HW4Solutions

HW4Solutions - ECEN 661 Modulation Theory Homework#4...

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ECEN 661 - Modulation Theory Homework #4 Solutions 1. (a) where . Hence, . (b) . Since the pulse shape does not have a null at , then we need the termin brackets to be zero at . This will happen if . The resulting spectrum will be . (c). To place nulls at , we would need to satisfy Φ ss f () Gf 2 T ---------------- = T 2 --sinc fT 2 e j π 2 e j 3 π 2 {} = jT –s i n c 2 π 2 sin e j π = Φ f T sinc 2 2 sin 2 π 2 = Φ f 2 T φ bb m [] e j 2 π m = φ m Eb n b nm + Ea n ka n 1 + a + 1 + + == 1 k 2 + φ aa m k φ m 1 + k φ m 1 ++ = 1 k 2 + m =0, km =1 , ± 0o t h e r w i s e . = Φ f 2 T ---------------- 1 k 2 2 k 2 π cos = f 1 T = f 1 T = k 1 = Φ f 2 T ---------------- 2 2 –2 π cos = 4 2 T ---------------- s i n 2 π = 4 T sinc 2 2 sin 2 π 2 sin 2 π = f 14 T =
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. This is not possible for any real . The desired null could be achieved by precoding as follows: . 2. (a). . (b). Writing, , this can be rewritten as: . This alternative form will be helpful for the next part. (c). . 3. (a). If 0 was sent, where and . In that case: . If was sent, then and 1 k 2 2 k π 2 --   cos ++ 1 k 2 +0 == k b n a n a n 4 = fab , () 1 2 πσ 2 ------------ am r 2 bm i 2 + 2 σ 2 -------------------------------------------------- exp 1 2 2 zm 2 2 σ 2 ----------------- exp p u φ , u 2 2 u φ cos m r 2 u φ sin m i 2 + 2 σ 2 ------------------------------------------------------------------------------------ exp = u 2 2 u 2 m 2 + 2 σ 2 ------------------ u
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HW4Solutions - ECEN 661 Modulation Theory Homework#4...

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