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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 $$\frac{a^{\lvert n\rvert}}{1-a^2}$$

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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 $$\phi_{xx}[n]=\frac{a^{|n|}}{1-a^2},\ |a|<1$$

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  • $\begingroup$ 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
    Commented Jul 15, 2017 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
    Commented Jul 15, 2017 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. ♦
    Commented Jul 15, 2017 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
    Commented Jul 15, 2017 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
    Commented Jul 15, 2017 at 15:47
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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?

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  • $\begingroup$ Oh gosh Peter, did I catch you with a "non answer" answer? Is there a SE.DSP badge for that :) ? $\endgroup$
    – Laurent Duval
    Commented Jul 18, 2017 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. ♦
    Commented Jul 18, 2017 at 17:42
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    $\begingroup$ OK then. Points added $\endgroup$
    – Laurent Duval
    Commented Jul 18, 2017 at 18:11

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