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If we give a input $x(t)=u(t)$ to a system $\mathcal{S}$ we get an output $y(t) = e^{-t} u(t)$.
After we Laplace-transform both the input and the output we get the transfer function $$H(s) = 1-\frac{1}{s+1}$$ which in the time domain is: $$h(t) = \delta(t)-e^{-t}u(t)$$

But why do we need to transform everything in the Laplace domain to find the transfer function? What does the Laplace domain describe fundamentally? Why aren't we able to find the transfer function by division of output with input in the time domain?

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    $\begingroup$ Given the questions that you have been asking recently: it might be a good idea to go through text book or online class on linear time invariant systems (including Laplace & Fourier Transform ). Without having a more formal foundation, it will be very difficult to get the hang of this. $\endgroup$
    – Hilmar
    Sep 8, 2022 at 16:18
  • $\begingroup$ "Yeah but WHY is the Laplace domain so important?" This is probably the question you should lead with. The short answer is that for linear, time-invariant (LTI) systems, it takes a lot of really tedious, difficult, and disconnected bits of math surrounding analyzing differential equations, and it expresses all of it in a unified, (fairly) easy to understand manner. It does it so well, that sometimes folks forget that it doesn't work for nonlinear systems, or that there are problems in LTI systems that are easier solved in the time domain. $\endgroup$
    – TimWescott
    Sep 8, 2022 at 19:19

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First of all it's important to understand that this is all about linear and time-invariant (LTI) systems. Otherwise, you can't generally use a transfer function to characterize a system. So if you have some input $x(t)$ and some corresponding system output $y(t)$, you generally can't say much about the system, unless you know that it is LTI.

The transfer function of an LTI system is defined in the frequency domain, not in the time domain. The transfer function $H(s)$ relates the Laplace transforms of the output and input signals:

$$Y(s)=H(s)X(s)\tag{1}$$

where $X(s)$ and $Y(s)$ are the Laplace transforms of the input and output signal, respectively, and $H(s)$ is the system's transfer function. The nice thing about Eq. $(1)$ is that input and output are related by multiplication. The reason for this is that the Laplace transform converts convolution into multiplication, and convolution is the process which relates the input signal to the output signal of an LTI system in the time domain:

$$y(t)=(x\star h)(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\tag{2}$$

Eq. $(2)$ is just the inverse Laplace transform of $(1)$. Note that the impulse response $h(t)$ and the transfer function are related by the Laplace transform. Both functions completely the characterize the relation between input and output of an LTI system.

Given input and output, it is usually much simpler to determine the transfer function, and from it the impulse response, than computing the impulse response directly in the time domain. Determining the impulse response in the time domain would mean to solve $(2)$ for $h(t)$, given $x(t)$ and $y(t)$. This is obviously much harder than to compute $H(s)$ from $(1)$ (assuming that the Laplace transforms $X(s)$ and $Y(s)$ are easy to compute).

And, answering your last question as to why we can't find the transfer function by dividing in the time domain: because that's not what LTI systems do; their inputs and outputs are not related via multiplication in the time domain, but by convolution. Here's where the Laplace transform comes into play, it converts convolution into multiplication.

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  • $\begingroup$ Yeah but WHY is the Laplace domain so important? $\endgroup$
    – Miss Mulan
    Sep 8, 2022 at 15:24
  • $\begingroup$ I started writing and went from the time domain: what if the input is a sine and, when the input is zero you measure the output to be different than zero? Would you say that the t.f. has infinite gain? What if the output is zero while the input is not? Would that mean there is nothing in the black box? Then your answer came. :-) $\endgroup$ Sep 8, 2022 at 15:25
  • $\begingroup$ @MissMulan: Among other things, because it converts convolution to multiplication. And it converts differential equations into algebraic equation, which are much easier to solve. $\endgroup$
    – Matt L.
    Sep 8, 2022 at 15:25
  • $\begingroup$ @aconcernedcitizen but what does the Laplace transform physically mean then? $\endgroup$
    – Miss Mulan
    Sep 8, 2022 at 15:29
  • $\begingroup$ @MattL. but why does it convert convolution to multiplication and differential equations to algebraic equations? $\endgroup$
    – Miss Mulan
    Sep 8, 2022 at 15:29
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(I was going to leave @Matt L.'s answer but, given the line of comments, I'll try, too)

Let's say you have a 1st order lowpass prototype and you feed it a sine:

$$\begin{align} &H(s)=\dfrac{1}{s+1} \tag{1} \\\\ &h(t)=u(t)\mathrm{e}^{-t} \tag{2} \\\\ &x(t)=u(t)\sin(t) \tag{3} \end{align}$$

Applying the convolution integral you get the output to be (omitting the $u(t)$ for a clearer view):

$$y(t)=\dfrac{\mathrm{e}^{-t}+\sin(t)-\cos(t)}{2}=\dfrac{\mathrm{e}^{-t}}{2}+\cfrac{\sin\left(t-\dfrac{\pi}{4}\right)}{\sqrt2} \tag{4}$$

After the transient has decayed enough that its value doesn't influence the output enough ($5t$ or more), every time the input is zero (at every $n\pi/2$) the output will be:

$$y\left(n\dfrac{\pi}{2}\right)=\cfrac{\sin\left(n\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)}{\sqrt2}$$

As long as the input stays as $\sin(t)$, that makes the output have alternating values $\pm 1/2$ for each input's zero crossing. What could you infer from this: that the output has infinite gain, that it creates something out of nothing? What will the transfer function be?

Conversely, every time the input is $\pm 1/\sqrt2$, the output will be zero. Would you say that the transfer function is zero?

This is why you can't rely on the time domain response in order to get the transfer function: all the results you'll get will vary from a fraction of a second to another.

And if you use a different frequency then you will get different results, each time, for each frequency. It is because of this that the Laplace transform is used (as long as it's an LTI, see @Matt L.'s answer): not only it converts from time domain into frequency domain thus, allowing us to determine the exact response of the system for each input, but it does so while managing to transform the difficult differential equations that govern the time domain into simple algebraic, linear equations, much easier to solve.

Note that this Laplace transform is nothing but a mathematical abstraction, it does not exist in the real world. It's a very useful tool to have but, in practice, even frequency response analyzers are oblivious to the existence of Laplace -- they do it by injecting a small signal with a swept frequency, then measuring the system for each of these frequencies. Then they collate the results and you get to see the frequency response. In practice. In theory, that's where you have the luxury of Laplace and we're ever so grateful for it.

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    $\begingroup$ Good answer +1. Just one remark: it might make things even clearer if you added the step functions to Eqs (1) and (2). For me, $\sin(t)$ extends from $-\infty$ to $\infty$, in which case there wouldn't be any transient. Also, people sometimes don't realize that $\sin(t)$ (without a step function) doesn't have a Laplace transform, but it does have a Fourier transform. $\endgroup$
    – Matt L.
    Sep 8, 2022 at 17:05
  • $\begingroup$ @MattL. Thank you for returning the favour. >:-) You're right about the $u(t)$, so I've added it. At the time I was thinking about the mathematical example, not the Laplace\<-\>time application. $\endgroup$ Sep 8, 2022 at 20:16

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