Autocorrelation

A signals’ self similarity, $y[n]=x[n]$. See Crosscorrelation An autocorrelation of 1 means that the signals are identical, 0 means that they are completely different.

$$r_{xx}[l]=\sum _{n=-\infty }^{\infty }x[n-l]\cdot x[n]=\sum _{n=-\infty }^{\infty }x[n+l]\cdot x[n]$$

^fc34ea Or alternatively, if we have stochastic variables

$$r_{xx}=E[X[n-l]X[n]]=E[X[n+l]X[n]]$$

Where $l$ is the lag. $r_{xx}$ is often denoted as ${\gamma }_{xx}$ in digital signal processing.

Properties

• Energy of a sequence

$$E_k=\sum _{n=-\infty }^{\infty }x[n{]}^2=r_{xx}[0]\ge 0$$

• Autocorrelation is maximum at $l=0$, $\to$ $E_k=r_{xx}[0]\ge \mid r_{xx}[l]\mid$

• If it is even $\to$ We only compute for $l\ge 0$

• We care about a signals’ normalized autocorrelation because: ?

• Given

$$y[n]=x[n]+Rx[n-D]$$

Where $R$ is attenuation, and D is delay

Then

$$\frac{r_{yy}[0]}{r_{yy}[D]}=\frac{1+R^2}{R}$$

And the reverse filter we use to get $y[n]=x[n]$ is $H(z{)}^{-1}=\frac{1}{1+R{z}^{-D}}$ Which means that both the poles and the zeroes of H(z) needs to be within the unit circle in order to be stable (Since the zeroes of $H(z)$ are the poles of $H(z{)}^{-1}$)

Examples

Examples 2, lecture 8

Given

$$y[n]=x[n]+Rx[n-D]$$

Estimate R and D using the autocorrelation of y

Solution

Finding $r_{yy}[l]$

r_{yy}[l] &= \sum_{n=-\infty}^{\infty}y[n]y[n-l] \\ r_{yy}[l] &= \sum_{n}(x[n]+Rx[n-D])*(x[n-l] + Rx[n-D-l]) \\ r_{yy}[l] &= r_{xx}[l] + Rr_{xx}[D+l] + Rr_{xx}[l-D] + R²r_{xx}[l] \\ r_{yy}[l] &= (1+R²)r_{xx}[l] + Rr_{xx}[l+D] + Rr_{xx}[l-D] \end{align}

Which will give us peaks at $0$, $-D$ and $D$

$$sexyman\int \sum \alpha dx$$