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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. 11 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.
 Spring '03
 WOODS

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