Convert half width half max to standard deviation for Gaussian
s = ieHwhm2SD(h,gDim)
Example:
We calculate the SD for a Gaussian with hwhm of 10; then we plot to show
that it reaches a value of 0.5 at 10 units from the center
s = ieHwhm2SD(10,2); g = fspecial('gauss',50,s);
x = 1:50; x = x - mean(x(:)); mesh(x,x,g/max(g(:))); axis equal
By default, we assume a bivariate Gaussian (gDim = 2)
g2 = (1/(2*pi*sx*sy))*exp(-(1/2)(x/sx)^2 + (y/sy)^2)
http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Bivariate_ca
se
N.B. The 1D case has not been tested significantly.
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Notes:
In the 2D case there is an elliptical curve(hx,hy) where the function is
1/2. The ellipse is defined from
0.5 = exp(-(1/2)*(hx/sx)^2 + (hy/sy)^2)
ln(2) = (1/2)*(hx/sx)^2 + (hy/sy)^2
We know that (hx,0) is on the contour, so
ln(2) = (1/2)*(hx/sx)^2
sx = hx / sqrt(2*ln(2))
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In the 1D case: g1 = (1/s sqrt(2pi)) exp(-(x/2s)^2)
The max is M = 1/s*sqrt(2*pi)
The half max occurs at a value, h, where
0.5 = exp(-(h/2s)^2)
ln(0.5) = -(h/2s)^2
sqrt(-ln(0.5)) = h/2s
2s sqrt(ln(2)) = h
We calculate the standard deviation in the 1D Gaussian case using a
slightly different formula
s = h/(2*sqrt(ln(2));
Copyright ImagEval Consultants, LLC, 2007.