Could it be possible that there is some offset when computing the sample frequency by calling FftSharp.Transform.FFTfreq()?

These are the steps I've followed and the results obtained:

• Create a 1-cycle 2-Hz-sinus wave at a sampling rate of 16 Hz (see 16 points 2 Hz sinus .txt) (The offset is less noticeable the higher the sampling frequency).
• Compute the FFT using:

signalFFT = FftSharp.Transform.FFTmagnitude(signalWindowed);
freq = FftSharp.Transform.FFTfreq(sampleFreq, signalFFT.Length);
plotFFT.Plot.AddScatterLines(freq, signalFFT);

• The plot shows the frequency of the signal at 1.78 Hz instead of 2 Hz.
  

Possible explanation:

• The array returned by FftSharp.Transform.FFTmagnitude() has one more point than half its size: new double[input.Length / 2 + 1] (2^n-1 +1).
• Inside function FftSharp.Transform.FFTfreq(), the frequency is computed as:

if (OneSided)
    double fftPeriodHz = sampleRate / pointCount / 2;
else
    double fftPeriodHz = sampleRate / pointCount;

• The problem is that pointCount is passed as the length of the array returned by FftSharp.Transform.FFTmagnitude() (2^n-1 +1), so it has one more point than what is required to compute fftPeriodHz.

So, i can think of two options:

1. Subtract 1 from the array's length when calling FftSharp.Transform.FFTfreq(). This might be counterintuitive and not obvious.
2. Perform the subtraction inside FftSharp.Transform.FFTfreq():

if (OneSided)
    double fftPeriodHz = sampleRate / (pointCount-1) / 2;
else
    double fftPeriodHz = sampleRate / (pointCount-1);

This last option produces the correct result for the sinus wave mentioned at the beginning. I hesitated before doing any PR since I'm not sure which one would you consider more appropriate.