## Abstract

Free-space laser communication systems use optical-fiber-based technology such as optical amplifiers, receivers, and high-speed modulators. In these systems using single-mode fibers, the fiber coupling efficiency is one of the most significant issues to be solved. Optimum relationships between a focused optical beam and mode field size of the optical fiber in the presence of random angular jitter are discussed in relation to fiber-coupled optical systems. Maximum fiber coupling efficiency is analytically derived with the optimum Airy disk radius normalized by the mode field radius as a function of random angular jitter. The fade level of fiber-coupled signals at desired fade probability is investigated. It is shown that the average bit error ratio significantly degrades with the random angular jitter normalized by the mode field radius larger than about 0.3 when the Airy disk size is optimally selected.

© 2006 Optical Society of America

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### Equations (28)

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(1)
$$U\left(r\right)=\mathrm{exp}\left[jk(f+\frac{{r}^{2}}{2f})\right]\frac{\pi {D}^{2}}{4\lambda f}\left[2\frac{{J}_{1}(kDr\u22152f)}{kDr\u22152f}\right],$$
(2)
$$A\left(r\right)=\frac{\pi {D}^{2}}{4\lambda f}\left[2\frac{{J}_{1}(3.83r\u2215{w}_{1})}{(3.83r\u2215{w}_{1})}\right],$$
(3)
$${w}_{0}\u2215a=0.65+1.619\u2215{V}^{3\u22152}+2.879\u2215{V}^{6},$$
(4)
$$M\left(r\right)=\sqrt{\frac{2}{\pi {w}_{0}^{2}}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{r}^{2}}{{w}_{0}^{2}}).$$
(5)
$$M(r,\rho )=\frac{2\sqrt{2\pi}r}{{w}_{0}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{r}^{2}+{\rho}^{2}}{{w}_{0}^{2}}){I}_{0}\left(\frac{2r\rho}{{w}_{0}^{2}}\right).$$
(6)
$$\eta =\frac{{\mid \int A\left(r\right)M\left(r\right)\mathrm{d}r\mid}^{2}}{\int {\mid A\left(r\right)\mid}^{2}\mathrm{d}r}.$$
(7)
$$\eta \left(\frac{\rho}{{w}_{0}}\right)={\mid \int \mathrm{d}r\frac{2\sqrt{2}}{{w}_{0}}{J}_{1}\left(\frac{3.83r}{{w}_{1}}\right)\times \mathrm{exp}(-\frac{{r}^{2}+{\rho}^{2}}{{w}_{0}^{2}}){I}_{0}\left(\frac{2r\rho}{{w}_{0}^{2}}\right)\mid}^{2}.$$
(8)
$$p\left(r\right)=\frac{r}{{\sigma}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{r}^{2}}{2{\sigma}^{2}}).$$
(9)
$$\u27e8\eta \u27e9=\frac{\u27e8{\mid \int A\left(r\right)M\left(r\right)\mathrm{d}r\mid}^{2}\u27e9}{\u27e8\int {\mid A\left(r\right)\mid}^{2}\mathrm{d}r\u27e9}=\frac{{\mid \int \int \mathrm{d}r\mathrm{d}\rho A\left(r\right)M(r,\rho )p\left(\rho \right)\mid}^{2}}{\int {\mid A\left(r\right)\mid}^{2}\mathrm{d}r}.$$
(10)
$$\u27e8\eta \u27e9={\mid \int \int \mathrm{d}r\mathrm{d}\rho \frac{2\sqrt{2}}{{w}_{0}}{J}_{1}\left(\frac{3.83r}{{w}_{1}}\right)\mathrm{exp}(-\frac{{r}^{2}+{\rho}^{2}}{{w}_{0}^{2}})\times {I}_{0}\left(\frac{2r\rho}{{w}_{0}^{2}}\right)\frac{\rho}{{\sigma}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{\rho}^{2}}{2{\sigma}^{2}})\mid}^{2}.$$
(11)
$$\u27e8\eta \u27e9={\mid \int \int \mathrm{d}r\mathrm{d}\rho \frac{2\sqrt{2}}{{w}_{0}}{J}_{1}\left(\frac{3.83r}{{w}_{1}}\right)\mathrm{exp}(-\frac{{r}^{2}}{{w}_{0}^{2}})\times \left\{\frac{\rho}{{\sigma}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}[-{\rho}^{2}(\frac{1}{{w}_{0}^{2}}+\frac{1}{2{\sigma}^{2}})]{I}_{0}\left(\frac{2r\rho}{{w}_{0}^{2}}\right)\right\}\mid}^{2}.$$
(12)
