2 ms·
The Laplace Transform is an easy(ish) way to transform differential equations into complex numbers. The response of a capacitor or inductor is a differential e
by TheOtherHobbes 2y ago
The Laplace Transform is an easy(ish) way to transform differential equations into complex numbers.
The response of a capacitor or inductor is a differential equation with an exponential solution. The Laplace transform replaces the exponentials with s = iω, where ω just means "a sine wave at some frequency".
(It's actually jω in EE, because reasons, but it means the same thing.)
So if you build a network of capacitors, resistors, and inductors, or springs and masses, or anything else that uses the same maths, you can turn it into a simple algebraic equation in s.
This is called the transfer function. It literally defines the frequency and phase response with some simple algebra.
In practice you don't bother with the calculus and just go straight to s. A single capacitor filter is 1/(1+sRC). Add more stages - the rules are pretty simple - and you get a more complex equation.
Then you can put ω = 200Hz or whatever into the transfer function, and you get a complex number out. This tells how big the output is at that frequency, and how much it's phase shifted.
If you graph ω, you get the complete frequency and phase response.
Really all you're doing is replacing a static DC test voltage with a sine wave "probe" across the frequencies you're interested in.
There's a bit more to it - which is where poles and zeroes appear - but not much.
(It's also related to the Fourier transform, but that's a whole other thing.)
It's incredibly impressive that all of this was invented in the early 19th century by a handful of French aristocrats and a couple of Germans, more or less as a hobby.
- aeronerd91 2y ago[dead]
- HeyLaughingBoy 2y ago> There's a bit more to it - which is where poles and zeroes appear - but not much. As someone who has bad memories of spending hours calculating Root Locus plots by hand, I'm going to have to take offense at that statement :-)
- eitland 2y agoI think you misunderstand. I love the Laplace transform. I just don't like how school decide to use three lecture hours to derive it and only afterwards explain what it is for and that this was only a demo of how one could arrive at the same result. We also derived the relativistic equations but then we got to know up front what was going on.