For purposes of understanding the process - not for any practical purpose for which I could use imresize - I wanted to show that 2x upsampling followed by convolution with an appropriate kernel (presumably the 2d equivalent of a "triangle") is equivalent to imresize(x,scalingFactor,'bilinear'), However I cannot do it perfectly. It "almost" works in the sense that qualitatively it looks very similar but it is as if it is slightly "shifted" cf. code below. Moreover for upsampling higher than ×2 I don't know how to build the correct kernel. I tried below, with zero "stuffing" but that is not correct. Any help and explanations/corrections are greatly appreciated.

N = 64;
%x = phantom(N)+0.25;
x = imresize(im2double(imread('cameraman.tif')),[N N]);
s = 2;

x_matlab_bilinear = imresize(x,s,'bilinear');

kernel1D = [0.5, 0.5*ones(1,s-2), 1, 0.5*ones(1,s-2), 0.5];
bilinearKernel =  kernel1D.'*kernel1D;

x_up = zeros(s*N,s*N);
x_up(1:s:end,1:s:end) = x;
x_custom_bilinear = conv2(x_up, bilinearKernel,'same');


imagesc(x_matlab_bilinear);colormap('gray');title('x up matlab bilinear');
imagesc(x_custom_bilinear);colormap('gray');title('x up custom bilinear');

enter image description here

For e.g. x8 upsampling I would use the following kernel (generated by my formula):

enter image description here

above is my kernel for interpolating a 2 times upsampled image.

For the delta-impulse image the results of 2 times upsampling are the following for the same above code but just replacing the first lines by

N=6; x=zeros(N); x(3,3)=1;


enter image description here

  • $\begingroup$ For 2x resampling, the only difference is that MATLAB uses a different output coordinate system, its output is effectively shifted by half a pixel compared to yours. I recommend that you experiment with a small delta-pulse image, which shows you perfectly what a linear process does. $\endgroup$ Commented Mar 30, 2022 at 20:28
  • $\begingroup$ For other scales, you also need to fix your kernel: it should be a triangle! $\endgroup$ Commented Mar 30, 2022 at 20:28
  • $\begingroup$ Try this test image: N=6; x=zeros(N); x(3,3)=1;. $\endgroup$ Commented Mar 30, 2022 at 20:29
  • $\begingroup$ ok will post the result with this delta pulse image, thanks alot. Also I posted the 8 times upsampling kernel I would use (just for the sake of example) is that correct? $\endgroup$
    – SheppLogan
    Commented Mar 30, 2022 at 20:31
  • $\begingroup$ No, it needs to be a triangle, so that when you’re closer to one sample than the other, you weigh that sample more than the other. $\endgroup$ Commented Mar 30, 2022 at 20:33

1 Answer 1


The linear interpolation kernel should be a triangle. You can construct it like this:

kernel1D = [(1:s)/s, (s-1:-1:1)/s];

The difference between MATLAB's imresize and your method for s=2 is a half-pixel shift. This is equivalent to shifting the kernel by half a pixel. So instead of a kernel [1/2, 1, 1/2], you'd use a kernel [1/4,3/4,3/4,1/4]. I was able to reproduce MATLAB's result like this (up to differences at the boundary, where MATLAB pads the input by repeating values, rather than adding zeros like conv2 does):

kernel1D = [1/4,3/4,3/4,1/4];  % kernel shifted by half a pixel!
bilinearKernel =  kernel1D.'*kernel1D;

x_up = zeros(s*N,s*N);
x_up(2:s:end,2:s:end) = x;   % note the shift here!
x_custom_bilinear = conv2(x_up, bilinearKernel,'same');

I'm guessing that the half-pixel shift makes the operation more symmetric: note that in your original result, you have a black column of pixels on the right. Shift by half a pixel, and now both the left and right columns of pixels contain the same fraction of "invented" (extrapolated) data.

  • $\begingroup$ very interesting thanks, but I wonder why is it not rather >> imtranslate([0.5 1 0.5],[0.5 0]) ans = 0.2500 0.7500 0.7500 for the 0.5 pixel shift rahter than 4 values you get [1/4,3/4,3/4,1/4]? $\endgroup$
    – SheppLogan
    Commented Mar 30, 2022 at 21:22
  • $\begingroup$ I hope I am not abusing of your kindness but maybe you will be interested to look at my other main question about this topic, a even more challenging one : dsp.stackexchange.com/questions/82295/… $\endgroup$
    – SheppLogan
    Commented Mar 30, 2022 at 21:41
  • $\begingroup$ @SheppLogan: It is the continuous-domain triangle that you translate, and then apply point-sampling. So there are 4 samples that hit the triangle, not 3. You only get 3 if the origin of the triangle hits a sample. $\endgroup$ Commented Mar 30, 2022 at 21:54
  • $\begingroup$ oh ok, thanks for the explanation, you are a very knowledgeable person! $\endgroup$
    – SheppLogan
    Commented Mar 30, 2022 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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