# CGH: Consolidate Expressions Regarding Parallel Apertures

## One-dimensional Apertures

From our accompanying discussion of the Utility of FFT Techniques, we start with the most general expression for the amplitude at one point on an image screen, namely,

 $~A(y_1)$ $~=$ $~\sum_j a_j e^{i(2\pi D_j/\lambda + \phi_j)} \, ,$

and, assuming that $~|Y_j/L| \ll 1$ for all $~j$, deduce that,

 $~A(y_1)$ $~\approx$ $~\sum_j a_j e^{i[ 2\pi L/\lambda + \phi_j]}\biggl[ \cos\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) - i \sin\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \biggr] \, ,$

where,

 $~L$ $~\equiv$ $~ Z \biggl[1 + \frac{y_1^2}{Z^2} \biggr]^{1 / 2} \, .$

Note that $~L$ is formally a function of $~y_1$, but in most of what follows it will be reasonable to assume, $~L \approx Z$. Notice, as well, that this last approximate expression for the (complex) amplitude at the image screen may be rewritten in the form that will be referred to as our,

Focal-Point Expression

$~A(y_1)$

$~\approx$

$~ e^{i 2\pi L/\lambda } \sum_j a_j e^{i \phi_j} \cdot e^{-i \Theta_j } \, ,$

where,

 $~\Theta_j$ $~\equiv$ $~\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \, .$

### Case 1

In a related accompanying derivation titled, Analytic Result, we made the substitution,

 $~a_j$ $~\rightarrow$ $~a_0(Y) dY = a_0(\Theta) \biggl[ \frac{w}{2\beta_1} \biggr] d\Theta \, ,$

where,

 $~\frac{1}{\beta_1}$ $~\equiv$ $~\frac{\lambda L}{\pi y_1w} \, ,$

and changed the summation to an integration, obtaining,

 $~A(y_1)$ $~\approx$ $~ e^{i 2\pi L/\lambda }\biggl[ \frac{w}{2\beta_1} \biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot e^{-i \Theta } d\Theta \, .$

If we assume that both $~a_0$ and $~\phi$ are independent of position along the aperture, and that the aperture — and, hence the integration — extends from $~Y_2 = -w/2$ to $~Y_1 = +w/2$, we have shown that this last expression can be evaluated analytically to give,

 $~A(y_1)$ $~\approx$ $~ e^{i [2\pi L/\lambda + \phi] }\biggl[ \frac{a_0 w}{2\beta_1} \biggr] \int_{\Theta_2}^{\Theta_1} e^{-i \Theta } d\Theta$ $~=$ $~ e^{i [2\pi L/\lambda + \phi] } \cdot a_0 w ~\mathrm{sinc}(\beta_1) \, .$

We need to explicitly demonstrate that an evaluation of our Focal-Point Expression with $~a_j = 1$, gives this last sinc-function expression, to within a multiplicative factor of, something like, $~j_\mathrm{max}$.

### Case 2

In our accompanying discussion of the Fourier Series, we have shown that a square wave can be constructed from the expression,

 $~f(x)$ $~=$ $~ \frac{c}{L} + \sum_{n=1}^{\infty} \biggl( \frac{2}{n\pi} \biggr) \sin \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr)$ $~=$ $~ \frac{2c}{L}\biggl\{\frac{1}{2} + \sum_{n=1}^{\infty} \mathrm{sinc} \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr) \biggr\} \, .$

Can we make this look like our above, Focal-Point Expression?

Let's start by setting

 $~Y_j$ $~=$ $~\frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \, ,$

for $~0 \le j \le (j_\mathrm{max}-1)$, in which case,

 $~\Theta_j$ $~\equiv$ $~ \frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \biggr] = \frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} \biggr] - \frac{2\pi y_1}{\lambda L} \biggl[ \frac{w}{2} \biggr]$ $~=$ $~ j \biggl[ \frac{2\pi y_1 w}{(j_\mathrm{max}-1) \lambda L} \biggr] - \frac{\pi y_1 w }{\lambda L} = \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L} \, ,$ $~=$ $~ j \cdot \Delta\Theta - \frac{(j_\mathrm{max} -1)}{2} \Delta\Theta \, ,$

where,

$~\Delta\Theta \equiv \frac{\pi y_1}{\mathfrak{L}} \, ,$     and     $~\mathfrak{L} \equiv \biggl[ \frac{(j_\mathrm{max}-1) \lambda L}{2w} \biggr] \, .$

This means that $~\Theta_{i} = - \Theta_{( j_\mathrm{max} - 1 - i )}$.

