Find autocorrelation of exponential signal $a^nu[n]$ 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]&amp;=\sum_{m=-\infty}^\infty x[m]x[m-n]\\
&amp;=\sum_{m=-\infty}^\infty a^m u[m]a^{m-n}u[m-n]\\
&amp;=\sum_{m=n}^\infty a^m a^{m-n}\\
&amp;=a^{-n}\sum_{m=n}^\infty  a^{2m}\\
&amp;=a^{-n}\sum_{l=0}^\infty  a^{2(l+n)}\quad\tag{with $m-n=l$}\\
&amp;=a^{-n}a^{2n}\sum_{l=0}^\infty  a^{2l}\\
&amp;=a^{n}\sum_{l=0}^\infty  \left(a^2\right)^l\\
&amp;=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 $$\frac{a^{\lvert n\rvert}}{1-a^2}$$
Let $x[n]=a^nu[n], |a|&lt;1$. 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 $n&gt;0$. In this case, we have 

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



For $n&lt;0$:

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

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

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