Discrete autocorrelation function

Question

I'm trying to solve following and already tryed some approaches, but seems to be I not really understand how I should start to solve it.

The following relationship is given x[n]=v[n-1]+w[n]+v[n+1], v[n] is a mean free white noise with variance one and w[n] is also a mean free white noise with variance one.

Note that v[n] and w[n] are jointly uncorrelated.

Find the discrete autocorrelation function $r_x[k]$

Could anybody help me?

Thanks in advance.

Answer 1

Today I've tried again to solve it and here is a solution.

Signal x[n] $x[n]=v[n-1]+w[n]+v[n+1]$

$r_x(k) = E\{x(k) x(l)\} = E\{(v[k-1]+w[k]+v[k+1]) (v[l-1]+w[l]+v[l+1])\} = $

$ = E\{v[k-1] v[l-1]\} + E\{v[k-1] w[l]\} + E\{v[k-1] v[l+1]\} + $

$ + E\{w[k] v[l-1]\} + E\{w[k] w[l]\} + E\{w[k] v[l+1]\} + $

$ + E\{v[k+1] v[l-1]\} + E\{v[k+1] w[l]\} + E\{v[k+1] v[l+1]\} $

$ E\{v[k+1] w[l]\} = 0 $

$ E\{v[k-1] w[l]\} = 0 $

$ E\{w[k] v[l-1]\} = 0 $

$ E\{w[k] v[l+1]\} = 0 $

$ E\{v[k-1] v[l+1]\} = 0 $

$ E\{v[k+1] v[l-1]\} = 0 $

$r_x(k) = E\{v[k-1] v[l-1]\} + E\{w[k] w[l]\} + E\{v[k+1] v[l+1]\} = $

$ = \sigma_v\delta(k+1) + \sigma_w\delta(k) + \sigma_v\delta(k-1) $