HW4Solutions

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

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ECEN 661 Modulation Theory Homework #4 Solutions Spring 2008 1. (a) where . Hence, . (b) . Since the pulse shape does not have a null at , then we need the term in brackets to be zero at . This will happen if . The resulting spectrum will be . 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 ----------------22 –2 π cos = 4 2 T ----------------sin 2 π = 4 T sinc 2 2 sin 2 π 2 sin 2 π =
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(c). To place nulls at , we would need to satisfy . This is not possible for any real . The desired null could be achieved by precoding as follows: . 2. Let be the output (low-pass equivalent) of the filter . The instantaneous SNR at the output of the filter is . The denominator of this expression works out to be . Using the Cauchy-Schwartz inequality which states that , then the SNR can be upper bounded by .
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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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HW4Solutions - ECEN 661 Modulation Theory Homework #4...

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