## Abstract

The chief emphasis of the paper is given to the calculation of the information capacity of a beam of light for on-off signal modulation. A preliminary and approximate calculation in Sec. 2 is based on Gaussian modulation. On-off modulation permits an exact calculation of the information capacity; the calculation is based on the Poisson statistics of nondegenerate ensembles of Bose entities.

Most of the results are given in terms of the information capacity per transmitted photon, measured in bits per photon, and called the information efficiency *I*_{t}. When there is no ambient light, and when the probability
$\mathcal{p}$ that the gate is open is equal to one-half, the maximum possible information efficiency is just one bit per transmitted photon; when, however, the probability
$\mathcal{p}$ of an open gate is less than one-half, the information efficiency *I*_{t} may be higher, with the upper bound
${\text{log}}_{2}\left(1/\mathcal{p}\right)$.

The calculation of the information efficiency in Sec. 3 supposes that no light reaches the receiver except that from the gated source. In Sec. 4 ambient light also is permitted to reach the receiver; extensive numerical results were obtained with a high-speed digital computer and are presented in tables and graphs.

The term *light* is used throughout the paper instead of radiation, in order to emphasize that the results hold only for nondegenerate beams of radiation. The output of laser oscillators is therefore excluded.

© 1962 Optical Society of America

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### Equations (33)

