WAV-file analysis C (libsndfile, fftw3)

Question

I'm trying to develop a simple C application that can give a value from 0-100 at a certain frequency range at a given timestamp in a WAV-file.

Example: I have frequency range of 44.1kHz (typical MP3 file) and I want to split that range into n amount of ranges (starting from 0). I then need to get the amplitude of each range, being from 0 to 100.

What I've managed so far:

Using libsndfile I'm now able to read the data of a WAV-file.

infile = sf_open(argv [1], SFM_READ, &sfinfo);

float samples[sfinfo.frames];

sf_read_float(infile, samples, 1);

However, my understanding of FFT is rather limited. But I know it's required inorder to get the amplitudes at the ranges I need. But how do I move on from here? I found the library FFTW-3, which seems to be suited for the purpose.

I found some help here: https://stackoverflow.com/a/4371627/1141483

and looked at the FFTW tutorial here: http://www.fftw.org/fftw2_doc/fftw_2.html

But as I'm unsure about the behaviour of the FFTW, I don't know to progress from here.

And another question, assuming you use libsndfile: If you force the reading to be single channeled (with a stereo file) and then read the samples. Will you then actually only be reading half of the samples of the total file? As half of them being from channel 1, or does automaticly filter those out?

Thanks a ton for your help.

EDIT: My code can be seen here:

double blackman_harris(int n, int N){
double a0, a1, a2, a3, seg1, seg2, seg3, w_n;
a0 = 0.35875;
a1 = 0.48829;
a2 = 0.14128;
a3 = 0.01168;

seg1 = a1 * (double) cos( ((double) 2 * (double) M_PI * (double) n) / ((double) N - (double) 1) );
seg2 = a2 * (double) cos( ((double) 4 * (double) M_PI * (double) n) / ((double) N - (double) 1) );
seg3 = a3 * (double) cos( ((double) 6 * (double) M_PI * (double) n) / ((double) N - (double) 1) );

w_n = a0 - seg1 + seg2 - seg3;
return w_n;
}

int main (int argc, char * argv [])
{   char        *infilename ;
SNDFILE     *infile = NULL ;
FILE        *outfile = NULL ;
SF_INFO     sfinfo ;


infile = sf_open(argv [1], SFM_READ, &sfinfo);

int N = pow(2, 10);

fftw_complex results[N/2 +1];
double samples[N];

sf_read_double(infile, samples, 1);


double normalizer;
int k;
for(k = 0; k < N;k++){
    if(k == 0){

        normalizer = blackman_harris(k, N);

    } else {
        normalizer = blackman_harris(k, N);
    }

}

normalizer = normalizer * (double) N/2;



fftw_plan p = fftw_plan_dft_r2c_1d(N, samples, results, FFTW_ESTIMATE);

fftw_execute(p);


int i;
for(i = 0; i < N/2 +1; i++){
    double value = ((double) sqrtf(creal(results[i])*creal(results[i])+cimag(results[i])*cimag(results[i]))/normalizer);
    printf("%f\n", value);

}



sf_close (infile) ;

return 0 ;
} /* main */

Answer 1

Well it all depends on the frequency range you're after. An FFT works by taking 2^n samples and providing you with 2^(n-1) real and imaginary numbers. I have to admit I'm quite hazy on what exactly these values represent (I've got a friend who has promised to go through it all with me in lieu of a loan I made him when he had financial issues ;)) other than an angle around a circle. Effectively they provide you with an arccos of the angle parameter for a sine and cosine for each frequency bin from which the original 2^n samples can be, perfectly, reconstructed.

Anyway this has the huge advantage that you can calculate magnitude by taking the euclidean distance of the real and imaginary parts (sqrtf( (real * real) + (imag * imag) )). This provides you with an unnormalised distance value. This value can then be used to build a magnitude for each frequency band.

