VB FFT - stuck understanding relationship of results to frequency

Question

Trying to understand an fft (Fast Fourier Transform) routine I'm using (stealing)(recycling)

Input is an array of 512 data points which are a sample waveform. Test data is generated into this array. fft transforms this array into frequency domain. Trying to understand relationship between freq, period, sample rate and position in fft array. I'll illustrate with examples:

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Sample rate is 1000 samples/s. Generate a set of samples at 10Hz.

Input array has peak values at arr(28), arr(128), arr(228) ... period = 100 sample points

peak value in fft array is at index 6 (excluding a huge value at 0)

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Sample rate is 8000 samples/s Generate set of samples at 440Hz

Input array peak values include arr(7), arr(25), arr(43), arr(61) ... period = 18 sample points

peak value in fft array is at index 29 (excluding a huge value at 0)

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How do I relate the index of the peak in the fft array to frequency ?

Answer 1

If you ignore the imaginary part, the frequency distribution is linear across bins:

Frequency@i = (Sampling rate/2)*(i/Nbins).

So for your first example, assumming you had 256 bins, the largest bin corresponds to a frequency of 1000/2 * 6/256 = 11.7 Hz. Since your input was 10Hz, I'd guess that bin 5 (9.7Hz) also had a big component. To get better accuracy, you need to take more samples, to get smaller bins.

Your second example gives 8000/2*29/256 = 453Hz. Again, close, but you need more bins. Your resolution here is only 4000/256 = 15.6Hz.

Answer 2

It would be helpful if you were to provide your sample dataset.

My guess would be that you have what are called sampling artifacts. The strong signal at DC ( frequency 0 ) suggests that this is the case.

You should always ensure that the average value in your input data is zero - find the average and subtract it from each sample point before invoking the fft is good practice.

Along the same lines, you have to be careful about the sampling window artifact. It is important that the first and last data point are close to zero because otherwise the "step" from outside to inside the sampling window has the effect of injecting a whole lot of energy at different frequencies.

The bottom line is that doing an fft analysis requires more care than simply recycling a fft routine found somewhere.

Here are the first 100 sample points of a 10Hz signal as described in the question, massaged to avoid sampling artifacts

> sinx[1:100]
  [1]  0.000000e+00  6.279052e-02  1.253332e-01  1.873813e-01  2.486899e-01  3.090170e-01  3.681246e-01  4.257793e-01  4.817537e-01  5.358268e-01
 [11]  5.877853e-01  6.374240e-01  6.845471e-01  7.289686e-01  7.705132e-01  8.090170e-01  8.443279e-01  8.763067e-01  9.048271e-01  9.297765e-01
 [21]  9.510565e-01  9.685832e-01  9.822873e-01  9.921147e-01  9.980267e-01  1.000000e+00  9.980267e-01  9.921147e-01  9.822873e-01  9.685832e-01
 [31]  9.510565e-01  9.297765e-01  9.048271e-01  8.763067e-01  8.443279e-01  8.090170e-01  7.705132e-01  7.289686e-01  6.845471e-01  6.374240e-01
 [41]  5.877853e-01  5.358268e-01  4.817537e-01  4.257793e-01  3.681246e-01  3.090170e-01  2.486899e-01  1.873813e-01  1.253332e-01  6.279052e-02
 [51] -2.542075e-15 -6.279052e-02 -1.253332e-01 -1.873813e-01 -2.486899e-01 -3.090170e-01 -3.681246e-01 -4.257793e-01 -4.817537e-01 -5.358268e-01
 [61] -5.877853e-01 -6.374240e-01 -6.845471e-01 -7.289686e-01 -7.705132e-01 -8.090170e-01 -8.443279e-01 -8.763067e-01 -9.048271e-01 -9.297765e-01
 [71] -9.510565e-01 -9.685832e-01 -9.822873e-01 -9.921147e-01 -9.980267e-01 -1.000000e+00 -9.980267e-01 -9.921147e-01 -9.822873e-01 -9.685832e-01
 [81] -9.510565e-01 -9.297765e-01 -9.048271e-01 -8.763067e-01 -8.443279e-01 -8.090170e-01 -7.705132e-01 -7.289686e-01 -6.845471e-01 -6.374240e-01
 [91] -5.877853e-01 -5.358268e-01 -4.817537e-01 -4.257793e-01 -3.681246e-01 -3.090170e-01 -2.486899e-01 -1.873813e-01 -1.253332e-01 -6.279052e-02

And here is the resulting absolute values of the fft frequency domain

 [1] 7.160038e-13 1.008741e-01 2.080408e-01 3.291725e-01 4.753899e-01 6.653660e-01 9.352601e-01 1.368212e+00 2.211653e+00 4.691243e+00 5.001674e+02
[12] 5.293086e+00 2.742218e+00 1.891330e+00 1.462830e+00 1.203175e+00 1.028079e+00 9.014559e-01 8.052577e-01 7.294489e-01

Answer 3

I'm a little rusty too on math and signal processing but with the additional info I can give it a shot.

