I always wondered why it didn’t work to compute the dual Gabor atom by using the image-to-vector methods I explained previously [1,2,3]. Dr. Kaiblinger showed me that the correct way was to use that special isomorphism that walks along the diagonal of the image. Because width and height have to be relatively prime, that path spans the whole image space. And because there are no jumps over pixels, 2D-modulations stay 1D-modulations. This is not yet proved formally, but I can already show first experiments:
> p=64; q=75; idx=linind(p,q); %index vector
> img=zeros(p,q); N=p*q
N = 4800
> img(idx(1:50))=1; imagesc(img); %step 50
> img(idx(1:100))=1; imagesc(img); %step 100
> img(idx(1:1000))=1; imagesc(img); %step 1000
> img(idx(1:4000))=1; imagesc(img); %step 4000
A 2D-frequency is given as a tensor product of two 1D-frequencies with signal lengths p and q, respectively. If their modulation parameters are given as kp and kq, then the corresponding 2D-modulation is given by a 1D-modulation of length N and parameter kN = kq p − kp q (mod N) ; as already said, a proof is not given yet.
> kp=4; frp = exp(2*pi*i*(0:p-1) * kp/p);
> kq=3; frq = exp(2*pi*i*(0:q-1) * kq/q);
> frpq = frp’ * frq; size(frpq)
ans =
64 75
> imagesc(real(frpq))
> kN = mod(kq*p-kp*q, N)
kN = 4692
> frN = exp(2*pi*i*(0:N-1) * kN/N);
> plot(real(frN(1:1000)))
> frN2=zeros(p,q);
> frN2(idx)=frN;
> compnorm(frpq, frN2)
quotient of norms: norm(x)/norm(y) = 1
difference of normalized versions = 1.478e-12
ans = 1.4780e-12
So those two 2D-frequencies are really identical. The plot of the second one is identical to the first one, so we skip it here.
Now we want to see if the 2D-dual of a separable 2D atom obtained by that isomorphism is identical to the tensor product of the two 1D-duals.