Shouldn't imfilter and conv2 output the same image here?

Question

So I am applying a Gaussian kernel to an ultrasound image, made up of integers values in the range of 0 to 255, like this

filteredImage = imfilter(image,kernel,'conv','same')

using imshow(filteredImage) I get a nicely blurred image:

Then, instead using

convImage = conv2(image,kernel,'same')

I get the following image

Aren't these two functions, used this way, supposed to produce the same kind of output?

Answer 1

imfilter and conv2 aren't exactly the same (imfilter is like conv2 with the filter flipped). If you use 'conv' then they are the same.

First you can check that you are using the correct image types using imfilter in the other way. Check this:

out1=conv2(double(image),kernel,'same');  
out_conv=uint8(out1); 
old_imfilter=imfilter(image,kernel,'same');
new_imfilter=imfilter(image,kernel(end:-1:1,end:-1:1),'same');

new_imfilter and out_conv should be the same. If so, you can do it with your method:

convImage = uint8(conv2(double(image),kernel,'same'));
filteredImage = imfilter(image,kernel,'conv','same');

Now filteredImage and convImage should be the same.

Answer 2

The difference between imfilter and conv2 when used in this way is that conv2 performs a conversion to double...

varargin{k} = double(varargin{k});

...while imfilter does not.

Images of type uint8 have a range of [0,255] while those of type double typically have range [0,1]. imshow works under these assumptions and so, if passed a double image will display any value greater than 1 as white.

As your original image was uint8, the result after calling conv2 will still have values in the range [0,255] (despite the conversion to double) and so will not be displayed correctly by imshow. A number of fixes (some suggested already) are:

• Allow imshow to detect the range: imshow(convImage,[])
• Normalize your result: imshow(convImage/255)
• Convert your result to uint8 before display: imshow(uint8(convImage))