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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?

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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.

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  • $\begingroup$ This one is very similar to the definition @Fat32 uses. I'm asking about the inverse though, which is defining the unit step first approximated as a linear function between 0 and $\Delta$ and then deriving that to get the unit impulse. $\endgroup$ – chilliefiber Aug 25 '20 at 14:47
  • $\begingroup$ I just gave you the formula, for the case where $\delta_\Delta$ and $u_\Delta$ are symmetrical around $t = 0$. Why can't you shift that over by $\Delta/2$? $u_\Delta$ is perfectly piecewise linear and continuous. $\endgroup$ – TimWescott Aug 25 '20 at 15:03
  • $\begingroup$ Sorry I was misinterpreting the graph. I've accepted your answer. $\endgroup$ – chilliefiber Sep 6 '20 at 18:04
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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.

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  • $\begingroup$ OK I had also seen that definition and that one I understand and makes sense when defining the unit step as its integral. But it doesn't help me with the actual question, which is the derivative of the unit step drawn as a linear function. $\endgroup$ – chilliefiber Aug 25 '20 at 14:45

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