quiz3soln08 - EE 230B Quiz 3, Spring 2008 Solutions Prof....

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EE 230B Quiz 3, Spring 2008 Solutions Prof. G. Pottie For the channel f ( z ) = 1 2 + 1 2 z 1 : 1/ 2 2 X Σ 2 X z -1 a) Show X ( e j ω T ) = 1 + cos T . Why is this channel problematic? Solution: Since this is a minimum phase channel (no precursors), we can easily compute the folded spectrum using the factorization of the equivalent noise-whitened matched filter channel: X ( z ) = f ( z ) f *( z 1 ) = (1/ 2 + z 1 /2 )(1/ 2 + z ) = 1 2 (1 + z + z 1 + 1) = 1 + 1 2 ( z + z 1 ) X ( e j T ) = 1 + 1 2 ( e j T + e j T ) = 1 + cos T This channel has a null at the bandedges, and so equalizers will considerably enhance the noise. b) Compute J min for the ideal MMSE linear equalizer and DFE and compare, using the trapezoid rule for integration with Δ = π /4 T and N 0 =0.05. You may use the following data: ω - π / T -3 π /4 T - π /2 T - π /4 T 0 π / T π /2 T 3 π /4 T π / T X ( e j T ) 0 0.293 1 1.707 2 1.707 1 0.293 0 Solution: For the linear equalizer, J = T 2 No X ( e j T ) + No d T 2 N 0 4 T 1 0.05 + 2 .298 + 2 1.05 + 2 1.757 + 1 2.05 / T / T J = .05 8 1 0.05 + 2 .298 + 2 1.05 + 2 1.757 + 1 2.05 = 0.189 Similarly, for the DFE
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This note was uploaded on 03/29/2011 for the course EECS 100 taught by Professor Dun during the Spring '11 term at CSU Channel Islands.

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quiz3soln08 - EE 230B Quiz 3, Spring 2008 Solutions Prof....

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