Discrete Time Convolution Evaluation I'm trying to solve a problem on convolution from Alan V.Oppenheim:
Find the convolution output $y[n]$ for the following signals:
$$x[n]= u[n]\quad\text{and}\quad h[n]=a^{n}u[-n-1], \ a&gt;1 $$
I started the evaluation:
$$y[n]=\sum_{k=-\infty}^{+\infty} u[k]a^{n-k}u[-n+k-1]$$
considering that $u[k]=1$ for $k\ge0$ and $u[-n+k-1]=1$ for $k\ge n+1$
which I evaluated to
$$y[n]=a^{n}\sum_{k=m}^{+\infty} a^k$$ where $m=n+1$ could be $&lt;0$ or $&gt;0$
and I tried to evaluate for $m&gt;0$ which is the same as $n&gt;-1$, which evaluated as:
\begin{align}
y[n]&amp;=a^{n}\sum_{k=m}^{+\infty} a^m\\
&amp;=a^{n}\left[\left(\sum_{k=0}^{+\infty} a^m\right)-\left(\sum_{k=0}^{m-1} a^m\right)\right]\\
&amp;=a^{n}\left[\left(\frac{1}{1-a^{-1}}\right)-\left(\frac{1-a^{-m}}{1-a^{-1}}\right)\right]\\
&amp;=\frac{a^{n-m}}{1-a^{-1}}\\
&amp;=\frac{a^{-1}}{1-a^{-1}}\tag{since $n-m=-1$}
\end{align}
but when I evaluated for $m&lt;0$  which is $n\le -1$ I am facing a problem:
$$y[n]= a^{n}\left[\left(\sum_{k=m}^{-1} a^m\right)+\left(\sum_{k=0}^{+\infty} a^m\right)\right]$$
How do I evaluate the first summation? I mean am I to consider $k=-m$ since $m&lt;0$?
The multiplication of the two unit step sequences $u[k]\cdot u[-n+k-1]$ is only non-zero if both sequences are non-zero. This means that the condition $k\ge 0$ as well as the condition $k\ge n+1$ must be satisfied. So you have two cases: for $n&lt;=-1$ you have to evaluate the sum with the lower limit $k=0$, and for $n&gt;-1$, you have to evaluate the sum with a lower limit $k=n+1$.