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delay delta Jmin equalizer f 0 832 { 0 . 33 , 0 . 027 , 0 . 070 , 0 . 01 } 1 134 { 0 . 66 , 0 . 36 , 0 . 16 , 0 . 08 } 2 30 { - 0 . 28 , 0 . 65 , 0 . 30 , 0 . 14 } 3 45 { 0 . 1 , - 0 . 27 , 0 . 64 , 0 . 3 }

The best equalizer is the one corresponding to a delay of 2, since this Jmin is the smallest. In this case, however, any of the last three open the eye.Observe that the number of errors (as reported in err ) is zero when the eye is open.

Plot the frequency response (using freqz ) of the channel b in LSequalizer.m . Plot the frequency response of each of the fourequalizers found by the program. For each channel/equalizer pair, form the product of the magnitude of thefrequency responses. How close are these products to unity?

Add (uncorrelated, normally distributed) noise into the simulation using the command r=filter(b,1,s)+sd*randn(size(s)) .

  1. For the equalizer with delay 2, what is the largest sd you can add and still have no errors?
  2. Make a plot of Jmin as a function of sd .
  3. Now try the equalizer with delay 1. What is the largest sd you can add, and still have no errors?
  4. Which is a better equalizer?

Use LSequalizer.m to find an equalizer that can open the eye for the channel b= [1 1 -0.8 -.3 1 1].

  1. What equalizer length n is needed?
  2. What delays delta give zero error at the output of the quantizer?
  3. What is the corresponding Jmin ?
  4. Plot the frequency response of this channel.
  5. Plot the frequency response of your equalizer.
  6. Calculate and plot the product of the two.

Modify LSequalizer.m to generate a source sequence from the alphabet ± 1 , ± 3 . For the default channel [0.5 1 -0.6] , find an equalizer that opens the eye.

  1. What equalizer length n is needed?
  2. What delays delta give zero error at the output of the quantizer?
  3. What is the corresponding Jmin ?
  4. Is this a fundamentally easier or more difficult task than when equalizing a binary source?
  5. Plot the frequency response of the channel and of the equalizer.

There is a way to convert the exhaustive search over all the delays δ in the previous approach into a single matrix operation.Construct the ( p - α ) × ( α + 1 ) matrix of training data

S ¯ = s [ α + 1 ] s [ α ] s [ 1 ] s [ α + 2 ] s [ α + 1 ] s [ 2 ] s [ p ] s [ p - 1 ] s [ p - α ] ,

where α specifies the number of delays δ that will be searched, from δ = 0 to δ = α . The ( p - α ) × ( n + 1 ) matrix of received data is

R ¯ = r [ α + 1 ] r [ α ] r [ α - n + 1 ] r [ α + 2 ] r [ α + 1 ] r [ α - n + 2 ] r [ p ] r [ p - 1 ] r [ p - n ] ,

where each column corresponds to one of the possible delays. Note that α > n is required to keep the lowest index of r [ · ] positive. In the ( n + 1 ) × ( α + 1 ) matrix

F ¯ = f 00 f 01 f 0 α f 10 f 11 f 1 α f n 0 f n 1 f n α ,

each column is a set of equalizer parameters, one corresponding to each of the possible delays. The strategy is to use S ¯ and R ¯ to find F ¯ . The column of F ¯ that results in the smallest value of the cost J L S is then the optimal receiver at the optimal delay.

The j th column of F ¯ , corresponds tothe equalizer parameter vector choice for δ = j - 1 . The product of R ¯ with this j th column from F ¯ is intended to approximate the j th column of S ¯ . The least-squares solution of S ¯ R ¯ F ¯ is

F ¯ = ( R ¯ T R ¯ ) - 1 R ¯ T S ¯ ,

where the number of columns of R ¯ (i.e.,  n + 1 ) must be less than or equal to the number of rows of R ¯ (i.e.,  p - α ) for ( R ¯ T R ¯ ) - 1 to exist. Consequently, p - α n + 1 implies that p > n + α . If so, the minimum value associated with a particularcolumn of F ¯ (e.g.,  F ¯ ) is, from [link] ,

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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