## Abstract

In traditional three-dimensional (3D) active imaging methods, the detection depth
range is observed to increase linearly with the detection time, and the intensity
information was not fully utilized. However, by encoding the relative values into
pseudovalues, the intensity information was fully utilized, and we found the maximum
detection depth range increases exponentially with the detection time. Furthermore,
we present a 3D imaging system capable of exponentially expanding the detection
depth range. A 3D scene reconstruction was undertaken with the targets placed at a
distance of $600\u20131100\text{\hspace{0.17em}}\mathrm{m}$. Experimental results indicate that the method expands the detection
depth range exponentially without distance resolution loss as compared with the
conventional method.

© 2011 Optical Society of America

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### Equations (22)

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(1)
$$Z={Z}_{0}+\mathrm{\Lambda}{I}_{2}/{I}_{1};$$
(2)
$$Z={Z}_{0}+\mathrm{\Lambda}(1+{I}_{3}/{I}_{2});$$
(3)
$$Z={Z}_{0}+\mathrm{\Lambda}(3-{I}_{2}/{I}_{3})\mathrm{.}$$
(4)
$$Z={Z}_{0}+\mathrm{\Lambda}{I}_{2}/{I}_{3};$$
(5)
$$Z={Z}_{0}+\mathrm{\Lambda};$$
(6)
$$Z={Z}_{0}+\mathrm{\Lambda}(2-{I}_{3}/{I}_{2});$$
(7)
$$Z={Z}_{0}+2\mathrm{\Lambda};$$
(8)
$$Z={Z}_{0}+\mathrm{\Lambda}(2+{I}_{1}/{I}_{2});$$
(9)
$$Z={Z}_{0}+3\mathrm{\Lambda};$$
(10)
$$Z={Z}_{0}+\mathrm{\Lambda}(4-{I}_{2}/{I}_{1});$$
(11)
$$Z={Z}_{0}+4\mathrm{\Lambda};$$
(12)
$$Z={Z}_{0}+\mathrm{\Lambda}(4+{I}_{2}/{I}_{1});$$
(13)
$$Z={Z}_{0}+5\mathrm{\Lambda};$$
(14)
$$Z={Z}_{0}+\mathrm{\Lambda}(6-{I}_{2}/{I}_{1});$$
(15)
$$Z={Z}_{0}+6\mathrm{\Lambda};$$
(16)
$$Z={Z}_{0}+\mathrm{\Lambda}(7-{I}_{1}/{I}_{3})\mathrm{.}$$
(17)
$${a}_{i,N}=[{b}_{1},{b}_{2},\dots {b}_{N}{]}^{T},\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}\text{where \hspace{0.17em} \hspace{0.17em}}i=\sum _{j=1}^{N}{b}_{j}{2}^{N-j},\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\phantom{\rule{0ex}{0ex}}{b}_{j}=0\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}\text{or}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}1,$$
(18)
$${A}_{3}=\left[\begin{array}{l}00011110\\ 01110100\\ 11000111\end{array}\right]=[{a}_{1,3},{a}_{3,3},{a}_{2,3},{a}_{6,3},{a}_{4,3},{a}_{7,3},{a}_{5,3},{a}_{1,3}]\mathrm{.}$$
(19)
$${A}_{N}=[{a}_{1,N},{A}_{N}^{\prime},{a}_{1,N}]\mathrm{.}$$
(20)
$${A}_{3}^{\prime}=[{a}_{3,3},{a}_{2,3},{a}_{6,3},{a}_{4,3},{a}_{7,3},{a}_{5,3}],\phantom{\rule{0ex}{0ex}}{A}_{3}=[{a}_{1,3},{A}_{3}^{\prime},{a}_{1,3}]\mathrm{.}$$
(21)
$${A}_{N+1}^{\prime}=\left[\begin{array}{cccc}{a}_{1,N},& {A}_{N}^{\prime},& {a}_{1,N},& {A}_{N}^{\prime}\\ 1,& {o}_{N},& 0,& {e}_{N}\end{array}\right],$$
(22)
$${A}_{N+1}=[{a}_{1,N+1},{A}_{N+1}^{\prime},{a}_{1,N+1}],$$