Exam1Soln - University of Illinois Spring 2011 ECE 361:...

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University of Illinois Spring 2011 ECE 361: Solutions to the First MidSemester Exam Tuesday February 22, 2011, 9:30 a.m. – 10:50 a.m. 1. Consider a binary digital communication system operating over an AWGN channel using signals s 0 ( t ) = ( A, 0 t < T, 0 , otherwise, and s 1 ( t ) = ( 2 A cos( πt/T ) , 0 t < T, 0 , otherwise. The receiver consists of a linear time-invariant filter whose output is sampled at T 0 = T and is compared to the maximum-likelihood threshold. (a) Suppose that the receiver filter has impulse response h ( t ) = s 0 ( t ). What is the error probability P e achieved by this receiver? Solution: Note that h ( t ) = s 0 ( t ) = s 0 ( T - t ) so that the filter is matched to s 0 ( t ). We have ˆ s 0 ( T ) = s 0 ? h ± ± t = T = Z -∞ s 0 ( t ) h ( T - t ) dt = Z T 0 A 2 dt = A 2 T = E 0 , ˆ s 1 ( T ) = s 1 ? h ± ± t = T = Z -∞ s 1 ( t ) h ( T - t ) dt = Z T 0 2 A 2 cos( πt/T ) dt = 0 = h s 0 ,s 1 i , σ 2 = N 0 2 Z -∞ h 2 ( t ) dt = N 0 2 Z T 0 A 2 dt = A 2 TN 0 2 , and so P e = Q ² ˆ s 0 ( T ) - ˆ s 1 ( T ) 2 σ ³ = Q A 2 T 2 p A 2 0 / 2 ! = Q A r T 2 N 0 ! (b) Find the error probability P * e achieved by the optimum (matched-filter) receiver for these signals and show that it is smaller than the P e you found in part (a). Solution: Since h s 0 1 i = 0 and s 1 ( t ) is a half cycle of a sinusoid of period 2 T and rms amplitude A , we have orthogonal signals of energy E 0 = E 1 = A 2 T , and thus P * e = Q s E 0 + E 1 - h s 0 1 i 2 N 0 = Q r E 0 + E 1 2 N 0 ! = Q A r T N 0 ! < Q A r T 2 N 0 ! = P e since the argument of Q ( · ) is larger by a factor of 2. Note that it is not necessary to find or specify the matched filter impulse, or the sampling time, or the signal outputs, or the noise variance, etc. in order to find P * e which is solely dependent on the signal energies E 0 and E 1 , the inner product h s 0 1 i , and the given value of N 0 .
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Exam1Soln - University of Illinois Spring 2011 ECE 361:...

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