# Delta Function and Unit Step Formula

I have come across the following formula:

$$u(n)=\sum_{m=-\infty}^{n}\delta(m)$$

where $u(n)$ is the Unit Step and $\delta(m)$ is the Delta Function:

What I can't understand is how this formula "works".

Expanding the formula we have:

$$u(n)=...+\delta(0)+\delta(1)+\delta(2)+...+\delta(n-1)+\delta(n)=\delta(0)=1$$

So expanding it, no matter what, gives us the same result which obviously is not a Unit Step, so I can't understand how that formula can produce the Unit Step.

I know that I have made a mistake somewhere but I don't know where. Can someone explain to me that formula and my mistake?

• Note that if $n<0$ then $u(n) = 0$ since the sum only contains terms $\delta(m)$ with $m$ negative. If $n>0$ we have a term $\delta(0)$ in the sum so $u(n) = 1$. Jun 5, 2015 at 1:51
• @Adam Just curious ... why are you calling the terms $\delta(n)$ "Delta Functions?" That term seems reserved for the Generalized Function, which for any "test function" $f$ we have $\int \delta (x)f(x)dx=f(0)$. On the other hand, the Kronecker Delta is an indicator function that is $1$ only when its argument is $0$. And this is the case here, I believe. Make sense? Jun 5, 2015 at 3:08
• @Dr.MV I have stumbled upon it on numerous circumstances in Control Systems and in Signal Processing where $\delta(n)$ is called Delta function. I know that it is also called Kronecker Delta but most non-mathematical books I guess don't make a distinction. :)
Jun 5, 2015 at 3:16
• @adam Yes. In a discete time system's course, the "Kronecker" label is often dropped. Similarly, in a continuous time system's course, the label "Dirac" is ofter dropped. Jun 5, 2015 at 13:56

If $n<0$, then $\delta(m) = 0$ for all $m \le n$, and so $u(n) = 0$.

if $n \ge 0$, then $\delta(0) = 0$ and $\delta(m) = 0$ for all $n \neq 0$, hence $u(n) = 1$.

• Thank you. It seems I was just confussed.
This is because $u(n)=1$ if $n \geq 0$ and $u(n)=0$ if $n \leq 0$. In this last cases, $\delta (0)$ never appears, so $u(n)=0$.