## Abstract

The Fourier transform method is applied to analyze the initial phase of linear and equispaced Fizeau fringes. We develop an algorithm for high-precision phase measurement by using the Fourier coefficient that corresponds to the spatial frequency of the Fizeau fringes, and we describe methods for determining the fringe carrier frequency. Errors caused by carrier frequency fluctuation and data truncation are studied theoretically and by computer simulation. To demonstrate the method we apply it to the real-time calibration of a piezoelectric transducer mirror in a Twyman–Green interferometer.

© 1994 Optical Society of America

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### Equations (29)

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(1)
$$I\phantom{\rule{0.1em}{0ex}}(x)=a\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}[1+\gamma (x)cos(2\pi fx+\varphi )\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}},$$
(2)
$$\begin{array}{cc}\hfill \xce\phantom{\rule{0.1em}{0ex}}(\upsilon )& ={\mathit{\int}}_{-\infty}^{\infty}I\phantom{\rule{0.1em}{0ex}}(x)exp(-2\pi i\upsilon x)\text{d}x\hfill \\ & =\xe2\phantom{\rule{0.1em}{0ex}}(\upsilon )+\u0109\phantom{\rule{0.1em}{0ex}}(\upsilon -f)exp(i\varphi )+\u0109\phantom{\rule{0.1em}{0ex}}(\upsilon +f)exp(-i\varphi ),\hfill \end{array}$$
(3)
$$\xe2\phantom{\rule{0.1em}{0ex}}(\upsilon )={\mathit{\int}}_{-\infty}^{\infty}a\phantom{\rule{0.1em}{0ex}}(x)exp(-2\pi i\upsilon x)\text{d}x\phantom{\rule{0.2em}{0ex}},$$
(4)
$$\u0109\phantom{\rule{0.1em}{0ex}}(\upsilon )=\xbd{\mathit{\int}}_{-\infty}^{\infty}a\phantom{\rule{0.1em}{0ex}}(x)\gamma (x)exp(-2\pi i\upsilon x)\text{d}x\phantom{\rule{0.2em}{0ex}},$$
(5)
$$log[\xce\phantom{\rule{0.1em}{0ex}}(f)\phantom{\rule{0.1em}{0ex}}]=log[\u0109\phantom{\rule{0.1em}{0ex}}(0)\phantom{\rule{0.1em}{0ex}}]+i\varphi \phantom{\rule{0.2em}{0ex}},$$
(6)
$$\phi =\varphi +\text{Im}\{log[\u0109\phantom{\rule{0.1em}{0ex}}(0)\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.1em}{0ex}}\}\phantom{\rule{0.2em}{0ex}},$$
(7)
$$I\phantom{\rule{0.1em}{0ex}}(x)=a\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\{1+\gamma (x)cos[2\pi fx+\omega (x)+\varphi ]\phantom{\rule{0.1em}{0ex}}\}\phantom{\rule{0.2em}{0ex}},$$
(8)
$$\u0109\phantom{\rule{0.1em}{0ex}}(\upsilon )=\xbd{\mathit{\int}}_{-\infty}^{\infty}a\phantom{\rule{0.1em}{0ex}}(x)\gamma (x)exp[-i\omega (x)\phantom{\rule{0.1em}{0ex}}]exp(-2\pi i\upsilon x)\text{d}x\phantom{\rule{0.1em}{0ex}}.$$
(9)
$$H\phantom{\rule{0.1em}{0ex}}(n/NT)=\sum _{k=0}^{N-1}h\phantom{\rule{0.1em}{0ex}}(kt)exp(-2\pi ikn/N)\phantom{\rule{0.2em}{0ex}},$$
(10)
$$h\phantom{\rule{0.1em}{0ex}}(k)=\frac{1}{N}\sum _{n=0}^{N-1}H\phantom{\rule{0.1em}{0ex}}(n/NT)exp(2\pi ikn/N)\phantom{\rule{0.2em}{0ex}},$$
(11)
$$\begin{array}{cc}\hfill I\phantom{\rule{0.1em}{0ex}}(k)& =a\phantom{\rule{0.1em}{0ex}}(k)\phantom{\rule{0.1em}{0ex}}[1+\gamma (k)cos(2\pi fk+\varphi )\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}},\hfill \\ \\ & \phantom{\rule{11.5em}{0ex}}(k=0,1,\dots ,N-1)\phantom{\rule{0.2em}{0ex}}.\hfill \end{array}$$
(12)
$$\begin{array}{cc}\hfill \xce\phantom{\rule{0.1em}{0ex}}(n/N)& =\xe2\phantom{\rule{0.1em}{0ex}}(n/N)+\u0109\phantom{\rule{0.1em}{0ex}}(n/N-f)exp(i\varphi )\hfill \\ & \phantom{\rule{1.2em}{0ex}}+\u0109\phantom{\rule{0.1em}{0ex}}(n/N+f)exp(-i\varphi )\phantom{\rule{0.2em}{0ex}},\hfill \\ & \hfill \phantom{\rule{4.5em}{0ex}}(n=0,1,\dots ,N-1)\phantom{\rule{0.2em}{0ex}},\hfill \end{array}$$
