# 4.12 Unknown signal delay  (Page 2/2)

We have argued the doubtfulness of assuming that the noise is white in discrete-time detection problems. The approach forsolving the colored noise problem is to use spectral detection. Handling the unknown delay problem in this way is relativelystraightforward. Since a sequence can be represented equivalently by its values or by its DFT, maximization can becalculated in either the time or the frequency domain without affecting the final answer. Thus, the spectral detector'sdecision rule for the unknown delay problem is (from this equation )

$\max\{ , \sum_{k=0}^{L-1} \frac{\Re (\overline{R(k)}S(k)e^{-\left(\frac{i\times 2\pi k}{L}\right)})}{{}_{k}^{2}}-\frac{1}{2}\frac{\left|S(k)\right|^{2}}{{}_{k}^{2}}\}\underset{{}_{0}}{\overset{{}_{1}}{}}$
where, as usual in unknown delay problems, the observationinterval captures the entire signal waveform no matter what the delay might be. The energy term is a constant and can beincorporated into the threshold. The maximization amounts to finding the best linear phase fit to the observations' spectrumonce the signal's phase has been removed. A more interesting interpretation arises by noting that the sufficient statistic isitself a Fourier Transform; the maximization amounts to finding the location of the maximum of a sequence given by $\Re (\sum_{k=0}^{L-1} \frac{\overline{R(k)}S(k)}{{}_{k}^{2}}e^{-\left(\frac{i\times 2\pi k}{L}\right)})$ The spectral detector thus becomes a succession of two Fourier Transforms with the final result determined by themaximum of a sequence!

Unfortunately, the solution to the unknown-signal-delay problem in either the time or frequency domains is confounded when twoor more signals are present. Assume two signals are known to be present in the array output, each of which has an unknown delay: $r(l)={s}_{1}(l-{}_{1})+{s}_{2}(l-{}_{2})+n(l)$ . Using arguments similar to those used in the one-signal case, the generalized likelihood ratio test becomes $\max\{{}_{1}, , {}_{2} , \sum_{l=0}^{L-1} r(l){s}_{1}(l-{}_{1})+r(l){s}_{2}(l-{}_{2})-{s}_{1}(l-{}_{1}){s}_{2}(l-{}_{2})\}\underset{{}_{0}}{\overset{{}_{1}}{}}^{2}\ln +\frac{{E}_{1}+{E}_{2}}{2}$ Not only do matched filter terms for each signal appear, but also a cross-term between the two signals. It is this latterterm that complicates the multiple signal problem: if this term is not zero for all possible delays, a non-separable maximization process results and both delays mustbe varied in concert to locate the maximum. If, however, the two signals are orthogonal regardless of the delay values, thedelays can be found separately and the structure of the single signal detector (modified to include matched filters for eachsignal) will suffice. This seemingly impossible situation can occur, at least approximately. Using Parseval's Theorem, thecross term can be expressed in the frequency domain. $\sum_{l=0}^{L-1} {s}_{1}(l-{}_{1}){s}_{2}(l-{}_{2})=\frac{1}{2\pi }\int_{-\pi }^{\pi } {S}_{1}()\overline{{S}_{2}()}e^{i({}_{2}-{}_{1})}\,d$ For this integral to be zero for all ${}_{1}$ , ${}_{2}$ , the product of the spectra must be zero. Consequently, if the two signals have disjoint spectralsupport, they are orthogonal no matter what the delays may be.

We stated earlier that this situation happens "at least approximately." Why the qualification?
Under these conditions, the detector becomes $(l, , \max\{{}_{1} , (r(l), {s}_{1}(D-1-l))\})+(l, , \max\{{}_{2} , (r(l), {s}_{2}(D-1-l))\})\underset{{}_{0}}{\overset{{}_{1}}{}}$ with the threshold again computed independently of the received signal amplitudes.
Not to be boring, but we emphasize that ${E}_{1}$ and ${E}_{2}$ are the energies of the signals ${s}_{1}(l)$ and ${s}_{2}(l)$ used in the detector, not those of their received correlates ${A}_{1}{s}_{1}(l)$ and ${A}_{2}{s}_{2}(l)$ .
${P}_{F}=Q(\frac{}{\sqrt{({E}_{1}+{E}_{2})^{2}}})$ This detector has the structure of two parallel, independently operating, matched filters, each of which is tuned to thespecific signal of interest.

Reality is insensitive to mathematically simple results. The orthogonality condition on the signals that yielded therelatively simple two-signal, unknown-delay detector is often elusive. The signals often share similar spectral supports,thereby violating the orthogonality condition. In fact, we may be interested in detecting the same signal repeated twice (or more) within the observation interval.Because of the complexity of incorporating inter-signal correlations, which are dependent on the relative delay, theidealistic detector is often used in practice. In the repeated signal case, the matched filter is operated over the entireobservation interval and the number of excursions above the threshold noted. An excursion is defined to be a portion of the matched filter's output that exceeds thedetection threshold over a contiguous interval. Because of the signal's non-zero duration, the matched filter's response tojust the signal has a non-zero duration, implying that the threshold can be crossed at more than a single sample. When onesignal is assumed, the maximization step automatically selects the peak value of an excursion. As shown in lower panels of this figure , a low-amplitude excursion may have a peak value less than anon-maximal value in a larger excursion. Thus, when considering multiple signals, the important quantities are the times atwhich excursion peaks occur, not all of the times the output exceeds the threshold.

This figure illustrates the two kinds of errors prevalent in multiple signal detectors. In the left panel, we find two excursions, the firstof which is due to the signal, the second due to noise. This kind of error cannot be avoided; we never said that detectorscould be perfect! The right panel illustrates a more serious problem: the threshold is crossed by four excursions, all of which are due to a single signal. Hence, excursions must besorted through, taking into account the nature of the signal being sought. In the example, excursions surrounding a largeone should be discarded if they occur in close proximity. This requirement means that closely spaced signals cannot bedistinguished from a single one.

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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20/(×-6^2)
Salomon
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Salomon
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Salomon
I got X =-6
Salomon
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Abhi
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