## Abstract

By the use of analytic continuation, the correct spectrum of an undersampled analog
input signal *f*_{a}(*t*) of a true bandwidth *B* is recovered
from an aliased Fourier spectrum that is computed directly from a data set consisting
of sinusoid-crossing locations {*t*_{i}}, where the signal *f*_{a}(*t*) intersects with a reference sinusoid
*r*(*t*) with a frequency of *W* <
*B*/2 and an amplitude of *A*. If *A*
≥ |*f*_{a}(*t*)| within the sampling period *T*, then a
crossing exists within each time interval Δ = 1/2*W*, and a total of
2*WT* = 2*M* sinusoid crossings are detected, where
*M* is a positive integer. The cut-off frequency for sampling is
*W* = ±*M/T*. In a crossing detector, a trade-off
exists between the size of Δ and the accuracy with which a crossing can be located
within it because the detector has a finite response time. Low-accuracy detection of
the crossing positions degrades the detection limit of the detector and results in a
computed Fourier spectrum that contains spurious wideband frequencies. We show
however that, if *f*_{a}(*t*) has a known compact support within *T*,
then sampling at a frequency of *W* < *B*/2 may
still be possible because the correct *f*_{a}(*t*) spectrum can be recovered from the aliased spectrum by
means of analytic continuation. The technique is demonstrated for an interferogram
test signal in both the absence and presence of additive Gaussian noise.

© 1996 Optical Society of America

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### Equations (9)

