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If the variance \sigma_x^2 of d(n)= x (n) - x(n-1) is one-tenth the variance \sigma_x^2 of a stationary zero mean discrete time signal x(n), then the normalized auto-correlation function
\frac{R_{xx}(k) }{\sigma_x^2} at k = 1 is
(a) 0.95
(b) 0.90
(c) 0.10
(d) 0.05Answer is ( a )
Let's understand the definition of variance \sigma^2 = Var [A]
Concept
, where A is any random process
Then , \sigma^2 = E [ (A-E(A))^2]
= E[(A^2 - 2AE(A) + (E(A))^2)]
Taking Expectation E within the brackets:
=E[A^2] -2E[A]E[A] + (E(A))^2
= E[A^2] -(E[A])^2
A mnemonic for the above expression is "mean of square minus square of mean"
The process is defined by :
d(n) = x(n) - x(n-1)
And also given that process is zero-mean.E[x(n)] =0 \implies (E[x(n)-x(n-1)])^ 2 = 0
Hence,Variance of d(n) = \sigma_d^2 = E[(x(n)-x(n-1))^2] - (E[x(n)-x(n-1)])^2
\sigma_d^2 = E[(x(n)-x(n-1))^2] - 0
\sigma_d^2 = E[(x(n))^2 +(x(n-1))^2 - 2x(n)x(n-1)]
\sigma_d^2 = E[(x(n))^2] +E[(x(n-1))^2] - 2 E[x(n)x(n-1)]
\sigma_d^2 = \sigma_x^2 \times \frac{1}{10}
The expression E[x(n)x(n-1)] is auto-correlation by lag 1.
In general , the autocorrelation function with lag k is defined by the following equation
R_{xx}(k) = E[x(n)x(n-k)]
\sigma_x^2 \times \frac{1}{10}= \sigma_x^2 + \sigma_x^2 - 2R_{xx}(1)
\implies \frac{ R_{xx}(1)}{\sigma_x^2 } = 19/20 = 0.95
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