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# Tutorial5_ans - EE 3008 Bandpass DM Solutions Q1) (a) Q 0...

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EE 3008 Bandpass DM Solutions Q1) (a) () 7 2.3 1.7 52 . 8 7 1 0 0.0144 QQ ⎛⎞ =≈ × ⎜⎟ ⎝⎠ (b) 7 11 . 6 5 2.87 10 × (c) This proportion for men is 1.7 1.7 1.6 1.7 1 1 0.5 0.202 0.298 −− −−≈ = This proportion for women is 1.6 1.6 1.7 1.6 1 = Thus, for all grow-up humans, this proportion is 0.298. Q2) We can use an envelope detector followed by an integral detector for detection. How to determine the error rate for this detection system? Q3) Baseband: NRZ and on-off spectra (1/ τ =10k) BPSK On-off ASK Q4) Refer to lecture notes: 9 4 2 2 2 10 5 2 10 ) 10 ( 2 = = = x A E b τ 0 2 / 5 0.013 b BER Q E N Q == Average power = Eb × (bit rate); new ( 2.5) 0.057 BER Q f 10k f 10k NRZ on-off 0 0

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Q5) (a) If A=+1, the integrator output is B = =+ ≈× ∫∫ cos ( ) . ( cos( )) . 29 0 10 9 0 4 210 4 05 1 4 10 4 05 10 ππ td t t d t The 4 levels are -3B, -B, B, 3B. (b) The energy levels are At d t 22 9 0 21 0 4 ) π = 9B, B, B, 9B. The average energy per symbol is (9B+B+B+9B)/4=5B The average energy per bit is E
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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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Tutorial5_ans - EE 3008 Bandpass DM Solutions Q1) (a) Q 0...

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