Autocorrelation of signal with offset

Question

If I have a stationary signal $x(t)$ with zero mean and with an auto correlation $r_{xx}(\tau)$, then what is the auto correlation of $y(t)=x(t)+b$ ?

My calculations so far:

$$\begin{align} r_{yy}(\tau) &= E\{ [x(t+\tau) + b] [x(t) + b] \} \\ &= E\{ x(t+\tau)x(t) + b x(t+\tau) + b x(t) + b^2 \} \\ &=r_{xx}(\tau) + b^2 \\ \end{align}$$

The problem is that the autocorrelation of a constant is a triangular pulse and not just a squared offset. What am I missing here?

Answer 1

I think what you're doing is using matlab or R or some other tool to find the autocorrelation numerically. In that case, you will get a triangular pulse because you're effectively doing the correlation of a noisy, short-duration pulse of height $b$ with itself.

In that case, the expectations in Mark's answer don't hold and you'll get a triangular pulse.

For example, if I do the following in R:

x <- rnorm(100)
acf(x+10, demean=FALSE, lag=100)
acf(x+10, demean=TRUE, lag=100)

then the first plot shows what you are suggesting (a triangular pulse) and the second plot is what Mark says (the de-meaned version).

Answer 2

Let me just write an answer because what I wrote above about the expectation is not correct and extremely confusing because I didn't show how the $b^2$ drops out.

By definition:

$$\operatorname{cov}(X, Y) \triangleq E(XY) - E(X)E(Y)$$

So, $$\begin{align} \operatorname{cov}\big((X_{t+\tau} + b),(X{_t} + b)\big) &= E\big((X_{t+\tau} + b)(X_{t} + b)\big) - E\big(X_{t+\tau} + b\big)\times E\big(X_{t} + b\big) \\ &= E\big(X_{t+\tau} \times X_{t}\big) + b^2 - \Big(E\big(X_{t+\tau}\big) \times E\big(X_{t}) + b^2\Big) \\ &= E\big(X_{t+\tau} \times X_{t}\big) - \Big(E\big(X_{t+\tau}\big) \times E\big(X_{t}\big)\Big) \\ &= \operatorname{cov}\big(X_{t+\tau}, X_{t}\big) \\ &= \gamma(\tau) \\ \end{align}$$

Apologies for confusion.