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This graph contains two waveforms. One wave is identified by a thin solid line and the other is identified by a dotted line. The x-axis is labeled ω/π and the y-axis is labeled |H(ω)|. Both wave forms follow the same general path. They start high on the y-axis and proceed to the right. The waves drop drastically to the x-axis and bounce along the x-axis until the end of the graph. The upper portion of the waves are bound vertically by two sets of horizontal line. Solid line has a greated amplitude and contained within these bounds are two dotted lines which mark the bounds of the dotted line. The dotted line has a smaller amplitude. After the drastic descent, the upper bound of the solid line wave is marked by a solid line. The dotted line wave's upper bound is marked by a horizontal dotted line. The amplitude of the solid line is greater than the dotted line.
Effects of transition bands in l filters.

These two results together illustrate the importance of the transition bandwidth for a CLS design. Clearly one can decrease maximum error tolerances by widening the transition band. Yet finding the perfect balance between a transition bandwidth and a given tolerance can prove a difficult task, as will be shown in [link] . Hence the relevance of a CLS method that is not restricted by two types of specifications competing against each other. In principle, one should just determine how much error one can live with, and allow an algorithm to find the optimal transition band that meets such tolerance.

Two problem solutions

[link] introduced some important remarks regarding the behavior of extrema points and transition bands in l 2 and l filters. As one increases the constraints on an l 2 filter, the result is a filter whose frequency response looks more and more like an l filter.

[link] introduced the frequency-varying problem and an IRLS-based method to solve it. It was also mentioned that, while the method does not solve the intended problem (but a similar one), it could prove to be useful for the CLS problem. As it turns out, in CLS design one is merely interested in solving an unweighted, constrained least squares problem. In this work, we achieve this by solving a sequence of weighted, unconstrained least squares problems, where the sole role of the weights is to "constraint" the maximum error of the frequency response at each iteration. In other words, one would like to find weights w such that

min h D ( ω ) - H ( ω ; h ) 2 subject to D ( ω ) - H ( ω ; h ) τ ω [ 0 , ω p b ] [ ω s b , π ]

is equivalent to

min h w ( ω ) · ( D ( ω ) - H ( ω ; h ) ) 2

Hence one can revisit the frequency-varying design method and use it to solve the CLS problem. Assuming that one can reasonably approximate l by using high values of p , at each iteration the main idea is to use an l p weighting function only at frequencies where the constraints are exceeded. A formal formulation of this statement is

w ϵ ( ω ) = | ϵ ( ω ) | p - 2 2 if | ϵ ( ω ) | > τ 1 otherwise

Assuming a suitable weighting function existed such that the specified tolerances are related to the frequency response constraints, the IRLS method would iterate and assign rather large weights to frequencies exceeding the constraints, while inactive frequencies get a weight of one. As the method iterates, frequencies with large errors move the response closer to the desired tolerance. Ideally, all the active constraint frequencies would eventually meet the constraints. Therefore the task becomes to find a suitable weighting function that penalizes large errors in order to have all the frequencies satisfying the constraints; once this condition is met, we have reached the desired solution.

This image contains three graphs. The first graph represents Polynomial weighting (p=100). The second graph represents Linear-log view (p=100), and the third graph represents Linear-log view (p=500). The x-axis is labeled ε for all of these graphs, and the y-axis is labeled w(ε).
CLS polynomial weighting function.

One proposed way to find adequate weights to meet constraints is given by a polynomial weighting function of the form

w ( ω ) = 1 + ϵ ( ω ) τ p - 2 2

where τ effectively serves as a threshold to determine whether a weight is dominated by either unity or the familiar l p weighting term. [link] illustrates the behavior of such a curve.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
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
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Abhi
is it a question of log
Abhi
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salma
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Tamia
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Uday
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salma
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a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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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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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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what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
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
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
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
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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
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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.
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, Iterative design of l_p digital filters. OpenStax CNX. Dec 07, 2011 Download for free at http://cnx.org/content/col11383/1.1
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