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solns_final_s11

# solns_final_s11 - UCSB Spring 2011 ECE 146B Solutions to...

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Unformatted text preview: UCSB Spring 2011 ECE 146B: Solutions to Final Examination 1 2 2 2 s1 s2 s3 s4 s5 s6 Figure 1: ML regions and distances between neighbors for Problem 1 Problem 1: (a) The ML regions for two typical points, s 1 and s 2 , are drawn in Figure 1. (b) Since there are 4 points like s 1 and 4 like s 2 , the average error probabillity equals P e = 4 8 P e | 1 + 4 8 P e | 2 . For the intelligent union bound, note that the ML boundaries for s 1 are determined by neighbors s 2 , s 3 and s 4 , so that P e | 1 ≤ Q parenleftbigg d 12 2 σ parenrightbigg + Q parenleftbigg d 13 2 σ parenrightbigg + Q parenleftbigg d 14 2 σ parenrightbigg The boundaries of the ML region for s 2 are determined by neighbors s 1 , s 5 and s 6 , so that P e | 2 ≤ Q parenleftbigg d 12 2 σ parenrightbigg + Q parenleftbigg d 25 2 σ parenrightbigg + Q parenleftbigg d 26 2 σ parenrightbigg As usual, we express the Q function arguments as d 2 σ = radicalBigg d 2 E b radicalbigg E b 2 N For r = 1, R = 2, we have (see Figure 1) d 12 = 1, d 13 = d 14 = √ 2, and d 25 = d 26 = 2 √ 2. The energy per symbol is given by E s = ( r 2 + R 2 ) / 2 = (1 2 + 2 2 ) / 2 = 5 / 2 The energy per bit is given by E b = E s log 2 8 = 5 / 6 Thus, d 2 12 E b = 6 / 5, d 2 13 E b = 12 / 5, d 2 25 E b = 24 / 5 Plugging in the numbers, we get P e | 1 ≤ Q parenleftBigg radicalbigg 3 E b 5 N parenrightBigg + 2 Q parenleftBigg radicalbigg 6 E b 5 N parenrightBigg P e | 2 ≤ Q parenleftBigg radicalbigg 3 E b 5 N parenrightBigg + 2 Q parenleftBigg radicalbigg 12 E b 5 N parenrightBigg Averaging, we get the following bound on the average error probability P e ≤ Q parenleftBigg radicalbigg 3 E b 5 N parenrightBigg + Q parenleftBigg radicalbigg 6 E b 5 N parenrightBigg + Q parenleftBigg radicalbigg 12 E b 5 N parenrightBigg (c) We would like to set d 13 = d 12 = d min in order to optimize power efficiency. Noting that d 13 = √ 2 r and d 12 = R − r , we get R = ( √ 2 + 1 ) r . Setting r = 1 without loss of generality, we have d min = √ 2 and E s = r 2 + R 2 2 = 1 2 + ( √ 2 + 1) 2 2 = 2 + √ 2 E b = E s log 2 8 = 2 + √ 2 3 so that d 2 min E b = 6 2 + √ 2 Comparing with (a)-(b), for which d min = 1, E b = 5 / 6, and d 2 min E b = 6 / 5, we obtain that the ratio of the power efficiencies is given by 5 / (2 + √ 2), corresponding to a gain of 10 log...
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solns_final_s11 - UCSB Spring 2011 ECE 146B Solutions to...

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