How to get spatial frequency from FFT?

The Fourier transform neither knows nor cares whether the units of the independent variable are time, space, or anything else. It will do whatever you ask it to do (within limits, of course)

Try this:

I1=0.7;
I2=0.5;
I3=0.3;
L1=200;
L2=170;
n1=1;
n2=1.444;
lam=(1.52:0.0001:1.56);
Q12=(4*pi*n1*L1)./lam;
Q23=(4*pi*n2*L2)./lam;
Q13=Q12+Q23;
I=I1+I2+I3+2*sqrt(I1*I2).*cos(Q12)+2*sqrt(I2*I3).*cos(Q23)+2*sqrt(I1*I3).*cos(Q13);
figure
plot(lam*1000,I)
L = numel(lam);
Ts = mean(diff(lam));
Fs = 1/Ts;
Fn = Fs/2;
FTI = fft(I)/L;
Fv = linspace(0, 1, fix(L/2)+1)*Fn * 1E-3;
Iv = 1:numel (Fv);
[pks,locs] = findpeaks(abs(FTI(Iv)));
figure
plot(Fv, abs(FTI(Iv)))
xlim([0  0.5])
xlabel('Spatial Frequency (nm^{-1})')
ylabel('Amplitude')
text(Fv(locs), abs(FTI(locs)), sprintfc('Peak %d',(1:numel(locs))), 'HorizontalAlignment','center', 'VerticalAlignment','bottom')

producing:

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Star Strider 2020-11-26

在 MATLAB Online 中打开

Since you have the Signal Processing Toolbox, make the appropriate changes to this code for each filter you want to create and use:

Fs = 44000;                                                         % Sampling Frequency
Fn = Fs/2;                                                          % Nyquist Frequency
Wp = [898 899]/Fn;                                                  % Stopband Frequency (Normalised)
Ws = [0.98 1.02].*Wp;                                               % Passband Frequency (Normalised)
Rp =  1;                                                            % Passband Ripple
Rs = 90;                                                            % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs);                                     % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp);                                        % Elliptic Filter Design: Zero-Pole-Gain 
[sos,g] = zp2sos(z,p,k);                                            % Second-Order Section For Stability
figure
freqz(sos, 2^18, Fs)                                                % Filter Bode Plot
set(subplot(2,1,1), 'XLim',Ws*Fn)                                   % Optional
set(subplot(2,1,2), 'XLim',Ws*Fn)                                   % Optional
signal_filt = filtfilt(sos, g, signal);                             % Filter Signal

You will need to create a different filter (using this code for each one separately) for each frequency band, then use it with the original signal to get the desired result.

All that is necessary is to define the passband frequencies, inside the square brackets [], in the ‘Wp’ assignment, and specify the appropriate sampling frequency ‘Fs’. This code will design an appropriate bandpass filter by default.

All the frequencies are in the same units. The code normalises them appropriately so the functions will work. If you modify it, remember that for a bandpass filter, the stopband frequencies ‘Ws’ need to include the limits of the passband frequencies, ‘Wp’. This code calculates them appropriately automatically.

Use the filtfilt function to do the actual filtering, as I do here. Supply your vector as the ‘signal’ argument.

Chueng 2024-7-3，13:27

hello，can you explain how wavelength is converted to spatial frequency during the FFT processing? Do you have any relevant formulas?

Star Strider 2024-7-3，14:09

编辑：Star Strider 2024-7-3，14:25

The Fourier transform converts time, distance, or other variables to frequency units of cycles-per-original uint. So for time domain signals with the sampling frequency in seconds, the resulting frequency uints are cycles-per-second, or Hertz (Hz). Here, with the original units being nanometres, the resulting frequency uints are in cycles-per-nanometre, or more simply,

or

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EDIT — (3 Jul 2024 at 14:25)

In terms of your other question (How to use fft to analyse the refelction specturm? that I just now saw), simply replace ‘cycles’ with ‘wavenumber’ since that is how you choose to express it, instead labelling it

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