Tutorial4_ans - EE 3008. Receiver Design Tutorial Solutions...

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EE 3008. Receiver Design Tutorial Solutions Q1) 2 2 2 1 (1 ) 0.5 0.5/ 2 2 2 11 Pr{1 0} 2 2 x y x y e dx e dy σ π πσ = −− −∞ →= = ∫∫ Pr{0 1} Pr{1 Q2) 2 2 2 1 ) 01 / 2 2 2 -1} 22 x y x y e dx e dy = −∞ −∞ = Pr{-1 1} Q3) Refer to lecture notes: 4 8 (10 ) 10 10 b E == () 0 2 / 5 0.013 b BER Q E N Q Average power = ) ( rate bit E b × The new ( 2.5) 0.057 BER Q =≈ x 0 1 1/2 error region for d=1 correct region for d=1 p.d.f. of x 2 2 2 ) 1 ( 2 2 1 ) ( = x x e x p
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Q4) (a) If A=+1, the output at the integrator output is 4 0.5 10 4 0 B 10 . 5 1 0 dt × =⋅ = × The 4 levels are -3B, -B, B and 3B. (b) The energy levels are 4 2 0 1 Ad t × = 9B, B, B, 9B. The average energy per symbol is (9B+B+B+9B)/4=5B The average energy per bit is E b =5B/2=2.5B (c) First suppose the transmitter only transmit the two smaller symbols (+1 and –1). The situation of A= –1 is shown
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This note was uploaded on 01/11/2011 for the course EE 3008 taught by Professor Pingli during the Fall '08 term at City University of Hong Kong.

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Tutorial4_ans - EE 3008. Receiver Design Tutorial Solutions...

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