solution07

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

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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 0 | v = - 1] = P [ v + N 0 | v = - 1] = P [ - 1 + N 0] = P [ N 1] = 1 - Φ(1) = Q (1) = 0 . 159 b. Similarly, if a 1 is sent P [error | v = 1] = P [ Y 0 | v = 1] = P [ v + N 0 | v = 1] = P [1 + N 0] = P [ N ≤ - 1] = Φ( - 1) = 1 - Q ( - 1) = Q (1) = 0 . 159 Figure 1 shows the areas of interest. 0 1 -1 y Figure 1: Regions corresponding to error in reception 2. Let T be the random variable denoting the lifetime of the transistor. Since T is gaussian, its pdf is given by f T ( t ) = 1 9 2 π e 1 2 ( t - 200) 2 9 2 (1) a. The probability that a particular transistor will last for more than 220 hours is P [ T > 220] = Q ± 220 - 200 9 ² 1

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Q (2 . 2) = 1 . 39 × 10 - 2 b. Here we want the probability that a particular transistor will last more than 220 hours given that the transistor lasts for 210 hours. This conditional probability is given by P [ T > 220 | T > 210] = P [ { T > 220 } ∩ { T > 210 } ] P [ { T > 210 } ] = P [ { T > 220 } ] P [ { T > 210 } ] = Q ± 220 - 200 9 ² Q ± 210 - 200 9 ² Q (2 . 2) Q (1 . 1) = 1 . 39 × 10 - 2 1 . 36 × 10 - 1 = 0 . 1022 3. X is a uniform random variable between 0 and 1. Now
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## This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.

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

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