# 0.10 Constrained least squares (cls) problem  (Page 3/5)

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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 ${l}_{\infty }$ 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 ${l}_{\infty }$ 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

$\begin{array}{cc}\underset{h}{\text{min}}\hfill & {\parallel D\left(\omega \right)-H\left(\omega ;h\right)\parallel }_{2}\hfill \\ \text{subject}\phantom{\rule{4.pt}{0ex}}\text{to}\hfill & {\parallel D\left(\omega \right)-H\left(\omega ;h\right)\parallel }_{\infty }\phantom{\rule{-0.166667em}{0ex}}\le \phantom{\rule{-0.166667em}{0ex}}\tau \phantom{\rule{1.em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}\omega \in \left[0,{\omega }_{pb}\right]\cup \left[{\omega }_{sb},\pi \right]\hfill \end{array}$

is equivalent to

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\parallel w\left(\omega \right)·\left(D\left(\omega \right)-H\left(\omega ;h\right)\right)\parallel }_{2}$

Hence one can revisit the frequency-varying design method and use it to solve the CLS problem. Assuming that one can reasonably approximate ${l}_{\infty }$ 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

$w\left(ϵ,\left(,\omega ,\right)\right)=\left\{\begin{array}{cc}{|ϵ\left(\omega \right)|}^{\frac{p-2}{2}}\hfill & \text{if}\phantom{\rule{4.pt}{0ex}}|ϵ\left(\omega \right)|\phantom{\rule{-0.166667em}{0ex}}>\phantom{\rule{-0.166667em}{0ex}}\tau \hfill \\ 1\hfill & \text{otherwise}\hfill \end{array}\right)$

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.

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

$w\left(\omega \right)=1+{\left|\frac{ϵ\left(\omega \right)}{\tau }\right|}^{\frac{p-2}{2}}$

where $\tau$ 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.

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I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
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20/(×-6^2)
Salomon
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Salomon
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Salomon
I got X =-6
Salomon
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oops. ignore that.
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Abhi
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Abhi
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Abhi
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salma
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