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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?

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  • \$\begingroup\$ Your reasoning and your answer look correct to me. Is there something about it you find confusing? \$\endgroup\$
    – Tanner Swett
    Commented May 5, 2023 at 10:20
  • \$\begingroup\$ @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 ? \$\endgroup\$
    – Homer Jay Simpson
    Commented May 5, 2023 at 10:22
  • \$\begingroup\$ Do you know the definition of \$\delta(t)\$? \$\endgroup\$
    – Tim Williams
    Commented May 5, 2023 at 10:28
  • \$\begingroup\$ @TimWilliams $$ \delta(t)= \begin{cases} 0 &\text{if ($t \neq 0$\ })\\ singular &\text{if \(t=0)\)}\\ \end{cases} $$ \$\endgroup\$
    – Homer Jay Simpson
    Commented May 5, 2023 at 10:35
  • \$\begingroup\$ @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 \$\endgroup\$
    – Marcus Müller
    Commented May 5, 2023 at 14:04

1 Answer 1

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

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

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  • \$\begingroup\$ so only the part where the Dirac function is involved I take care of $t=0$ \$\endgroup\$
    – Homer Jay Simpson
    Commented May 5, 2023 at 10:45
  • \$\begingroup\$ 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 \$\endgroup\$
    – Marcus Müller
    Commented May 5, 2023 at 14:10
  • \$\begingroup\$ 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. \$\endgroup\$
    – Marcus Müller
    Commented May 5, 2023 at 14:11
  • \$\begingroup\$ 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. \$\endgroup\$
    – edwillys
    Commented May 6, 2023 at 9:55

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