How to find the autocorrelation function

Question

If $X(t)$ and $Y(t)$ are independent $WSS$ processes with zero mean,find the autocorrelation function of $Z(t)= a + bX(t) + cY(t)$.

Answer 1

The autocorrelation sequence $R_X(t,\tau)$ of a process $x(t)$ is defined as $$ R_X(t,\tau) = \mathbf E \left[ x(t) \cdot x(t+\tau) \right]. $$

Write out that definition for your sequence $Z$ in terms of $X$ and $Y$ and expand it, you'll be left with the expected value of terms that are mixed products of $X$ and $Y$ and terms that have each of the sequences on their own.

In this case, you are also told that $X$ and $Y$ are WSS (so $R_X(t, \tau) = R_X(\tau)$ and $R_Y(t,\tau) = R_Y(\tau)$) and independent, so the expected value of mixed products is zero.

If you are having problems with that post what you have tried so we can help you from there.

The solution:

The idea behind this problem is to find an expression for the autocorrelation of $Z$, $R_Z$, in terms of the autocorrelations of $X$ and $Y$. We don't know what the distribution of $X$ or $Y$ is, so our job is to somehow find $R_X$ and $R_Y$ "hidden" in the definition of $R_Z$. Writing it out in terms of $X$ and $Y$ we find:

$$ R_Z(t,\tau) = \mathbf E[ z(t) \cdot z(t+\tau) ] $$

$$ = \mathbf E[ \left( a + b X(t) + c Y(t) \right) \cdot \left( a + b X(t+\tau) + c Y(t + \tau) \right) ] $$

$$ = \mathbf E[ a^2 + a b X(t + \tau) + a c Y(t + \tau) + a b X(t) + b^2 X(t) X(t + \tau) + b c X(t) Y(t + \tau) + a c Y(t) + b c X(t + \tau) Y(t) + c^2 Y(t) Y(t + \tau) ] $$

Knowing that expectation is a linear operator, we may distribute it among the linear terms:

$$ = \mathbf E[a^2] + a b \mathbf E[X(t+\tau)] + a c \mathbf E[Y(t + \tau)] + a b \mathbf E[X(t)] + b^2 \mathbf E[X(t) X(t+\tau)] + b c \mathbf E[X(t) Y(t + \tau)] + a c \mathbf E[Y(t)] + b c \mathbf E[X(t + \tau) Y(\tau)] + c^2 \mathbf E[Y(t) Y(t+\tau)]$$

Now we know that $X$ and $Y$ are zero mean, so $\mathbf E[X(\cdot)] = \mathbf E[Y(\cdot)] = 0$. Also $X$ and $Y$ are independent, therefore they are uncorrelated and $\mathbf E[X(\cdot)Y(\cdot)] = 0$:

$$ = \mathbf E[a^2] + b^2 \mathbf E[X(t) X(t + \tau)] + c^2 \mathbf E[Y(t) Y(t+\tau)] $$

Note here that the term with $a^2$ does not disappear - it is not a random variable, but all this means is that $\mathbf E[a^2] = a^2$ since it always has the same value. This is important since it has the effect of adding a mean to $Z$, as opposed to $X$ and $Y$ which have no mean.

Finally, in that last expression we recognize the definition for $R_X(t,\tau) = \mathbf E[X(t) X(t + \tau)]$ and similarly for $R_Y$, so:

$$ R_Z(t,\tau) = a^2 + b^2 R_X(t,\tau) + c^2 R_Y(t,\tau). $$

Note that up to this point we have never used the fact that $X$ and $Y$ are WSS. This added condition means that in reality $R_X$ doesn't depend on $t$ - the "time" at which the autocorrelation is evaluated - but only on $\tau$ - the "displacement" or difference in time between the samples that we wish to compute the correlation for. All of this we imply when we write $R_X(t,\tau) = R_X(\tau)$.

Finally, since both $X$ and $Y$ depend only on $\tau$ it follows that $Z$ also only depends on $\tau$:

$$ R_Z(t,\tau) = R_Z(\tau) = a^2 + b^2 R_X(\tau) + c^2 R_Y(\tau). $$

So $Z$ is a WSS process (but not zero mean!)