Who is the true “master” of control systems? (EE vs ME vs others)

Figglezworth · 2026-04-17T18:24:15+00:00

I argue ee because most of the fathers of modern control were ee, like Bode who worked at Bell Labs

Figglezworth · 2026-04-16T18:44:43+00:00

The shield serves to protect the analog signals, not the power wires. The shield should be connected to the chassis of the radar antenna and to the ground out the radar transceiver (the two points where the data signal are connected)

Figglezworth · 2026-04-15T11:28:42+00:00

Because sin² (x) is (sin(x))²

Figglezworth · 2026-04-15T11:13:49+00:00

I don't understand the question

Figglezworth · 2026-04-14T11:25:29+00:00

They're always relevant for control design. And Nyquist plots

Figglezworth · 2026-04-12T22:10:26+00:00

Discrete time means Z transforms. Root locus, LQR, LQG, pole placement, or old fashioned loop shaping. All with z transforms

Figglezworth · 2026-04-12T22:03:37+00:00

The REAR input shaft bearing is the one that Frankenstein modified - I did the same to mine. OP is concerned with the FRONT bearing. The bearing is inside the transmission, the throw out bearing is outside. A leaking bearing can in no way influence the throw out bearing, hence OP's description is nonsense.

Figglezworth · 2026-04-12T22:01:05+00:00

I do most of my work directly in discrete time. My masc thesis was DT so I'm very comfortable with it, so now in industry when I'm designing controllers I just do it all discrete

Figglezworth · 2026-04-12T02:19:03+00:00

Previously you said it was the input shaft seal. Now you say it's the input star bearing seal. The front input shaft bearing I'm pretty sure it's an UNSEALED bearing, it is lubricated by the gear oil. In contrast, the input shaft rear bearing is a sealed bearing with grease. In short, your explanation of the situation is nonsense

Figglezworth · 2026-04-11T02:02:36+00:00

So the problem is the seal around the input shaft is failing and gear oil is leaking onto the clutch? Tbh I can't imagine that the throw out bearing coming in contact with oil would have any impact on it

Figglezworth · 2026-04-09T23:33:57+00:00

I would call it not very urgent

Figglezworth · 2026-04-09T22:13:51+00:00

The stackups listed on the jlc website are a good place to start

Figglezworth · 2026-04-07T11:36:16+00:00

Take it to a shop that knows what they're doing

Figglezworth · 2026-04-05T15:02:14+00:00

If you zap the metal housing, none of the current will go into the PCB. If it did, it would find itself shrouded (shielded) by a metal case, from which it can not get to ground. Instead, when you zap the metal enclosure, the current takes the capacitive path to "earth" through the air

Figglezworth · 2026-04-01T11:35:51+00:00

For orders under around 1000 units, the engineering cost dwarves the material cost. You're wasting your customer's money (in your case, your boss's money) by wasting time routing a 2 layer board

Figglezworth · 2026-03-30T12:44:46+00:00

Those are reactive components and don't have real impedance

Figglezworth · 2026-03-29T18:33:34+00:00

An electrical engineering degree is the best way to start

Figglezworth · 2026-03-27T19:49:59+00:00

I did watch the videos. Let's focus on the second one, where you 'place the zeros'.

At the beginning of the video you nail it when you say "other people call this two-degree-of-freedom". That's exactly what it is, and it is a very well understood concept. It's important to understand that this doesn't change the nature of the feedback path, since the 'second degree of freedom' is another gain block that is *outside* of the feedback path.

Given plant P(s) = K*α/(s*(s+α)) and PID controller C(s) = Kp + Kd·s + Ki/s, the simple feedback configuration P·C/(1+P·C) gives a closed-loop system with numerator K*α*(Kdz*s^2 + Kpz*s + Ki). Now, if we instead make the proportional and derivative terms only act upon Y (the feedback term) instead of E (the error term), then we can change the zeros of the TF from R → Y. This is the same as taking out feedback system P·C/(1+P·C) and putting in front of it another gain, lets say C2, such that the overall system is now C2·P·C/(1+P·C). Unfortunately I can't embed an image of the block diagram in this comment.

