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Course by: Don Johnson. E-mail the author

# Unknown Signal Delay

Module by: Don Johnson. E-mail the author

A uniformly most powerful decision rule may not exist when an unknown parameter appears in a nonlinear way in the signal model. Most pertinent to array processing is the unknown time origin case: the signal has been subjected to an unknown delay ( slΔ s l Δ , Δ=? Δ ? ) and we must determine the signal's presence. The likelihood ratio cannot be manipulated so that the sufficient statistic can be computed without having a value for ΔΔ. Thus, the search for a uniformly most powerful test ends in failure and other methods must be sought. As expected, we resort to the generalized likelihood ratio test.

More specifically, consider the binary test where a signal is either present ( 1 1 ) or not ( 0 0 ). The signal waveform is known, but its time origin is not. For all possible values of ΔΔ, the delayed signal is assumed to lie entirely in the observations (Figure 1). This signal model is ubiquitous in active sonar and radar, where the reflected signal's exact time-of-arrival is not known and we want to determine whether a return is present or not and the value of the delay. 1 Additive white Gaussian noise is assumed present. The conditional density of the observations made under 1 1 is p r | 1 Δ r=12πσ2L2e(12σ2l=0L1rlslΔ2) p r 1 Δ r 1 2 σ 2 L 2 1 2 σ 2 l 0 L 1 r l s l Δ 2

The exponent contains the only portion of this conditional density that depends on the unknown quantity ΔΔ. Maximizing the conditional density with respect to ΔΔ is equivalent to maximizing l=0L1rlslΔ12s2lΔ l 0 L 1 r l s l Δ 1 2 s l Δ 2 . As the signal is assumed to be contained entirely in the observations for all possible values of ΔΔ, the second term does not depend on ΔΔ and equals half of the signal energy EE. Rather than analytically maximizing the first term now, we simply write the logarithm of the generalized likelihood ratio test as max Δ Δ l =ΔΔ+D1rlslΔ 0 1 σ2lnη+E2 Δ l Δ Δ D 1 r l s l Δ 0 1 σ 2 η E 2 where the non-zero portion of the summation is expressed explicitly. Using the matched filter interpretation of the sufficient statistic, this decision rule is expressed by max Δ Δ rl*sD1l| l =D1+Δ 0 1 γ l D 1 Δ Δ r l s D 1 l 0 1 γ This formulation suggests that the matched filter having a unit-sample response equal to the zero-origin signal be evaluated for each possible value of ΔΔ and that we use the maximim value of the resulting output in the decision rule. In the known-delay case, the matched-filter output is sampled at the "end" of the signal; here, the filter, which has a duration DD less than the observation interval LL, is allowed to continue processing over the allowed values of signal delay with the maximum output value chosen. The result of this procedure is illustrated here. There two signals, each having the same energy, are passed through the appropriate matched filter. Note that the index at which the maximim output occurs is the maximim likelihood estimate of ΔΔ. Thus, the detection and the estimation problems are solved simultaneously. Furthermore, the amplitude of the signal need not be known as it enters in expression for the sufficient statistic in a linear fashion and an UMP test exists in that case. We can easily find the threshold γγ by establishing a criterion on the false-alarm probability; the resulting simple computation of γγ can be traced to the lack of a signal-related quantity or an unknown parameter appearing in 0 0 .

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

max Δ Δ k=0L1Rk¯Skei2πkΔL σ k 212|Sk|2 σ k 2 0 1 γ Δ k 0 L 1 R k S k 2 k Δ L σ k 2 1 2 S k 2 σ k 2 0 1 γ
(1)
where, as usual in unknown delay problems, the observation interval captures the entire signal waveform no matter what the delay might be. The energy term is a constant and can be incorporated into the threshold. The maximization amounts to finding the best linear phase fit to the observations' spectrum once the signal's phase has been removed. A more interesting interpretation arises by noting that the sufficient statistic is itself a Fourier Transform; the maximization amounts to finding the location of the maximum of a sequence given by k=0L1Rk¯Sk σ k 2ei2πkΔL k 0 L 1 R k S k σ k 2 2 k Δ L The spectral detector thus becomes a succession of two Fourier Transforms with the final result determined by the maximum of a sequence!

