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Unknown signal parameters

Applying the techniques described in the previous section may be difficult to justify when the signal and/or noise modelsare uncertain. For example, we must "know" a signal down to the precise value of every sample. In other cases, we mayknow the signal's waveform, but not the waveform's amplitude as measured by a sensor. A ubiquitous example of this uncertainty is propagation loss: the range of afar-field signal can only be lower-bounded, which leads to the known waveform, unknown amplitude detection problem. Anotheruncertainty is the signal's time origin : Without this information, we do not know when to start the matchedfiltering operation! In other circumstances, the noise may have a white power spectrum but its variance is unknown. Muchworse situations (from the point of view of detection theory) can be imagined: the signal may not be known at all and onemay want to detect the presence of any disturbance in the observations other than that of well-specified noise. These problems are very realistic, butthe detection schemes as presented are inadequate to attack them. The detection results we have derived to date need tobe extended to incorporate the presence of unknowns just as we did in hypothesis testing ( Detection in the Presence of Unknowns ).

Unknown signal amplitude

Assume that a signal's waveform is known exactly, but the amplitude is not. We need an algorithm to detect thepresence or absence of this signal observed in additive noise at an array's output. The models can be formallystated as 0 : r l n l , l 0 L 1 1 : r l A s l n l , l 0 L 1 , A ? As usual, L observations are available and the noise is Gaussian. This problem isequivalent to an unknown parameter problem described in Detection in the Presence of Unknowns . We learned there that the first step is to ascertain the existence of a uniformly most powerful test.For each value of the unknown parameter A , the logarithm of the likelihood ratio is written A r K s A 2 s K s 0 1 Assuming that A 0 , a typical assumption in array processing problems, we write this comparison as r K s 0 1 1 A A s K s As the sufficient statistic does not depend on the unknown parameter and one of the models ( 0 ) does not depend on this parameter, a uniformly most powerful test exists: the threshold term, despite itsexplicit dependence on a variety of factors, can be determined by specifying a false-alarm probability. If thenoise is not white, the whitening filter or a spectral transformation may be used to simplify the computation ofthe sufficient statistic.

Assume that the waveform, but not the amplitude, of a signal is known. The Gaussian noise is white with a variance of 2 . The decision rule expressed in terms of a sufficient statistic becomes r s 0 1 The false-alarm probability is given by P F Q E 2 where E is the assumed signal energy which equals s 2 . The threshold is thus found to be E 2 Q P F The probability of detection for the matched filter detector is given by P D Q A E E 2 Q Q P F A 2 E 2 where A is the signal's actual amplitude relative to the assumed signal having energy E . Thus, the observed signal, when it is present, has energe A 2 E . The probability of detection is shown in as a function of the observed signal-to-noise ratio. For any false-alarm probability, thesignal must be sufficiently energetic for its presence to be reliably determined.

The false-alarm probability of the detector was fixed at 10 -2 . The signal equaled A 0 l , l 0 L 1 , where 0 was 2 0.1 and L 100 ; the noise was white and Gaussian. The detection probabilities that result from a matched filter detector, asign detector, and a square-law detector are shown. These detectors make progressively fewer assumptions about thesignal, consequently yielding progressively smaller detection probabilities.

All too many interesting problems exist where a uniformly most powerful decision rule cannot be found. Suppose in theproblem just described that the amplitude is known ( A 1 , for example), but the variance of the noise is not. Writing the covariance matrix as 2 K , where we normalize the covariance matrix to have unit variance entries by requiring tr K L , unknown values of 2 express the known correlation structure, unknownnoise power problem. From the results just given, the decision rule can be written so that sufficient statisticdoes not depend on the unknown variance. r K s 0 1 2 s K s However, as both models depend on the unknown parameter, performance probabilities cannot becomputed and we cannot design a detection threshold.

Hypothesis testing ideas show the way out; estimate the unknown parameter(s) under each model separately and thenuse these estimates in the likelihood ratio ( Non-Random Parameters ). Using the maximum likelihood estimates for the parameters results in the generalized likelihoodratio test for the detection problem ( Kelly, et al , Kelly, et al , van Trees ). Letting denote the vector of unknown parameters, be they for the signal or the noise, thegeneralized likelihood ratio test for detection problems is expressed by r p n r s 1 p n r s 0 0 1 Again, the use of separate estimates for each model (rather than for the likelihood ratio as awhole) must be stressed. Unknown signal-related parameter problems and unknown noise-parameter problems have differentcharacteristics; the signal may not be present in one of the observation models. This simplification allows a thresholdto be established objectively. In contrast, the noise is present in each model; establishing a threshold valueobjectively will force new techniques to be developed. We first continue our adventure in unknown-signal-parameterproblems, deferring the more challenging unknown-noise-parameter ones to later .

Questions & Answers

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
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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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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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
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Prasenjit
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Damian
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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.
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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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