How can I correct inhomogeneous intensity in image?

Dear,

I am new to MATLAB, i need some help from you guys.The below image has inhomogeneous intensity distribution, the intensity is max at the top and it is decreasing linearly as it is going down.

• Idea was to consider the small portion of the images top corner and bottom corner and to loop through those pixels, get the intensity difference and set it to the max intensity value.
• for example if the intensity is getting reduced in step of 5 after reaching center of the image than see the difference and correct it by adding that difference

Thank you in advance

The image:

Answer 1

[Thorsten]

I = im2double(imread('blob.bmp'));

y = mean(I(:, 1:100), 2);

sigma = 30; % choosen by visual inspection
G = fspecial('gaussian', 3*sigma+1, sigma);
yb = imfilter(y, G, 'replicate');

Ic = bsxfun(@plus, I, yb(1) - yb);
subplot(1,2,1), imshow(I)
subplot(1,2,2), imshow(Ic)

[Rahul shetty]

Thanks Thorsten

[Image Analyst]

Note though that this is not as accurate as my method since he estimates the background based only on the average vertical profile in the left 100 lines, not over the whole image like I suggested you do. This method is good if your nonuniformity occurs only in the vertical direction, which is unlikely if you snapped your image with a camera, even one attached to a microscope.

[Rahul shetty]

@ Image analyst thanks for the comment. the non uniformity is occurring in vertical direction because the LED lights where not placed in the center or there was not enough light for the lower part of the image during capturing the image, so the intensity is getting decreased in steps, as we go vertically down.

can any one of you please explain me this code what exactly is going in there. Thanks

[Thorsten]

First we want to estimate the vertical illumination gradient, so we take the mean of the first 100 columns in the image. If we plot them, we realise that they are quite noisy. To get rid of the noise, we blurred the illumination gradient with a Gaussian. The difference along the gradient is the first point in the top line yb(1) minus the other points. Now to get rid of the gradient we have to add this correction to all columns of image. To add a the single column to a full RGB image, we use bsxfun.

[Rahul shetty]

Thanks

[Image Analyst]

Rahul, if you're using a lens, and I think you are, then there is shading across the lateral (x) direction. I took the code and added code to take profiles across the image above and below the spot:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;

rgbImage = im2double(imread('spot.bmp'));
[rows, columns, numberOfColorChannels] = size(rgbImage);

y = mean(rgbImage(:, 1:100), 2);

sigma = 30; % Chosen by visual inspection
G = fspecial('gaussian', 3*sigma+1, sigma);
yb = imfilter(y, G, 'replicate');

correctedImage = bsxfun(@plus, rgbImage, yb(1) - yb);
subplot(2,2,1)
imshow(rgbImage)
title('Original Image', 'fontSize', fontSize);
subplot(2,2,2);
imshow(correctedImage);
title('Corrected Image', 'fontSize', fontSize);
axis on;
% Let user mouse around and see RGB values
hp = impixelinfo();
for row = 1 : 20 : 150
  subplot(2,2,2);
  line([1, columns], [row, row]);
  hold on;
  thisLine = correctedImage(row, :, 2);
  subplot(2,2,3:4)
  plot(thisLine, '-');
  hold on;
end
% Get the means
topBand = correctedImage(1:150, :, 2);
meanTop = mean(topBand, 1);
plot(meanTop, 'k-', 'LineWidth', 3);
for row = 700 : 20 : rows
  hold on;
  subplot(2,2,2);
  line([1, columns], [row, row]);
  thisLine = correctedImage(row, :, 2);
  subplot(2,2,3:4)
  plot(thisLine, '-');
  hold on;
end
grid on;
title('Plots Across Image.  Black = top lines, Magenta = bottom lines', 'fontSize', fontSize);
% Get the means
bottomBand = correctedImage(700:end, :, 2);
meanBottom = mean(bottomBand, 1);
plot(meanBottom, 'm-', 'LineWidth', 3);

% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')

As you can see there is definitely a shading across the corrected image of about 2% from center to edges. Not only that but there is a difference in brightness from top to bottom even after correction of about 2%.

Now, maybe correcting out that last 2% doesn't matter to you, in which case don't worry about it. I'm not really sure what your plans for the corrected image are. If you're just getting a rough idea of the area of the spot, then it probably won't change the area much if you had a more accurate correction. But if you're doing high precision calibrated color measurement, then it could matter.

[Rahul shetty]

Image Analyst, thanks a lot, very well explained.

• This images are taken from DSLR camera
• I am writing a algorithm for automatic detection of circles, there is a one more small outer most circle in this image which can be seen by changing image attributes. The detection for the lower part of the circle is not accurate because of non homogeneous intensity.

Thanks for your replay.

[Image Analyst]

You might be interested in this paper: http://www.ecse.rpi.edu/homepages/qji/Papers/ellipse_det_icpr02.pdf

[Rahul shetty]

thanks Image Analyst

[Qing Wang]

Hi Thorsten,

What is the principle of your method?

Thanks,

Qiang

[Image Analyst]

@Qing Wang, click the link "Show 10 older comments" and read them all. It's discussed there.

Answer 2

[Image Analyst]

The BEST way to do it, and I do it all the time, is to take a "blank shot" and then divide your image by it.

In other words, take a picture of that white stuff (assuming it's truly flat and uniform, like a sheet of paper), without the yellow stuff. This will give you the pattern of exposure due to the combination of lens shading and lighting non-uniformity. So let's say some pixel in the corner had 90% as much light reaching the sensor as at the middle. Well, you'd want to divide by 90% to bring the value up to what it's supposed to be, wouldn't you? (Note, radiographs and fluorescence images might also need a subtraction). OK, now you need to get rid of small defects in the image like scratches, specks, video noise, etc. So you can either blur the image or fit it to a 2-D polynomial to get the perfect, noise-free background (I do the latter). Now you normalize your noise-free image so that it goes between 0 and 1. It will be 1 at the brightest point and the min, in the corners, will be whatever it is, like 0.6 or 0.8 or whatever, depending on how severe your shading it. The severity of lens shading varies greatly with manufacturers. We've tried stock lenses from manufacturers like Nikon and Canon and they show severe shading - they're junk. You'll get much flatter fields from specialty (non mass market) lens makers like Schneider (my favorite) and Linos. But their lenses cost twice as much. Obviously it's better to start with a better image than a crummy image if you're going to try to fix it. OK, now you just divide your actual image by the test image to get the flat field image:

correctedImage = double(testImage) ./ backgroundImage;

There are ways to improve upon that and be even more accurate, but I don't want to deliver my whole class on it here. They're more complicated and I think this may be enough for you.

If you can't get a blank shot, there are ways to estimate the background with that yellow spot still in there. You just have to remove it as best you can and fit the background to what's left. A simple way to do that is attached in the demo, but I'd actually do it a somewhat different way.

[Rahul shetty]

Thanks a lot Image Analyst, for detailed explanation.:)