So lets take an order 10 FFT (2^10). You input 1024 samples. You FFT those samples and you get 512 imaginary and real values back (the particular ordering of those values depends on the FFT algorithm you use). So this means that for a 44.1Khz audio file each bin represents 44100/512 Hz or ~86Hz per bin.

One thing that should stand out from this is that if you use more samples (from whats called the time or spatial domain when dealing with multi dimensional signals such as images) you get better frequency representation (in whats called the frequency domain). However you sacrifice one for the other. This is just the way things go and you will have to live with it.

Basically you will need to tune the frequency bins and time/spatial resolution to get the data you require.

First a bit of nomenclature. The 1024 time domain samples I referred to earlier is called your window. Generally when performing this sort of process you will want to slide the window on by some amount to get the next 1024 samples you FFT. The obvious thing to do would be to take samples 0->1023, then 1024->2047, and so forth. This unfortunately doesn't give the best results. Ideally you want to overlap the windows to some degree so that you get a smoother frequency change over time. Most commonly people slide the window on by half a window size. ie your first window will be 0->1023 the second 512->1535 and so on and so forth.

Now this then brings up one further problem. While this information provides for perfect inverse FFT signal reconstruction it leaves you with a problem that frequencies leak into surround bins to some extent. To solve this issue some mathematicians (far more intelligent than me) came up with the concept of a window function. The window function provides for far better frequency isolation in the frequency domain though leads to a loss of information in the time domain (ie its impossible to perfectly re-construct the signal after you have used a window function, AFAIK).

Now there are various types of window function ranging from the rectangular window (effectively doing nothing to the signal) to various functions that provide far better frequency isolation (though some may also kill surrounding frequencies that may be of interest to you!!). There is, alas, no one size fits all but I'm a big fan (for spectrograms) of the blackmann-harris window function. I think it gives the best looking results!

However as I mentioned earlier the FFT provides you with an unnormalised spectrum. To normalise the spectrum (after the euclidean distance calculation) you need to divide all the values by a normalisation factor (I go into more detail here).

this normalisation will provide you with a value between 0 and 1. So you could easily multiple this value by 100 to get your 0 to 100 scale.

This, however, is not where it ends. The spectrum you get from this is rather unsatisfying. This is because you are looking at the magnitude using a linear scale. Unfortunately the human ear hears using a logarithmic scale. This rather causes issues with how a spectrogram/spectrum looks.

To get round this you need to convert these 0 to 1 values (I'll call it 'x') to the decibel scale. The standard transformation is 20.0f * log10f( x ). This will then provide you a value whereby 1 has converted to 0 and 0 has converted to -infinity. your magnitudes are now in the appropriate logarithmic scale. However its not always that helpful.

At this point you need to look into the original sample bit depth. At 16-bit sampling you get a value that is between 32767 and -32768. This means your dynamic range is fabsf( 20.0f * log10f( 1.0f / 65536.0f ) ) or ~96.33dB. So now we have this value.

Take the values we've got from the dB calculation above. Add this -96.33 value to it. Obviously the maximum amplitude (0) is now 96.33. Now didivde by that same value and you nowhave a value ranging from -infinity to 1.0f. Clamp the lower end to 0 and you now have a range from 0 to 1 and multiply that by 100 and you have your final 0 to 100 range.

And that is much more of a monster post than I had originally intended but should give you a good grounding in how to generate a good spectrum/spectrogram for an input signal.

and breathe

Further reading (for people other than the original poster who has already found it):

Converting an FFT to a spectogram

Edit: As an aside I found kiss FFT far easier to use, my code to perform a forward fft is as follows:

CFFT::CFFT( unsigned int fftOrder ) :
    BaseFFT( fftOrder )
{
    mFFTSetupFwd    = kiss_fftr_alloc( 1 << fftOrder, 0, NULL, NULL );
}

bool CFFT::ForwardFFT( std::complex< float >* pOut, const float* pIn, unsigned int num )
{
    kiss_fftr( mFFTSetupFwd, pIn, (kiss_fft_cpx*)pOut );
    return true;
}