solution07 - Solutions to Homework 07 1. a. In this case we...

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Unformatted text preview: Solutions to Homework 07 1. a. In this case we have transmitted a 0 (i.e., v =- 1). The receiver makes an error if the received signal is greater than 0 (i.e., if v + N 0). Then P [error | v =- 1] = P [ Y | v =- 1] = P [ v + N | v =- 1] = P [- 1 + N 0] = P [ N 1] = 1- (1) = Q (1) = 0 . 159 b. Similarly, if a 1 is sent P [error | v = 1] = P [ Y | v = 1] = P [ v + N | v = 1] = P [1 + N 0] = P [ N - 1] = (- 1) = 1- Q (- 1) = Q (1) = 0 . 159 Figure 1 shows the areas of interest. 1-1 y Figure 1: Regions corresponding to error in reception 2. The pdf of the Gaussian random variable T is f T ( t ) = 1 9 2 e- ( t- 200) 2 2 9 2 . a. The probability that a particular transistor will last for more than 200 hours is P [ T > 220] = 1- P [ T 220] 1 = 1- parenleftbigg 220- 200 9 parenrightbigg = 1- parenleftbigg 1- Q parenleftbigg 220- 200 9 parenrightbiggparenrightbigg Q (2 . 2) 1 . 39 10- 2 b. This is the probability that a particular transistor will last more than 200 hours given the condition that the transistor lasts for 210 hours. We need to find the conditional probability. P [ T > 200 | T > 210] = P [( T > 220) AND ( T > 210)] P [ T > 210] = P [ T > 220] P [ T > 210] = 1- P [ T 220] 1- P [ T 210] = 1- parenleftBig 220- 200 9 parenrightBig 1- parenleftBig 210- 200 9 parenrightBig = 1- parenleftBig 1- Q parenleftBig 220- 200 9 parenrightBigparenrightBig 1- parenleftBig 1- Q parenleftBig 210- 200 9 parenrightBigparenrightBig = Q (20 / 9) Q (10 / 9)...
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This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.

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solution07 - Solutions to Homework 07 1. a. In this case we...

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