Unfortunately, the solution to the unknown-signal-delay problem in either the time or frequency domains is confounded when two or more signals are present. Assume two signals are known to be present in the array output, each of which has an unknown delay: rl= s 1 l Δ 1 + s 2 l Δ 2 +nl r l s 1 l Δ 1 s 2 l Δ 2 n l . Using arguments similar to those used in the one-signal case, the generalized likelihood ratio test becomes max Δ 1 , Δ 2 Δ 1 , Δ 2 l=0L1rl s 1 l Δ 1 +rl s 2 l Δ 2 s 1 l Δ 1 s 2 l Δ 2 0 1 σ2lnη+ E 1 + E 2 2 Δ 1 Δ 2 l 0 L 1 r l s 1 l Δ 1 r l s 2 l Δ 2 s 1 l Δ 1 s 2 l Δ 2 0 1 σ 2 η 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 latter term 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 must be varied in concert to locate the maximum. If, however, the two signals are orthogonal regardless of the delay values, the delays can be found separately and the structure of the single signal detector (modified to include matched filters for each signal) will suffice. This seemingly impossible situation can occur, at least approximately. Using Parseval's Theorem, the cross term can be expressed in the frequency domain. l=0L1 s 1 l Δ 1 s 2 l Δ 2 =12πππ S 1 ω S 2 ω¯eiω( Δ 2 Δ 1 )dω l 0 L 1 s 1 l Δ 1 s 2 l Δ 2 1 2 ω S 1 ω S 2 ω ω Δ 2 Δ 1 For this integral to be zero for all Δ 1 Δ 1 , Δ 2 Δ 2 , the product of the spectra must be zero. Consequently, if the two signals have disjoint spectral support, they are orthogonal no matter what the delays may be.2 Under these conditions, the detector becomes max Δ 1 Δ 1 rl* s 1 D1l| l =D1+ Δ 1 +max Δ 2 Δ 2 rl* s 2 D1l| l =D1+ Δ 2 0 1 γ l D 1 Δ 1 Δ 1 r l s 1 D 1 l l D 1 Δ 2 Δ 2 r l s 2 D 1 l 0 1 γ with the threshold again computed independently of the received signal amplitudes.3 P F =Qλ( E 1 + E 2 )σ2 P F Q λ E 1 E 2 σ 2 This detector has the structure of two parallel, independently operating, matched filters, each of which is tuned to the specific signal of interest.

Reality is insensitive to mathematically simple results. The orthogonality condition on the signals that yielded the relatively 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, the idealistic detector is often used in practice. In the repeated signal case, the matched filter is operated over the entire observation 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 the detection threshold over a contiguous interval. Because of the signal's non-zero duration, the matched filter's response to just the signal has a non-zero duration, implying that the threshold can be crossed at more than a single sample. When one signal 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 a non-maximal value in a larger excursion. Thus, when considering multiple signals, the important quantities are the times at which 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 first of which is due to the signal, the second due to noise. This kind of error cannot be avoided; we never said that detectors could 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 be sorted through, taking into account the nature of the signal being sought. In the example, excursions surrounding a large one should be discarded if they occur in close proximity. This requirement means that closely spaced signals cannot be distinguished from a single one.

## Footnotes

1. For a much more realistic (and harder) version of the active radar/sonar problem, see this problem.
2. We stated earlier that this situation happens "at least approximately." Why the qualification?
3. Not to be boring, but we emphasize that E 1 E 1 and E 2 E 2 are the energies of the signals s 1 l s 1 l and s 2 l s 2 l used in the detector, not those of their received correlates A 1 s 1 l A 1 s 1 l and A 2 s 2 l A 2 s 2 l .

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