The key expression under the summation therefore becomes,

 $~a_j e^{i \phi_j} \cdot e^{-i \Theta_j }$ $~=$ $~~a_j e^{i \phi_j} \cdot \biggl[ \cos \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0 \biggr) - i \sin \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0 \biggr) \biggr] \, ,$

where,

$~\Theta_0 \equiv \frac{(j_\mathrm{max} - 1)}{2} \cdot \pi y_1 \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] = \frac{\pi y_1 w}{\lambda L} \, .$

Now, what is the argument of the sinc function? By default, it needs to be something along the lines of,

 $~\frac{j \pi c}{\mathfrak{L}}$ $~=$ $~j \pi c \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] \, .$

Then, as $~j$ varies from $~0$ to $~(j_\mathrm{max} - 1)$, the argument goes from $~0$ to $~[2\pi w c/(\lambda L)]$. In an effort to make the function exhibit reflection symmetry as we move from one side of the aperture to the next, let's subtract half of this upper limit; that is, let's modify the argument of the sinc function to read,

 $~\frac{j \pi c}{\mathfrak{L}} - \frac{\pi w c}{\lambda L}$ $~=$ $~ j \pi c \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] - \frac{\pi w c}{\lambda L} = \biggl[ \frac{2j}{j_\mathrm{max}-1} - 1\biggr]\biggl[ \frac{\pi w c}{\lambda L} \biggr] \, .$

This means that in our above, Focal-Point Expression we want to set,

 $~a_j$ $~=$ $~ \mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1} - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] \, .$

This therefore gives the following,

Focal-Point Expression for a Square Wave

$~A(y_1)$

$~\approx$

$~ e^{i 2\pi L/\lambda } \sum_{j=0}^{j_\mathrm{max}-1} e^{i \phi_j} \cdot~ \mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1} - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] \biggl\{ \cos \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] \biggr\} \, .$

This exhibits a very desirable feature: Both the sinc function and the sine function — and, hence, also their product — have reflection symmetry about the summation index, $~j = (j_\mathrm{max}-1)/2$. As a result, if the overall phase factor, $~e^{i \phi_j}$, behaves in an appropriately simple way — for example, if it is zero everywhere — then under the summation the sine term will sum to zero and leave only the desired — and real — product, $~\mathrm{sinc} \times \cos$. Try this out in Excel to see if it works!

This could use a little more manipulation. Let's define the alternate summation index,

 $~n$ $~\equiv$ $\frac{1}{2} \biggl[ j_\mathrm{max}-1 \biggr] \biggl( \frac{2j}{j_\mathrm{max}-1} - 1 \biggr) \, ,$

in which case we can write,

 $~A(y_1)$ $~\approx$ $~ e^{i 2\pi L/\lambda } \sum_{n~=~-(j_\mathrm{max} - 1)/2}^{+(j_\mathrm{max} - 1)/2} e^{i \phi_j} \cdot~ \mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] \biggl\{ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] \biggr\}$ $~=$ $~ e^{i 2\pi L/\lambda } e^{i \phi_{j=0} } ~+~e^{i 2\pi L/\lambda } \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2} 2e^{i \phi_j} \cdot~ \mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] ~ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr]$ $~=$ $~ e^{i 2\pi L/\lambda } e^{i \phi_{j=0} } ~+~e^{i 2\pi L/\lambda } \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2} 2e^{i \phi_j} \cdot~ \mathrm{sinc} \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) ~ \cos \biggl( \frac{n \pi y_1 }{\mathfrak{L} } \biggr)$ $~=$ $~ e^{i 2\pi L/\lambda } \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{ e^{i \phi_{j=0} } \biggl(\frac{c}{\mathfrak{L}} \biggr) ~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2} e^{i \phi_j} \cdot~ \biggl(\frac{ 2 }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) ~ \cos \biggl( \frac{n \pi y_1 }{\mathfrak{L} } \biggr) \biggr\} \, .$

Finally, recalling that,

 $~L$ $~\equiv$ $~ Z \biggl[1 + \frac{y_1^2}{Z^2} \biggr]^{1 / 2} \approx Z \biggl[1 + \frac{1}{2}\frac{y_1^2}{Z^2} \biggr] = Z + \frac{y_1^2}{2Z} \, ,$

let's set …

 $~e^{i\phi_j}$ $~=$ $~ e^{-i2\pi Z/\lambda}$ $~\Rightarrow ~~~ e^{i2\pi L/\lambda} \cdot e^{i\phi_j}$ $~=$ $~ e^{i2\pi (L-Z)/\lambda} \approx e^{i\pi y_1^2/(\lambda Z)} = \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \, .$

As a result, we have,

 $~A(y_1)$ $~\approx$ $~ \biggl[ \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \biggr] \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{ \biggl(\frac{c}{\mathfrak{L}} \biggr) ~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2} \biggl(\frac{ 2 }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) ~ \cos \biggl( \frac{n \pi y_1 }{\mathfrak{L} } \biggr) \biggr\} \, .$

Therefore, a clean square wave will appear only if $~[\pi y_1^2/(\lambda Z)] \ll 1$.