Hi Ahmed

I have applied some changes to your code but perhaps you would like to consider using a different filter.

attached test567.m

1.

reference image

clear all;clc

img = imread( ('eight.tif') ); % read image, use gray-level images here. figure(1);imshow(img);title('Original Image') % it's useful to mark references as early as possible

2.

really contaminated image

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A = imnoise(img,'Gaussian',0,1); figure(2);imshow(A)title('Noisy Image')

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IMG2 = fft2 ( img ); % Fourier of img sz = size( img )

G = fspecial('gaussian' ,[5,5]) % create a filter with std sigma same size as img %# Filter it h = imfilter(A,G,'same') H = fft2 ( h ) % Fourier of filter

F = IMG2.*H % filter in Fourier space f = ifft2( F ) % back to spatial domain. f=f-min(min(f)); f=f*255/max(max(f));

figure(3); imagesc(f);title('Gaussian filter in Frequency Domain')

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Please note that min max values of the filter itself are really high, and the range between such min max is really narrow compared to the values of literally any pixel of the filter.

You may want a higher range.

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figure(6); imshow(uint8(f)) % Calculate MSE, mean square error.

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Mean_Square_Error =immse(img,f) Mean_Square_Error = 1.8836e+04

apparently command immse suggested by Image Analyst yields an error assessment way larger than applying Ahmed's error formula.

either **im2double()** on unit8 type or the other way round, or **uint8()** on double type both ways get to same error figure.

[M N] = size(img) M = 242 N = 308 Mean_Square_Error2 = sum(sum(error0 .* error0)) / (M * N) Mean_Square_Error2 = 208.7750 error02 = im2double(img) - f; Mean_Square_Error3 = sum(sum(error0 .* error0)) / (M * N) Mean_Square_Error3 = 208.7750

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Ahmed

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thanks in advance

John BG