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ENC3250 Solutions pg 23

# ENC3250 Solutions pg 23 - s(t sinw ct = d 1(t coswct sinwct...

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-23- s(t) sinw c t = d 1 (t) cosw c t sinw c t + d 2 (t)sin 2 w c t = (1/2)d 1 (t) sin2w c t + (1/2)d 2 (t) - (1/2)d 2 (t) cos2w c t All terms at 2 w c are filtered out by the low-pass filter, yielding: y 1 (t) = (1/2)d 1 (t); y 2 (t) = (1/2)d 2 (t) 6.2 T s = signal element period; T b = bit period; A = amplitude = 0.005 a. T s = T b = 10 -5 sec P = 1 T s s 2 t ( ) 0 T s = A 2 2 E b = P × T b = P × T s = A 2 2 × T s ; N 0 = 2 . 5 × 10 8 × T s E b N 0 = A 2 2 ( ) × T s 2 . 5 × 10 8 × T s = 500 ; E b N 0 ( ) dB = 10 log 500 = 27 dB b. T b = T s 2 ; E b = P × T s 2 ; N 0 = 2 . 5 × 10 8 × T s E b N 0 ( ) = 250 ; E b N 0 ( ) dB = 10 log 250 = 24 dB 6.3 Each signal element conveys two bits. First consider NRZ-L. It should be clear that in this case, D = R/2. For the remaining codes, one must first determine the average number of pulses per bit. For example, for Biphase-M, there is an average
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