Autocorrelation of given signal.

Question

I don't understand how the autocorrelation of a given input signal is determined. I've got the answers, but don't understand how they were derived.

Given is the following input signal $u(t)$

$u(t) = 1$ $\qquad$ if $\quad$ $t=1$ and $2$

$u(t) = -1$ $\quad$if $\quad$ $t=3$ and $4$

and $u(t+4)=u(t)$

I understand that the autocorrelation function is equal to:

$R_u(\tau)=\bar{E}u(t)u(t-\tau)=lim_{N\rightarrow \infty}\frac{1}{N}\Sigma^N_{t=1}Eu(t)u(t-\tau)$

The answers state that for $\tau = 0, 1, 2$ we get:

$R_u(0)=lim_{N\rightarrow \infty}\frac{1}{N}\Sigma^N_{t=1}u^2(t)=1$

$R_u(1)=lim_{N\rightarrow \infty}\frac{1}{N}\Sigma^N_{t=1}u(t)u(t-1)=0$

$R_u(2)=lim_{N\rightarrow \infty}\frac{1}{N}\Sigma^N_{t=1}u(t)u(t-2)=-1$

I don't understand how the $0 , 1$ and $-1$ result from the limits to $\infty$

Any help will be greatly appreciated.

Answer 1

If a signal is periodic so is its autocorrelation, and with the same period.

Given $$ \eqalign{ & u(t + T) = u(t) \cr & R(\tau ) = E\left( {u(t)u(t + \tau )} \right) = \mathop {\lim }\limits_{N\, \to \,\infty } {1 \over N}\sum\limits_{t = 1}^N {u(t)u(t + \tau )} \cr} $$ we have in fact $$ R(\tau + T) = E\left( {u(t)u(t + \tau + T)} \right) = E\left( {u(t)u(t + \tau + T)} \right) = E\left( {u(t)u(t + \tau )} \right) $$

Then if $R(\tau)$ is periodic, we can limit the sum to only the period, because $$ \eqalign{ & \mathop {\lim }\limits_{N\, \to \,\infty } {1 \over N}\sum\limits_{t = 1}^N {u(t)u(t + \tau )} = \mathop {\lim }\limits_{N\, \to \,\infty } {1 \over {nT + q}}\left( {\sum\limits_{t = 1}^T {u(t)u(t + \tau )} + \sum\limits_{t = T + 1}^{2T} {u(t)u(t + \tau )} + \cdots + \sum\limits_{t = (n - 1)T + 1}^{nT} {u(t)u(t + \tau )} + \sum\limits_{t = nT + 1}^{nT + q} {u(t)u(t + \tau )} } \right) = \cr & = \mathop {\lim }\limits_{N\, \to \,\infty } {1 \over {nT + q}}\left( {n\sum\limits_{t = 1}^T {u(t)u(t + \tau )} + \sum\limits_{t = 1}^q {u(t)u(t + \tau )} } \right) = \cr & = \mathop {\lim }\limits_{N\, \to \,\infty } {n \over {nT + q}}\sum\limits_{t = 1}^T {u(t)u(t + \tau )} + \mathop {\lim }\limits_{N\, \to \,\infty } {1 \over {nT + q}}\sum\limits_{t = 1}^q {u(t)u(t + \tau )} = \cr & = {1 \over T}\sum\limits_{t = 1}^T {u(t)u(t + \tau )} \cr} $$

That fixed I think you can clear your doubts.

And, by the way, if you take $t$ and $\tau$ to start from $0$, you will have an easier task.