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# 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. The pdf of the Gaussian random variable T is f T ( t ) = 1 9 2 π 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

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= 1 - Φ 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 - Φ parenleftBig 220 - 200 9 parenrightBig 1 - Φ 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) Q (2 . 2) Q (1 . 1) 1 . 39 × 10 - 2 1 . 36 × 10 - 1 0 . 10221 3. X is a uniform random variable between 0 and 1. Now Y = 2 X + 3. Thus it has a range from 2 × 0 + 3 = 3 to 2 × 1 + 3 = 5. Now, F Y ( y ) = P [ { Y y } ] = P [ { 2 X + 3 y } = P [ { X 1 2 ( y - 3) } ] = F X
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