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Unformatted text preview: ECE 146B/Spring 2011: Solutions to Problem Set 5 Problem 6.7: First some general observations. We know that samples of WGN through an LTI system with impulse response h ( t ) are Gaussian with zero mean and variance 2 || h || 2 (the statistics of such a sample does not depend on the sampling time, since the input and output random processes are each WSS and Gaussian, and hence stationary). When we send a signal s ( t ) through the same LTI system, the sample at time t is ( s h )( t ), which does depend on the sampling time. Thus, when we send signal plus noise through the system, we want to choose the sampling time to maximize the signal contribution. To make the signal scaling explicit, let us set the maximum amplitude of s ( t ) as A (the figure shows A = 1). (a) Let Z = ( y h )( t ), where h ( t ) = s ( t ). Then we are in the setting of Example 6.1.3: Z N ( m,v 2 ) if 1 sent, and Z N (0 ,v 2 ) if 0 sent, where v 2 = 2 || h || 2 = 2 4 A 2 1 t 2 dt = 4 3 2 A 2 For sampling time t = 1, we have m = ( s h )( t ) = s ( t ) s ( t t ) dt = A 2 1 t (1 t ) dt = A 2 6 Thus, for the ML decision rule (comparing Z with threshold m/ 2) P e = Q ( | m | 2 v ) = Q ( 3 A 24 ) In order to express this in terms of E b /N , note that E b = 1 2 || s || 2 = 2 A 2 3 and N = 2 2 . Plugging this in, we obtain P e == Q ( 1 8 E b N ) (b) We can improve the error probability by sampling at t = 0, which will maximize the signal contribution to the matched filter output. (see Theorem 5.10.2 and the discussion on matched filter in Chapter 5). We then have m = || s || 2 = 4 A 2 3 , while v 2 is as before. This gives, reasoning as in (a), P e = Q ( | m | 2 v ) = Q ( E b N ) which is a well-known formula for the performance of optimal reception of on-off keying in AWGN.which is a well-known formula for the performance of optimal reception of on-off keying in AWGN....
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This note was uploaded on 09/10/2011 for the course ECE 146B taught by Professor Upamanyumadhow during the Spring '11 term at UC Merced.
- Spring '11