HW3Solutions

# HW3Solutions - ECEN 661 - Modulation Theory Homework #3...

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ECEN 661 - Modulation Theory Homework #3 Solutions 1. (a) Since the shifted pulses in each series are non-overlapping, all cross terms will be zero and hence this simplifies to: . Taking the data symbols to be , this simplifies to . Focus attention on the interval , and we get . Note, an identical result is obtained for any other interval of length we look at. Hence, for all . (b) Write , where , . Note that since both and are zero-mean and independent, then . ut () 2 a n gt 2 nT n 2 b n 2 T n 2 + = 2 a n 2 g 2 t 2 n b n 2 g 2 t 2 T n + = a n b n , 1 ± {} 2 g 2 t 2 g 2 t 2 T + [] n = 0 tT < 2 g 2 t g 2 + +s i n 2 π t 2 T ------   sin 2 π + 2 T ------------------- + == sin 2 π t 2 T cos 2 π t 2 T +1 T 2 1 = t xt jy t + = a n 2 n = yt b n 2 T n = φ uu τ 1 2 -- Ex t τ + τ + + = 1 2 Ext τ + 1 2 Eyt τ + + = 1 2 φ xx τ 1 2 φ yy τ + φ τ

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Note in the last step above we take advantage of the fact that is of the same nature as (just delayed in time by ) and hence has the same autocorrelation. From the above result we find
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## HW3Solutions - ECEN 661 - Modulation Theory Homework #3...

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