Reconstructed Pulse Out of Phase
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In my research I'm having a problem reconstructing some pulses, I've replicated the problem I'm having with a simple gaussian pulse. Essentially when I take the IFFT of the pulse, the reconstructed time domain signal is out of phase. I've included my code with the plots, I think the problem is within how I'm reconstructing the time space. Any help is appreciated, thank you!
clear variables
%% FFT Test on Gaussian Pulse
Fs = 100; %Sampling Frequency
t = -0.5:1/Fs:0.5; %Time Vector
L = length(t);
x = 1/(4*sqrt(2*pi*0.01))*(exp(-t.^2/(2*0.01))); %Gaussian pulse
plot(t,x) %plotting pulse
title('Gaussian Pulse')
xlabel('time(t)')
ylabel('Pulse')
n = 2^nextpow2(L);
Y = fft(x,n); %plotting frequency spectrum
P = abs(Y);
f = Fs*(0:(n/2))/n;
plot(f,P(1:n/2+1)
%% Reconstructing Pulse
recon = ifft(Y);
t2 = linspace(-0.5,0.5,128); %Reconstructing Time Space
figure()
plot(t,x)
hold on
plot(t2,recon,'--')
hold off
legend('Original','Reconstruction')
0 Comments
Accepted Answer
Paul
on 6 Sep 2022
Hi Doug,
I think the short answer is, don't use nextpow2.
Fs = 100; %Sampling Frequency
t = -0.5:1/Fs:0.5; %Time Vector
L = length(t);
x = 1/(4*sqrt(2*pi*0.01))*(exp(-t.^2/(2*0.01))); %Gaussian pulse
Y = fft(x);
xr = ifft(Y);
figure;
plot(t,x,t,xr,'o')
However, that answer seems too obvious.
If desired to use more points in the fft for some reason, then we just need to account for what that means on the time vector
N = 2^nextpow2(numel(x));
tr = (0:(numel(x)-1))/Fs + t(1);
xr = ifft(fft(x,N));
xr = xr(1:numel(tr));
In the plot below, I'm also including the what is effectively t2 and recon from the Question. We see that t2 isn't really the correct time vector becasue it effectively redefined the sampling frequency.
figure;
plot(t,x,tr,xr,'o',.1+linspace(-0.5,0.5,128),ifft(fft(x,128)))
legend('Original','Reconstructed','recon from Question')
2 Comments
Paul
on 6 Sep 2022
Good luck. Keep in mind that
fft(x)
Only give the correct magnitude, but not the correct phase, because the first element of x corresponds to a negative time. In this case, since we're going to frequency and back to time, it doesn't matter, but keep it in mind.
More Answers (1)
Cris LaPierre
on 6 Sep 2022
I'm not an FFT expert, but it appears to be related to your value of n. At least the shift goes away when I use L instead (seethis example)
%% FFT Test on Gaussian Pulse
Fs = 100; %Sampling Frequency
t = -0.5:1/Fs:0.5; %Time Vector
L = length(t);
x = 1/(4*sqrt(2*pi*0.01))*(exp(-t.^2/(2*0.01))); %Gaussian pulse
plot(t,x) %plotting pulse
title('Gaussian Pulse')
xlabel('time(t)')
ylabel('Pulse')
% n = 2^nextpow2(L);
Y = fft(x); %plotting frequency spectrum
P = abs(Y/L);
f = Fs*(0:(L/2))/L;
plot(f,P(1:L/2+1))
%% Reconstructing Pulse
recon = ifft(Y);
t2 = linspace(-0.5,0.5,L); %Reconstructing Time Space
figure()
plot(t,x)
hold on
plot(t2,recon,'--')
hold off
legend('Original','Reconstruction')
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