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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 )

k 0 L 1 R k S k 2 k L k 2 1 2 S k 2 k 2 0 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 k 0 L 1 R k S k k 2 2 k L 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 1 2 l 0 L 1 r l s 1 l 1 r l s 2 l 2 s 1 l 1 s 2 l 2 0 1 2 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. l 0 L 1 s 1 l 1 s 2 l 2 1 2 S 1 S 2 2 1 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 D 1 1 1 r l s 1 D 1 l l D 1 2 2 r l s 2 D 1 l 0 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 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
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
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linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Statistical signal processing. OpenStax CNX. Dec 05, 2011 Download for free at http://cnx.org/content/col11382/1.1
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