Super- resolution: Warp Matrix

The super resolution model is as such: Y_k = DHFX + N , where Y_k is one of the N measured low resolution image, D is the Decimation Matrix, H is the blur matrix , F is the warp matrix, N is the noise and X is the high resolution image. Lets say the X is [L *L] or we can say [L^2 * 1], F is [L^2*L^2], and H is [L^2*L^2] and D is [M_k^2*L^2] where M_k is the size of Y. In the super resolution problem we need to find X but for that we need to find the F(warp matrix). The shifting is done to get the sub pixel information as far as I understand. I am also able to find the [3*3 ] warp matrix which transform the coordinate space but that is not required. So if anyone can help me to find the F matrix. Thank you

• it was better to put your comment in the comment section! I taught you meant " can move over the coordinate system". Aug 16 '17 at 20:58
• i need you to explain more about your question. I don't get it what a projective transform (your 3*3 matrix) got to do with super-resolution algorithms. Aug 16 '17 at 21:00
• The equation Y=DHX is a model for signal sampling where H is a operator for a low-pass filter or anti-aliasing filter and D maps the filtered signal to a lower dimension space in other words sample the signal (images are 2D extension of signals). Aug 16 '17 at 21:12
• so you want to reconstruct a super resolution image using multiple low resolution image from different view. Aug 18 '17 at 10:50