0.12 Linear equalization  (Page 11/17)

Use LMSequalizer.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 in the output of the quantizer?
3. How does the answer compare with the design in [link] ?

Modify LMSequalizer.m and EqualizerTest.m to generate a source sequence from the alphabet $\pm 1$ , $\pm 3$ . 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 in the output of the quantizer?
3. Is this a fundamentally easier or more difficult task than when equalizing a binary source?
4. How does the answer compare with the design in [link] ?

The trained adaptive equalizer updates its impulse response coefficients $f_i\left[k\right]$ via [link] based on a gradient descent of the performance function $J_{LMS}$ [link] . With channels that have deep nulls in their frequency responseand a high SNR in the received signal, the minimization of (the average) of $J_{LMS}$ results in a large spike in the equalizer frequency response (in order for their product to be unity at thefrequency of the channel null). Such large spikes in the equalizer frequency responsealso amplify channel noise. To inhibit the resulting large values of $f_i$ needed to create a frequency response with segments of high gain, consider acost function that also penalizes the sum of the squares of the $f_i$ , e.g.

Derive the associated adaptive element update law corresponding to this performance function.

Show that the decision-directed LMS algorithm [link] can be derived as an adaptive element with performance function $(1/2)\text{avg}\left\{{(\mathrm{sign}\left\{y\left[k\right]\right\}-y\left[k\right])}^2\right\}$ . Hint: Suppose that the derivative of the sign function $\frac{d\mathrm{sign}\left\{x\right\}}{dx}$ is zero everywhere.