## Abstract

In a recent published work we proposed a technique to recover the absolute phase maps of two fringe patterns with different spatial frequencies. It is demonstrated that a number of selected frequency pairs can be used for the proposed approach, but the published work did not provide a guideline for frequency selection. In addition, the performance of the proposed technique in terms of its anti-noise capability is not addressed. In this paper, the rules for selecting the two frequencies are presented based on theoretical analysis of the proposed technique. Also, when the two frequencies are given, the anti-noise capability of technique is formulated and evaluated. These theoretical conclusions are verified by experimental results.

© 2012 OSA

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### Equations (31)

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(1)
$$\{\begin{array}{c}{\Phi}_{1}(x)=2\pi {m}_{1}(x)+{\varphi}_{1}(x)\\ {\Phi}_{2}(x)=2\pi {m}_{2}(x)+{\varphi}_{2}(x)\end{array}$$
(2)
$${m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}=\Psi (x)$$
(3)
$${\Phi}_{1}(x)={f}_{1}{\Phi}_{0}(x),{\Phi}_{2}(x)={f}_{2}{\Phi}_{0}(x)$$
(4)
$${m}_{1}(x)=\{\begin{array}{cc}\lfloor {f}_{1}/2\rfloor & [{f}_{1}-({f}_{1}\mathrm{mod}2+1)]\pi /{f}_{1}\le {\Phi}_{0}(x)<\pi \\ \mathrm{...}& \mathrm{...}\\ 1& \pi /{f}_{1}\le {\Phi}_{0}(x)<3\pi /{f}_{1}\\ 0& -\pi /{f}_{1}<{\Phi}_{0}(x)<\pi /{f}_{1}\\ -1& -3\pi /{f}_{1}<{\Phi}_{0}(x)\le -\pi /{f}_{1}\\ \mathrm{...}& \mathrm{...}\\ -\lfloor {f}_{1}/2\rfloor & -\pi {<}_{1}{\Phi}_{0}(x)\le -[{f}_{1}-({f}_{1}\mathrm{mod}2+1)]\pi /{f}_{1}\end{array}$$
(5)
$${m}_{2}(x)=\{\begin{array}{cc}\lfloor {f}_{2}/2\rfloor & [{f}_{2}-({f}_{2}\mathrm{mod}2+1)]\pi /{f}_{2}\le {\Phi}_{0}(x)<\pi \\ \mathrm{...}& \mathrm{...}\\ 1& \pi /{f}_{2}\le {\Phi}_{0}(x)<3\pi /{f}_{2}\\ 0& -\pi /{f}_{2}<{\Phi}_{0}(x)<\pi /{f}_{2}\\ -1& -3\pi /{f}_{2}<{\Phi}_{0}(x)\le -\pi /{f}_{2}\\ \mathrm{...}& \mathrm{...}\\ -\lfloor {f}_{2}/2\rfloor & -\pi <{\Phi}_{0}(x)\le -[{f}_{2}-({f}_{2}\mathrm{mod}2+1)]\pi /{f}_{2}\end{array}$$
(6)
$$\underset{{m}_{1}(x),{m}_{2}(x)}{Min}\left\{\left|{m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}-\Phi (x)\right|\right\}$$
(7)
$$N={N}_{1}+{N}_{2}-1=2\lfloor {f}_{1}/2\rfloor +2\lfloor {f}_{2}/2\rfloor +1$$
(8)
$$\frac{{f}_{2}{\varphi}_{1}({x}_{a})-{f}_{1}{\varphi}_{2}({x}_{a})}{2\pi}\ne \frac{{f}_{2}{\varphi}_{1}({x}_{b})-{f}_{1}{\varphi}_{2}({x}_{b})}{2\pi}or{m}_{2}({x}_{a}){f}_{1}-{m}_{1}({x}_{a}){f}_{2}\ne {m}_{2}({x}_{b}){f}_{1}-{m}_{1}({x}_{b}){f}_{2}$$
(9)
$${m}_{2}({x}_{a}){f}_{1}-{m}_{1}({x}_{a}){f}_{2}={m}_{2}({x}_{b}){f}_{1}-{m}_{1}({x}_{b}){f}_{2}$$
