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Expanding the dot product, r s i l 0 L 1 r l s i l another signal processing interpretation emerges. The dot product now describes a finite impulse response (FIR) filteringoperation evaluated at a specific index. To demonstrate this interpretation, let h l be the unit-sample response of a linear, shift-invariant filterwhere h l 0 for l 0 and l L . Letting r l be the filter's input sequence, the convolution sum expresses the output. r k h k l k L 1 k r l h k l Letting k L 1 , the index at which the unit-sample response's last value overlaps the input's value at the origin, we have k L 1 r k h k l 0 L 1 r l h L 1 l If we set the unit-sample response equal to the index-reversed, then delayed signal h l s i L 1 l , we have k L 1 r k s i L 1 k l 0 L 1 r l s i l which equals the observation-dependent component of the optimal detector's sufficient statistic. depicts these computations graphically.

The detector for signals contained in additive, white Gaussian noise consists of a matched filter, whose output is sampled atthe duration of the signal and half of the signal energy is subtracted from it. The optimum detector incorporates amatched filter for each signal compares their outputs to determine the largest.

The sufficient statistic for the i th signal is thus expressed in signal processing notation as k L 1 r k s i L 1 k E i 2 . The filtering term is called a matched filter because the observations are passed through a filter whose unit-sample response "matches" that of the signalbeing sought. We sample the matched filter's output at the precise moment when all of the observations fall within thefilter's memory and then adjust this value by half the signal energy. The adjusted values for the two assumed signals aresubtracted and compared to a threshold.

To compute the performance probabilities, the expressions should be simplified in the ways discussed in the hypothesis testingsections. As the energy terms are known a priori they can be incorporated into the threshold with the result l 0 L 1 r l s 1 l s 0 l 1 0 2 E 1 E 0 2 The left term constitutes the sufficient statistic for the binary detection problem. Because the additive noise is presumed Gaussian,the sufficient statistic is a Gaussian random variable no matter which model is assumed. Under i , the specifics of this probability distribution are l 0 L 1 r l s 1 l s 0 l s i l s 1 l s 0 l 2 s 1 l s 0 l 2 The false-alarm probability is given by P F Q 2 E 1 E 0 2 s 0 l s 1 l s 0 l s 1 l s 0 l 2 1 2 The signal-related terms in the numerator of this expression can be manipulated with the false-alarm probability (and thedetection probability) for the optimal white Gaussian noise detector succinctly expressed by P F Q 1 2 2 s 1 l s 0 l 2 1 s 1 l s 0 l 2 1 2 P F Q 1 2 2 s 1 l s 0 l 2 1 s 1 l s 0 l 2 1 2

Note that the only signal-related quantity affecting this performance probability (and all of the others)is the ratio of energy in the difference signal to the noise variance. The larger this ratio, the better (smaller) the performance probabilities become. Note that thedetails of the signal waveforms do not greatly affect the energy of the difference signal. For example, consider the case wherethe two signal energies are equal ( E 0 E 1 E ); the energy of the difference signal is given by 2 E 2 s 0 l s 1 l . The largest value of this energy occurs when the signals are negatives of each other, with the difference-signalenergy equaling 4 E . Thus, equal-energy but opposite-signed signals such as sine waves, square-waves, Bessel functions,etc. all yield exactly the same performance levels. The essential signal properties that do yield goodperformance values are elucidated by an alternate interpretation. The term s 1 l s 0 l 2 equals s 1 s 0 2 , the L 2 norm of the difference signal. Geometrically, the difference-signal energy is the same quantity as the square ofthe Euclidean distance between the two signals. In these terms, a larger distance between the two signals will mean betterperformance.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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I'm not sure why it wrote it the other way
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Commplementary angles
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
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Kristine 2*2*2=8
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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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. 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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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
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how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
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I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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?
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Source:  OpenStax, Signal and information processing for sonar. OpenStax CNX. Dec 04, 2007 Download for free at http://cnx.org/content/col10422/1.5
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