2

I am self studying linear circuits and I came to a very basic problem.

The problem: $$y(t) = e^{-t} u(t)$$ where \$u(t)\$ is the step function.

Calculate the derivative of \$y(t)^{\prime}\$.

My take is:

$$ \begin{align} y(t)^{\prime} & =\left( e^{-t} \right)^{\prime} u(t) + e^{-t} u(t)^{\prime} \\ & = e^{-t} (-t)^{\prime} u(t) + e^{-t} \left( u(t) \right)^{\prime} \\ & = -e^{-t} u(t) + e^{-t} \delta(t) \end{align} $$

But this gives a solution, $$-e^{-t}u(t)+\delta(t)$$ Why?

Tim Williams
  • 22,874
  • 1
  • 20
  • 71
Homer Jay Simpson
  • 231
  • 6
  • Your reasoning and your answer look correct to me. Is there something about it you find confusing? – Cassie Swett May 05 '23 at 10:20
  • @CassieSwett yes why in the book the second term becomes from $$ -e^{-t}u(t)+e^{-t}\delta(t) = -e^{-t}u(t)+\delta(t)$$.Where did the $$e^{-t}$$ went ? – Homer Jay Simpson May 05 '23 at 10:22
  • Do you know the definition of \$\delta(t)\$? – Tim Williams May 05 '23 at 10:28
  • @TimWilliams $$ \delta(t)= \begin{cases} 0 &\text{if \($t \neq 0$\ })\\ singular &\text{if \(t=0)\)}\\ \end{cases} $$ – Homer Jay Simpson May 05 '23 at 10:35
  • @HomerJaySimpson "singular" is a lack of the property "boundedness" at a point, not a definition. Coming from a mathematical point of view, \$y(t)\$ simply has no derivative, as it's not differentiable. So, from a mathematical point of view, the statement "The derivative of \$e^{-t}u(t)\$ is \$-e^{-t} u(t) + \delta(t)\$" is hence wrong; that's twice as true as \$\delta\$ really isn't a function. What your pseudo-derivative is good for is being a *test function* if you will if you multiply something else with it in a integrative context. Doing math with things involving Dirac deltas and acting – Marcus Müller May 05 '23 at 14:04
  • as if it's a function that you can then deal like with any other function has led to very many tears amongst students. I'd be *super* careful with the material you're learning from! – Marcus Müller May 05 '23 at 14:06

1 Answers1

0

The Dirac function only has non-zero value at t=0, where your exponential has a value of 1. Therefore:

$$ −e^{−t}u(t)+e^{−t}δ(t) = −e^{−t}u(t)+δ(t) $$

edwillys
  • 96
  • 6
  • so only the part where the Dirac function is involved I take care of $t=0$ – Homer Jay Simpson May 05 '23 at 10:45
  • The delta dirac is not a function. The function \$y(t)\$ is not differentiable – it's not even continuous. The value of \$y'(0)\$ is simply undefined. OP's exercise was badly worded – similar problems *do* appear when you're asking yourself questions like "I saw how the system reacted to my input signal, I wonder how it reacts to the derivative of the input signal", and that very quickly leads to the difference between step and impulse response, but claiming you can essentially derive \$u(t)\$ and get a function just gets you through all kinds of hell. So, while it's clear where whoever – Marcus Müller May 05 '23 at 14:10
  • stated the problem wanted to take things, actually doing the introduction to step and impulse responses that way will lead to a lot of confusion. And the Dirac delta (not) being a function is one of them. – Marcus Müller May 05 '23 at 14:11
  • I would rather say that the Dirac is a _special_ kind of function. There are many functions that are not differentiable at some points and not continuous. Both of them are pretty common (and useful) in signal analysis, though in "real" life they are of course approximations. – edwillys May 06 '23 at 09:55