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 ( ) G f ( ) 2 T ---------------- = G f ( ) T 2 --sinc fT 2 ( ) e j π fT 2 e j 3 π fT 2 { } = jT sinc fT 2 ( ) π fT 2 ( ) sin e j π fT = Φ ss f ( ) T sinc 2 fT 2 ( ) sin 2 π fT 2 ( ) = Φ ss f ( ) G f ( ) 2 T ---------------- φ bb m [ ] e j 2 π fT m = φ bb m [ ] E b n b n m + [ ] E a n ka n 1 + ( ) a n m + ka n m 1 + + ( ) [ ] = = 1 k 2 + ( aa m [ ] k φ aa m 1 + [ ] k φ aa m 1 [ ] + + = 1 k 2 + m =0, k m = 1, ± 0 otherwise. = Φ ss f ( ) G f ( ) 2 T ---------------- 1 k 2 2 k 2 π fT ( ) cos + + { } = f 1 T = f 1 T = k 1 = Φ ss f ( ) G f ( ) 2 T ---------------- 2 2 2 π fT ( ) cos { } = 4 G f ( ) 2 T ----------------sin 2 π fT ( ) = 4 T sinc 2 fT 2 ( ) sin 2 π fT 2 ( ) sin 2 π fT ( ) =
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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 .
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