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In that formula :

$$ u(t)=K_{\mathrm{P}} e(t)+K_{\mathrm{I}} \int_{0}^{t} e(\tau) \mathrm{d} \tau+K_{\mathrm{D}} \frac{\mathrm{d} e(t)}{\mathrm{d} t} $$

I know from the formulae collection we refer to that : $$ \mathcal{L} \left\{ \int_{0}^{t} e(\tau) \mathrm{d} \tau \right\} = \frac{E(s)}{s} $$

But I don't manage to find a proof. I watched a few videos on convolution theorem with two functions, but they are general, I can't find this exact example. I assume one of the functions must be 1 and integrate as $\frac{1}{s}$ of course but I have difficulties with the limits of integration and reconstitute the proof.

Also and maybe more importantly, why is integration helping with the error? (just started PID controllers and would appreciate some insights)

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  • $\begingroup$ I do not know your background in engineering but have you ever taken a course on differential equations ? Do you know the Laplace transform integral? Btw. your equation uses an inverse Laplace transform, you should use just $\mathcal{L}$ instead. $\endgroup$ – Fat32 Feb 17 at 11:09
  • $\begingroup$ No I know the basics and I read about Laplace transforms. I know it's a basis but I struggle to find it in relevant literature. I will edit, thanks! $\endgroup$ – Dovendyr Feb 17 at 11:11
  • $\begingroup$ * I read on my own I meant! I just know the absolut basics $\endgroup$ – Dovendyr Feb 17 at 11:18
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Considering the Laplace transform of a function $x(t)$ as : $$X(s) = \mathcal{L}\{x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-st} dt \tag{0}$$

then to find the following Laplace transform $$\mathcal{L} \{ \int_{-\infty}^{t} e(\tau) d\tau \} \tag{1} $$

you can write the integral as a convolution of $e(t)$ with the unit-step function $u(t)$ as $$\mathcal{L}\{ \int_{-\infty}^{t} e(\tau)d\tau \} = \mathcal{L}\{ e(t) \star u(t) \} \tag{2} $$

Then by using the Convolution Theorem of the Laplace transform (which states that Laplace transform of a convolution is equivalent to multiplication of individual Laplace transforms)

then you have: $$\mathcal{L}\{ \int_{-\infty}^{t} e(\tau)d\tau \} = \mathcal{L}\{ e(t) \star u(t) \} = E(s) U(s) \tag{3} $$

Where, $E(s)$ and $U(s)$ are the Laplace transforms of the functions $e(t)$ and $u(t)$, repectively.

It can be shown that $U(s)$ = $\frac{1}{s}$, hence the result follows: $$\mathcal{L}\{ \int_{-\infty}^{t} e(\tau)d\tau \} = \mathcal{L}\{ e(t) \star u(t) \} = \frac{E(s)}{s} \tag{4} $$

Regarding, why there's an integration part in the PID, one could argue that it provides a means to correct (or control) steady-state errors; i.e., errors that remain as $t \to \infty$. These errors are better captured by their integral.

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