2*.159^2a communication channel accepts an arbitrary voltage input v and outputs a voltage y = v + n, where n is a gaussian random variable with mean 0 and variance 1. suppose that the channel is used to transmit binary information as follows: to transmit 0, the input v = −1, to transmit 1, the input v = 1. the receiver decides a 0 was sent if the voltage is negative and a 1 otherwise. find the probability of the receiver making an error if a 0 was sent; if a 1 is sent.

a) The probability of the receiver making an error if 0 was sent is 15.86%

b) The probability of the receiver making an error if 1 was sent is also 15.86%

Step-by-step explanation:

a) If 0 was sent to the channel, the input v=-1 and the output voltage becomes:

Y= v + N = -1 + N

The receiver makes an error if Y is positive, i.e. Y= -1 + N > 0, i.e N > 1.

Since N is a Gaussian random variable with mean=0 and variance=1, the probability N>1 = P(N>1) = 15.86%

b) If 1 was sent, the input v=1 and the output voltage becomes:

Y= v + N = 1 + N

The receiver makes an error if Y is negative, i.e. Y= 1 + N < 0, i.e N < -1.

Since N is picked randomly from standard normal distribution, the probability N < -1  = P(N<-1) = 15.86%

Below is the percentage probabilities of values N in the standard distribution with respect to their distances from the mean.