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This question may seem slightly unnecessary but I'm really curious about the answer. For a system (LTI ones at least not sure about the rest) the output can be described as some function of the input: $$y(t) = f(x(t))$$ So surely the impulse response of a system can be thought of the same way: $$h(t) = f(\delta (t))$$ Now let's say we have the impulse response: $$h(t) = e^{-t}u(t) = f(\delta (t))$$ This could be the sound of a clap decaying in a room or something.

What I'm interested in is understanding what kind of function $f$ is able turn the delta function, which only exists for the briefest moment, into a decaying function that exists for a period of time.

When I think of it in terms of the decaying sound in a room, it makes perfect sense that the response would be a decaying function because its just the sound bouncing off the walls creating an echo that decays away, but I can't see mathematically how any function could take the delta function and somehow create this kind of a response.

I'm not sure if this question has a definite answer, I'm sorry if it doesn't, I just really hope it does. The delta function always seems like some math tool that was created because it's useful as opposed to representing real life.

Thanks, Richard

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  • $\begingroup$ I'm not sure what this would represent physically, but the following works (it's rather basic though): $$f(x(t))=\int_\Bbb R e^{-t'}u(t')x(t-t')\,dt'$$ $\endgroup$ – John Doe Dec 11 '17 at 0:15
  • $\begingroup$ Sorry about this but what does the prime symbol on some of the t's represent, that its negated? Or just a different t? Or maybe something else? Thanks $\endgroup$ – richard davies Dec 11 '17 at 0:28
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    $\begingroup$ Just that its a different $t$. If it makes it clearer, you can write it as $$f(x(t))=\int_\Bbb R e^{-p}u(p)x(t-p)\,dp$$ It is just a dummy variable over which you integrate. Then it gives $$f(\delta(t))=\int_\Bbb R e^{-p}u(p)\delta(t-p)\,dp=e^{-t}u(t)$$ as required. $\endgroup$ – John Doe Dec 11 '17 at 0:29
  • $\begingroup$ Ahh thank you so much, just knowing its possible this way is so relieving, I never would of though of this kind of a solution. $\endgroup$ – richard davies Dec 11 '17 at 0:40
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By the definition of impulse response, $f()$ is the convolution operator (as shown in @John Doe's comments)

$$f(x(t)) = h(t) * x(t) $$

And now note that

$$f(\delta(t)) = h(t) * \delta(t) = h(t)$$

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  • $\begingroup$ So quick and simple, I love it, it's kind of melted my brain a little, I'm going in circles of thought now with this but it'll sink in soon $\endgroup$ – richard davies Dec 11 '17 at 0:41
  • $\begingroup$ Haha, I get it now, the function that turns the impulse into the impulse response is the impulse response. Such madness $\endgroup$ – richard davies Dec 11 '17 at 0:44
  • $\begingroup$ Yeah, as an engineer, I understand the Dirac delta function as a mathematical "thingy" that can only really be evaluated in the context of being an integral's integrand. It is defined/designed to pluck out the function with which it is being convolved with. Delayed delta functions, $\delta(t - t_k)$, will produce a delayed copy of the impulse response, $h(t-t_k)$. $\endgroup$ – Andy Walls Dec 11 '17 at 0:49

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