Solution Found!
Answer: Let X denote the voltage at the output of a
Chapter 4, Problem 18E(choose chapter or problem)
Problem 18E
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<– .5 .
a. What is P( Y = .5)?
b. Obtain the cumulative distribution function ofY and graph it.
Questions & Answers
QUESTION:
Problem 18E
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<– .5 .
a. What is P( Y = .5)?
b. Obtain the cumulative distribution function ofY and graph it.
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) =
P(Y=0.5) =
P(Y=0.5) =
P(Y=0.5) =
P(Y=0.5) =
Therefore, P(Y=0.5) is 0.25.