## Abstract

A method of diffraction calculation between tilted planes with variable sampling rates is proposed. The proposed method is based on the Fourier spectrum rotation from a tilted plane to a parallel plane. The nonuniform fast Fourier transformation (NUFFT) is used to calculate the nonuniform sampled Fourier spectrum on the tilted plane with variable sampling rates, which overcomes the sampling restriction of FFT in the conventional method. Both of the computer simulation and the optical experiment shows the feasibility of our method in calculating the hologram of polygon-based object with scalable size, which can be considered as an important application in the holographic three-dimensional display.

© 2014 Optical Society of America

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### Equations (10)

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(1)
$$\left[\begin{array}{c}{u}_{t}\\ {v}_{t}\\ {w}_{t}({u}_{t},{v}_{t})\end{array}\right]=T\cdot \left[\begin{array}{c}{u}_{r}\\ {v}_{r}\\ {w}_{r}({u}_{r},{v}_{r})\end{array}\right]=\left[\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {a}_{4}& {a}_{5}& {a}_{6}\\ {a}_{7}& {a}_{8}& {a}_{9}\end{array}\right]\cdot \left[\begin{array}{c}{u}_{r}\\ {v}_{r}\\ {w}_{r}({u}_{r},{v}_{r})\end{array}\right]$$
(2)
$$\begin{array}{l}{u}_{t}=\alpha ({u}_{r},{v}_{r})={a}_{1}{u}_{r}+{a}_{2}{v}_{r}+{a}_{3}{w}_{r}({u}_{r},{v}_{r})\\ {v}_{t}=\beta ({u}_{r},{v}_{r})={a}_{4}{u}_{r}+{a}_{5}{v}_{r}+{a}_{6}{w}_{r}({u}_{r},{v}_{r})\end{array}$$
(3)
$$\begin{array}{c}{G}_{t}({u}_{t},{v}_{t})={G}_{t}\left[{m}_{u}\Delta \left({u}_{t}\right),{m}_{v}\Delta \left({v}_{t}\right)\right]\\ =NUFFT2\left[{f}_{t}({m}_{x}\Delta x,{m}_{y}\Delta y)\right]\\ =NUFFT2\left[{f}_{t}({x}_{t},{y}_{t})\right]\end{array}$$
(4)
$${G}_{r}({u}_{r},{v}_{r})={G}_{r}({\alpha}^{-1}({u}_{t},{v}_{t}),{\beta}^{-1}({u}_{t},{v}_{t}))={G}_{t}({u}_{t},{v}_{t})$$
(5)
$${G}_{h}({u}_{h},{v}_{h})={G}_{r}({u}_{r},{v}_{r})\cdot \mathrm{exp}\left(2\pi iz\sqrt{\frac{1}{{\lambda}^{2}}-{u}_{r}{}^{2}-{v}_{r}{}^{2}}\right)$$
(6)
$$\begin{array}{c}{f}_{h}({x}_{h})=NUFFT1\left\{{G}_{r}({u}_{r})\cdot \mathrm{exp}\left(2\pi iz\sqrt{\frac{1}{{\lambda}^{2}}-{u}_{r}{}^{2}}\right)\right\}\\ =NUFFT1\left\{{G}_{t}({u}_{t})\cdot \mathrm{exp}\left(2\pi iz\sqrt{\frac{1}{{\lambda}^{2}}-{u}_{r}{}^{2}}\right)\right\}\\ =NUFFT1\left\{NUFFT2\left[{f}_{t}({x}_{t})\right]\cdot \mathrm{exp}\left(2\pi iz\sqrt{\frac{1}{{\lambda}^{2}}-{u}_{r}{}^{2}}\right)\right\}\end{array}$$
(7)
$$F(n)=NUFFT1\left[f({x}_{m})\right]={\displaystyle \sum _{m}f({x}_{m})\mathrm{exp}\left(-in{x}_{m}\right)}$$
(8)
$$F({x}_{n})=NUFFT2\left[f(m)\right]={\displaystyle \sum _{m}f(m)\mathrm{exp}\left(-i{x}_{n}m\right)}$$
(9)
$$\begin{array}{c}{G}_{t}({u}_{n})={\displaystyle \sum {f}_{t}({x}_{m})\cdot \mathrm{exp}\left(-i2\pi mR\Delta T{u}_{n}\right)}\\ ={\displaystyle \sum {f}_{t}({x}_{t})\cdot \mathrm{exp}\left(\frac{-2i\pi Rm{u}_{n}}{{L}_{t}}\right)}\end{array}$$
(10)
$$\begin{array}{c}{f}_{h}(m)={\displaystyle \sum {G}_{h}({u}_{h})\cdot \mathrm{exp}\left(i2\pi m\Delta Hn\Delta {u}_{h}\text{'}\right)}\\ ={\displaystyle \sum {G}_{h}({u}_{h})\cdot \mathrm{exp}\left(\frac{i2\pi m\Delta Hn{L}_{h}\text{'}}{N}\right)}\\ ={\displaystyle \sum {G}_{h}({u}_{h})\cdot \mathrm{exp}\left(\frac{i2\pi m\Delta Hn{L}_{h}}{RN}\right)}\\ ={\displaystyle \sum {G}_{h}({u}_{h})\cdot \mathrm{exp}\left(\frac{i2\pi mn}{RN}\right)}\end{array}$$