$$\u27e8\eta \u27e9={\mid \int \mathrm{d}r\frac{2\sqrt{2}}{{w}_{0}}{J}_{1}\left(\frac{3.83r}{{w}_{1}}\right)\mathrm{exp}(-\frac{{r}^{2}}{{w}_{0}^{2}})\times \left\{\frac{{w}_{0}^{2}}{{w}_{0}^{2}+2{\sigma}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}\left[\frac{2{\sigma}^{2}{r}^{2}}{{w}_{0}^{2}({w}_{0}^{2}+2{\sigma}^{2})}\right]\right\}\mid}^{2}.$$
(13)
$$\u27e8\eta \u27e9={\mid \int \mathrm{d}r\frac{2\sqrt{2}{w}_{0}}{{w}_{0}^{2}+2{\sigma}^{2}}{J}_{1}\left(\frac{3.83r}{{w}_{1}}\right)\mathrm{exp}(-\frac{{r}^{2}}{{w}_{0}^{2}+2{\sigma}^{2}})\mid}^{2}.$$
(14)
$$\u27e8\eta \u27e9={\mid \frac{\sqrt{2\pi}}{\sqrt{1+2{(\sigma \u2215{w}_{0})}^{2}}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{3.83}^{2}}{8}\frac{1+2{(\sigma \u2215{w}_{0})}^{2}}{{({w}_{1}\u2215{w}_{0})}^{2}})\times {I}_{1\u22152}\left(\frac{{3.83}^{2}}{8}\frac{1+2{(\sigma \u2215{w}_{0})}^{2}}{{({w}_{1}\u2215{w}_{0})}^{2}}\right)\mid}^{2},$$
(15)
$$\u27e8\eta \u27e9=\frac{2\pi}{1+2{(\sigma \u2215{w}_{0})}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-2\frac{{3.83}^{2}}{8}\frac{1+2{(\sigma \u2215{w}_{0})}^{2}}{{({w}_{1}\u2215{w}_{0})}^{2}})\times {I}_{1\u22152}^{2}\left(\frac{{3.83}^{2}}{8}\frac{1+2{(\sigma \u2215{w}_{0})}^{2}}{{({w}_{1}\u2215{w}_{0})}^{2}}\right).$$
(16)
$$\u27e8\eta \u27e9=\frac{2\pi}{1+2{(\sigma \u2215{w}_{0})}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-2z){I}_{1\u22152}^{2}\left(z\right),$$
(17)
$$z=\frac{{3.83}^{2}}{8}\frac{1+2{(\sigma \u2215{w}_{0})}^{2}}{{({w}_{1}\u2215{w}_{0})}^{2}}.$$
(18)
$$\frac{\mathrm{d}\u27e8\eta \u27e9}{\mathrm{d}{w}_{1}}=\frac{2\pi}{1+2{(\sigma \u2215{w}_{0})}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-2z)(-2{z}^{\prime}){I}_{1\u22152}\left(z\right)[{I}_{1\u22152}\left(z\right)-{I}_{1\u22152}^{\prime}\left(z\right)],$$
(19)
$${I}_{1\u22152}\left(z\right)-{I}_{1\u22152}^{\prime}\left(z\right)=0.$$
(20)
$$2{I}_{1\u22152}\left(z\right)-{I}_{-1\u22152}\left(z\right)-{I}_{3\u22152}\left(z\right)=0.$$
(21)
$$\frac{{w}_{1}}{{w}_{0}}=1.709\sqrt{1+\frac{2{\sigma}^{2}}{{w}_{0}^{2}}}.$$
(22)
$$\text{\hspace{0.17em}}{\phantom{\mid}\u27e8\eta (\frac{\sigma}{{w}_{0}},\frac{{w}_{1}}{{w}_{0}})\u27e9\mid}_{\mathrm{opt}}=\frac{0.8145}{1+2{(\sigma \u2215{w}_{0})}^{2}}.$$
(23)
$${P}_{F}={\int}_{\rho}^{\infty}p\left(r\right)\mathrm{d}r={\int}_{\rho}^{\infty}\frac{r}{{\sigma}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\frac{{r}^{2}}{2{\sigma}^{2}})\mathrm{d}r,$$
(24)
$$\frac{\rho}{{w}_{0}}=\frac{\sigma \sqrt{2(-\mathrm{ln}\phantom{\rule{0.2em}{0ex}}{P}_{F})}}{{w}_{0}}.$$
(25)
$${F}_{T}=10\phantom{\rule{0.2em}{0ex}}\mathrm{log}\left[\frac{\eta (\rho \u2215{w}_{0})}{{\phantom{\mid}\u27e8\eta (\sigma \u2215{w}_{0},{w}_{1}\u2215{w}_{0})\u27e9\mid}_{\mathit{opt}}}\right].$$
(26)
$${F}_{T}=10\phantom{\rule{0.2em}{0ex}}\mathrm{log}\left[\frac{\eta (\sigma \sqrt{2(-\mathrm{ln}\phantom{\rule{0.2em}{0ex}}{P}_{F})}\u2215{w}_{0})}{{\phantom{\mid}\u27e8\eta (\sigma \u2215{w}_{0},{w}_{1}\u2215{w}_{0})\u27e9\mid}_{\mathit{opt}}}\right].$$
(27)
$$\mathrm{BER}\left(Q\right)=\frac{1}{2}\mathrm{erfc}\left(\frac{Q}{\sqrt{2}}\right),$$
(28)
$$\mathrm{BER}=\int \mathrm{d}\rho p\left(\rho \right)\mathrm{BER}\left[Q{\phantom{\mid}\u27e8\eta (\frac{\rho}{{w}_{0}},\frac{{w}_{1}}{{w}_{0}})\u27e9\mid}_{\mathit{opt}}\right]=\int \mathrm{d}\rho p\left(\rho \right)\frac{1}{2}\mathrm{erfc}\left[\frac{Q}{\sqrt{2}}{\phantom{\mid}\u27e8\eta (\frac{\rho}{{w}_{0}},\frac{{w}_{1}}{{w}_{0}})\u27e9\mid}_{\mathit{opt}}\right].$$