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(1)
$$\begin{array}{ll}f(t)\hfill & =\text{\u2211}_{m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-M}^{M}c\phantom{\rule{0.1em}{0ex}}(m)exp(\phantom{\rule{0.1em}{0ex}}j\phantom{\rule{0.1em}{0ex}}m2\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}{f}_{0}t)\phantom{\rule{0.2em}{0ex}}=\text{\u2211}_{m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-M}^{M}c\phantom{\rule{0.1em}{0ex}}(m)\phantom{\rule{0.1em}{0ex}}{Z}^{m}\hfill \\ \hfill & ={Z}^{-M}[c\phantom{\rule{0.1em}{0ex}}(-M\phantom{\rule{0.1em}{0ex}})+c\phantom{\rule{0.1em}{0ex}}(-M+1)\phantom{\rule{0.1em}{0ex}}Z+\dots +c\phantom{\rule{0.1em}{0ex}}(M)\phantom{\rule{0.1em}{0ex}}{Z}^{2M}]\hfill \\ \hfill & ={Z}^{-M}\prod _{i=1}^{2M}(Z-{Z}_{i}\phantom{\rule{0.1em}{0ex}})={Z}^{-M}\phantom{\rule{0.1em}{0ex}}p\phantom{\rule{0.1em}{0ex}}(Z)\hfill \\ \hfill & ={Z}^{-M}\phantom{\rule{0.1em}{0ex}}[a\phantom{\rule{0.1em}{0ex}}(0)+a\phantom{\rule{0.1em}{0ex}}(1)\phantom{\rule{0.1em}{0ex}}Z+a\phantom{\rule{0.1em}{0ex}}(2)\phantom{\rule{0.1em}{0ex}}{Z}^{2}+\dots +a\phantom{\rule{0.1em}{0ex}}(2M)\phantom{\rule{0.1em}{0ex}}{Z}^{2M}]\phantom{\rule{0.1em}{0ex}},\hfill \end{array}$$
(2)
$$a\phantom{\rule{0.1em}{0ex}}(M+m)=-\frac{1}{(M+m)}\sum _{n\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1}^{M+m}{s}_{n}\phantom{\rule{0.1em}{0ex}}a\phantom{\rule{0.1em}{0ex}}(M+m-n)\phantom{\rule{0.1em}{0ex}},$$
(3)
$$\text{NMSE}=\frac{\text{\u2211}_{k\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}0}^{M-1}{|\phantom{\rule{0.2em}{0ex}}{f}_{a}\phantom{\rule{0.1em}{0ex}}(k)-f\phantom{\rule{0.1em}{0ex}}(k)\phantom{\rule{0.2em}{0ex}}|}^{2}}{\text{\u2211}_{k\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}0}^{M-1}{|\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.1em}{0ex}}(k)\phantom{\rule{0.2em}{0ex}}|}^{2}}\phantom{\rule{0.1em}{0ex}},$$
(4)
$$f(k\phantom{\rule{0em}{0ex}}\Delta \phantom{\rule{0em}{0ex}})=\frac{A\phantom{\rule{0.1em}{0ex}}(0)}{2}+\sum _{m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.1em}{0ex}}1}^{2M\phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1}\phantom{\rule{0.1em}{0ex}}\left[A\phantom{\rule{0.1em}{0ex}}(m)cos\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.3em}{0ex}}\frac{k}{M}\right)+B\phantom{\rule{0.1em}{0ex}}(m)sin\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\frac{k}{M}\right)\right]\phantom{\rule{0.1em}{0ex}},$$
(5)
$${f}_{u}(k\phantom{\rule{0em}{0ex}}\Delta )=\frac{A\phantom{\rule{0.1em}{0ex}}(0)}{2}+\sum _{m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.1em}{0ex}}1}^{\beta \phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1}\phantom{\rule{0.1em}{0ex}}\left[A\phantom{\rule{0.1em}{0ex}}(m)\phantom{\rule{0em}{0ex}}cos\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\frac{k}{M}\right)+B\phantom{\rule{0.1em}{0ex}}(m)\phantom{\rule{0em}{0ex}}sin\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\frac{k}{M}\right)\right]\phantom{\rule{0.1em}{0ex}}.$$
(6)
$${f}_{e}\phantom{\rule{0.1em}{0ex}}(k\phantom{\rule{0em}{0ex}}\Delta )={f}_{u}\phantom{\rule{0.1em}{0ex}}(k\phantom{\rule{0em}{0ex}}\Delta )+\sum _{m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\beta}^{2M\phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1}\phantom{\rule{0.1em}{0ex}}\left[A\phantom{\rule{0.1em}{0ex}}(m)\phantom{\rule{0em}{0ex}}cos\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\frac{k}{M}\right)+B\phantom{\rule{0.1em}{0ex}}(m)\phantom{\rule{0em}{0ex}}sin\phantom{\rule{0em}{0ex}}\left(\phantom{\rule{0em}{0ex}}\pi \phantom{\rule{0em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\frac{k}{M}\right)\right]\phantom{\rule{0.1em}{0ex}}.$$
(7)
$$\varphi =\sum _{k<{k}_{1}\phantom{\rule{0.1em}{0ex}},\phantom{\rule{0.1em}{0ex}}k>{k}_{2}}\phantom{\rule{0.1em}{0ex}}{\left|{f}_{e}\phantom{\rule{0.1em}{0ex}}(k\phantom{\rule{0em}{0ex}}\Delta \phantom{\rule{0em}{0ex}})\phantom{\rule{0.1em}{0ex}}\right|}^{2}\phantom{\rule{0.1em}{0ex}},$$
(8)
$$\sum _{k<{k}_{1}\phantom{\rule{0.1em}{0ex}},\phantom{\rule{0.1em}{0ex}}k>{k}_{2}}\phantom{\rule{0.1em}{0ex}}\left|{f}_{e}\phantom{\rule{0.1em}{0ex}}(k\phantom{\rule{0em}{0ex}}\Delta )\phantom{\rule{0.1em}{0ex}}\right|cos\phantom{\rule{0em}{0ex}}\left(\frac{\pi \phantom{\rule{0em}{0ex}}mk}{M}\right)=0,m=\beta \phantom{\rule{0.1em}{0ex}},\beta +1\phantom{\rule{0.1em}{0ex}},\dots \phantom{\rule{0.1em}{0ex}},2M-1\phantom{\rule{0.1em}{0ex}},$$
(9)
$$\sum _{k<{k}_{1}\phantom{\rule{0.1em}{0ex}},\phantom{\rule{0.1em}{0ex}}k>{k}_{2}}\phantom{\rule{0.1em}{0ex}}\left|{f}_{e}\phantom{\rule{0.1em}{0ex}}(k\phantom{\rule{0em}{0ex}}\Delta )\phantom{\rule{0.1em}{0ex}}\right|sin\phantom{\rule{0em}{0ex}}\left(\frac{\pi \phantom{\rule{0em}{0ex}}mk}{M}\right)=0,m=\beta ,\beta +1\phantom{\rule{0.1em}{0ex}},\dots ,2M-1\phantom{\rule{0.1em}{0ex}},$$