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Unformatted text preview: ECE 562 Fall 2011 October 10, 2011 SOLUTIONS TO HW 4 1. SemiOrthogonal Signal Set. Consider the signal set with M = 2 N signals given by: s m ( t ) = E g m ( t ) m = 1 ,...,N j E g m N ( t ) m = N + 1 ,...,M. where { g k ( t ) } N k =1 are realvalued orthonormal functions. Clearly this signal set satis fies: Re [ k,` ] = 0, for k 6 = ` . (a) Argue that R k = h r ( t ) ,g k ( t ) i ,k = 1 ,...,N , form sufficient statistics for optimal decision making at the receiver for an AWGN channel. (b) Now define the M realvalued statistics y m = r m,I m = 1 ,...,N r ( m N ) ,Q m = N + 1 ,...,M. Show that the MPE decision rule is given by m MPE = arg max m y m (c) Find an expression for P e for the MPE decision rule Ans: (a) The functions { g k ( t ) } form a basis for our complex signal space. By the principle of irrelevance we get a sufficient statistic if we project our received signal onto each g k . Therefore the R k are sufficient. (b) Assume equal priors so the MPE detector is the ML detector. Let W CN (0 ,N o I ). Then for 1 m N : R = E > + W and for N + 1 m 2 N : R = j E > + W The conditional pdf for 1 m N is given by p m ( r ) = 1 ( N o ) N exp k r m k 2 N o = 1 ( N o ) N exp ( r m,I E ) 2 + N o = 1 ( N o ) N exp 2 E N o y m k r k 2 N o E N o c V.V. Veeravalli, 2011 1 Similarly for N + 1 m 2 N : p m ( r ) = 1 ( N o ) N exp 2 E N o y m k r k 2 N o E N o Only the y m term is a function of the message and the exponential is monotonic increasing, so the MPE detector is given by m MPE = arg max m { 1 , 2 ,..., 2 N } y m (c) By symmetry P e = P e , 1 , so we only need to calculate the probability of error given that we sent 1. Given that we sent 1, Y 1 = E + W 1 ,I and Y k = W k,I or Y k = W k M,Q for k > 1. Thus Y k is a set of i.i.d. N (0 ,N / 2) random variables for k 6 = 1, and W 1 ,I is independent of them. Therefore the probability of correct decision making when 1 is sent is the same as in the case of coherent demodulation in the completely orthogonal case (see notes 9). P c , 1 = 1 N Z  " 1 Q E + x p N / 2 !# M 1 e x 2 N dx = 1 2 Z  [1 Q ( t )] M 1 e 1 2 ( t 2 s ) 2 dt where s = E s N = E N . And P e = 1 1 2 Z  [1 Q ( t )] M 1 e 1 2 ( t 2 s ) 2 dt 2. In this problem you will show the following result for Mary orthogonal modulation lim M P c = 1 b > ln 2 b < ln 2 where P c is the probability of correct decision making....
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 Fall '09

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