HW7Solutions

HW7Solutions - ECEN 661 Modulation Theory Homework#7...

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ECEN 661 - Modulation Theory Homework #7 Solutions 1. (a) First, we start with the trellis diagram for The trellis diagram for was presented in class and is repeated below. h 13 = θ n π 3 -- = θ n π 3 = θ n 0 = θ n 2 π 3 ------ = θ n 2 π 3 = θ n π = . . . . . . I 1 1 = pt () I 1 1 = p * t I 2 1 / e j π 3 p * t = I 2 1 / e j π 3 = I 2 1 / e j π 3 p * t = I 2 e j π 3 = I 3 e j 2 π 3 = I 3 e j 2 π 3 p * t = I 3 = I 3 1 = p * t I 3 e j –2 π 3 = I 3 e j π 3 p * t = e j π t 3 T s = h 12 = θ n π 2 = θ n π 2 = θ n 0 = θ n π = . . . . . . e j π t 2 T s = I 1 = I 1 1 / p * t = I 2 1 / jp * t = I 2 j = I 2 jp t = I 2 1 / - * t = I 3 = I 3 1 / p * t = I 3 1 / p * t = I 3 1 / - =

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Finally, here is the trellis diagram for . (b) Consider the two data sequences and where the x’s indicate arbitrary data symbols (as long as they are the same for both sequences. These two sequences cause the following sequences of transmitted waveforms: , . The distance between these two waveforms is . Hence the distance is . The energy per bit is . Hence the asymptotic efficiency is θ n 2 π 3 ------ = θ n 2 π 3 = θ n 0 = I 1 1 / pt () = I 1 1 / p * t = I 2 e j 2 π 3 = I 2 1 / e j 2 π 3 p * t = I 3 e j –2 π 3 = I 3 1 / e j π 3 p * t = e j 2 π t 3 T s = . . . . . . h 23 = I 11 xxx ,,,, , = I '1 –1 , = st I ; e jh π p * t ()… ,, [] = I ' ; p * t e j h π = d 2 II ' , p * t 2 0 T s dt e π p * t e j h π 2 0 T s + = p * t 2 0 T s e π tT s e j h π s 2 0 T s = 22 2 π ht T s -----------   cos 0 T s 2 T s 1s i n c 2 h == e π p * t e j h π 2 0 T s e π e j h π s e j h π e π s 2 0 T s = 2 π T s 2 π h cos t d 0 T s 2 T s i n c 2 h d 2 ' , 4 T s i n c 2 h = E b e j θ 2 0 T s 0 T s T s =
. 2. (a) Following the same procedure as in Problem 1(b), we calculate the distance between the paths that correspond to the sequences and . As before, these two sequences cause the following transmitted waveforms: , , where now . The distance between these two waveforms is . . The remaining integrand is odd symmetric about and hence the integral is zero. Thus . As with the previous problem, and hence . This is the same as MSK. (b) The plots of the PSD of the CPM modulation format given and of MSK are shown in the fig- η d min 2 4 E b ---------- d 2 II ' , () 4 E b ------------------- 1s i n c 2 h [] 0.59 h =1/3 1 h =1/2 1.21 h =2/3 == = I 11 xxx ,,,, , = I '1 –1 , = st I ; pt jp * t ()… ,, = I ' ; p * t j = p t j π 4 ----1 c o s π t T s ----   exp j π 2 ----sin 2 π t 2 T s -------- exp d 2 ' , p * t 2 0 T s dt * t jp t 2 0 T s +2 p * t 2 0 T s 2 j π 2 2 π t 2 T s exp j π 2 –s i n 2 π t 2 T s exp 2 t d 0 T s = 8c o s 2 π 2 --sin 2 π t 2 T s t d 0 T s = 16 π ----- c o s 2 π 2 2 θ θ d 0 π 2 -- = 8 π 1 π 2 π 2 2 θ cos cos + θ d 0 π 2 = 4 π 1 π 2 u cos sin + u d 0 π = 4 4 π π 2 u cos sin u d 0 π + = u π 2 = d 2 ' , 4 = E b 1 = η d 2 ' , 4 E b 1 =

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ure below. The -30dB bandwidths are very similar.
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This note was uploaded on 02/21/2012 for the course ECEN 611 taught by Professor Miller during the Spring '08 term at Texas A&M.

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HW7Solutions - ECEN 661 Modulation Theory Homework#7...

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