# Into the sub-pixel realm: Optical simulation

Started Jan 1, 2013 | Discussions thread
Re: Into the sub-pixel realm: Optical simulation

Joofa wrote:

Joofa wrote:

Leo360 wrote:

Joofa wrote:

Also here we are dealing with two successive convolutions applied one after the other: first the lens then the sensor. Y= Hs*Hl*x. In this case variance of the result does not seem to equal sum of the variances.

Take the variance as the second central moment.

Variances add up for a sum of independent random variables. Here, on the other hand we have successive convolution -- a multiplicative relationship. Here is a simple 1D example I have in mind regarding two successive convolutions. One is for lens another for the sensor. For illustration purposes I choose convolution with a simple square kernel which mixes together central point and two of its nearest neighbors like this:

y(n) = x(n-1) + x(n) + x(n+1)

Start with a unit point source which is all zeros except for a single point x(0)=1. The resulting Y-sequence will have y(-1)=y(0)=y(1)= 1, the rest zero. "Blur disk" diameter is two, radius is 1 unit. Intensity -- constant. Now apply the second convolution (for the sensor) to the output Y. For simplicity, I choose the same coefficients:

z(n)= y(n-1) + y(n) + y(n+1)

The resulting sequence is all zeros except for the following elements: z(-2)=1; z(-1)=2; z(0)=3; z(1)=2; z(2)=1. "Blur disk" diameter is 4, radius is 2 and the intensity if higher in the center and less on the edges. Thus, "Blur Energy" is concentrated in the effective area smaller than the geometrical sum of radii. It seems that the "effective" radius with intensity variations factored in should be smaller than 2, might be even sqrt(2) for that matter. This way one can justify the Marianne's formula, I think. Of course, for a proper analysis one should perform 2-D convolution which is very similar but involves much more typing

As I mentioned before, we need to take variance as the second central moment. Taking your example, var(x[n]) = 0, var(y[n]) = 2/3, var(z[n]) = 2/3, and var (output[n]) = 4/3. Hence, since 4/3 = 2/3 + 2/3 + 0, the formula for additive variances under convolution satisfies.

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Furthermore, the blur "radius" for y[n] and z[n] would be sqrt(2/3) = 0.81 (smaller than 1 in your example above), and for output[n], sqrt(4/3) = 1.15 (smaller than 2).

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In 2D the radii will be somewhat bigger and closer to sqrt(2).

Leo

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