# 0.12 Linear equalization  (Page 13/17)

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## Dispersion-minimizing linear equalization

This section considers an alternative performance function that leads to another kind of blind equalizer.Observe that for a binary $±1$ source, the square of the source is known, even when the particular values of the source are not.Thus ${s}^{2}\left[k\right]=1$ for all $k$ . This suggests creating a performance function that penalizes thedeviation from this known squared value $\gamma =1$ . In particular, consider

${J}_{DM}=\frac{1}{4}\text{avg}\left\{{\left(\gamma -{y}^{2}\left[k\right]\right)}^{2}\right\},$

which measures the dispersion of the equalizer output about its desired squared value $\gamma$ .

The associated adaptive element for updating the equalizer coefficients is

${f}_{i}\left[k+1\right]={f}_{i}\left[k\right]-\mu {\left(\frac{\partial {J}_{DM}}{\partial {f}_{i}}|}_{{f}_{i}={f}_{i}\left[k\right]}.$

Mimicking the derivation in [link] through [link] yields the dispersion-minimizing algorithm This is also known as the constant modulus algorithm and as the Godard algorithm. (DMA) for blindly adapting the coefficientsof a linear equalizer. The algorithm is

${f}_{i}\left[k+1\right]={f}_{i}\left[k\right]+\mu \text{avg}\left\{\left(1-{y}^{2}\left[k\right]\right)y\left[k\right]r\left[k-i\right]\right\}.$

Suppressing the averaging operation, this becomes

${f}_{i}\left[k+1\right]={f}_{i}\left[k\right]+\mu \left(1-{y}^{2}\left[k\right]\right)y\left[k\right]r\left[k-i\right],$

which is shown in the block diagram of [link] .

When the source alphabet is $±1$ , then $\gamma =1$ . When the source is multilevel, it is still useful to minimize the dispersion, but theconstant should change to $\gamma =\frac{\text{avg}\left\{{s}^{4}\right\}}{\text{avg}\left\{{s}^{2}\right\}}$ .

While DMA typically may converge to the desired answer from a worse initialization than decision-directed LMS,it is not as robust as trained LMS. For a particular delay $\delta$ , the (average) squared recovery errorsurface being descended (approximately) along the gradient by trained LMS is unimodal (i.e., it has only one minimum).Therefore, no matter where the search is initialized, it finds the desired sole minimum, associated with the $\delta$ used in computing the source recovery error. The dispersion performance function is multimodal withseparate minima corresponding to different achieved delays and polarities.To see this in the simplest case, observe that an answer in which all $+1$ 's are swapped with all $-1$ 's has the same value at the optimal point. Thus, the convergent delay and polarity achieved depend onthe initialization used. A typical initialization for DMA is a single nonzero spikelocated near the center of the equalizer. The multimodal nature of DMA can be observed in theexamples in the next section.

A simple M atlab program that implements the DMA algorithm is given in DMAequalizer.m . The first few lines define the channel, createthe binary source, and pass the input through the channel. The last few lines implement the equalizerand calculate the error between the output of the equalizer and the source as a way of measuring theperformance of the equalizer. These parts of the code are familiar from LSequalizer.m . The new part of the code is in the center,which defines the length n of the equalizer, the stepsize mu of the algorithm, and the initialization of the equalizer (which defaultsto a “center spike” initialization). The coefficients of the equalizer are updatedas in [link] . As with the other equalizers, thecode in EqualizerTest.m can be used to test the operation of the converged equalizer.

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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)
hmm well what is the answer
Abhi
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can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
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
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
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Abhi
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salma
Commplementary angles
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a perfect square v²+2v+_
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. 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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Cied
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Porter
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Cesar
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Stotaw
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Prasenjit
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Azam
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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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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