transfer function of camera !!!!!!!!!!!!! what is this

hi today i here about what is called a transfer function of camera

its matrix with 8 unknown parameter

can any help about what is this!!!???

Answer 1

[Image Analyst]

You'd have to give more info. Usually I think transfer function refers to the plot of output gray level versus input gray level, often called "gamma" or the opto-electronic conversion function. Basically it's a curve that usually expands (brightens) dark parts of the photo while compressing brighter parts of the photo. See the Gamma FAQ here: http://www.poynton.com/ Poynton also has a color FAQ here: http://www.poynton.com/ColorFAQ.html for those who are interested. But I've never heard of the gamma being a matrix with 8 unknown parameters. I've heard of it as a 256 by 1 lookup table. The index into the table is the input gray level and the value of the table at that element is the gray level the output image will be mapped to. Go here: http://www.image-engineering.de/ for tools for measuring the gamma of cameras.

[mangood UK]

dear image thank you for replay

i will told you about my main problem that make me search around to find the solution and some told me about the T.F

i hava camera to capture a image for object on the ground

i want the area of object in pixel to be constant what ever why change the position of camera (height)

if camera have height 1 meter above object and area of object='100'

then if camera have height equal 0.5 meter above object the area must still=100

what is the processing to do this operation??!!!!

best reggards

[Image Analyst]

As the camera moves closer or farther away, a fixed size object will look larger or smaller, respectively. In other words, it will cover more or fewer pixels. If you want it to always be 100 pixels, then you must find it, determine the current magnification, and then use imresize. For example, you move the camera farther away and the object is now only 25 pixels. So you must do

newImage = imresize(originalImage, 100/25);

[Walter Roberson]

Better make that sqrt(100/25) as the magnification occurs along x and y.

[Image Analyst]

You're right. When it said "B = imresize(A, scale) returns image B that is scale times the size of A. " I thought size meant area (like the "size" of the image in pixels), but later down in the help it says: "If you specify a scalar, imresize uses the value as the scale factor for each dimension" so you need the square root.

[mangood UK]

dear:

is this same thing can be done for other feature of the object like Perimeter and circularity and other feature ???

best regards

[Image Analyst]

You have to look at the units. Area is a squared thing so you need to multiple pixels by the calibration factor squared when the pixels represent area. When the pixels represent a linear thing, like distance or perimeter, then multiply by the calibration factor (not squared). For a unitless ratio like circularity, you don't need to multiply by anything since it's = perimeter^2/(4*pi*area) so the calibration factor may already be in the perimeter or area and it just cancels out and you don't want to introduce it again. You can also use perimeter and area in pixels without converting to some other scale or calibration.

[mangood UK]

ok ,it clear now but i need to know about last problem

if i calculate the area or perimeter of object in pixel

dose the Nonlinear distortion caused by camera lens effect on object area or perimeter??? if so how can we solve the problem

best regards

[Image Analyst]

Yes, it would. Optical distortion (in the strict optical definition) takes two forms: pincushion (lines bow inwards) and barrel (lines bow outward). You can tell which you have by imaging a grid of lines or dots and seeing what kind of pattern they exhibit. A square area on the graph paper you image will not land on square block of pixels in your image - it will be some weird shape. So you need to measure the grid intersection points and define where they should be. To do this you use imtransform(). Pass in the actual coordinates, and the ideal, true, desired coordinates into maketform(), and get a transformation which you will pass along with your image into imtransform() and it will warp/fix your image.

[mangood UK]

dear Image Analyst:

you replay is always perfect yes it perfect

for Optical distortion i think also affect on the center of the object or center of mass of object is this right!!!!

for the imtransform() do you any know any good tutorial i mean with som figure for what you say and for the solving this problem

best regards

[Image Analyst]

Since distortion redistributes the light, it could affect the location of the center of mass, and the intensity in the scene. The help has some examples for imtransform, though not for distortion. It wouldn't be too hard to make one for distortion. Give it a try.

[mangood UK]

thank

Answer 2

[mangood UK]

thank you

alot

 a lot

 a lot

best regards