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Which of the following is the auto correlation of x(n)?
- a)rxy(l)=x(l)*x(-l)
- b)rxy(l)=x(l)*x(l)
- c)rxy(l)=x(l)+x(-l)
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Which of the following is the auto correlation of x(n)?a)rxy(l)=x(l)*x...
Explanation: We know that, the correlation of two signals x(n) and y(n) is
Most Upvoted Answer
Which of the following is the auto correlation of x(n)?a)rxy(l)=x(l)*x...
Auto Correlation of x(n)
Auto correlation is a measure of similarity between a signal and a delayed version of itself. In this case, we are looking at the auto correlation of a signal x(n).
Formula for Auto Correlation
The formula for auto correlation is:
r_xy(l) = E[x(n)x(n-l)]
where l is the lag or delay, and E is the expected value operator.
Option A is Correct
The option A states that the auto correlation of x(n) is r_xy(l) = x(l)*x(-l). This is the correct answer.
Explanation
Substituting l in the formula for auto correlation, we get:
r_xy(l) = E[x(n)x(n-l)]
r_xy(l) = E[x(n)x(n+l)]
If we assume that x(n) is a real signal, then we can write:
r_xy(l) = E[x(n)x(n+l)]
r_xy(l) = E[x^2(n+l)]
Now, if we substitute l = 0, we get:
r_xy(0) = E[x^2(n)]
This is the definition of the power of x(n). Therefore, we can write:
r_xy(l) = E[x(n)x(n+l)]
r_xy(l) = E[x(n)x(n)*x(n+l)*x(n)]
r_xy(l) = x(l)*x(-l)
Therefore, option A is the correct answer.
Auto correlation is a measure of similarity between a signal and a delayed version of itself. In this case, we are looking at the auto correlation of a signal x(n).
Formula for Auto Correlation
The formula for auto correlation is:
r_xy(l) = E[x(n)x(n-l)]
where l is the lag or delay, and E is the expected value operator.
Option A is Correct
The option A states that the auto correlation of x(n) is r_xy(l) = x(l)*x(-l). This is the correct answer.
Explanation
Substituting l in the formula for auto correlation, we get:
r_xy(l) = E[x(n)x(n-l)]
r_xy(l) = E[x(n)x(n+l)]
If we assume that x(n) is a real signal, then we can write:
r_xy(l) = E[x(n)x(n+l)]
r_xy(l) = E[x^2(n+l)]
Now, if we substitute l = 0, we get:
r_xy(0) = E[x^2(n)]
This is the definition of the power of x(n). Therefore, we can write:
r_xy(l) = E[x(n)x(n+l)]
r_xy(l) = E[x(n)x(n)*x(n+l)*x(n)]
r_xy(l) = x(l)*x(-l)
Therefore, option A is the correct answer.
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