How to find energy of the image

7 次查看（过去 30 天）

aarti sawant 2014-2-3

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/114626-how-to-find-energy-of-the-image

编辑： Image Analyst 2014-2-3

   Please help me to find energy of the image given by equation,
   E(x) = for i=1 to x (p(x))

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Image Analyst 2014-2-3

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/114626-how-to-find-energy-of-the-image#answer_123068

在 MATLAB Online 中打开

What is p? The units of images are gray levels, and gray levels are units of energy (joules). Why? Well because you have radiation falling on a sensor. That's watts per square meter or joules per second per square meter. But the sensor has an area and it collects photons for a limited time so you multiply by the area of the pixel and the integration time of the pixel and you get joules. So to get the energy in an image you have to sum up all the gray levels.

totalEnergy = sum(imageArray(:));

What is x and p in your equation? Why is E a function of x (whatever x is)?

7 个评论
显示 5更早的评论隐藏 5更早的评论

Dishant Arora 2014-2-3

h you calculated above in turn comes out to be PDF, it's quite obvious that you will get 1 as answer by summing up h. As mentioned earlier by Image analyst you can simply add up all the pixel values in image to get the energy, why complicate it

Image Analyst 2014-2-3

编辑：Image Analyst 2014-2-3

在 MATLAB Online 中打开

aarti, please read this: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

Plus those for loops aren't needed and nullifying zeros in h isn't needed. You could do this:

zeroPixelLocations = (a == 0); % Find all zeros.
a(zeroPixelLocations ) = 1; % Set to 1 instead of 0.
h = imhist(a, 256);
h = h / numel(a); % a must be grayscale if you use numel.
E = sum(h(:))

请先登录，再进行评论。

请先登录，再回答此问题。