FFT and IFFT : filtering frequencies

Question

I am sampling at $30 Hz$.

I have a sample of $N=150$.

Here is an example of how my sample looks like, which starts at 54 and ends on 203 on the X Axis

I am interested in observing frequencies between 15 and 20 per minute. ( Cycles per minute ) Hence, I would derive the cycles per second ( Hz ) like this.

$15*0.0166666666667=0.25 \ Hz$

$20*0.0166666666667=0.33 \ Hz$

When I take the FFT, my data looks as follows. ( X axis are the bins to N / 2, not frequency )

To derive the frequency for each bin, I am using this stack answer to construct this logic:

Frequency = ( bin * sampleRate ) / dft.Points

So I might have some output as follows:

Frequency = 0 * 30/300
Frequency = 1 * 30/300
Frequency = 2 * 30/300
Frequency = 3 * 30/300

If I re-plot the same data, where the X axis now represents frequency, this is what I get :

Here is where my question kicks in: ( Note, if something above is wrong in my logic, please point it out, as it could be the reason why I am getting weird results )

I now want to take the IFFT of the dataset, given that the frequency of the bin is less than . Here is some java code to help illustrate how I am doing this with JTransforms:

final double[] ifft = new double[fft.length];
for (int idx = 0; idx < fft.length; idx++) {
    final double fftPoint = fft[idx];
    final Double sampleCount = Double.valueOf(idx * 30);
    final Double binToFrequency = Double.valueOf(sampleCount / fft.length);
    ifft[idx] = binToFrequency <= 0.33333334 ? fftPoint : 0;
}

analyzer.realInverseFull(ifft, false);

Variable fft above is the DFT/FFT array data points. Since I have 150 samples. fft.length should equate to 300. I am then checking the frequency of the bin, and if it is less than 0.33333334 I am using it, otherwise, it is getting zeroed out.

When I plot my graph, the odd part is that I get two waves:

If I zoom in on the top one, here is what I see:

I also notice that the numbers on the x axis are increments of 2. For example, the wave where the amplitude is around 400K on the Y axis starts at 1, and has points at 3,5,7,9, etc. The lower one starts at 0, and has points 2,4,6,8,etc.

Question:

Should I be seeing two waves? If yes, how do I combine it into one? If no, what step might have I destroyed?

From my cloudy knowledge, the only thing I can think of that can explain this is the notion of imaginary and real numbers. But, I do not see why they should be shown in a time domain, after an iFFT occurs.

Answer 1

FloatFFT has a funny array layout. Real and Imaginary components of each datapoint/bin have to be interleaved in the data array you provide. The FFT plot you show it is not a common FFT plot: is the real and imaginary part of the FFT interleaved: $[ Re\{X[0]\}, Im\{X[0]\}, Re\{X[1]\}, Im\{X[1]\}, ... ]$.

Handle two consecutive points everywhere accordingly. For example: $bin[0] = abs( sqrt( a[0]*a[0] + a[1]*a[1] )$ and so on... Also, follow that layout when you perform the inverse.

Documentation is not the best and you will have to read it carefully.

Bonus: the reason you have to provide a $2 \cdot n$ array to some functions is not that the FFT has to double the data set length, it is because each bin is stored in two consecutive array positions.

Answer 2

The first problem might be that you need to sample more than twice the rate of the highest frequency in your signal. Otherwise your samples will be contaminated by aliasing.

The second problem in plotting your spectrum is that the lowest bins of an FFT or DFT of strictly real data can be mixed (contaminated) by the nearby complex conjugate negative frequency image. This can be solved by sampling longer (more samples covering more time), which will move the spectrum of interest to higher numbered FFT result bins.

Third, attempting to filter a signal by zeroing FFT bins does not work well. See this answer for details: Why is it a bad idea to filter by zeroing out FFT bins?