[Random phases in a communication system] A particular communication system tran

Question

[Random phases in a communication system] A particular communication system transmits a perfectly sinusoidal carrier wave, X = sin(T), where T is measured in microseconds. A particular receiver samples the signal at a random time T, and measures just one output voltage, X = sin(T). T has the following pmf: pT(u) = Find E[T] and sigma 2T. Find E[X], E[X2], and sigma 2X without finding the pmf of X. Suppose that a different receiver samples the signal at time S, where ps(v) = where N is a constant positive integer. Let the measured signal sample be Y = sin (S). Find E[Y], E[Y2], and sigma 2Y. Be sure to describe the outcome in the case when N is even, as well as the case when N is odd.

Explanation / Answer

a) E(T) =/2* 0.2 + 3/2 * 0.2 + 5/2 * 0.2 + 7/2 * 0.2 + 9/2 * 0.2= /2* 0.2*[1+3+5+7+9]= /2*0.2*25= 2.5* = 5/2= 7.854

T2 = Var( T)

E(T^2)= ^2/4* 0.2 + 9^2/4 * 0.2 + 25^2/4 * 0.2 + 49^2/4 * 0.2 + 81^2/4 * 0.2= 0.2* ^2/4[ 1+9+25+49+81]= 165* ^2*0.2/4= 81.4242

hence T2 = Var( T)= E(T^2)- (E(T))^2= 19.739

b) for T= /2, 5/2 and 9/2, X= 1 and for T= 3/2 and 7/2, X= -1

hence E(X)= 0.2-0.2+0.2-0.2+0.2= 0.2

X^2= 1 with probability 1 as X^2 is always 1.

Hence E(X^2)= 1

x2= Var(X)= 1- 0.2^2= 1-0.04= 0.96

c) Case 1: N is even

when n is odd, Y= 1 and when n is even Y= -1 for n belonging to {1,2,...N} and the probability of each outcome is 1/N.

since N is even, there are equal no. of odd and even n's, hence E(Y)= 0 as no of times Y is 1 is same as the no. of times Y is -1 and each outcome has same probability.

Y^2= 1 with probability 1 however.

hence E(Y^2)=1

Thus y2= Var(Y)=1

Case 2: N is odd

when n is odd, Y= 1 and when n is even Y= -1 for n belonging to {1,2,...N} and the probability of each outcome is 1/N.

since N is odd, no of odd numbers in {1,2..,N} is 1 more than the no. of even numbers, hence E(Y)= -1*1/N as no of times Y is 1 is 1 less than the no. of times Y is -1 and each outcome has same probability 1/N.

Y^2= 1 with probability 1 however.

hence E(Y^2)=1

Thus y2= Var(Y)=1-(-1/N)^2= 1-1/N^2

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.