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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Problem Set 3: Solutions Mary Amplitude Shift Keying; Repetition Coding 1. [Average energy in Mary amplitude shift keying] The total energy in the M signals is 2(1 2 + 3 2 + ··· + ( M 1) 2 ) E if M is even, and 2(2 2 + 4 2 + ··· + ( M 1) 2 ) E if M is odd. Now, 1 2 + 2 2 + 3 2 + ··· + n 2 = n ( n + 1)(2 n + 1) 6 and so, with M = 2 n +1 being an odd number, and using M 1 = 2 n and ( M +1) / 2 = n +1, we have 2 2 + 4 2 + ··· + ( M 1) 2 = 2 2 + 4 2 + ··· + (2 n ) 2 = 4(1 2 + 2 2 + 3 2 + ··· + n 2 ) = 2 n ( n + 1)(2 n + 1) 3 = ( M 1)( M + 1) M 6 = M ( M 2 1) 6 . On the other hand, if M = 2 n + 2 is even, then using M 1 = 2 n + 1, 1 2 + 3 2 + ··· + ( M 1) 2 = (1 2 + 2 2 + 3 2 + ··· + (2 n + 1) 2 ) (2 2 + 4 2 + ··· + (2 n ) 2 ) = (2 n + 1)(2 n + 2)(4 n + 3) 6 2 n ( n + 1)(2 n + 1) 3 = (2 n + 1)(2 n + 2)(4 n + 3 2 n ) 6 = (2 n + 1)(2 n + 2)(2 n + 3) 6 = ( M 1) M ( M + 1) 6 = M ( M 2 1) 6 . The claimed result follows straightforwardly from this. 2. [Bit error probabilities in 4ASK systems] (a) The receiver has thresholds set at ± 2 √ E and 0; that is, the decision regions are Γ 3 = (∞ , 2 √ E ) , Γ 2 = ( 2 √ E , 0) , Γ 1 = (0 , 2 √ E ) , Γ = (2 √ E , ∞ ). (Remember that the receiver decides that i was trans mitted if the matched filter output Y ∈ Γ i ). Thus, the error probabilities are P ( E  A ) = P ( E  A 3 ) = Q ( p 2 E /N ) (end points); P ( E  A 1 ) = P ( E  A 2 ) = 2 Q ( p 2 E /N ) (interior points). Hence, P ( E ) = ∑ 3 i =0 P ( E  A i ) P ( A i ) = 1 4 (1 + 2 + 2 + 1) Q ( p 2 E /N ) = 1 . 5 Q ( p 2 E /N )....
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This document was uploaded on 01/24/2012.
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