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(1)
$$\gamma =1/\left({e}^{h\nu /k{T}_{s}}-1\right).$$
(2)
$$\mathrm{\lambda}{T}_{s}\ll 14\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}\mu -deg,$$
(3)
$$N\equiv {\u3008{\left(M-\overline{M}\right)}^{2}\u3009}_{\text{av}}=\overline{M}.$$
(4)
$$S={K}^{2}{\overline{M}}^{2}.$$
(5)
$$C={\text{log}}_{2}{\left(1+S/N\right)}^{\frac{1}{2}},$$
(6)
$$C=\frac{1}{2}{\text{log}}_{2}\left(1+{K}^{2}\overline{M}\right),$$
(7)
$$I=C/\overline{M}=\left(1/2\overline{M}\right)\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}\left(1+{K}^{2}\overline{M}\right),$$
(8)
$$I=\left[\frac{1}{2}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}e\right]{K}^{2}=0.72{K}^{2}$$
(9)
$$N=\overline{M}+A,$$
(10)
$$I=\left(1/2\overline{M}\right)\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}\left[1+{K}^{2}{\overline{M}}^{2}/\left(\overline{M}+A\right)\right].$$
(11)
$$\begin{array}{l}{P}_{0}=q\\ {P}_{1}=\mathcal{p}\\ {{P}_{0}}^{\prime}=1-\mathcal{p}\left(1-{e}^{-M}\right)\\ {{P}_{1}}^{\prime}=\mathcal{p}\left(1-{e}^{-M}\right).\end{array}$$
(12)
$$\begin{array}{l}{\mathcal{p}}_{00}=1\\ {\mathcal{p}}_{01}=0\\ {\mathcal{p}}_{10}={e}^{-M}\\ {\mathcal{p}}_{11}=1-{e}^{-M}.\end{array}$$
(13)
$$C={\displaystyle {\sum}_{i}{P}_{i}}{\displaystyle {\sum}_{j}{\mathcal{p}}_{ij}}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}{\mathcal{p}}_{ij}-{\displaystyle {\sum}_{j}{{P}_{j}}^{\prime}}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}{{P}_{j}}^{\prime}.$$
(14)
$$\begin{array}{l}C=\mathcal{p}\mathcal{M}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}{\mathcal{p}}^{-1}\\ \phantom{\rule{1em}{0ex}}+\left(1-\mathcal{p}\mathcal{M}\right)\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}{\left(1-\mathcal{p}\mathcal{M}\right)}^{-1}-\mu \mathcal{p}M{e}^{-M},\end{array}$$
(15)
$$\mathcal{M}\equiv 1-{e}^{-M}$$
(16)
$$C=\mathcal{p}M\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}\left(1/\mathcal{p}\right).$$
(17)
$${I}_{s}=C/M=\mathcal{p}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}\left(1/\mathcal{p}\right).$$
(18)
$$\text{max}\phantom{\rule{0.2em}{0ex}}\text{re}\phantom{\rule{0.2em}{0ex}}\mathcal{p}\phantom{\rule{0.2em}{0ex}}\text{of}\phantom{\rule{0.2em}{0ex}}{I}_{s}={e}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}e=0.53\phantom{\rule{0.2em}{0ex}}\text{bit}\phantom{\rule{0.2em}{0ex}}\text{per}\phantom{\rule{0.2em}{0ex}}\text{photon}.$$
(19)
$${I}_{s}=0.5\phantom{\rule{0.2em}{0ex}}\text{bit}\phantom{\rule{0.2em}{0ex}}\text{per}\phantom{\rule{0.2em}{0ex}}\text{photon}$$
(20)
$${I}_{t}=C/\mathcal{p}M={\text{log}}_{2}1/\mathcal{p}.$$
(21)
$${I}_{t}=1.0\phantom{\rule{0.2em}{0ex}}\text{bit}\phantom{\rule{0.2em}{0ex}}\text{per}\phantom{\rule{0.2em}{0ex}}\text{photon}.$$
(22)
$$\begin{array}{cc}{I}_{t}\cong {1/\left(1+{M}^{2}/4\right)}^{\frac{1}{2}},& \mathcal{p}=\frac{1}{2},\end{array}$$
(23)
$$\begin{array}{cc}{C\cong \frac{1}{2}M/\left(1+{M}^{2}/4\right)}^{\frac{1}{2}},& \mathcal{p}=\frac{1}{2}.\end{array}$$
(24)
$${I}_{t}\cong {{\text{log}}_{2}{\mathcal{p}}^{-1}/\left(1+{M}^{2}/{{M}_{c}}^{2}\right)}^{\frac{1}{2}}$$
(25)
$$C\cong {\mathcal{p}M\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}{\mathcal{p}}^{-1}/\left(1+{M}^{2}/{{M}_{c}}^{2}\right)}^{\frac{1}{2}},$$
(26)
$$\begin{array}{l}\begin{array}{lll}{M}_{c}=2,& \text{for}& \mathcal{p}=\frac{1}{2}\end{array}\\ \begin{array}{ccc}{M}_{c}=1+\left({\text{log}}_{2}e\right)/\left({\text{log}}_{2}{\mathcal{p}}^{-1}\right),& \text{for}& {\mathcal{p}}^{-1}\gg 1\end{array}\\ \begin{array}{ccc}{M}_{c}=1,& \text{for}& {\text{log}}_{2}{\mathcal{p}}^{-1}\gg 1.\end{array}\end{array}$$
(27)
$$U\left(B,W\right)={e}^{-B}{B}^{W}/W!$$
(28)
$$S\left(B,R\right)={\displaystyle \sum _{W=R}^{\infty}{e}^{-B}{B}^{W}/W!}.$$
(29)
$$\begin{array}{l}{P}_{0}=1-\mathcal{p}\\ {P}_{1}=\mathcal{p}\\ {{P}_{0}}^{\prime}=\mathcal{p}\left\{1-S\left(A+M,R\right)\right\}+\left(1-\mathcal{p}\right)\left\{1-S\left(A,R\right)\right\}\\ {{P}_{1}}^{\prime}=\mathcal{p}S\left(A+M,R\right)+\left(1-\mathcal{p}\right)S\left(A,R\right)\end{array}$$
(30)
$$\begin{array}{l}{\mathcal{p}}_{00}=1-S\left(A,R\right)\\ {\mathcal{p}}_{01}=S\left(A,R\right)\\ {\mathcal{p}}_{10}=1-S\left(A+M,R\right)\\ {\mathcal{p}}_{11}=S\left(A+M,R\right).\end{array}$$
(31)
$$\begin{array}{r}C=F\left[\mathcal{p}T+\left(1-\mathcal{p}\right)Q\right]+F\left[\mathcal{p}\left(1-T\right)+\left(1-\mathcal{p}\right)\phantom{\rule{0.2em}{0ex}}\left(1-Q\right)\right]\\ -\mathcal{p}\left(F\left[T\right]+F\left[1-T\right]\right)-\left(1-\mathcal{p}\right)\phantom{\rule{0.2em}{0ex}}(F\left[Q\right]\\ +F\left[1-Q\right]),\end{array}$$
(32)
$$\begin{array}{c}Q\equiv S\left(A,R\right)\\ T\equiv S\left(A+M,R\right)\\ F\left[X\right]\equiv X\phantom{\rule{0.2em}{0ex}}{\text{log}}_{2}\left(1/X\right).\end{array}$$
(33)
$${I}_{t}=C/\mathcal{p}M.$$