Texts p418 consider the case where a random process xn is...

Texts: P4.18 Consider the case where a random process {x(n)} is generated as in Figure P4.18 I[(6-u)x ... (I-u)x (u)x]=(u)x (i) Find and present the correlation matrix R = E[x(n)x(n)]. (ii) Find and present the eigenvalues, λ, and eigenvectors, qᵢ, of R. (iii) Present a plot of the power spectral density X(eᵢ). (iv) Add the plots of the magnitude responses of the eigenfilters of R to the power spectral density plot in Part (iii). (v) Make an attempt to relate the plots in Part (iv) to the eigenvalues in Part (ii). Unit variance H(z) = 1 + 2z⁻¹ - z⁻² white process Figure P4.18 Unit variance white process H Figure P4.20

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0:00 Hello students.

00:02 Today we will discuss about this question.

00:04 In this question, we need to consider a random process that is given by z of t is equal to a cost of omega 0 t plus teta, where theta is a uniform random variable over closed interval minus pi divide by 2 to plus pi divide by 2.

00:24 Assume z of t goes through a squirting device to produce a new random process.

00:31 That is x of t is equals to z square of t now here in the part a we need to show that or we need to prove that the auto correlation function of x of t auto correlation function of x of t that can be given as r x of t that is equal to a square divide by two one plus two cos of two omega zero tau so we need to show this or we need to prove this.

01:06 So here first of all we will we know that the let the consider a function that is z of t is equals to a cos omega 0 plus theta.

01:17 So therefore the probability density function that can be given as f theta that is equals to here one multi one divide by pi where minus pi divide by two less than teta less than pi divide by two and zero otherwise.

01:35 So now consider a random per portion that is x of t that is equals to z square of t.

01:44 So therefore here x of t that can be given as a square, cost square of omega 0 t plus theta.

01:52 So therefore x of t that is equals to a square divide by 2 1 plus cos of 2 omega 0t plus 2 theta.

02:02 So here, rx of tau, that can be given as x of t multiplied by t plus tau whole bar, that is equals to integration minus pi divide by 2 ,2 plus pi divide by 2 f theta, multiplied by a square divided by 4 in bracket 1 plus cos of 2 omega 0 t plus 2 theta multiplied by 1 plus cos of 2 omega 0t plus 2 theta multiplied by 1 plus cos of 2 .0...

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