1
$\begingroup$

Is there an accurate way to create an aliased image from the Fourier transform of the original image?

in other words, i have the Fourier coefficients of an image and i want to make down-sample in frequency domain.

$\endgroup$
1
$\begingroup$

I am trying to construct an aliasing image.

I=rgb2gray(imread('Liver.png')); % a 512*512 image 
subplot(1,2,1),imshow(I),title('Original image')  % shown on the left side of figure below

when you implement F=fftshift(fft2(I));, the corresponding axis represent the frequency range of [-255:256]/512; and the cut-off frequency U = V = 256/512 = 0.5;

based on Nyquist Sampling Theorem, the sampling periods in spatial domain should satisfy

delta_x<=1/(2*U)=1;
delta_y<=1/(2*V)=1;

to avoid the aliasing.

So if your bandwidth of recorder is 64 * 64, the corresponding point interval in spatial domain is 1 / (64/256) = 4. Thus,

Ia=imresize(I(1:4:end,1:4:end),[512 512]);
subplot(1,2,2),imshow(Ia),title('Aliased image') % shown on the right side of the figure below

enter image description here

You can observe the aliasing effect at the edge.

$\endgroup$
0
$\begingroup$

Yes, there is a (simple) way to do this.

Aliasing comes from the fact that the bandwidth of the recording device is smaller than the bandwidth of the signal. Out-of-bandwidth frequencies are not zeroed-out however, but injected into the captured part of the frequency domain.

You can find into which frequency coefficient they are injected without any computation by using the periodicity of the DFT. If you suppose that the bandwidth of the recorder is some rectangle in the frequency domain, then this rectangle is replicated (translated) in each direction. So, you just have to translate mentally this rectangle to see which (in-bandwidth) position corresponds to an aliased frequency.

$\endgroup$
  • $\begingroup$ please, can you start with say 256*256 frequency coeff. and show me the steps of creating aliased image when the BW of the recorder is 64*64 rectangle. $\endgroup$ – HforHesham Dec 16 '13 at 19:25

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.