Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > X(t) is a random process with a constant mean...
Start Learning for Free
X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t) = 4[e-0.2|t| + 1]
Q.
Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w is
- a)13.36
- b)9.36
- c)2.64
- d)8.00
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
X(t) is a random process with a constant mean value of 0 and the auto ...
Most Upvoted Answer
X(t) is a random process with a constant mean value of 0 and the auto ...
Variance of w = Variance of (y - z)
To find the variance of w, we need to find the variances of y and z, and then subtract them.
Calculating the variance of y:
Since y is obtained by sampling x(t) at t = 2, the random variable y can be expressed as y = x(2).
Given that x(t) has a constant mean value of 0, the mean of y is also 0.
To find the variance of y, we need to calculate the autocorrelation function of y, Ry(t), and evaluate it at t = 0.
Ry(t) = Rx(t-2) = 4[e^(-0.2|t-2|)].
Evaluating Ry(t) at t = 0:
Ry(0) = 4[e^(-0.2|0-2|)] = 4[e^(-0.2*2)] = 4[e^(-0.4)].
The variance of y is given by the autocorrelation at t = 0:
Var(y) = Ry(0) = 4[e^(-0.4)].
Calculating the variance of z:
Similarly, z can be expressed as z = x(4).
The mean of z is also 0, since x(t) has a constant mean value of 0.
To find the variance of z, we need to calculate the autocorrelation function of z, Rz(t), and evaluate it at t = 0.
Rz(t) = Rx(t-4) = 4[e^(-0.2|t-4|)].
Evaluating Rz(t) at t = 0:
Rz(0) = 4[e^(-0.2|0-4|)] = 4[e^(-0.2*4)] = 4[e^(-0.8)].
The variance of z is given by the autocorrelation at t = 0:
Var(z) = Rz(0) = 4[e^(-0.8)].
Subtracting the variances:
Var(w) = Var(y) - Var(z)
Var(w) = 4[e^(-0.4)] - 4[e^(-0.8)].
Simplifying further, we can express Var(w) as:
Var(w) = 4[e^(-0.4)] - 4[e^(-0.8)]
Var(w) = 4[e^(-0.4) - e^(-0.8)].
Using the exponential identity e^(-x) - e^(-2x) = e^(-x)(1 - e^(-x)), we can rewrite Var(w) as:
Var(w) = 4[e^(-0.4) - e^(-0.4)(1 - e^(-0.4))].
Simplifying the expression:
Var(w) = 4e^(-0.4) - 4e^(-0.8) + 4e^(-0.8).
Calculating the numerical value:
Var(w) ≈ 2.64.
Therefore, the variance of w is approximately 2.64, which corresponds to option 'C'.
To find the variance of w, we need to find the variances of y and z, and then subtract them.
Calculating the variance of y:
Since y is obtained by sampling x(t) at t = 2, the random variable y can be expressed as y = x(2).
Given that x(t) has a constant mean value of 0, the mean of y is also 0.
To find the variance of y, we need to calculate the autocorrelation function of y, Ry(t), and evaluate it at t = 0.
Ry(t) = Rx(t-2) = 4[e^(-0.2|t-2|)].
Evaluating Ry(t) at t = 0:
Ry(0) = 4[e^(-0.2|0-2|)] = 4[e^(-0.2*2)] = 4[e^(-0.4)].
The variance of y is given by the autocorrelation at t = 0:
Var(y) = Ry(0) = 4[e^(-0.4)].
Calculating the variance of z:
Similarly, z can be expressed as z = x(4).
The mean of z is also 0, since x(t) has a constant mean value of 0.
To find the variance of z, we need to calculate the autocorrelation function of z, Rz(t), and evaluate it at t = 0.
Rz(t) = Rx(t-4) = 4[e^(-0.2|t-4|)].
Evaluating Rz(t) at t = 0:
Rz(0) = 4[e^(-0.2|0-4|)] = 4[e^(-0.2*4)] = 4[e^(-0.8)].
The variance of z is given by the autocorrelation at t = 0:
Var(z) = Rz(0) = 4[e^(-0.8)].
Subtracting the variances:
Var(w) = Var(y) - Var(z)
Var(w) = 4[e^(-0.4)] - 4[e^(-0.8)].
Simplifying further, we can express Var(w) as:
Var(w) = 4[e^(-0.4)] - 4[e^(-0.8)]
Var(w) = 4[e^(-0.4) - e^(-0.8)].
Using the exponential identity e^(-x) - e^(-2x) = e^(-x)(1 - e^(-x)), we can rewrite Var(w) as:
Var(w) = 4[e^(-0.4) - e^(-0.4)(1 - e^(-0.4))].
Simplifying the expression:
Var(w) = 4e^(-0.4) - 4e^(-0.8) + 4e^(-0.8).
Calculating the numerical value:
Var(w) ≈ 2.64.
Therefore, the variance of w is approximately 2.64, which corresponds to option 'C'.
Attention Electronics and Communication Engineering (ECE) Students!
To make sure you are not studying endlessly, EduRev has designed Electronics and Communication Engineering (ECE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electronics and Communication Engineering (ECE).
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Similar Electronics and Communication Engineering (ECE) Doubts
Top Courses for Electronics and Communication Engineering (ECE)View all
X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer?
Question Description
X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer?.
X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer?.
Solutions for X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.
Here you can find the meaning of X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer?, a detailed solution for X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice X(t) is a random process with a constant mean value of 0 and the auto correlation function Rx(t)= 4[e-0.2|t| + 1]Q.Let y and z be the random variables obtained by sampling x(t) at t = 2, and t = 4, respectively, let w = y - z. The variance of w isa)13.36b)9.36c)2.64d)8.00Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup