simple test for nufft not working

3 vues (au cours des 30 derniers jours)
Pieter van Wyk
Pieter van Wyk le 27 Mar 2023
I have a very simple test code for using the nufft function:
%% define signal
t = [-1.0/(2*pi),1.0/(2*pi)];
X = [1.0,1.0];
%% compute the NFFT
f_k = -10:10;
Y_k = nufft(X,t,f_k);
Accroding to the equation given in the documentation the result of this test should be , but the result is just a real constant across . How do I need to modify my program to deliver the correct result, or is this a bug?

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Paul
Paul le 27 Mar 2023
Modifié(e) : Paul le 27 Mar 2023
Hi Pieter,
I don't think that expression for Y(k) is correct. We can check symbolically
%% define signal
t = sym([-1.0,1.0]);
X = [1.0/(2*sym(pi)),1.0/(2*sym(pi))];
%% compute the NFFT
%f_k = sym(-10:10);
%Y_k = nufft(X,t,f_k)
syms f_k
% expression of Y(k) from the doc page for nufft
Yk = sum(X.*exp(-1j*2*sym(pi)*t*f_k))
Yk = 
Yk = simplify(Yk)
Yk = 
Subbing back in the values of f_k from the original problem
cos(2*pi*(-10:10))/pi
ans = 1×21
0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183
which matches the result from nufft.
nufft(double(X),double(t),-10:10)
ans = 1×21
0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183 0.3183

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