## Abstract

We present a method that allows the reconstruction of smooth phase distributions from their laterally sheared representation. The proposed approach is efficient in the sense that only one sheared distribution is needed to completely restore the signal. A mandatory requirement is that the phase distribution is spatially limited. The method is exemplified by means of a synthetic signal, and in addition a practical algorithm is given. Finally,
experimental results are presented. The deformation of a metallic surface is investigated by both speckle shearography and electronic speckle pattern interferometry (ESPI)
respectively. To give proof of the proposed technique, the phase distribution reconstructed from the shearographic measurement is shown to match the results obtained by the ESPI.

© 2007 Optical Society of America

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### Equations (13)

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(1)
$$\Delta g\left(x\right)=g\left(x-\frac{s}{2}\right)-g\left(x+\frac{s}{2}\right)=g\left(x\right)\otimes \left[\delta \left(x-\frac{s}{2}\right)-\delta \left(x+\frac{s}{2}\right)\right]=g\left(x\right)\otimes h\left(x,s\right),$$
(2)
$$F\left\{\Delta g\left(x\right)\right\}\left(v\right)=G\left(v\right)H\left(v,s\right).$$
(3)
$$H(v,s\text{)}=2i\text{\hspace{0.17em} sin}\left(\pi vs\right).$$
(4)
$$g\left(x\right)={F}^{-1}\left\{F\left\{\Delta g\left(x\right)\right\}\left(v\right)T\left(v,s\right)\right\}.$$
(5)
$$\Delta {g}_{n}=\Delta g\left(x\right){\displaystyle \sum _{{n}_{i}\mathrm{,}{n}_{j}=0}^{N-1}\delta \left({x}_{i}-{n}_{i}\Delta x,{y}_{i}-{n}_{j}\Delta x\right)\text{\hspace{1em}}}$$
(6)
$${n}_{i},{n}_{j}=0,1\text{, \hspace{0.17em}\u2026\hspace{0.17em}},N-1.$$
(7)
$$\Delta {g}_{n}={g}_{n-s/2}-{g}_{n+s/2}.$$
(8)
$${F}_{k}\left\{\Delta {g}_{n}\right\}={\displaystyle \sum _{{n}_{i}\text{,}{n}_{j}=0}^{N-1}\Delta {g}_{n}\text{exp}\left(i\text{\hspace{0.17em}}\frac{2\pi}{N}\text{\hspace{0.17em}}nk\right)}.$$
(9)
$$F\left\{\Delta {{g}_{n}}^{\left(0\right)}\right\}\left(v\right)={\displaystyle \sum _{k=-\infty}^{\infty}{F}_{k}\left\{\Delta {g}_{n}\right\}}\text{sinc}\left[\pi \left(\frac{v}{\Delta v}-k\right)\right],$$
(10)
$${F}_{k}\left\{{g}_{n}\right\}=F\left\{\Delta {{g}_{n}}^{\left(0\right)}\right\}\left(v\right)T\left(v,s\right){\displaystyle \sum _{{k}_{i}\text{,}{k}_{j}=0}^{N-1}\delta \left({v}_{i}-{k}_{i}\Delta v,{v}_{j}-{k}_{j}\Delta v\right)}\text{.}$$
(11)
$$\beta {s}_{i}=N\Delta x.$$
(12)
$${F}_{k+0.5}\left\{{g}_{n}\right\}=F\left\{\Delta {{g}_{n}}^{\left(0\right)}\right\}\left(v\right)T\left(v\right)\times {\displaystyle \sum _{{k}_{i}\text{,}{k}_{j}=0}^{N-1}\delta \left({v}_{i}+\frac{\Delta v}{2}-{k}_{i}\Delta v,{v}_{j}-{k}_{j}\Delta v\right)}.$$
(13)
$${g}_{n}\text{exp}\left(i\text{\hspace{0.17em}}\frac{\pi}{N}\text{\hspace{0.17em}}{n}_{i}\right)={F}^{-1}\left\{{F}_{k+0.5}\left\{{g}_{n}\right\}\right\}.$$