# hw4_soln - ECE 562 Fall 2011 SOLUTIONS TO HW 4 1...

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Unformatted text preview: ECE 562 Fall 2011 October 10, 2011 SOLUTIONS TO HW 4 1. “Semi-Orthogonal” 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 real-valued orthonormal functions. Clearly this signal set satis- fies: Re [ ρ 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 real-valued 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 ˆ 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 ( πN o ) N exp- k r- m k 2 N o = 1 ( πN o ) N exp- ( r m,I- √ E ) 2 + ··· N o = 1 ( πN 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 ( πN 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 ˆ 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 √ πN Z ∞-∞ " 1- Q √ E + x p N / 2 !# M- 1 e- x 2 N dx = 1 √ 2 π Z ∞-∞ [1- Q ( t )] M- 1 e- 1 2 ( t- √ 2 γ s ) 2 dt where γ s = E s N = E N . And P e = 1- 1 √ 2 π Z ∞-∞ [1- Q ( t )] M- 1 e- 1 2 ( t- √ 2 γ s ) 2 dt 2. In this problem you will show the following result for M-ary orthogonal modulation lim M →∞ P c = 1 γ b > ln 2 γ b < ln 2 where P c is the probability of correct decision making....
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hw4_soln - ECE 562 Fall 2011 SOLUTIONS TO HW 4 1...

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