0
$\begingroup$

I need to find the autocorrelation of the following discrete signal $$x[n]=a^nu[n] $$ So I tried finding the convolution of $x[n]$ and $x[-n]$. \begin{align} \phi_{xx}[n]&=\sum_{m=-\infty}^\infty x[m]x[m-n]\\ &=\sum_{m=-\infty}^\infty a^m u[m]a^{m-n}u[m-n]\\ &=\sum_{m=n}^\infty a^m a^{m-n}\\ &=a^{-n}\sum_{m=n}^\infty a^{2m}\\ &=a^{-n}\sum_{l=0}^\infty a^{2(l+n)}\quad\tag{with $m-n=l$}\\ &=a^{-n}a^{2n}\sum_{l=0}^\infty a^{2l}\\ &=a^{n}\sum_{l=0}^\infty \left(a^2\right)^l\\ &=a^n\frac{1}{1-a^2} \end{align}

However the result should be an even function of $n$ and instead of $$\frac{a^n}{1-a^2}$$ I should have found $$\frac{a^{\lvert n\rvert}}{1-a^2}$$

$\endgroup$
3
$\begingroup$

Let $x[n]=a^nu[n], |a|<1$. Autocorrelation is

$$\phi_{xx}[n]=\sum_{m=-\infty}^{\infty}x[m]x[m-n]=\sum_{m=-\infty}^{\infty}a^mu[m]a^{m-n}u[m-n]$$ First assume that $n>0$. In this case, we have

$$u[m]u[m-n]=\begin{cases}0,& \forall m<n\\ 1,& \forall m\ge n\end{cases}$$ Therefore, $$\begin{align} \phi_{xx}[n]&=\sum_{\color{red}{m=n}}^{\infty}x[m]x[m-n]\\ &=\sum_{m=n}^{\infty}a^ma^{m-n}\\ &=a^n(1 + a^2 + a^4 +\cdots )\\ &=\frac{a^n}{1-a^2}\end{align}$$


For $n<0$:

$$u[m]u[m-n]=\begin{cases}0,& \forall m<0\\ 1,& \forall m\ge 0\end{cases}$$

$$\begin{align} \phi_{xx}[n]&=\sum_{\color{red}{m=0}}^{\infty}x[m]x[m-n]\\ &=\sum_{m=0}^{\infty}a^ma^{m-n}\\ &=a^{-n}(1 + a^2 + a^4 +\cdots )\\ &=\frac{a^{-n}}{1-a^2}\end{align}$$

and since $\phi_{xx}[n]=\phi_{xx}[-n]$, we can write it for all $n$ as $$\phi_{xx}[n]=\frac{a^{|n|}}{1-a^2},\ |a|<1$$

$\endgroup$
  • $\begingroup$ This doesn't really answer the question: Where does the OP's calculation break down if his $n$ (a.k.a. your $m$) is a negative number? $\endgroup$ – Dilip Sarwate Jul 15 '17 at 14:27
  • $\begingroup$ Won't I get a^n for every m if I don't assume that m>0? I could assume that m>0 and then add the mod since autocorrelation is an even function but my problem is it's still a^n for m<0 if we don't start with the m>0 assumption. $\endgroup$ – John Katsantas Jul 15 '17 at 14:29
  • $\begingroup$ @JohnKatsantas Doesn't $m$ have to be greater than 0 going from your first line to your second because you aren't taking account of the $u(m)$ term? $\endgroup$ – Peter K. Jul 15 '17 at 14:31
  • $\begingroup$ @PeterK. The sum starts from 0 in that case and the a^-n remains intact. That solves it, you can add the answer $\endgroup$ – John Katsantas Jul 15 '17 at 14:36
  • $\begingroup$ I am late in the party! Unfortunately, the image was not loaded when I left the answer. Hence is the change of notation (will fix it). DilipSarwate I didn't see that explicitly in the question. But yes, as PeterK commented earlier, we have two terms in the sum and the other one "controls" the sum for negative $m$ values. $\endgroup$ – msm Jul 15 '17 at 15:47
3
$\begingroup$

Doesn't $m$ have to be greater than 0 going from your first line to your second because you aren't taking account of the $u(m)$ term?

$\endgroup$
  • $\begingroup$ Oh gosh Peter, did I catch you with a "non answer" answer? Is there a SE.DSP badge for that :) ? $\endgroup$ – Laurent Duval Jul 18 '17 at 17:27
  • $\begingroup$ @LaurentDuval The OP said in the comments on the other answer "That solves it, you can add the answer" so I did. :-) $\endgroup$ – Peter K. Jul 18 '17 at 17:42
  • 1
    $\begingroup$ OK then. Points added $\endgroup$ – Laurent Duval Jul 18 '17 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.