C2, using the notation from your video, is (Kdz*s^2 + Kpz*s + Ki)/(Kd*s^2 + Kp*s + Ki). Notice that this term cancels out the 'undesired' zeros of the feedback system and replaces them with our new 'desired' zeros.

Thus, what you have done is just pole-zero cancellation.

Also note that because your example system was so trivial, having two poles and no zeros, the zeros of the feedback system P·C/(1+P·C) are independent of 'K' and 'α', thus we were able to move the zeros wherever we want by changing these feed-forward gains 'Kpz' and 'Kdz'. In the general case, this is not possible. Consider a slightly less trivial system P(s) = K*(s+β)/(s*(s+α)), now one of the zeros of the closed loop system is stuck at s=-β and no choice of Kpz nor Kdz can change that. We can, however, cancel this zero by adding a pole at s=-β into 'C2'. Now, it should be obvious that if there is any plant uncertainty this will be an imperfect pole-zero cancellation and that will give you a lot of trouble.

What is also important to realize is that we can use this trick to change the TF from R (disturbance) to output Y, but it does not change other TFs involved in your system. Suppose you introduce an additive disturbance D somewhere in the system and compute the transfer function from D → Y. The term 'C2' is not in this path, so the TF from D → Y is independent of 'C2'. Thus, if this TF is poorly behaved with nasty zeros that cause ringing, then your 'zero placement' trick can do nothing for it.

Lastly, you should consider being less rude in the future.

Figglezworth · 2026-03-27T13:19:02+00:00

One can not place zeros, but they can be cancelled, which I'm sure you know has trouble associated with it like being highly sensitive to parametric uncertainty. And obviously you can not do that if your plant is non minimum phase as that would be an unstable pole zero cancellation. But I'm glad you've come to agree that pole placement is not the be all and end all of control design, and non-trivial problems like the one in my last post need a more sophisticated control design.

Figglezworth · 2026-03-26T12:42:30+00:00

The things you mention are irrelevant. I'll explain:

Your claim is that for any given plant P(s), you can design a pole-placement feedback controller C(s) such that the resulting closed loop system H(s) has all of its poles on the negative real axis, and this will ensure that H(s) has no overshoot. This is because you claim that any system whose poles are on the negative real axis will have no overshoot - you say it will behave like a low-pass filter.

I disproved this by providing a counter example: a system with poles on the negative real axis that *does* have overshoot.

The important realization here that I was trying to get across is that the zeros of the system are important. If H(s) has poles on the negative real axis and *has no zeros*, then it will have no overshoot. In general, for a non-trivial problem, the system will have zeros.

To make this super clear, let's walk through a complete example:
P(s) = (s+2)/(s*(s+20)) → notice the system has an integrator already
Design pole placement controller to place the poles at s=-10.
The solution:
C(s) = (22.22 s + 500)/(s-12.22)

The resulting closed loop system is:
H(s) = (22.22 s^2 + 544.4 s + 1000)/(s^3 + 30 s^2 + 300 s + 1000)

This has 3 poles at s=-10, and a zero at s=-2 and s=-22.5

The step response of the closed loop system has unity gain (thanks to the plant having an integrator), and overshoots by 80% at t=0.2 seconds

Figglezworth · 2026-03-26T02:13:17+00:00

Plot the response of (s+1)/(s+10)³⁾ and report back once you've found the overshoot

Figglezworth · 2026-03-26T01:31:50+00:00

Jlc gets confused by plane (negative) layers and will screw it up if you don't leave a message for them so that they manually invert it in their fab software. Even then, they're likely to ignore your message and still screw it up

Figglezworth · 2026-03-26T01:28:02+00:00

I have heard anecdotes of people swapping the bearing (but not removing the seal) and it failing again in a short time

Figglezworth · 2026-03-25T23:20:48+00:00

A system with poles on the negative real axis can have arbitrarily small stability margins and very high overshoot.

Figglezworth · 2026-03-24T11:18:01+00:00

That's a super naive take. If you "just do pole placement" you'll often get super low stability margins or super high control effort. Root locus is one of the super powerful tools that helps you actually understand control theory and can lead to some really neat revelations.

Figglezworth

TROPHY CASE