MATLAB: Can u hhelp me by giving the syntax for making an intensity matrix from the given image…

image processingImage Processing Toolboxintensity matrix..

Image=imread('D:\semester 3\matlab work\images\hand.jpeg');
imshow(image);
   disp(image);

after executing it the command window shows its intensity values….i want to make the matrix from these intensity values… how can i do it????can u help me for the same???

i want to use these intensity values in my code…dats y……

Best Answer

You read the intensity values into an array named "Image" with a capital-I. And after that you used "image" with a lower-case-I. image with a lower-case-I happens to be the name of the MATLAB routine to create a graphics image. It happens to be valid to call image() with no parameters, in which case it displays a default image and returns the handle-graphics handle of the image object.

You then apply imshow() to the handle of the image object; imshow is probably going to automatically erase your current figure contents, causing the default image object to be deleted, and then creating a new image object that displays based upon the chance numeric value the handle had.

You then disp(image) which is going to create a new default image object and output the handle graphics handle of the image object.

What you need to do is go back and use consistent variable names in your code, preferably avoiding using "image" or "Image" as the variable name.

Related Solutions

MATLAB: How to find intensity of a gray image

It depends on what you mean by intensity. I know you've been working, and instructing students at your university, long enough in the image processing field that you know how to take a histogram:

% Let's compute and display the histogram.
[pixelCount, grayLevels] = imhist(grayImage);
subplot(2, 2, 2); 
bar(pixelCount);
grid on;
title('Histogram of original image', 'FontSize', fontSize);
xlim([0 grayLevels(end)]); % Scale x axis manually.

and I know that you know how to get the overall mean of the image:

meanGrayLevel = mean2(grayImage);

I know you already know all that. So now we need to have you explain what you mean by "intensity". What you (and most others) may not know is that the units of a standard optical image are NOT intensity. They're lumens-seconds, a form of visible energy. Why? Think of lumens like watts, but just in the visible wavelength. So it's like energy per second. Because you have Lux (lumens per square meter) at the sensor (CCD), and you integrate that over the area (square meters) of a pixel and over time (seconds), then the units are lumen-seconds, kind of like a visible energy. See this for further discussion.

Now, intensity is an SI base unit along with meter, second, ampere, kelvin, mole, and kilogram. Its units are candela. You can not get the luminous intensity of an image - it doesn't make sense. It is a characteristic of the light source, not a surface with light incident on it. Wikipedia goes into a great amount of detail on all these units.

Anyway, we're back to what you mean by intensity. The mean gray level and histogram are too trivial (and you already know those), so please explain exactly what you mean.

MATLAB: Relation between “Image” intensity and “Light” Intensity

Welcome to the very confusing field of optical units. An image does not have units of intensity. See the table of the bottom of this page: https://en.wikipedia.org/wiki/Candela I'm surprised you couldn't find anything because there is tons of information out there, unfortunately it will make your head spin unless you have a Ph.D. in optics (sometimes even if you do, speaking from personal experience). The units of an image are like joules, or in photometric units lux*m^2*seconds, which is lumens*seconds, which is candela*steradian*second (to get it into all base SI units). Anyway, think of it as a measure of energy (or luminous energy). Let's use regular radiometric units instead of photometric (luminous) units (which are restricted to the human visual range and a lot more complicated). So you have optical power hitting your sensor. Like 10 watts over an area of 1 cm by 1 cm. Now the CCD well integrates those photons. Each pixel might be 5 microns by 5 microns. So now we have watts per area multiplied by the area. That's how many watts are integrated by that pixel. But the pixel only integrates for a certain number of seconds, and watts is joules per second, so you have watts*seconds = (joules/second)*second = joules. That's why I say it's like joules or energy.

OK, that's more than you wanted to know, so I won't even bother to go into the "intensity" of a light source which is even more complicated. Even the "experts" don't agree. For example the American Institute of Physics says that the "intensity" of a light source is W/steradian, yet intensity is an SI Base quantity (like meter, kg, second, ampere, kelvin, and mole) and it's units are candela. My late optics professor, Jim Palmer of the College of Optics at the University of Arizona, got so worked up on the sloppy usage that he wrote a paper on it: "Getting intense about intensity", Metrologia, 1993, vol 30, pp. 371-372. I attach a partial screenshot of his paper, for educational purposes.

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Related Solutions

MATLAB: How to find intensity of a gray image

MATLAB: Relation between “Image” intensity and “Light” Intensity

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