discrete Fourier transform of Gaussian
4 ビュー (過去 30 日間)
古いコメントを表示
I'm trying to understand the following code, where we calculate the discrete Fourier transform of the Gaussian kernel:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
n1=64;
n2=64;
n3=64;
i = sqrt(-1);
wx=exp(i*2*pi/n1); % nth root of unity.
wy=exp(i*2*pi/n2); % nth root of unity.
wz=exp(i*2*pi/n3); % nth root of unity.
[x,y,z]=meshgrid([1:n1],[1:n2],[1:n3]);
x=permute(x,[2 1 3]);
y=permute(y,[2 1 3]);
z=permute(z,[2 1 3]);
KERNEL = n1*n1*(2-wx.^(x-1)-wx.^(1-x)) + n2*n2*(2-wy.^(y-1)-wy.^(1-y)) + ...
n3*n3*(2-wz.^(z-1)-wz.^(1-z));
KERNEL = exp( -dt*KERNEL );
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The Gaussian kernel is defined as follows: .
Let and and grid points .
The discrete Fourier transform (1D) of a grid function is the coefficient vector with .
But here in the code we compute the kernel in a different way. Does somebody know why the Fourier transform of the kernel can be computed this way?
What I mean is, why do we get constants (I write n for , ) because I thought it should be since we are in the 3D case. And I don't understand why we compute , because we then compute the kernel in a range since x takes the values instead of and this isn't the range from the discrete Fourier transform definition. I'd really appreciate if someone could give me a hint to understand this computation.
0 件のコメント
回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Fourier Analysis and Filtering についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!