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Consider the following recursive difference equation of a LTI system, where $v[n]$ is a white noise, zero-mean process with $\sigma_v^2 = 1$.

$ x[n] = v[n] + 0.75x[n-1]-0.25x[n-2] $

I want to calculate the auto-correlation sequence $r_{xx}[k]$ for $k=0,1,2$. I started as follows:

$ r_{xx}[k] = E\{(v[n] + 0.75x[n-1]-0.25x[n-2])(v[n-k] + 0.75x[n-k-1]-0.25x[n-k-2])\} \\ r_{xx}[0] = E\{ v[n]^2 \} + 0.75^2 E\{ x[n-1]^2 \} - 2 \times 0.25 \times 0.75 E\{ x[n-1] x[n-2] \} + 0.25^2 E\{ x[n-2]^2 \} = \sigma_v^2 + \frac{5}{8} r_{xx}[0] - \frac{3}{8}r_{xx}[1] $

As you can see, in the equation for $r_{xx}[0]$, a term with $r_{xx}[1]$ remains. If I then want to determine $r_{xx}[1]$, a term with $r_{xx}[2]$ remains, and so on. How is it possible to solve this problem?

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With

$$\begin{align}r_k=E\{x_nx_{n+k}\}&=E\{(v_n+\alpha x_{n-1}+\beta x_{n-2})x_{n+k}\}\\&=E\{v_nx_{n+k}\}+\alpha E\{x_{n-1}x_{n+k}\}+E\{x_{n-2}x_{n+k}\}\\&=\sigma_v^2+\alpha r_{k+1}+\beta r_{k+2}\tag{1}\end{align}$$

and with

$$r_k=r_{-k}$$

you can obtain $3$ equations

$$\begin{align}r_0&=\sigma_v^2+\alpha r_1+\beta r_2\\ r_1&=\alpha r_0+\beta r_1\\r_2&=\alpha r_1+\beta r_0\end{align}$$

from which you get

$$\begin{align}r_0&=\frac{\sigma_v^2(1-\beta)}{(1+\beta)[(1-\beta)^2-\alpha^2]}\\r_1&=\frac{\sigma_v^2\alpha}{(1+\beta)[(1-\beta)^2-\alpha^2]}\\r_2&=\frac{\sigma_v^2[\alpha^2+\beta(1-\beta)]}{(1+\beta)[(1-\beta)^2-\alpha^2]}\end{align}$$

With $\sigma_v^2=1$, $\alpha=0.75$ and $\beta=-0.25$, this evaluates to

$$r_0=\frac53,\; r_1=1,\; r_2=\frac13$$

From these values you can compute all other values $r_k$ using the recursion $(1)$.

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We have \begin{equation} x[n] = v[n] + 0.75x[n-1]-0.25x[n-2] \end{equation} You started correct, you get \begin{equation} \begin{split} r_{xx}[k] = E\{(v[n] + 0.75x[n-1]-0.25x[n-2])(v[n-k] + 0.75x[n-k-1]-0.25x[n-k-2])\} \\ \end{split} \end{equation} You get 9 terms \begin{equation} \begin{split} r_{xx}[k] &= Ev[n]v[n-k] +0.75E x[n-k-1]v[n] - 0.25E x[n-k-2]v[n]\\ &+0.75 Ex[n-1]v[n-k] + 0.75^2Ex[n-1]x[n-k-1]+0.75(0.25)Ex[n-1]x[n-k-2]\\ &-0.25Ex[n-2]v[n-k] -0.75(0.25)Ex[n-2]x[n-k-1] + 0.25^2Ex[n-2]x[n-k-2] \end{split} \tag{1} \end{equation} Since the noise is white then \begin{equation} Ev[n]v[n-k] = \delta(k)\sigma_v^2 = \delta(k) \end{equation} Since $k\geq 0$, write down $x[n-k-1]$ and $x[n-k-2]$ do not contain lags that align with $v[n]$, so \begin{align} E x[n-k-1]v[n] &= 0\\ E x[n-k-2]v[n] &= 0 \end{align} On the other hand, $Ex[n-1]v[n-k]$ and $Ex[n-2]v[n-k]$, will agree for $k = 1$ and $k=2$ according to \begin{align} x[n-1]v[n-k] &= \overbrace{Ev[n-1]v[n-k]}^{\delta(k-1)} + 0.75\overbrace{ Ex[n-2]v[n-k] }^{\delta(k-2)}- 0.25\overbrace{Ex[n-3]v[n-k]}^{0} \\ x[n-2]v[n-k] &= \underbrace{Ev[n-2]v[n-k]}_{\delta(k-2)} + \underbrace{E(0.75x[n-3]-0.25x[n-4])v[n-k]}_{0} \end{align} So \begin{align} x[n-1]v[n-k] &= \delta(k-1) + 0.75\delta(k-2) \\ x[n-2]v[n-k] &= \delta(k-2) \end{align} The other four terms follow from the AR definition, i.e. \begin{align} E x[n-1]x[n-k-1] &= r_{xx}[n-1-n+k+1]= r[k] \\ E x[n-1]x[n-k-2] &= r_{xx}[n-1-n+k+2]= r[k+1] \\ E x[n-2]x[n-k-1] &= r_{xx}[n-2-n+k+1]= r[k-1] \\ E x[n-2]x[n-k-2] &= r_{xx}[n-2-n+k+2]= r[k] \end{align} Replacing all results in equation $(1)$, we get \begin{equation} \begin{split} r[k] &= \delta(k) +0.75\delta(k-1) + 0.75^2\delta(k-1) -0.25\delta(k-2)\\ &+0.75^2 r[k] +0.75(0.25)r[k+1]-0.75(0.25)r[k-1] + 0.25^2 r[k] \end{split} \end{equation} which suggests that \begin{equation} 0.375r[k] = \delta(k) + 1.3125\delta(k-1) - 0.25\delta(k-2) +0.1875(r[k+1] - r[k-1]) \end{equation} For $k = 0$, we get \begin{equation} 0.375r[0] = \delta(0) + 1.3125\delta(-1) - 0.25\delta(-2) +0.1875(r[+1] - r[-1]) \end{equation} we know that $\delta(0) = 1$ and $\delta(-1) = \delta(-2) = 0$. Also $r[1] = r[-1]$ due to the fact that $r[k]$ is even \begin{equation} r[0] =\frac{1}{0.375} \delta(0) = \frac{8}{3} \end{equation} For $k = 1$, we get \begin{equation} 0.375r[1] = 1.3125 + 0.1875(r[2] - r[0]) \end{equation} which is (again because $r[k]$ is an even function) \begin{equation} 0.375r[1] = 1.3125 + 0.1875(r[2] - \frac{8}{3}) \tag{*} \end{equation} and at $k = -1$ \begin{equation} 0.375r[-1] = 0.1875(r[0] - r[-2]) \end{equation} which is (again because $r[k]$ is an even function) \begin{equation} 0.375r[1] = 0.1875(\frac{8}{3} - r[2])\tag{**} \end{equation} Now solve the system of two equations in two unknowns $(*),(**)$ to get $r[1],r[2]$, i.e. \begin{equation} \begin{bmatrix} 0.3750 & -0.1875\\ 0.3750 & 0.1875 \end{bmatrix} \begin{bmatrix} r[1]\\ r[2] \end{bmatrix} = \begin{bmatrix} 0.8125\\ 0.5000 \end{bmatrix} \end{equation}

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    $\begingroup$ There must be a mistake somewhere because the end result is wrong. $\endgroup$ – Matt L. Sep 21 '18 at 11:29

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