 |
| Ask |
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) 0.05
Answer is ( a )
 |
| Concept |
Let's understand the definition of variance $$\sigma^2 = Var [A] $$, 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
$$R_{xx}(k) = E[x(n)x(n-k)]$$
$\sigma_x^2 \times \frac{1}{10}= \sigma_x^2 + \sigma_x^2 - 2R_{xx}(1)$
$\implies \frac{ R_{xx}(1)}{\sigma_x^2 } = 19/20 = 0.95$
