## Abstract

We propose a novel scheme for suppressing stimulated Brillouin scattering in optical fibers. The scheme makes use of a single or a sampled Bragg grating fabricated within the fiber used for transmitting intense Q-switched pulses. The grating is designed such that the spectrum of the Stokes pulse generated through stimulated Brillouin scattering falls entirely within its stop band. We show numerically that 15-ns pulses with 2-kW peak power can be transmitted though a 1-m-long fiber with little energy loss using this scheme. A sampled grating can be used for longer fibers but its coupling coefficient should be higher. The proposed scheme should prove useful for double-clad fiber lasers and amplifiers.

©2003 Optical Society of America

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### Equations (10)

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(1)
$$n\left(z\right)=\overline{n}+2{n}_{1}\left(z\right)\mathrm{cos}\left(2{k}_{B}z\right)+{n}_{2}{\mid E\left(z\right)\mid}^{2},$$
(2)
$$E(z,t)=\mathrm{Re}\{{E}_{p}(z,t)\mathrm{exp}\left[i\left({k}_{p}z-{\omega}_{p}t\right)\right]$$
(2)
$$\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}\phantom{\rule{.2em}{0ex}}+{E}_{s}^{-}(z,t)\mathrm{exp}[-i\left({k}_{s}z+{\omega}_{s}t\right)]+{E}_{s}^{+}(z,t)\mathrm{exp}\left[i\left({k}_{s}z-{\omega}_{s}t\right)\right]\}$$
(3)
$$\rho (z,t)={\rho}_{0}+\mathrm{Re}\left\{\rho (z,t)\mathrm{exp}\left[i\left({k}_{A}z-{\Omega}_{B}t\right)\right]\right\}.$$
(4)
$$\frac{\partial {E}_{p}}{\partial z}+\frac{1}{{v}_{g}}\frac{\partial {E}_{p}}{\partial t}=-{g}_{B}^{e}{E}_{s}^{-}Q+i\Gamma \left({\mid {E}_{p}\mid}^{2}+2{\mid {E}_{s}^{+}\mid}^{2}+2{\mid {E}_{s}^{-}\mid}^{2}\right){E}_{p}$$
(5)
$$-\frac{\partial {E}_{s}^{-}}{\partial z}+\frac{1}{{v}_{g}}\frac{\partial {E}_{s}^{-}}{\partial t}={g}_{B}^{e}{E}_{p}{Q}^{*}+i\delta {E}_{s}^{-}+i{\kappa}^{*}{E}_{s}^{+}+i\Gamma \left({\mid {E}_{s}^{-}\mid}^{2}+2{\mid {E}_{p}\mid}^{2}+2{\mid {E}_{s}^{+}\mid}^{2}\right){E}_{s}^{-}$$
(6)
$$\frac{\partial {E}_{s}^{+}}{\partial z}+\frac{1}{{v}_{g}}\frac{\partial {E}_{s}^{+}}{\partial t}=i\delta {E}_{s}^{+}+i\kappa {E}_{s}^{-}+i\Gamma \left({\mid {E}_{s}^{+}\mid}^{2}+2{\mid {E}_{p}\mid}^{2}+2{\mid {E}_{s}^{-}\mid}^{2}\right){E}_{s}^{+}$$
(7)
$${\tau}_{A}\frac{\partial Q}{\partial t}+Q={E}_{p}{E}_{s}^{-*}+{Q}_{0},$$
(8)
$$\kappa \left(z\right)=\frac{\pi {n}_{1}}{\lambda},\phantom{\rule{.9em}{0ex}}\Gamma =\frac{2\pi}{\lambda}{n}_{2},\phantom{\rule{.9em}{0ex}}\delta ={k}_{s}-{k}_{B\dot{-}}$$
(9)
$$\u3008f(z,t)\u3009=0,\u3008f(z,t){f}^{*}(z\text{'},t\text{'})\u3009=F\delta \left(z-z\text{'}\right)\delta \left(t-t\text{'}\right).$$