## Abstract

I consider the far-field (or focal plane) irradiance distribution of a Gaussian beam that is truncated by a circular aperture in the presence of random angular jitter. First, for absence of jitter, I derive an accurate analytic approximation for the irradiance distribution within the main lobe of the beam for the case in which the beam diameter is less than the aperture diameter. Then I obtain the corresponding mean irradiance distribution in the presence of circularly symmetric normally distributed jitter. By maximizing the on-axis intensity I obtain the optimum ratio of the beam diameter to the aperture diameter in the presence of jitter and present results for the corresponding maximum on-axis intensity and encircled power as a function of the jitter statistics.

© 1995 Optical Society of America

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### Equations (21)

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(1)
$$I(\theta )=I(0)G(\theta ),$$
(2)
$$I(0)={I}_{U}{F}_{\text{trunc}},$$
(3)
$${I}_{U}=\frac{PA}{{(\lambda R)}^{2}},$$
(4)
$${F}_{\text{trunc}}=\frac{2{[1-\text{exp}(-{\mu}^{2})]}^{2}}{{\mu}^{2}};$$
(5)
$$G(\theta )={\left[\frac{2{\int}_{0}^{\mu}\text{d}xx\hspace{0.17em}\text{exp}(-{x}^{2}){J}_{0}(ux/\mu )}{1-\text{exp}(-{\mu}^{2})}\right]}^{2},$$
(6)
$$u=\frac{\theta}{(\lambda /\pi D)},$$
(7)
$$\mu =\frac{D}{d},$$
(8)
$${I}_{A}(\theta )=I(0){G}_{A}(\theta ),$$
(9)
$${G}_{A}(\theta )=\text{exp}\left[-\frac{1}{2}({\theta}^{2}/{{\theta}_{B}}^{2})\right],$$
(10)
$${{\theta}_{B}}^{2}={\left(\frac{\lambda}{\pi d}\right)}^{2}+{\left(\frac{\lambda b}{\pi D}\right)}^{2},$$
(11)
$${P}_{T}=[1-\text{exp}(-2{\mu}^{2})]P,$$
(12)
$$b=\frac{\sqrt{2}\mu}{{[\text{exp}({\mu}^{2})-1]}^{1/2}}.$$
(13)
$${\theta}_{1{e}^{2}}=\frac{2\lambda}{\pi {D}_{\text{eff}}},$$
(14)
$${D}_{\text{eff}}=d{[\text{tanh}({D}^{2}/2{d}^{2})]}^{1/2}.$$
(15)
$$\begin{array}{l}\u3008I(0)\u3009=I(0){F}_{\text{jitter}}\\ ={I}_{U}{F}_{\text{trunc}}{F}_{\text{jitter}},\end{array}$$
(16)
$${F}_{\text{jitter}}=\frac{1}{1+{{\sigma}_{j}}^{2}/{{\theta}_{B}}^{2}}$$
(17)
$${G}_{Aj}(0)=\text{exp}\left[-\frac{1}{2}\left(\frac{{\theta}^{2}}{{{\theta}_{B}}^{2}+{{\sigma}_{j}}^{2}}\right)\right].$$
(18)
$$2{{\mu}_{o}}^{2}-[\text{exp}({{\mu}_{o}}^{2})-1]\left\{1-\frac{2{\rho}^{2}}{{[\text{exp}({{\mu}_{o}}^{2})+1]}^{2}}\right\}=0,$$
(19)
$$\rho ={\sigma}_{j}/(\lambda /\pi D)$$
(20)
$${(d/D)}_{\text{opt}}=\{\begin{array}{ll}0.892-0.0536{\rho}^{2}+0.0089{\rho}^{4}\hfill & \text{for}\hspace{0.17em}0\le \rho \le 1.3\hfill \\ 0.468+\frac{0.592}{\sqrt{\rho}}-\frac{0.085}{\sqrt{\text{ln}\hspace{0.17em}\rho}}\hfill & \text{for}\hspace{0.17em}1.3<\rho \le 10\hfill \end{array}.$$
(21)
$$\begin{array}{l}\frac{\mathrm{\Delta}P({\theta}_{0})}{{P}_{T}}={\int}_{0}^{{\theta}_{0}}{I}_{A}(\theta )2\pi \theta \text{d}\theta ,\\ =1-\text{exp}\left(-\frac{1}{2}\frac{{{\theta}_{0}}^{2}}{{{\theta}_{B}}^{2}}\right).\end{array}$$