## Abstract

The article deals with a method of calculation of off-axis light propagation between parallel planes using discretization of the Rayleigh-Sommerfeld integral and its implementation by fast convolution. It analyses zero-padding in case of different plane sizes. In case of memory restrictions, it suggests splitting the calculation into tiles and shows that splitting leads to a faster calculation when plane sizes are a lot different. Next, it suggests how to calculate propagation in case of different sampling rates by splitting planes into interleaved tiles and shows this to be faster than zero-padding and direct calculation. Neither the speedup nor memory-saving method decreases accuracy; the aim of the proposed method is to provide reference data that can be compared to the results of faster and less precise methods.

©2010 Optical Society of America

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### Equations (14)

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(1)
$$t(x)={\displaystyle {\int}_{Source}s}(\xi )p(x-\xi )\text{d}\xi $$
(2)
$$t[n]={\displaystyle \sum _{m=0}^{M-1}s}[m]p[n-m]\text{\hspace{1em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} for \hspace{0.17em}}0\le m\le M-1,\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.05em}}0\le n\le N-1.$$
(3)
$$t[n]={\displaystyle \sum _{m=0}^{C-1}s}[m]p[(n-m)\text{\hspace{0.17em} mod \hspace{0.17em}}C]\text{\hspace{1em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} for \hspace{0.17em}}0\le m\le C-1,\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.05em}}0\le n\le C-1,$$
(4)
$$t=IDFT(DFT(s)\times DFT(p)),$$
(5)
$$U(x,y,z)=\frac{-1}{2\pi}\underset{-\infty}{\overset{\infty}{{\displaystyle \int \text{}}\text{}{\displaystyle \int \text{}}\text{}}}U(\xi ,\eta ,0)\frac{\partial}{\partial z}\frac{\mathrm{exp}(\text{j}kr)}{r}\text{d}\xi \text{d}\eta $$
(5)
$$U(x,y,z)=U(x,y,0)\otimes \left[\frac{-z}{2\pi}\left(\text{ j}k-\frac{1}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\right)\frac{\mathrm{exp}(\text{j}k\sqrt{{x}^{2}+{y}^{2}+{z}^{2}})}{{x}^{2}+{y}^{2}+{z}^{2}}\right]$$
(6)
$$\text{(}DFT\text{\hspace{0.05em} 'scount})\text{\hspace{0.05em}}C\mathrm{log}C$$
(8)
$${t}_{\text{basic}}(M,N)=3{(M+N)}^{2}\mathrm{log}{(M+N)}^{2}\equiv 6{(M+N)}^{2}\mathrm{log}(M+N)$$
(9)
$${t}_{\text{commontiles}}(M,N,T)=\left(1+2{T}^{2}\right)2{\left(M+\frac{N}{T}\right)}^{2}\mathrm{log}\left(M+\frac{N}{T}\right)$$
(10)
$${s}_{1}(\omega ,N,T)=\frac{{t}_{\text{basic}}(\omega N,N)}{{t}_{\text{commontiles}}(\omega N,N,T)}$$
(11)
$${t}_{\text{upscaled}}(M,\tau )=6{(\tau M+\tau M)}^{2}\mathrm{log}(\tau M+\tau M)\equiv 24{\tau}^{2}{M}^{2}\mathrm{log}(2\tau M)$$
(12)
$${t}_{\text{interleavedtiles}}(M,\tau )=\left(1+2{\tau}^{2}\right)2{\left(M+M\right)}^{2}\mathrm{log}\left(M+M\right)\equiv \left(1+2{\tau}^{2}\right)8{M}^{2}\mathrm{log}\left(2M\right)$$
(13)
$${s}_{2}(M,\tau )=\frac{{t}_{\text{upscaled}}(M,\tau )}{{t}_{\text{interleavedtiles}}(M,\tau )}\equiv \frac{3{\tau}^{2}\mathrm{log}(2\tau M)}{\left(1+2{\tau}^{2}\right)\mathrm{log}\left(2M\right)}$$
(14)
$$({M}_{x}+{N}_{x}-1)({M}_{y}+{N}_{y}-1)\le C$$