How does nufft function work in matlab?

2021 3 2
1 个回答
更新时间：2024 6 29
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I am trying to understand how to use nufft function. For this, I started by comparing the results of the fft and nufft transforms of a Gaussian to see if they yield the same result. I find that for a spacing in t space that is not unity, they yield different results. The exmple in MATLAB for nufft also has unit spacing between the t coordinates. Does nufft function not work for non-unit spacing? This would be a problem for me since the next step is to get the fourier transform of a gaussian with different spacings in the t space, i.e, dt varies for each element.
Thank you.
tmax = 12;
n = 2^11;
tau = 1; % Width of the gaussian pulse
dt = 2*tmax/n;
t = (-tmax:dt:tmax-dt);
fmax = 1/(2*dt);
df = 2*fmax/n;
f = -fmax:df:fmax-df;
Pulse = exp(-(t/tau).^2);
%Pulse = sin(t/tau);
fftPulse = ifftshift(fft(Pulse));
subplot(2,1,1)
plot(t,Pulse)
subplot(2,1,2)
plot(f,abs(fftPulse)); hold on;
nufftPulse = ifftshift(nufft(Pulse,t));
plot(f,abs(nufftPulse))

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dpb
dpb 2021-3-2
Of course they're not the same when t is not uniformly sampled; see the "More About" section and the references for what it means.
I'll agree the examples are somewhat limited in their utility for understanding -- the first is simply a gap in the time vector; the sampling rate is uniform.
If you do not sample with a uniform sample rate, however, then the waveform frequency is modified when you generate the time vector as well -- we would have to see what you're generating with the time-varying sample rate.
What need to compare would be the same signal sample uniformly and nonuniformly; the nufft analysis should be able to account for the latter and reproduce closely the correct frequencies.
Ghanasyam Remesh
Ghanasyam Remesh 2021-3-2
Thank you. I think I figured out the right code to yield the same results. We need to scale the frequencies differently when using nufft. This is a direct result of its definition.
Once again, thank you so much.
Ghanasyam Remesh
Ghanasyam Remesh 2021-3-2
tmax = 20;
n = 2^11;
tau = 0.5;
dt = 2*tmax/n;
t = (-tmax:dt:tmax-dt);
fmax = 1/(2*dt);
df = 2*fmax/n;
f = (-fmax:df:fmax-df);
Pulse = exp(-(t/tau).^2);
%Pulse = sin(t/tau);
fftPulse = ifftshift(fft(Pulse));
subplot(2,1,1)
plot(t,Pulse)
xlabel('t');ylabel('f(t)')
subplot(2,1,2)
plot(f,abs(fftPulse)); hold on;
nufftPulse = (nufft(Pulse,t*2)); % The 2 factor here and next line allows us to make the freqency range [0-2]
plot((f+fmax)*dt*2,abs(nufftPulse)) % PLEASE NOTE THAT WE HAVE TO SCALE AND SHIFT
xlabel('\nu');ylabel('f(\nu)')
evelyn
evelyn 2024-6-29
how to define the new time index, like 't*2', you mention?

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回答（1 个）

Paul
Paul 2024-6-29
Ran in:

0 个投票

Ran in:
To get fft and nufft to yield the same results ....
tmax = 12;
n = 2^11;
tau = 1; % Width of the gaussian pulse
dt = 2*tmax/n;
t = (-tmax:dt:tmax-dt);
fmax = 1/(2*dt);
df = 2*fmax/n;
f = -fmax:df:fmax-df;
Pulse = exp(-(t/tau).^2);
%fftPulse = ifftshift(fft(Pulse));
fftPulse = fftshift(fft(ifftshift(Pulse)));
nufftPulse = nufft(Pulse,t,f);
figure
subplot(2,1,1)
% plot the phase of the transform instead of the original signal
%plot(t,Pulse)
plot(f,angle(fftPulse),f,angle(nufftPulse));
subplot(2,1,2)
plot(f,abs(fftPulse)); hold on;
plot(f,abs(nufftPulse))
The hair on the angle plot is just numerical noise that results from taking the angle of complex number that has very small magnitude. At low frquencies where the magnitude isn't small we see that the angle is essentially zero as would be expected.

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