If you want to know the signal energy per bin you need the magnitude of the complex output. So just looking at the real output is not enough. Even when the input is only real numbers. For every bin the magnitude of the output is sqrt(real^2 + imag^2), just like pythagoras :-)

bins 0 to 449 are positive frequencies from 0 Hz to 500 Hz. bins 500 to 1000 are negative frequencies and should be the same as the positive for a real signal. If you process one buffer every second frequencies and array indices line up nicely. So the peak at index 6 corresponds with 6Hz so that's a bit strange. This might be because you're only looking at the real output data and the real and imaginary data combine to give an expected peak at index 10. The frequencies should map linearly to the bins.

The peaks at 0 indicates a DC offset.

Answer 4

It's been some time since I've done FFT's but here's what I remember

FFT usually takes complex numbers as input and output. So I'm not really sure how the real and imaginary part of the input and output map to the arrays.

I don't really understand what you're doing. In the first example you say you process sample buffers at 10Hz for a sample rate of 1000 Hz? So you should have 10 buffers per second with 100 samples each. I don't get how your input array can be at least 228 samples long.

Usually the first half of the output buffer are frequency bins from 0 frequency (=dc offset) to 1/2 sample rate. and the 2nd half are negative frequencies. if your input is only real data with 0 for the imaginary signal positive and negative frequencies are the same. The relationship of real/imaginary signal on the output contains phase information from your input signal.

Answer 5

The frequency for bin i is i * (samplerate / n), where n is the number of samples in the FFT's input window.

If you're handling audio, since pitch is proportional to log of frequency, the pitch resolution of the bins increases as the frequency does -- it's hard to resolve low frequency signals accurately. To do so you need to use larger FFT windows, which reduces time resolution. There is a tradeoff of frequency against time resolution for a given sample rate.

You mention a bin with a large value at 0 -- this is the bin with frequency 0, i.e. the DC component. If this is large, then presumably your values are generally positive. Bin n/2 (in your case 256) is the Nyquist frequency, half the sample rate, which is the highest frequency that can be resolved in the sampled signal at this rate.

If the signal is real, then bins n/2+1 to n-1 will contain the complex conjugates of bins n/2-1 to 1, respectively. The DC value only appears once.

Answer 6

The samples are, as others have said, equally spaced in the frequency domain (not logarithmic).

For example 1, you should get this:

alt text http://home.comcast.net/~kootsoop/images/SINE1.jpg

For the other example you should get

alt text http://home.comcast.net/~kootsoop/images/SINE2.jpg

So your answers both appear to be correct regarding the peak location.

What I'm not getting is the large DC component. Are you sure you are generating a sine wave as the input? Does the input go negative? For a sinewave, the DC should be close to zero provided you get enough cycles.

Answer 7

Another avenue is to craft a Goertzel's Algorithm of each note center frequency you are looking for.

Once you get one implementation of the algorithm working you can make it such that it takes parameters to set it's center frequency. With that you could easily run 88 of them or what ever you need in a collection and scan for the peak value.

The Goertzel Algorithm is basically a single bin FFT. Using this method you can place your bins logarithmically as musical notes naturally go.

Some pseudo code from Wikipedia:

s_prev = 0
s_prev2 = 0
coeff = 2*cos(2*PI*normalized_frequency);
for each sample, x[n],
  s = x[n] + coeff*s_prev - s_prev2;
  s_prev2 = s_prev;
  s_prev = s;
end
power = s_prev2*s_prev2 + s_prev*s_prev - coeff*s_prev2*s_prev;

The two variables representing the previous two samples are maintained for the next iteration. This can be then used in a streaming application. I thinks perhaps the power calculation should be inside the loop as well. (However it is not depicted as such in the Wiki article.)

In the tone detection case there would be 88 different coeficients, 88 pairs of previous samples and would result in 88 power output samples indicating the relative level in that frequency bin.

Answer 8

WaveyDavey says that he's capturing sound from a mic, thru the audio hardware of his computer, BUT that his results are not zero-centered. This sounds like a problem with the hardware. It SHOULD BE zero-centered.

When the room is quiet, the stream of values coming from the sound API should be very close to 0 amplitude, with slight +- variations for ambient noise. If a vibratory sound is present in the room (e.g. a piano, a flute, a voice) the data stream should show a fundamentally sinusoidal-based wave that goes both positive and negative, and averages near zero. If this is not the case, the system has some funk going on!

-Rick