Conceptualising the continuous time unit impulse function as derivative of unit step

Question

This a very newbie question.

I just watched Lecture 3 of Oppenheim's Signals course and he defines here the continuous time function as the derivative of the unit step function like so:

$$ \delta_\Delta(t)=\frac {du_\Delta(t)}{dt}$$

and that $ \delta(t)= \delta_\Delta(t) $ as $\Delta \to 0$

He claims that the derivative is equal to 1 no matter the value of $\Delta$, because that derivative can be interpreted as the area of a rectangle with sides $\Delta$ and $\frac 1 \Delta$

I can't conceptualise this the way the function $u_\Delta(t)$ is drawn at all. If the function is linear, that is, $y = mx + b$ passes through the origin, meaning $b=0$, and we can see that it has the point $(\Delta, 1)$ we can easily tell that $m= \frac 1 \Delta$ and that should be the derivative.

Can someone explain to me the error in my line of thought? Why is the derivative the area of a square?

Answer 1

Let $\delta_\Delta(t) = \begin{cases}\frac{1}{\Delta} && -\frac{1}{2\Delta} \le t \le \frac{1}{2\Delta} \\ 0 && \mathrm{otherwise}\end{cases}$.

Then just integrate it! $u_\Delta(t)$ must be zero for $t < -\frac{1}{2\Delta}$, it must be 1 for $t > \frac{1}{2\Delta}$, and it must be a straight line in between:

$$u_\Delta(t) = \begin{cases} 0 && t < -\frac{1}{2\Delta} \\ \Delta(t + \frac{1}{2\Delta}) && -\frac{1}{2\Delta} \le t \le \frac{1}{2\Delta} \\ 1 && t > \frac{1}{2\Delta} \end{cases}$$

Note that you can womp up almost any $\delta_\Delta(t)$, to suit the problem at hand. Just choose a function of $\Delta$ that's zero outside of some bounds and that integrates to 1. It can be triangular, a half-sine, raised sine, etc. It's generally easier to not bother, but if you feel compelled to go back to basics and find things in the limit as $\Delta \to 0$, you can.

Answer 2

I didn't look at how the function $\delta_\Delta(t)$ is defined in that lecture, but the convention is that it's a rectangular pulse of area being $1$. The definition is :

$$\delta_\Delta(t) = \begin{cases}{ \frac{1}{\Delta} ~~~, ~~~ 0 < t < \Delta \\ ~0~ ~~~,~~~\text{otherwise} } \end{cases} $$

As you can see the area of that pulse, $A(\Delta) = \frac{1}{\Delta} \times \Delta = 1$ for any finite value of $\Delta$. According to calculus, that area is still $1$ even when the parameter $\Delta$ takes on inadmissible values of $0$ or $\infty$ through a limiting process. In particular the continuous Dirac delta function is defined as $\delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t)$ ; when the limit of $\Delta$ goes to zero.

The rectangular pulse of unit area, $\delta_{\Delta}(t)$, is only one of many other possible ways of defining $\delta(t)$ through a limit of functions process.