Frequency scale in fft after zero filling (odd/even number of data)

Product: Matlab 2017b

If we have data set "A" which consists of time versus light absorbance.There is another set of data "B". Both data are sampled at the same rate Fs.

The two sets are to be divided in the frequency domain. C= fft(A)/fft(B). Prior to division, in order to make the length of A and B equal, zero filling in done or one can use the fft(X, n). Sometimes there are odd number of data points and sometimes, even in A and B after zero filling.

In such a case, how do we make a frequency axis for C in MATLAB? The number of original data points have changed after zero filling in either A or B.

Example: The key question is, assume we had initially 1000 data points in A, and 800 data points in B. Let us use C= fft (A, 1024) ./ fft(B, 1024). We get a new vector C. How should we make a frequency scale for C now given that the number of data points were changed.

Thanks.

Answer 1

[Star Strider]

I am not certain what you are doing. One way to make the frequencies equal would be to use the second argument, ‘n’, in the fft function, to be the same for all of them (and preferably relatively large, so that all your data have the same frequency vector). This has the effect of ‘zero-padding’ the argument vector, increasing the frequency resolution and equalising the frequencies of the various signals.

The only way to determine if this approach is appropriate to your data is to try it with them.

[FW]

Thank you for the suggestion. Division in frequency domain is being done for deconvolution purposes. I have added an example so that the question becomes clear.

[Star Strider]

My pleasure.

The frequency vector for ‘C’ is necessarily the same as it is for ‘fft(A,1024)’ and ‘fft(B,1024)’, because it would be the same for both (given the same sampling frequencies for both). Note that if the lengths of ‘A’ and ‘B’ in the time domain are different, and that the sampling for both is continuous and at the same sampling frequency, this could affect the spectral content for the low frequency signals, so lower-frequency signals could be present in the longer recording that are not present (or not detectable) in the shorter recording. This is a theoretical concern and would be the case only if there was a significant difference in the recording lengths. I doubt it would be a concern given that one has 800 samples and the other has 1000 samples.

I would also subtract the mean of each signal from each signal, to eliminate the d-c (0 Hz) offset from them.

This simulation demonstrates that no information is lost in the shorter signal, even at low frequencies:

t = linspace(0, 250, 1000);
y = sum(sin([0.05; 1.0; 1.5]*t*2*pi)) + 0.1*randn(size(t));

FTy1 = fft(y, 1024);
FTy2 = fft(y(1:800), 1024);

Fs = 1/t(2);
Fn = Fs/2;
Fv = linspace(0, 1, fix(numel(t)/2)+1)*Fn;
Iv = 1:numel(Fv);

figure
plot(Fv, abs(FTy1(Iv))*2, 'LineWidth',1)
hold on
plot(Fv, abs(FTy1(Iv))*2)
hold off
grid

Experiment with it to determine if there are any effects that might be relevant to the time-domain length difference.

[FW]

Thank you. Actually, I am interested in low frequency content (usually below 10 Hz). The data consists of spaced Gaussian peaks (due to light absorption) and light absorbing molecules pass through a flow cell. The data sampling rate is usually 250 Hz.

[Star Strider]

As always, my pleasure.

Longer time-domain record lengths will be best to detect low-frequency signals, particularly if your data are noisy.

With a 250 Hz sampling frequency, a record of 1000 samples would be 4 seconds in length. A signal that would complete 1 cycle in 4 seconds would have a frequency of 0.25 Hz. For a record of 800 samples, that frequency would be about 0.3125 Hz. The lowest frequency signal that would complete 1 cycle in a given record length would then be:

at least as I calculate it.

[FW]

Thanks. Just a final confirmation, whenever we do zero padding of the original data by using fft( X, n) option, our frequency axis should be made as per our chosen "n" rather than original length of t.

So let say original data A had 1000 points sampled at 250 Hz. We did FFT (A, 1024). This essentially means that we sampled for a longer time. Should our t be changed in the equation below to accomodate the zero padding. This is exactly causing confusion about zero padding and making a frequency axis now.

Fv = linspace(0, 1, fix(numel(t)/2)+1)*Fn

e.g. in your code

length (t) = 1000

however length(FFTy1) is 1024.

How does the frequency axis take care of the new number of data points?

Thanks.

[Star Strider]

As always, my pleasure.

‘... whenever we do zero padding of the original data by using fft( X, n) option, our frequency axis should be made as per our chosen "n" rather than original length of t.’

Yes.

The ‘Fv’ vector should be:

Fv = linspace(0, 1, fix(numel(FTy1)/2)+1)*Fn;

My apologies for the error. (Since ‘FTy1’ and ‘FTy2’ are the same lengths here, there is no need to compute the ‘Fv’ vector separately for each one.) This also assumes the two are vectors. If not, use the length function instead of numel to calculate the correct length of ‘Fv’.