exp function is returning zeros

Hi,

i am trying to compute a normal function for all pixels on the image ,

m = mean(I(:));
segma = std (I(:));
fb1=(exp(-abs((I(i)-m).^2)./2.*segma.^2))% Feature based1
fb2=(exp(-abs((I(j)-m).^2)./2.*segma.^2))% Feature based 2

it returns zeros,am i clculating the mean and the std in wrong way !!

Thanks

Answer 1

[Steven Lord]

Using Symbolic Math Toolbox, we can see what exp(-850) and exp(-950) would be, displayed to 20 digits.

>> vpa(exp(sym([-850; -950])), 20)

ans =

   7.0744125465834323713e-370
   2.6317352159005409842e-413

The smallest positive double precision number is not quite that small.

>> eps(0)

ans =

   4.9407e-324

[sbei arafet]

thanks for the precision

Answer 2

[Star Strider]

Assuming that ‘I’ is single or double, and not uint8 (that would throw an error for std), the most likely possibility is that the exp argument is very large.

[sbei arafet]

   I=double(dicomread('C:\Users\USER\Desktop\DDN250454 - 0815_002\Unnamed\WB MAC PSF ON - 8265\IM-0001-0001-0001.dcm'))./255;

yes it is, the arguments are in the range of -950 and -850

[Star Strider]

On my computer, the realmin function returns 2.225073858507201e-308, corresponding to exp(-708.396418532264). The exponential function of any value less than -708.396418532264 will return zero. (Even the Symbolic Math Toolbox with its extended resolution returns zero.)

Answer 3

[Torsten]

fb1=exp(-(I(i)-m).^2./(2.*segma.^2))% Feature based1
fb2=exp(-(I(j)-m).^2./(2.*segma.^2))% Feature based 2

By the way: "segma" is "sigma".

Best wishes

Torsten.

[sbei arafet]

thanks