iPhone Accelerate Framework FFT vs Matlab FFT

accelerate-framework, fft, iphone, matlab, objective-c

Solution

One big problem: C arrays are indexed from 0, unlike MATLAB arrays which are 1-based. So you need to change your loop from

for(int i = 1; i <= 16; i++)

to

for(int i = 0; i < 16; i++)

A second, big problem - you're mixing single precision (`float`) and double precision (`double`) routines. Your data is `double` so you should be using `vDSP_ctozD`, not `vDSP_ctoz`, and `vDSP_fft_zripD` rather than `vDSP_fft_zrip`, etc.

Another thing to watch out for: different FFT implementations use different definitions of the DFT formula, particularly in regard to scaling factor. It looks like the MATLAB FFT includes a 1/N scaling correction, which most other FFTs do not.

Here is a complete working example whose output matches Octave (MATLAB clone):

#include <stdio.h>
#include <stdlib.h>
#include <Accelerate/Accelerate.h>

int main(void)
{
    const int log2n = 4;
    const int n = 1 << log2n;
    const int nOver2 = n / 2;

    FFTSetupD fftSetup = vDSP_create_fftsetupD (log2n, kFFTRadix2);

    double *input;

    DSPDoubleSplitComplex fft_data;

    int i;

    input = malloc(n * sizeof(double));
    fft_data.realp = malloc(nOver2 * sizeof(double));
    fft_data.imagp = malloc(nOver2 * sizeof(double));

    for (i = 0; i < n; ++i)
    {
       input[i] = (double)(i + 1);
    }

    printf("Input\n");

    for (i = 0; i < n; ++i)
    {
        printf("%d: %8g\n", i, input[i]);
    }

    vDSP_ctozD((DSPDoubleComplex *)input, 2, &fft_data, 1, nOver2);

    printf("FFT Input\n");

    for (i = 0; i < nOver2; ++i)
    {
        printf("%d: %8g%8g\n", i, fft_data.realp[i], fft_data.imagp[i]);
    }

    vDSP_fft_zripD (fftSetup, &fft_data, 1, log2n, kFFTDirection_Forward);

    printf("FFT output\n");

    for (i = 0; i < nOver2; ++i)
    {
        printf("%d: %8g%8g\n", i, fft_data.realp[i], fft_data.imagp[i]);
    }

    for (i = 0; i < nOver2; ++i)
    {
        fft_data.realp[i] *= 0.5;
        fft_data.imagp[i] *= 0.5;
    }

    printf("Scaled FFT output\n");

    for (i = 0; i < nOver2; ++i)
    {
        printf("%d: %8g%8g\n", i, fft_data.realp[i], fft_data.imagp[i]);
    }

    printf("Unpacked output\n");

    printf("%d: %8g%8g\n", 0, fft_data.realp[0], 0.0); // DC
    for (i = 1; i < nOver2; ++i)
    {
        printf("%d: %8g%8g\n", i, fft_data.realp[i], fft_data.imagp[i]);
    }
    printf("%d: %8g%8g\n", nOver2, fft_data.imagp[0], 0.0); // Nyquist

    return 0;
}

Output is:

Input
0:        1
1:        2
2:        3
3:        4
4:        5
5:        6
6:        7
7:        8
8:        9
9:       10
10:       11
11:       12
12:       13
13:       14
14:       15
15:       16
FFT Input
0:        1       2
1:        3       4
2:        5       6
3:        7       8
4:        9      10
5:       11      12
6:       13      14
7:       15      16
FFT output
0:      272     -16
1:      -16 80.4374
2:      -16 38.6274
3:      -16 23.9457
4:      -16      16
5:      -16 10.6909
6:      -16 6.62742
7:      -16  3.1826
Scaled FFT output
0:      136      -8
1:       -8 40.2187
2:       -8 19.3137
3:       -8 11.9728
4:       -8       8
5:       -8 5.34543
6:       -8 3.31371
7:       -8  1.5913
Unpacked output
0:      136       0
1:       -8 40.2187
2:       -8 19.3137
3:       -8 11.9728
4:       -8       8
5:       -8 5.34543
6:       -8 3.31371
7:       -8  1.5913
8:       -8       0

Comparing with Octave we get:

octave-3.4.0:15> x = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 ]
x =

    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16

octave-3.4.0:16> fft(x)
ans =

 Columns 1 through 7:

   136.0000 +   0.0000i    -8.0000 +  40.2187i    -8.0000 +  19.3137i    -8.0000 +  11.9728i    -8.0000 +   8.0000i    -8.0000 +   5.3454i    -8.0000 +   3.3137i

 Columns 8 through 14:

    -8.0000 +   1.5913i    -8.0000 +   0.0000i    -8.0000 -   1.5913i    -8.0000 -   3.3137i    -8.0000 -   5.3454i    -8.0000 -   8.0000i    -8.0000 -  11.9728i

 Columns 15 and 16:

    -8.0000 -  19.3137i    -8.0000 -  40.2187i

octave-3.4.0:17>

Note that the outputs from 9 to 16 are just a complex conjugate mirror image or the bottom 8 terms, as is the expected case with a real-input FFT.

Note also that we needed to scale the vDSP FFT by a factor of 2 - this is due to the fact that it's a real-to-complex FFT, which is based on an N/2 point complex-to-complex FFT, hence the outputs are scaled by N/2, whereas a normal FFT would be scaled by N.

Problem

Original source