0
$\begingroup$

I have a image that I need to shift with less than a pixel.

My plan was to do a Fourier transform and multiply the signal with $e^{-aiu-biv}$ where $a, b$ are the shifts in x and y direction. This might make sense in theory but in practice I don't really know how to represent $e^{-aiu-biv}$.

I am using numpy to do a 2d FFT of the image which, ofcourse, is represented as a matrix of complex values. From this point of I'm kinda lost, how do I multiply this with $e^{-aiu-biv}$?

It seems like fundamentally something is wrong as I need to temporarily represent the image in a higher resolution, shift is slightly and then collapse it back to the original resolution.

$\endgroup$
0
$\begingroup$

I don't do much image processing but maybe the simpler 1D case will help:

A delay in the time domain is a phase shift in the frequency domain, this notion can be used to shift your FFT in the frequency domain by a rotation much smaller than a sample

$$x(n-n_o) = W_N^{kn_o}X(k)$$

$$W_N^{kn_o} = e^{\frac{-j 2 \pi k n_o}{N}}$$

You need to apply this principle row wise and column wise. I tried giving it a go by building a 2D matrix of the x y shift and then applying it to the image. For integer shift this works, when you get into fractional shift (0-.99) for $n_o$ it gets weird.

Non-fractional shift of a "cat eating fancy ice cream" to prove that it does what I think it does: enter image description here

Matlab code:

close all 
clear all

%load image
I = imread('cat_eating_fancy_ice_cream.jpg');


I = im2double(I);

%make phase shift matrix for frequency domain shifting
sz = size(I)
y = e.^((-j*2*pi*[0:sz(1)-1]*100)/sz(1)).';
x = e.^((-j*2*pi*[0:sz(2)-1]*200)/sz(2));
shift = y*x;

%convert image to grayscale for easier manipulation
pF1 =(fft2(rgb2gray(I)));

%phase shift frequency domain info
pF = pF1.*shift;

%plots
figure
subplot(121)
imshow(rgb2gray(I))
subplot(122)
imshow(ifft2(pF))
$\endgroup$
  • $\begingroup$ To clarify, a delay in the time domain is a linear phase shift vs frequency. (The higher the slope, the longer the delay) $\endgroup$ – Dan Boschen Oct 19 '18 at 18:29
0
$\begingroup$

The following Matlab line shows a 1D example of a fractional delay, implemented via DFT/FFT multiplication instead of a time-domain approach. Accuracy should be considered, noticeable ringing occurs.

You can extend it to 2D easily.

N = 64;                      % signal length
d = 0.5;                     % fractional shift

x = sin(2*pi*0.0323*[0:N-1]);  % signal
y = real( ifft( fftshift( exp(-j*(2*pi/N)*[-N/2:N/2-1]*d)).*fft(x,N) , N));

figure,stem(x,'g');
hold on
stem(y,'r+')
title('original x[n] and shifted y[n]');
legend('x[n]','y[n]');

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.