How to convolve $u(n)$ and $u(-n-5)$ without using $z$ transform

Question

As the title says, I wanted to test the property of the $z$ transform where:

$$z[x(n) \cdot x(h)] = z[x(n)] z [h(n)].$$

I have already solved the right hand part. All that remains to show is the left part, which I don't know how to do.

Thank you

Answer 1

I assume that you meant to type $u[n-5]$ rather than $u[-n-5]$ since the convolution that you have provided results in a non-convergent sum.

We compute $$ (x_1 * x_2)[n] = u[n-5] * u[n] = \sum_{k=-\infty}^\infty u[k-5]u[n-k] $$ Note that each product being added comes out to either $1$ or $0$.

Now, for a fixed $n$, we note that $n - k \geq 0$ means that $k \leq n$. On the other hand, $k -5 \geq 0$ means that $k \geq 5$. Thus, for any $n$, $(x_1*x_2)[n]$ is the number of integers $k$ that satisfy both $k \geq 5$ and $k \leq n$. That is, we have $$ (x_1 *x_2)[n] = \begin{cases} 0 & n < 5\\ (n-4) & n \geq 5 \end{cases} = (n-4)\cdot u[n-4] $$