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  1. While using FFT sample code from Apple documentation, what actually does the N, log2n, n and nOver2 mean?
  2. Does N refer to the window size of the fft or the whole number of samples in a given audio, and
  3. how do I calculate N from an audio file?
  4. how are they related to the audio sampling rate i.e. 44.1kHz?
  5. What would be the FFT frame size in this code?

Code:

/* Set the size of FFT. */
log2n = N;
n = 1 << log2n;

stride = 1;
nOver2 = n / 2;

printf("1D real FFT of length log2 ( %d ) = %d\n\n", n, log2n);

/* Allocate memory for the input operands and check its availability,
 * use the vector version to get 16-byte alignment. */
A.realp = (float *) malloc(nOver2 * sizeof(float));
A.imagp = (float *) malloc(nOver2 * sizeof(float));
originalReal = (float *) malloc(n * sizeof(float));
obtainedReal = (float *) malloc(n * sizeof(float));
Firdous
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2 Answers2

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  1. N or n typically refers to the number of elements. log2n is the base-two logarithm of n. (The base-two logarithm of 32 is 5.) nOver2 is n/2, n divided by two.
  2. In the context of an FFT, n is the number of samples being fed into the FFT.
  3. n is usually determined by a variety of constraints. You want more samples to provide a better quality result, but you do not want so many samples that processing takes up a lot of computer time or that the result is not available until so late that the user notices a lag. Usually, it is not the length of an audio file that determines the size. Rather, you design a “window” that you will use for processing, then you read samples from the audio file into a buffer big enough to hold your window, then you process the buffer, then you repeat with more samples from the file. Repetitions continue until the entire file is processed.
  4. A higher audio sampling rate means there will be more samples in a given period of time. E.g., if you want to keep your window under 1/30th of a second, then a 44.1 kHz sampling rate will have less than 44.1•1000/30 = 1470 samples. A higher sampling rate means you have more work to do, so you may need to adjust your window size to keep the processing within limits.
  5. That code uses N for log2n, which is unfortunate, since it may confuse people. Otherwise, the code is as I described above, and the FFT frame size is n.

There can be some confusion about FFT size or length when a mix of real data and complex data is involved. Typically, for a real-to-complex FFT, the number of real elements is said to be the length. When doing a complex-to-complex FFT, the number of complex elements is the length.

Eric Postpischil
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'N' is the number of samples, i.e., your vector size. Corresponding, 'log2N' is the logarithm of 'N' with the base 2, and 'nOver2' is the half of 'N'.

To answer the other questions, one must know, what do you want to do with FFT. This document, even it is written with a specific system in mind, can serve as an survey about the relation and the meaning of the parameters in (D)FFT.

Matthias
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  • Given that the code in question contains `log2n = N`, it does not seem that `log2N` is the logarithm of `N`, in this case. – Eric Postpischil Mar 20 '13 at 17:24
  • @EricPostpischil: Take a look at the next line: the shift by N corresponds to the power of 2, i.e., N is the dual logarithm of n. – Matthias Mar 20 '13 at 19:57
  • The text in your answer says “'N' is the number of samples”. In the code in the question, `n` is the number of samples, but `N` is not, nor is `log2N` the logarithm of `N`, nor is `nOver2` half of `N`. I understand that common usage in other code is that `N` is the number of samples, but it simply is not the case in the code in the question, which means that this answer contains incorrect statements. – Eric Postpischil Mar 20 '13 at 20:00