How to get spatial frequency from FFT?
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Sohel Rana
el 25 de Nov. de 2020
Comentada: Star Strider
el 26 de Nov. de 2020
Hi,
I have got the first graph based on the following code. How can I get the second graph after performing FFT?
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);
plot(lam*1000,I)
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Star Strider
el 25 de Nov. de 2020
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:
.
9 comentarios
Star Strider
el 26 de Nov. de 2020
As always, my pleasure!
You need to choose the passband frequencies from the centre frequency and the desirecd passband limits.
I am not certain what you are doing, so I must leave the details to you.
Remember the 1E-3 scaling factor in the ‘Fv’ vector in choosing the passbands with respect to your computed signals. It may be necessary to correct for that in the filter design.
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