EE567_Fall2010_HW3_Solutions

EE567_Fall2010_HW3_Solutions - Homework#3 Solutions EE 567...

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Unformatted text preview: Homework #3 Solutions EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 (a) (a) At the output of the receiver, the pdfs will be Gaussian with the means, ‐3Ep, ‐Ep, Ep and 3Ep, respectively, all with variance σ = N/2 Ep. The decision thresholds are at ‐2Eb, 0, and 2Eb. (b) P(e|m‐3) = P(e|m3) = Q( / σ) = Q( / / ) = Q( ) ) P(e|m‐1) = P(e|m1) = 2 × Q( P(e) = / σ) = Q( ) = 2 × Q( P(e) = Where Eb = Ep can be used to find EE 567 Communication Systems Fall 2010 11.2‐8 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 EE 567 Communication Systems Fall 2010 ...
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