2.47 In this problem, we will modify the communication system described in Exerc

Question

2.47 In this problem, we will modify the communication system described in Exercise 2.46 so that the detector at the receiver is allowed to make one of three possible decisions: "" the detector decides the received signal was a 0 "l"the detector decides the received signal was a "he detector is not sure and declares the received signal an erasure (i.e., the receiver chooses not to choose) The operation of the detector is described by the following set of conditional probabilities: Pr(0 received 10transmitted) = 0.90, Pr(0 received! I transmitted) = 0.04, Pil receivedjo transmitted) = 0.01 , Pr(I received! I transmitted) = 0.80, Pr(E received 10 transmitted) = 0.09, Pr(E received!! transmitted) = 0.16 Again, assume that Os and 1s are equally likely to be transmitted. (a) What is the probability that a symbol is erased at the receiver? (b) Given that a received symbol is declared an erasure, what is the probability that a 0 was actually transmitted? What is the probability of error of this receiver? That is, what is the probability that a 0 was transmitted and it is detected as a 1 ora l was transmitted and it is detected as a 0? (c) PROBLEM 2.46 for reference 2.46 A communication system sends binary data [O or 1) which is then detected at the receiver. The receiver occasionally makes mistakes and sometimes a 0 is sent and is detected as a 1 or a 1 can be sent and detected as a 0. Suppose the communication system is described by the following set of conditional probabilities: Pr(0 receivedjo transmitted) = 0.95, Pr(I receivedjo transmitted) = 0.05, Pr(0 received! l transmitted) = 0.10, Pr(1 receivedll transmitted) = 0.90

Explanation / Answer

Since 0 and 1 are equally likely so

P(0 transmitted) = P(1 transmitted) = 0.5

(a)

By the law of total probability, the probability that symbol will erase is

P(E received) = P(E received | 0 transmitted)P(0 transmitted)+ P(E received | 1 transmitted)P(1 transmitted)

=0.09 * 0.5 + 0.16 * 0.5 = 0.045 + 0.08 = 0.125

(b)

P(0 transmitted | E received) = [ P(E received | 0 transmitted)P(0 transmitted)] / P(E received) = [0.09 * 0.5] / 0.125 = 0.36

(c)

P(0 received and 1 transmitted) = P(0 received | 1 transmitted) P(1 transmitted) = 0.04 * 0.5 = 0.02

P(1 received and 0 transmitted) = P(1 received | 0 transmitted) P(0 transmitted) = 0.01 * 0.5 = 0.005

So required probability is

P(0 received and 1 transmitted) + P(1 received and 0 transmitted) =0.02 +0.005 = 0.025

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