Let X denote the voltage at the output of a microphone,

Chapter 5, Problem 18E

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QUESTION:

Let ?X ?denote the voltage at the output of a microphone, and suppose that ?X ?has a uniform distribution on the interval from to –1. The voltage is processed by a “hard limiter” with cutoff values and – .5, so the limiter output is a random variableY? ?related to ?X ?by Y = X if |X| ? .5 Y = .5 if X> .5, andY = – .5 if X

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QUESTION:

Let ?X ?denote the voltage at the output of a microphone, and suppose that ?X ?has a uniform distribution on the interval from to –1. The voltage is processed by a “hard limiter” with cutoff values and – .5, so the limiter output is a random variableY? ?related to ?X ?by Y = X if |X| ? .5 Y = .5 if X> .5, andY = – .5 if X

ANSWER:

Answer : Step 1 : Given, Let X denote the voltage at the output of a microphone, and suppose that X has a uniform distribution on the interval from to –1. The voltage is processed by a “hard limiter” with cutoff values and – .5, From the given information X is uniform distribution on interval -1 to +1. a). Now we have to find P( Y =0.5) P(Y=0.5) = f (x) dx X 0.5 P(Y=0.5) = 1 dx 2 0.5 P(Y=0.5) =1 4 P(Y=0.5) = 0.25

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