# Explanation of the result of a solved exercise concerning the unit step response of a system

I came across a solved exercise regarding Signals and Systems in which there is a system with impulse response $h(n)=u(n)$ and input $x(n)=u(n)$. The goal was to find the output of that system (unit step response).

(Note that $u(n)$ is the unit step)

So the solution begins saying that the output is the convolution between the input and the impulse response which is:

$$y(n)=x(n)\ast h(n)=\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m)=\sum_{m=0}^{n}1=n+1$$

After that the solution concludes that the output is:

$$y(n)=(n+1)\cdot u(n)$$

Everything is really elementary and I probably miss something here but how does the term $u(n)$ appeared in the output? Wasn't proved that the output is $n+1$?

Exectly the same happens in another one concerning a system with $x(n)=h(n)=a^{n} \cdot u(n)$ where although $y(n)=a^{n}\cdot (n+1)$ the result eventually is said to be $y(n)=a^{n}\cdot (n+1)\cdot u(n)$.

When they say:

$$y(n)=x(n)\ast h(n)=\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m)$$

they are correct, but in the next passage they oversimplify a bit.

Let's do it step by step:

\begin{align*} &\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = \sum_{m=-\infty}^{\infty}u(m)\cdot u(n-m) = \\ &= \sum_{m=0}^{\infty} 1 \cdot u(n-m) = \sum_{m=0}^{\infty} u(n-m) \\ \end{align*}

The function to be summed is $u(n-m)$, which is equal to 1 if and only if $n-m>0 \Leftrightarrow m < n$.

Since the sum is carried out only for $m>=0$, if $n>=0$ the result is actually $n+1$, but if $n<0$ the result is 0, therefore they should have written:

$\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = (n+1)u(n)$

rightaway.

The following image may help:

the red dots are the function values, the green shaded area highlights the value range that will be summed. The image was drawn with $n = 3$. You may infer that if $n<0$ the green area will disappear, indicating there is nothing to sum.

u(n) is like the Heaviside function in Laplace, it ensures that y(n)=0 for n<0, which must be the case. If you just wrote y(n)=n+1, then y(n) would have negative values for negative n