# Optics aperture modelling rect functions.

I am trying to model a shape T aperture through 2D rect functions. Both the horizontal and vertical ‘lines’ have length b and width a, and do not overlap. The origin should be taken to be the centre of the vertical line.

The question has hinted at the function describing the aperture to be cartesian seperable.

My issues, I think, seem to stem from not knowing how the rect functions can be combined through operations together…

Here are my thoughts on the vertical ‘line’ :

The RHS of it (to the right of the origin) I believe is: $\frac{a}{2}$ rect $\frac{y}{b}$ [1]
Similarly the LHS i think is : -$\frac{a}{2}$rect$\frac{y}{b}$. [2]

My problem is then to express these two together. I am not sure how you define [1] + [2].
Would this be zero?

Perhaps you should multiply them, in which case I get -$\frac{a^{2}}{4}$ rect$^{2}$ $\frac{y}{b}$; so to me it then makes more sense to look at $\frac{a}{2}$ rect$^{2}$ $\frac{y}{b}$

Again I’m not sure how you would define a rect$^{2}$ function.

Here are my thoughts on the horizontal ‘line’ :

First of all, it can not be a single rect function as either the top or bottom line would then be missing.

I think $\frac{b}{2}$ e$_{2}$ – $\frac{a}{2}$rect$\frac{x}{b}$ for the bottom half, and ($\frac{b}{2}$ + $\frac{a}{2}$ )e$_{2}$ + $\frac{a}{2}$rect $\frac{x}{b}$ for the top half.

BUT as said above, the question hints towards the function being Cartesian separable, but in describing the horizontal ‘line’ I have introduced e$_{2}$ – the unit vector in the y direction. This also doesn’t look right in general, as isn’t rect a scalar ?

Many thanks to anyone who can help shed some light .
!

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