## Abstract

A new two-step phase-shifting fringe projection profilometry is proposed. The slowly variable background intensity of fringe patterns is removed by the use of an intensity differential algorithm. The high-resolution differential algorithm is achieved based on global interpolation of fringe gray level on a subpixel scale. Compared with the traditional three- or four-step phase-shifting method, the profile measurement is sped up with this approach.
Computer simulation and experimental performance are evaluated to demonstrate the validity of the proposed measurement method. The experimental results compared with those of the four-step phase-shifting method are presented.

© 2007 Optical Society of America

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### Equations (14)

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(1)
$$I\left(x,y\right)=R\left(x,y\right)\left\{A\left(x,y\right)+B\left(x,y\right)\mathrm{sin}\left[2\pi fx+\varphi \left(x,y\right)\right]\right\}\text{,}$$
(2)
$$I\left(x,y\right)=a\left(x,y\right)+b\left(x,y\right)\mathrm{sin}\left[2\pi fx+\varphi \left(x,y\right)\right]\text{,}$$
(3)
$$\text{}{I}_{0}\left(x,y\right)=a\left(x,y\right)+b\left(x,y\right)\mathrm{sin}\left[2\pi fx+\varphi \left(x,y\right)\right]\text{,}$$
(4)
$${I}_{1}\left(x,y\right)=a\left(x,y\right)+b\left(x,y\right)\mathrm{sin}\left[2\pi fx+\varphi \left(x,y\right)+\frac{\pi}{2}\right]\text{,}$$
(5)
$${I}_{0}\prime \left(x,y\right)=\frac{\mathrm{d}{I}_{0}\left(x,y\right)}{\mathrm{d}x}=b\left(x,y\right)\left[2\pi f+\frac{\mathrm{d}\varphi \left(x,y\right)}{\mathrm{d}x}\right]\times \mathrm{cos}\left[2\pi fx+\varphi \left(x,y\right)\right]\text{,}$$
(6)
$${I}_{1}\prime \left(x,y\right)=\frac{\mathrm{d}{I}_{1}\left(x,y\right)}{\mathrm{d}x}=-b\left(x,y\right)\left[2\pi f+\frac{\mathrm{d}\varphi \left(x,y\right)}{\mathrm{d}x}\right]\times \mathrm{sin}\left[2\pi fx+\varphi \left(x,y\right)\right]\text{,}$$
(7)
$$\varphi \left(x,y\right)=\mathrm{arctan}\text{\hspace{0.17em}}\frac{-{I}_{1}\prime \left(x,y\right)}{{I}_{0}\prime \left(x,y\right)}-2\pi fx\text{,}$$
(8)
$$\Delta I\left(x,y\right)=I\left(x+1,y\right)-I\left(x,y\right)\text{,}$$
(9)
$$g\left(x,y\right)={\displaystyle \sum _{i=k}^{k+7}I\left({x}_{i},y\right)}{\displaystyle \prod _{\underset{j\ne i}{j\text{=}k}}^{k+7}[\left(x-{x}_{j}\right)/\left({x}_{i}-{x}_{j}\right)]}\text{,}$$
(10)
$$\frac{\mathrm{d}I\left(x,y\right)}{\mathrm{d}x}=\frac{I\left(x+\mathrm{d}x,y\right)-I\left(x,y\right)}{\mathrm{d}x}\text{,}$$
(11)
$${I}_{i}\left(x\right)=R\left(x\right)\left\{A+B\text{\hspace{0.17em}}\mathrm{sin}\left[2\pi fx+\varphi \left(x\right)+\frac{i\pi}{2}\right]\right\}\text{}\left(i=0,1\right)\text{,}$$
(12)
$${g}_{i}\left(x,y\right)=110\left(1-0.4\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{x}{50}\right)\times \left\{0.5+\mathrm{sin}\left[0.1\pi x+\varphi \left(x,y\right)+\frac{i\pi}{2}\right]\right\}+\mathrm{rand}\left(20\right)\text{}\left(i=0,1\right)\text{,}$$
(13)
$$\varphi \left(x,y\right)=\text{\hspace{1em}}\{\begin{array}{c}\phi \left(x,y\right)=1.5\pi -\left[{\left(x-256\right)}^{2}+{\left(y-256\right)}^{2}\right]/3276.8\\ 0\text{\hspace{1em}}\mathrm{if}\text{\hspace{0.17em}}\phi \left(x,y\right)<0\hfill \end{array}\text{,}$$
(14)
$$f\left(x,y\right)=g\left(x,y\right)+\lambda \left[-g\left(x-1,y\right)+2g\left(x,y\right)-g\left(x+1,y\right)-g\left(x,y-1\right)+2g\left(x,y\right)-g\left(x,y+1\right)\right]\text{,}$$