(13)
$$\xe2\phantom{\rule{0.1em}{0ex}}(n/N)=\sum _{k=0}^{N-1}a\phantom{\rule{0.1em}{0ex}}(k)exp(-2\pi ikn/N)\phantom{\rule{0.2em}{0ex}},$$
(14)
$$\u0109\phantom{\rule{0.1em}{0ex}}(n/N)=\xbd\sum _{k=0}^{N-1}a\phantom{\rule{0.1em}{0ex}}(k)\gamma (k)exp(-2\pi ikn/N)\phantom{\rule{0.2em}{0ex}}.$$
(15)
$$log[\xce\phantom{\rule{0.1em}{0ex}}(f)\phantom{\rule{0.1em}{0ex}}]=log[\u0109\phantom{\rule{0.1em}{0ex}}(n/N-f)\phantom{\rule{0.1em}{0ex}}]+i\varphi \phantom{\rule{0.2em}{0ex}},$$
(16)
$$f=({n}_{f}+\Delta n)/N\phantom{\rule{0.2em}{0ex}},$$
(17)
$$log[\xce\phantom{\rule{0.1em}{0ex}}(f={n}_{f}/N)\phantom{\rule{0.1em}{0ex}}]=log[\u0109\phantom{\rule{0.1em}{0ex}}(0)\phantom{\rule{0.1em}{0ex}}]+i\varphi \phantom{\rule{0.2em}{0ex}},$$
(18)
$$\varphi =\text{Im}[log\xce\phantom{\rule{0.1em}{0ex}}({n}_{f}/N)\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}}.$$
(19)
$$log[\xce\phantom{\rule{0.1em}{0ex}}(f)\phantom{\rule{0.1em}{0ex}}]=log[\u0109\phantom{\rule{0.1em}{0ex}}(\Delta n/N)\phantom{\rule{0.1em}{0ex}}]+i\varphi \phantom{\rule{0.2em}{0ex}},$$
(20)
$$\u0109\phantom{\rule{0.1em}{0ex}}(\Delta n/N)=\sum _{k=0}^{N-1}a\phantom{\rule{0.1em}{0ex}}(k)\gamma (k)exp(-2\pi ik\Delta n/N)\phantom{\rule{0.2em}{0ex}}.$$
(21)
$$\u0109\phantom{\rule{0.1em}{0ex}}(\Delta n/N)=a\gamma \frac{1-exp(-2\pi i\Delta n)}{1-exp(-2\pi i\Delta n/N)}\phantom{\rule{0.2em}{0ex}}.$$
(22)
$$\begin{array}{cc}\hfill 1-exp(-i2\pi \Delta n)& =1-cos(2\pi \Delta n)+isin(2\pi \Delta n)\hfill \\ & =2\phantom{\rule{0.1em}{0ex}}|\phantom{\rule{0.1em}{0ex}}sin(\pi \Delta n)\phantom{\rule{0.1em}{0ex}}|\phantom{\rule{0.1em}{0ex}}exp[i\phantom{\rule{0.1em}{0ex}}(\pi /2-\pi \Delta n)\phantom{\rule{0.1em}{0ex}}]\hfill \end{array}\phantom{\rule{0.2em}{0ex}}.$$
(23)
$$log[1-exp(-2\pi i\Delta n)\phantom{\rule{0.1em}{0ex}}]=log2\phantom{\rule{0.1em}{0ex}}|\phantom{\rule{0.1em}{0ex}}sin\pi \Delta n\phantom{\rule{0.1em}{0ex}}|\phantom{\rule{0.1em}{0ex}}+i(\pi /2-\pi \Delta n)\phantom{\rule{0.2em}{0ex}}.$$
(24)
$$log[\xce\phantom{\rule{0.1em}{0ex}}(f)\phantom{\rule{0.1em}{0ex}}]=loga\gamma \frac{|sin\pi \Delta n\phantom{\rule{0.1em}{0ex}}|}{|sin\pi \Delta n/N\phantom{\rule{0.1em}{0ex}}|}+i\phantom{\rule{0.1em}{0ex}}[\varphi +(1/N-1)\pi \Delta n]\phantom{\rule{0.2em}{0ex}}.$$
(25)
$$Im\{log[\xce\phantom{\rule{0.1em}{0ex}}(f)\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}}\}=\varphi -(1-1/N)\pi \Delta n\phantom{\rule{0.2em}{0ex}},$$
(26)
$$\begin{array}{cc}\hfill I(k)& =1+cos[\frac{2\pi (n\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\Delta n)}{N}k+\varphi ]\phantom{\rule{0.2em}{0ex}},\hfill \\ & \phantom{\rule{8.0em}{0ex}}(k=0,1,2,\dots ,N-1)\phantom{\rule{0.2em}{0ex}},\hfill \end{array}$$
(27)
$$\xce\phantom{\rule{0.1em}{0ex}}(f)=\sum _{-\infty}^{\infty}\xce\phantom{\rule{0.2em}{0ex}}(n/N)\frac{sin\pi N(f-n/N)}{\pi N(f-n/N)}\phantom{\rule{0.2em}{0ex}}.$$
(28)
$$\begin{array}{c}\xce\phantom{\rule{0.1em}{0ex}}(n/N+\Delta n/N)\hfill \\ \phantom{\rule{1.2em}{0ex}}=\text{\u2211}_{k=0}^{N-1}I\phantom{\rule{0.1em}{0ex}}(k)exp[-2\pi ik(n+\Delta n)/N]\hfill \\ \phantom{\rule{1.2em}{0ex}}=\text{\u2211}_{k=0}^{N-1}[I\phantom{\rule{0.1em}{0ex}}(k)exp(-2\pi ik\Delta n/N)\phantom{\rule{0.1em}{0ex}}]exp(-2\pi ikn/N)\phantom{\rule{0.2em}{0ex}}.\hfill \end{array}$$
(29)
$$\varphi =\frac{2\pi}{\lambda}2\phantom{\rule{0.1em}{0ex}}L\phantom{\rule{0.2em}{0ex}}.$$