Problem with matrix calculation

I have the following line of code that represents the inverse function of:

a+b*(100*10^-x)^c

this is the x:

x=double(gssq.(strcat(sheetNamesOECF{rgby2d},num2str(rgby2d2)))(:,:,1))/255

      0.1098    0.1137    0.1137    0.1137    0.1137    0.1137    0.1098    0.1098    0.1137
      0.1098    0.1176    0.1137    0.1098    0.1098    0.1098    0.1098    0.1098    0.1098
      0.1137    0.1176    0.1137    0.1098    0.1137    0.1137    0.1176    0.1137    0.1137
      0.1098    0.1137    0.1098    0.1098    0.1098    0.1098    0.1137    0.1098    0.1098
      0.1137    0.1098    0.1059    0.1098    0.1137    0.1098    0.1098    0.1098    0.1098

this is the inverse function:

-log10(((coeffr(1,1)-double(x)/100^coeffr(1,3)/coeffr(1,2))^(1/coeffr(1,3))))

now, when I fill it in like this:

-log10(((coeffr(1,1)-double(0.1)/100^coeffr(1,3)/coeffr(1,2))^(1/coeffr(1,3))))

I get:

2.2059 + 0.7575i

but when I fill in the matrix instead of the constant I get values like this:

1.1654 + 0.7560i

these are way smaller than what they are supposed to be.

does anyone know why filling in the same value gives me different results?

Kind regards,

Thomas Koelen

[Joep]

double(x) and double(0.11) are not equal. double(x) is in this cames almost 9*0.11. That's why the number is 2 times as high.

The second why use a second double(x)? the numbers are already double. If you want to use the first number of x. Use x(1) or x(1,1).

The third why you using double(...)/255 you can use instead of these im2double, does exact the same but the deference is this function already devides it by 255.

The forth point is are you sure you not reflecting the RGB coefficients with these coefficients? otherwise you should multiply this coefficients by 255.

[Thomas Koelen]

1. I don't understand why double x would be 9*0.11, could you explain yourself further?

2. This is my mistake, I just replaced the 0.1 with x here, in the code it's actually x.

3. Because I didn't know this function earlier on and already created a structute that has all the seperate channels.

4. because it is a normalised pixel value to calculate the OECF curve, so the pixels go from 0-1.

Kind regards

Answer 1

[Matt J]

Are you sure the cube root of -1 is also -1? This command thinks so

    >> nthroot(-1,3)

    ans =

        -1

but this command has a different opinion,

    >> (-1)^(1/3)

    ans =

       0.5000 + 0.8660i