Discrete Time Convolution Evaluation

Question

I'm trying to solve a problem on convolution from Alan V.Oppenheim:

Find the convolution output $y[n]$ for the following signals:

$$x[n]= u[n]\quad\text{and}\quad h[n]=a^{n}u[-n-1], \ a>1 $$

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 $<0$ or $>0$ and I tried to evaluate for $m>0$ which is the same as $n>-1$, which evaluated as:

\begin{align} y[n]&=a^{n}\sum_{k=m}^{+\infty} a^m\\ &=a^{n}\left[\left(\sum_{k=0}^{+\infty} a^m\right)-\left(\sum_{k=0}^{m-1} a^m\right)\right]\\ &=a^{n}\left[\left(\frac{1}{1-a^{-1}}\right)-\left(\frac{1-a^{-m}}{1-a^{-1}}\right)\right]\\ &=\frac{a^{n-m}}{1-a^{-1}}\\ &=\frac{a^{-1}}{1-a^{-1}}\tag{since $n-m=-1$} \end{align}

but when I evaluated for $m<0$ 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 $m<0$?

Answer 1

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 $n<=-1$ you have to evaluate the sum with the lower limit $k=0$, and for $n>-1$, you have to evaluate the sum with a lower limit $k=n+1$.