Explanation of binning in frequency analysis post-FFT

Question

I am creating an Android app which records sound for t seconds at a sampling rate of 44.1kHz with a buffer size of 8192 (16 bit mono samples). I need to plot a graph of amplitude against frequency, and so first I need to run an FFT over the buffer each time it is full (which happens multiple times every second, more so with an increasing sampling rate). However, after I run the FFT I understand I need to 'bin' the frequencies and then apply an A-weighting to each bin. From the Nyquist criterion, I understand that I only need to look at the first half of the frequencies, and disregard the first corresponding to the DC component. However, I am unsure of what 'binning' actually means and would like some clarification as to what it is, and how to choose what goes in what bin.

Please advise :) .

Answer 1

In its simplest form, the 'binning' process consists in summing the energies (squared magnitude) within groups of adjacent FFT values. This will give you the total energy in a set of disjoint frequency bands.

A more elaborate form consists in using overlapping triangle filter banks - you compute a weighted sum of the energy in a range of FFT bins to get a number which can be interpreted as the energy measured at the output of a band-pass filter of a given center frequency/width. For example, one can use this technique to build a visualization which uses a 1/3 octave scale on the frequency axis.

Answer 2

If you desire to plot a graph with other than 4096 points (half the FFT length) on the frequency axis, then you will have to resample the FFT result. This re-binning may have to composite several FFT magnitude bins into each plot point bin, depending on your desired x axis plot resolution.

Answer 3

If all you need to plot the FFT spectrum (ie plot amplitude vs frequency), then just get an array with the discrete FFT bin frequencies and plot it against the amplitudes.The frequencies only exist at the discrete locations given by

$$ f = n\frac{f_s}{N} - \frac{f_s}{2} \hspace{2mm} \text{where n } \hspace{0.5mm} = 0, 1, 2,...N $$

and the amplitudes A are given by

$$ A = \text{Magnitude(Data[n])} \hspace{2mm} \text{where n } \hspace{0.5mm} = 0, 1, 2,...N $$

If you have a 8192 long Audio signal for example, you would do something similar to what I have done in the small dirty pseudo code / C# code sample I have written below.

N = 8192;
Sample_Rate = 44100;
float[] myAudio = new Float[N];
myAudio = getAudio();
Complex[] myFFT_Audio = FFT(myAudio);  //Returns N complex values

//Get Array of frequency Values
float[] frequencies = new float[N]
for( i = 0 ; i < N; i++)
   { frequencies[i] = i * (Sample_Rate / N) - (Sample_Rate / 2); }

//Get Amplitudes
float[] Amplitudes = new float[N]
for( i = 0; i < N; i++)
  { Amplitudes[i] = myFFT_Audio[i].Magnitude(); }

//Plot Frequencies vs Amplitudes - will plot for frequencies -fs/2 to fs/2
PLOT(frequencies , Amplitudes);

If you want to only plot the values from $0$ to $f_s/2$ instead of from $-f_s/2$ to $f_s/2$ you can discard the first half of the values and just plot the second half.

You also need to make sure you know what sort of data FFT samples your FFT algorithm is returning as algorithms can either return FFT samples corresponding to either frequencies ranging from $-f_s / 2$ to $f_s/2$ or from $0$ to $f_s$.The code sample above assumes the former is true but that isn't always the case.

Matlab, for example, usually gives FFT samples corresponding to frequencies ranging from $0$ to $f_s$ so your frequencies in this case would just be

$$ f = n\frac{f_s}{N} $$

you can still get a plot from $-f_s/2$ to $f_s/2$ fairly easily though because the Fourier transform is periodic with period equal to $f_s$.