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# HW08 - University of Illinois Spring 2011 ECE 361 Problem...

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University of Illinois Spring 2011 ECE 361: Problem Set 8: Solutions Equalizers in Band-Limited Communications 1. [Zero-forcing transversal filter] The zero-forcing equalizer is a transversal filter designed to produce a signal that is as much like a unit pulse as possible when the input is the channel pulse response. For a three-tap equalizer, we need to find c [ - 1], c [0] and c [1] such that c [ - 1] h [0] + c [0] h [ - 1] + c [1] h [ - 2] = +1 . 00 c [ - 1] + 0 . 20 c [0] - 0 . 05 c [1] = 0 c [ - 1] h [1] + c [0] h [0] + c [1] h [ - 1] = +0 . 30 c [ - 1] + 1 . 00 c [0] + 0 . 20 c [1] = 1 c [ - 1] h [2] + c [0] h [1] + c [1] h [0] = - 0 . 07 c [ - 1] + 0 . 30 c [0] + 1 . 00 c [1] = 0 which set of three linear equations in three unknowns can be solved to get the equalizer coefficients c [ - 1] = - 0 . 215 0 . 8692 ≈ - 0 . 247354, c [0] = 0 . 9965 0 . 8692 1 . 146457, and c [1] = - 0 . 314 0 . 8692 ≈ - 0 . 361252. Thus, ˆ Y [ m ] ≈ - 0 . 247354 Y [ m + 1] + 1 . 146457 Y [ m ] - 0 . 361252 Y [ m - 1] or, substituting the X [ m + i ] for Y [ m + 1] , Y [ m ] , Y [ m - 1], - 0 . 247354 × ( - 0 . 05 X [ m + 3] +0 . 2 X [ m + 2] + X [ m + 1]+0 . 3 X [ m ] - 0 . 07 X [ m - 1]) +1 . 146457 × ( - 0 . 05 X [ m + 2] +0 . 2 X [ m + 1] + X [ m ] +0 . 3 X [ m - 1] - 0 . 07 X [ m - 2]) - 0 . 361252 × ( - 0 . 05 X [ m + 1]+0 . 2 X [ m ] + X [ m - 1] +0 . 3 X [ m - 2] - 0 . 07 X [ m - 3]) 0 . 02 X [ m + 3] - 0 . 11 X [ m + 2] + X [ m ] - 0 . 19 X [ m - 2] +0 . 03 X [ m - 3] showing that the contributions of X [ m ± 1] have been eliminated in the ISI affecting ˆ Y [ m ] (though contributions from X [ m ± 3] have been introduced). Note also that eliminating the ISI from X [ m ± 1] has increased the ISI from X [ m ± 2] which previously had coefficients - 0 . 07 and - 0 . 05 but now have coefficients - 0 . 11 and - 0 . 19. However, the maximum ISI is reduced from 0 . 62 E T in Y [ m ] to only 0 . 33 E T in ˆ Y [ m ]. Conversely, X [ m ] will not show up in ˆ Y [ m ± 1] but will show up with larger coefficients in ˆ Y [ m ± 2] than it did in Y [ m ± 2], and X [ m ] will show up in ˆ Y [ m ± 3] whereas it does not affect Y [ m ± 3] at all. 2. [Comparison of equalizers] (a) Y [ m ] = E T + 0 . 5 X [ m - 1] + N [ m ] ∼ N ( E T (1 ± 0 . 5) , σ 2 ). Hence, the maximum conditional error probability is Q (0 . 5 E T ) = Q (1 . 5) = 6 . 681 × 10 - 2 . the minimum conditional error probability is Q (1 . 5 E T ) = Q (4 . 5) = 3 . 398 × 10 - 6 . the average error probability P ( E ) is 1 2 Q (1 . 5) + Q (4 . 5) 1 2 Q (1 . 5) = 3 . 340 × 10 - 2 . In the absence of ISI, the error probability would be Q ( p E T 2 ) = Q (3) = 1 . 350 × 10 - 3 , an order of magnitude smaller than the maximum conditional error probability and the average error probability. Note that the SNR is 9 and SINR = E T 0 . 25 E T + σ 2 = SNR 1 + 0 . 25 SNR = 9 3 . 25 = 2 . 769.

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HW08 - University of Illinois Spring 2011 ECE 361 Problem...

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