(10)
$$\frac{{m}_{1}({x}_{a})-{m}_{1}({x}_{b})}{{m}_{2}({x}_{a})-{m}_{2}({x}_{b})}=\frac{{f}_{1}}{{f}_{2}}$$
(11)
$${m}_{1}({x}_{a})-{m}_{1}({x}_{b})=k{f}_{1}and{m}_{2}({x}_{a})-{m}_{2}({x}_{b})=k{f}_{2}$$
(12)
$$-2\lfloor \frac{{f}_{1}}{2}\rfloor \le {m}_{1}({x}_{a})-{m}_{1}({x}_{b})\le 2\lfloor \frac{{f}_{1}}{2}\rfloor and-2\lfloor \frac{{f}_{2}}{2}\rfloor \le {m}_{2}({x}_{a})-{m}_{2}({x}_{b})\le 2\lfloor \frac{{f}_{2}}{2}\rfloor $$
(13)
$${m}_{1}({x}_{a})-{m}_{1}({x}_{b})=\pm {f}_{1},{m}_{2}({x}_{a})-{m}_{2}({x}_{b})=\pm {f}_{2}$$
(14)
$$\lfloor \frac{{f}_{1}}{2}\rfloor =\frac{{f}_{1}}{2}and\lfloor \frac{{f}_{2}}{2}\rfloor =\frac{{f}_{2}}{2}$$
(15)
$${m}_{2}({x}_{a}){f}_{1}-{m}_{1}({x}_{a}){f}_{2}\ne {m}_{2}({x}_{b}){f}_{1}-{m}_{1}({x}_{b}){f}_{2}$$
(16)
$${m}_{2\mathrm{max}}(x)=\lceil \frac{{\Phi}_{0\mathrm{max}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}\rceil \le \frac{{\Phi}_{0\mathrm{max}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}+1$$
(17)
$${m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}\le {m}_{2\mathrm{max}}(x){f}_{1}-{m}_{1}(x){f}_{2}$$
(18)
$${m}_{2\mathrm{max}}(x){f}_{1}-{m}_{1}(x){f}_{2}\le (\frac{{\Phi}_{0\mathrm{max}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}){f}_{1}-{m}_{1}(x){f}_{2}=\frac{{f}_{1}+{f}_{2}}{2}$$
(19)
$${m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}\le \frac{{f}_{1}+{f}_{2}}{2}$$
(20)
$${m}_{2\mathrm{min}}(x)=\lceil \frac{{\Phi}_{0\mathrm{min}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}\rceil \ge \frac{{\Phi}_{0\mathrm{min}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}$$
(21)
$${m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}\ge {f}_{1}{m}_{2\mathrm{min}}(x)-{f}_{2}{m}_{1}(x)$$
(22)
$${f}_{1}{m}_{2\mathrm{min}}(x)-{f}_{2}{m}_{1}(x)\ge (\frac{{\Phi}_{0\mathrm{min}}(x)-\pi /{f}_{2}}{2\pi /{f}_{2}}){f}_{1}-{m}_{1}(x){f}_{2}=-\frac{{f}_{1}+{f}_{2}}{2}$$
(23)
$${m}_{2}(x){f}_{1}-{m}_{1}(x){f}_{2}\ge -\frac{{f}_{1}+{f}_{2}}{2}$$
(24)
$$-\frac{{f}_{1}+{f}_{2}}{2}\le {m}_{2}(x){f}_{1}-{f}_{2}{m}_{1}(x)\le \frac{{f}_{1}+{f}_{2}}{2}$$
(25)
$$-\frac{{f}_{1}+{f}_{2}}{2}\le \frac{{f}_{2}{\varphi}_{1}(x)-{f}_{1}{\varphi}_{2}(x)}{2\pi}\le \frac{{f}_{1}+{f}_{2}}{2}$$
(26)
$$N=2\lfloor {f}_{1}/2\rfloor +2\lfloor {f}_{2}/2\rfloor +1$$
(27)
$$N-1=2\lfloor {f}_{1}/2\rfloor +2\lfloor {f}_{2}/2\rfloor $$
(28)
$$G=\frac{{f}_{1}+{f}_{2}}{2\lfloor {f}_{1}/2\rfloor +2\lfloor {f}_{2}/2\rfloor}<2$$
(29)
$$\left|\frac{{f}_{2}\Delta {\varphi}_{1}(x)-{f}_{1}\Delta {\varphi}_{2}(x)}{2\pi}\right|<0.5$$
(30)
$$0\le \Delta {\varphi}_{\mathrm{max}}<\frac{\pi}{{f}_{1}+{f}_{2}}$$
(31)
$${f}_{1}+{f}_{2}<\frac{\pi}{\Delta {\varphi}_{\mathrm{max}}}$$