Signal Processing: Finding the auto-correlation

[Master1022]

Homework Statement

For the system below, where $e(t)$ is a zero mean white noise sequence with variance $\sigma_e ^2$ , determine the first two terms in the autocorrelation sequence $R_v (0)$ and $R_v (1)$

Relevant Equations

Autocorrelation

Hi,

I am working on the following problem from a textbook, but am getting stuck and am not sure how to proceed.

Question: For the system below:
$$v(t) = -cv(t - 1) + e(t)$$
where $e(t)$ is a zero mean white noise sequence with variance $\sigma_e ^2$ , determine the first two terms in the autocorrelation sequence $R_v (0)$ and $R_v (1)$

Attempt:
I am not sure which assumptions I ought to make to proceed with this question. At first, I thought about assuming that the signal $v(t)$ was a stationary signal, such that:
- $E[v(t)] = \mu = \text{constant}$
- The autocorrelation is only a function of the time difference $\tau$: $R(\tau) = E[v(t) v(t - \tau)]$

However, I don't think this assumption really makes sense as taking the expectation of the equation for $v(t)$ yields the fact that $E[v(t)] = -c E[v(t]$, which shows the mean is non-constant, unless the mean is 0.

Nonetheless, if I proceed with this assumption:
$$R_v (\tau) = E[v(t) v(t - \tau)] = E[\left( e(t) - c v(t - 1) \right) \left( e(t - \tau) - v(t - 1 - \tau) \right) ]$$
$$= E[e(t)e(t - \tau)] + c^2 E[v(t - 1) v(t - 1 - \tau)] - c E[e(t) v(t - 1 - \tau)] - c E[v(t - 1) e(t - \tau)]$$
Then by the stationarity assumption: $E[v(t - 1) v(t - 1 - \tau)] = R_v(\tau)$

$$R_v (\tau) = R_e (\tau) + c^2 R_v(\tau) - c E[e(t) v(t - 1 - \tau)] - c E[v(t - 1) e(t - \tau)]$$
Here is where I don't know how to deal with their cross terms. The first thought that comes to mind is that they are independent and thus I can split them up into a product of the expected values (which will yield zero). However, surely something like $v(t - 1)$ does depend on previous values of the error?

If I expand the expression, then I get: $y(t) = e(t) - c y(t - 1) = e(t) - c( e(t - 1) - c y(t - 2) ) = ...$ which then leads to something like:
$$y(t) = ( - c)^{t} + \sum_{k = 0}^{t} ( - c)^k e(t - k)$$

but am not sure how to use this to determine whether the terms $E[e(t) v(t - 1 - \tau)]$ and $E[v(t - 1) e(t - \tau)]$ are non-zero.

Any help would be greatly appreciated.

 

[FactChecker]

You are making things complicated by keeping $\tau$ as a variable in the equations. The problem is only asking for the cases of $\tau=0$ and $\tau=1$. You should only consider those cases and the equations should be fairly simple.

 

[Master1022]

Thanks for the reply.

You are making things complicated by keeping $\tau$ as a variable in the equations. The problem is only asking for the cases of $\tau=0$ and $\tau=1$. You should only consider those cases and the equations should be fairly simple.

Yes that is true - I was trying to generalize and then just substitute $\tau = 0, 1$, but that is more complicated than it needs to be.

Using $\tau = 0$, I get:
$$R_v (0) = E[v(t) v(t - 0)] = E[e(t)e(t)] - 2cE[v(t - 1)e(t)] + c^2 E[v(t - 1) v(t - 1)]$$
$$R_v (0) = R_e (0) + c^2 R_v (0) \rightarrow R_v (0) = \frac{\sigma_e ^2}{1 - c^2}$$

Then for $\tau = 1$:
$$R_v (1) = E[v(t) v(t - 1)] = E[e(t)e(t - 1)] - cE[e(t)v(t - 2)] - cE[e(t-1)v(t - 1)] + c^2 E[v(t - 1) v(t - 2)]$$
$$R_v (1) = c^2 R_y (1) - cE[e(t-1)v(t - 1)] = c^2 R_v(1) - c \left( E[e(t - 1) e(t - 1)] - c E[e(t - 1) v(t - 2)] \right)$$
$$\rightarrow R_v(1) = \frac{-c R_e (0)}{1 - c^2} = \frac{- c \sigma_e ^2}{1 - c^2}$$