PID and convolution

Question

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)

Answer 1

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.