solns6 - ECE 146B/Spring 2011: Solutions to Problem Set 6...

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Unformatted text preview: ECE 146B/Spring 2011: Solutions to Problem Set 6 Problem 6.17 (a) Signal space representations with respect to the given orthonormal basis are: Signal Set A: s 1 = (1 , , , 0) T , s 2 = (0 , 1 , , 0) T , s 3 = (0 , , , 1) T and s 4 = (0 , , , 1) T Signal Set B: s 1 = (1 , , , 1) T , s 2 = (0 , 1 , 1 , 0) T , s 3 = (1 , , 1 , 0) T and s 4 = (0 , 1 , , 1) T (b) For Signal Set A, the pairwise distance between any two points satisfies d 2 = d 2 min = 2, while the energy per symbol is E s = 1. Thus, E b = E s / (log 2 4) = 1 / 2, and d 2 min /E b = 4. The union bound on symbol error probability is therefore given by P e (signal set A) ≤ 3 Q parenleftbigg radicalBig d 2 min /E b radicalbig E b / 2 N parenrightbigg = 3 Q parenleftBig radicalbig 2 E b /N parenrightBig For signal set B, each signal has one neighbor at distance given by d 2 1 = 4 and two at distance given by d 2 2 = d 2 min = 2. The energy per symbol is E s = 2, so that E b = 1. The union bound is given by P e (signal set B) ≤ 2 Q parenleftBig radicalbig d 2 2 /E b radicalbig E b / 2 N parenrightBig + Q parenleftBig radicalbig d 2 1 /E b radicalbig E b / 2 N parenrightBig = 2 Q parenleftBig radicalbig E b /N parenrightBig + Q parenleftBig radicalbig 2 E b /N parenrightBig (c) For exact analysis of error probability for Signal Set B, suppose that the received signal in signal space is given by Y = ( Y 1 , Y 2 , Y 3 , Y 4 ). Condition on the signal s = (1 , , , 1) T being sent. Then Y 1 = 1 + N 1 , Y 2 = N 2 , Y 3 = N 3 , Y 4 = 1 + N 4 where N 1 , ..., N 4 are i.i.d. N (0 , σ 2 ) random variables. A correct decision is made if ( Y , s ) > ( Y , s k ) , k = 1 , 2 , 3. These inequalities can be written out as ( Y , s 1 ) = Y 1 + Y 4 > ( Y , s 2 ) = Y 2 + Y 3 ( Y , s 1 ) = Y 1 + Y 4 > ( Y , s 3 ) = Y 1 + Y 3 ( Y , s 1 ) = Y 1 + Y 4 > ( Y , s 4 ) = Y 2 + Y 4 The second and third inequalities give Y 1 > Y 2 and Y 4 > Y 3 , and imply the first inequality. Thus, the conditional probability of correct reception is given by P c | 1 = P [ Y 1 > Y 2 and Y 4 > Y 3 | s sent] = P [ Y 1 > Y 2 | s sent] P [ Y 4 > Y 3 | s sent] since Y k are conditionally independent given the transmitted signal. Noting that Y 1 − Y 2 and Y 4 − Y 3 are independent N (1 , 2 σ 2 ), we have P e | 1 = 1 − Pc | 1 = 1 − parenleftbigg 1 − Q parenleftbigg 1 √ 2 σ 2 parenrightbiggparenrightbigg 2 = 2 Q parenleftbigg 1 √ 2 σ 2 parenrightbigg...
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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.

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solns6 - ECE 146B/Spring 2011: Solutions to Problem Set 6...

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