## Abstract

We performed experiments to verify the theory of photorefractive pulse coupling
in the frequency domain. In particular, we confirm that the phase added to the
diffracted pulse depends quadratically on the relative delay between the two
input pulses.

© 1993 Optical Society of America

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### Equations (5)

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(1)
$${E}_{\text{shaped}}\left(t\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{E}_{\text{signal}}\left(t\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}G\left(\tau \right)\left[\text{exp}\left(\eta L\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}1\right]{E}_{\text{pump}}\left(t\right),$$
(2)
$$\begin{array}{cc}G\left(\tau \right)\phantom{\rule{0.2em}{0ex}}& =\phantom{\rule{0.2em}{0ex}}{\displaystyle {\int}_{-\infty}^{\infty}\overline{{E}_{\text{signal}}\left(t\right){E}_{\text{pump}}^{\ast}\left(t\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\tau \right)}}\mathrm{d}t/{I}_{0}\\ & \equiv \phantom{\rule{0.2em}{0ex}}\left|G\left(\tau \right)\right|\text{exp}\left[i{\varphi}_{\text{diff}}\left(\tau \right)\right].\end{array}$$
(3)
$$E\left(t\right)=\frac{1}{2\pi}{\displaystyle {\int}_{-\infty}^{\infty}F\left(\omega \right)\text{exp}\left(-i\omega t\right)\mathrm{d}\omega},$$
(4)
$$\begin{array}{l}\overline{\frac{{\left|{F}_{\text{shaped}}\left(\omega \right)\right|}^{2}}{{\left|{F}_{\text{signal}}\left(\omega \right)\right|}^{2}}}\\ \phantom{\rule{0.8em}{0ex}}\phantom{\rule{0.4em}{0ex}}=\phantom{\rule{0.2em}{0ex}}a\left(\tau \right)\left\{1\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}b\left(\tau \right)\text{cos}\left[{\varphi}_{\text{diff}}\left(\tau \right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\tau \omega \phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\beta {\omega}^{2}\right]\right\},\end{array}$$
(5)
$${\varphi}_{\text{diff}}\left(\tau \right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{tan}}^{-1}\frac{\beta}{\alpha}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{\beta}{4\left({\alpha}^{2}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}{\beta}^{2}\right)}{\tau}^{2}.$$