Timeline for Discretizing a Controller with the Backward Difference Method
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2019 at 21:19 | comment | added | TimWescott | I couldn't give you a super-rigorous mathematical proof of why It holds for every $G(s)$, but loosely speaking it holds because you can take $G(s)$ apart into a differential equation, then use the time-domain approximation for differentiation, then sample, then take the z transform. | |
| Nov 20, 2019 at 21:17 | comment | added | TimWescott | If you treat the sampling process as a multiplication by a train of unit impulses, then the result is that the sampled-time signal equals $\sum_{n=0}^\infty \delta(t - nT) x(nT)$. Take the Laplace transform of that and it equals the z transform with $z = e^{sT}$. This should be in the first or second chapter of any signal processing book that deals with discrete-time signals. | |
| Nov 20, 2019 at 21:15 | comment | added | TimWescott | It's an approximation. Those properties of the Laplace transform and the z transform tell us that we can approximate differentiation by substituting $s = (1 - z^{-1})/T$. | |
| Nov 20, 2019 at 21:03 | comment | added | AleWolf | Why the fact that the laplace transform of the derivative is $sX(s)$ and the zeta trasform of the discrete derivative is $\frac{1}{T}\frac{z-1}{z}X(z)$ tell us that, in order to discretize Laplace we have to substitute $s=\frac{1-z^{-1}}{T}$ ? Moreover, laplace and zeta trasform are different things to me ... And why it holds for every $G(s)$ ? There could be also second and third derivatives... | |
| Nov 20, 2019 at 0:29 | history | answered | TimWescott | CC BY-SA 4.0 |