The following C# NAudio code produces a different result to MATLAB by a factor of 4. Why does this occur and is one of them incorrect?
Complex[] tmp = new Complex[4];
tmp[0].X = 1.0f;
tmp[1].X = 0.5f;
tmp[2].X = 1.0f;
tmp[3].X = 0.25f;
tmp[0].Y = 0.0f;
tmp[1].Y = 0.0f;
tmp[2].Y = 0.0f;
tmp[3].Y = 0.0f;
FastFourierTransform.FFT(true, 2, tmp);
NAUDIO OUTPUT:
0.6875 + 0.0000i
0.0000 - 0.0625i
0.3125 + 0.0000i
0.0000 + 0.0625i
MATLAB OUTPUT:
2.7500 + 0.0000i
0.0000 - 0.2500i
1.2500 + 0.0000i
0.0000 + 0.2500i
The Discrete Fourier transform and its inverse require a certain normalization so that ifft(fft(x))==x. How this normalization is done changes from implementation to implementation.
It seems that, in this case, NAudio has chosen a different normalization than MATLAB.
MATLAB uses the most common normalization, where fft(x) at k=0 is equal to sum(x), and the inverse transform does the same thing but divides by n (number of samples). This is also the equation as described in the Wikipedia page for the DFT. In this case, the inverse transform matches the equation for the Fourier series.
NAudio seems to do the division by n in the forward transform, such that at k=0 you have mean(x).
Given the above, you can use the first frequency bin (the DC component) to verify what normalization is used (assuming there is a DC component, if the signal has a zero mean this will not work): If the DC component is equal to the sum of all the sample values, then the "common" normalization is used. It can also be equal to the sum divided by sqrt(n), in the case of a symmetric definition, where the forward and inverse transform carry the same normalization. In the case of NAudio it will be equal to the sum divided by n (i.e. the mean of the sample values). In general, take the DC component and divide it by the sum of the sample values. The result q is the normalization term used. The inverse transform should have a normalization term 1/qn.