Infinite sum of discrete unit-step signals

Question

Trying to sketch the following signal:

$$\sum_{k=-\infty}^\infty (u[k]-u[k-3])(u[n-k]-u[n-k-3])$$

Where $u[n]$ is the unit step signal (the Heaviside function, $1$ when $n\ge 0$ and $0$ otherwise).

My issue is that once $n$ gets involved I have no idea what to do. I believe that the first portion is just a DC signal of $1$ for all $n$ (when summed from $-\infty$ to $\infty$) but when multiplied by the second half I'm not sure what to do.

Answer 1

To begin with, consider each factor separately.

$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.

$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.

So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-k\in\{0,1,2\}$, which means $n\in \{k,k+1,k+2\}$.

So, you have:

• a bump at $0,1,2$ when $k=0$
• a bump at $1,2,3$ when $k=1$
• a bump at $2,3,4$ when $k=2$

and are adding these up.