0];
        [fftResult setValue:imagp Row:i*2+1 Column:0];
    }

    NSLog(@"FFT Result KW Energy: %@",fftResult);

    vDSP_destroy_fftsetup(fftSetup);

    // release memory
    if (f) {
       free(f);
        f = NULL;
    }
    if (A.realp) {
        free(A.realp);
        A.realp = NULL;
    }
    if (A.imagp) {
        free(A.imagp);
        A.imagp = NULL;
    }

    return PSD;
}

使用这个向量，我得到了

上面向量的结果：

如何使用 numpy fft 我得到这个结果：

   [65.22805000000001, 
0.0, 
-0.0085759487914117867,
 -0.015603049865923341,
 -0.0069788925854868157,
 0.015041138788651551,
 -0.010263059930525575,
 0.012321275391622862,
 -0.021832981634872219,
 -0.026161796455543486,
 -0.0018237156004342254,
 -0.013884928458450145,
 -0.0091755911990841002,
 -0.016388522893239089,
 -0.003204667797210365,
 -0.033305086561749964,
 -0.00048255506211134121,
 -0.015783138731002323,
 -0.0075337096722261337,
 -0.0071538868820418761, 
-0.00066829905037151319,
 -0.017467879005406146,
 0.0058655519004814438, 
0.011289870623693908, 
-0.021329700649694319,
 0.0018507094402767038,
 -0.0062391410740094827,
 -0.013248826390988894, 
0.0091139373764349274,
 -0.0076327732856707516,
 -0.003359005157638993,
 0.0064289012241005097, 
-0.013430000000003162, 
-0.0029959999999995546,
 -0.0065307135571379222,
 -0.006605572447413598,
 -0.0099422652522889073,
 -0.015080331241097611,
 -0.0016757131155144375,
 -0.010765855482435674,
 -0.0028346121922537357,
 -0.014091061276246888,
 -0.0014857985251928623,
 -0.0056375351611077946,
 -0.011988911430201157,
 -0.0049400354875454924,
 -0.0099028010321836699,
 -0.0045315332341146703,
 -0.018321444937884485,
 -0.0050991387310010752,
 -0.027683215530298805,
 0.0015174037330767526,
 -0.0075530851759402738,
 0.019037454812601148,
 0.00020298122212324523, 
0.0081699113196296563,
 -0.0011547055231802498,
 0.0043124328279300489,
 -0.0014283591672329676,
 0.0029302575544578394,
 -0.001462892683058185,
 0.0018357150212062624,
 -0.0014666841715850814,
 0.00088237880086386444, 
-0.0014699999999976399,
 0.0]

我已经尝试了很多东西，但我没有想法有人知道我可能做错了什么吗？在查看文档之前，我从未在 ios 上使用过 FFT 函数，在这种情况下似乎没有帮助。

Answer 1

发现得很好。

缩放因子 2 is a property of Apple's real FFT implementation。不知道为什么会这样，也许这样更有效率。

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