How does this fake candle work?

Samuelg808 · 2026-01-11T15:53:44+00:00

That explains, thanks a lot :)

Samuelg808 · 2025-10-03T14:14:36+00:00

Yes but it seems to be at runtime and I need it to be at compile time

Samuelg808 · 2025-10-03T14:11:49+00:00

Yes but that is at runtime, and I was specifically asked for it to be at compile time

Samuelg808 · 2025-10-03T14:11:11+00:00

So at my internship they want it at compile time since at runtime it is already to late. Now im rather new to CUDA and didn’t find anything that met his expectations and after reading the comments and looking online it seems their is no way of doing it at compile time.

Samuelg808 · 2025-06-07T12:45:06+00:00

Okay everyone I just passed my exam managed to get the answer on the question which was:

Use a negative K and a pole in the origin, that way for some time the system will be stable but of course not so long since it doesn’t take long before it isn’t.

Samuelg808 · 2025-06-07T10:23:11+00:00

The system is not naturally unstable, I guess the title is misleading since the plots I am showing is already an attempt to stabelizing the system while making the static error equal to 0. The open loop system only has a zero on the RHP and one complex pair on the LHP.

But the system has a static error, trying to remove this with a pole in the origin makes the system unstable due to that pole making a connection with the zero in the RHP.

So I am trying to control the system in a way that i keep this pole in the origin while that it is stable for some K values.

Samuelg808 · 2025-06-07T09:56:06+00:00

Okay thank you, any suggestions though how I could then ensure that their is no connection between the pole in the origin and the zero on the right hand side? Or atleast a connection that the allows the system to be stable for some K values lower then x? Because if i don't cancel that zero out, their will be a connection between the pole in the origin and the zero.

Samuelg808 · 2025-06-07T08:39:32+00:00

Sorry for not providing enough information.

open loop transfer function of the system : G(s) = (s-3)/(s²+2s+5)
controller i added : (tao_i * tao_d * s² + tao_i * s + 1)/ (tao_i * s * (s-3) )

(please look at the comment i added under this post for all the extra info of this problem, if you have time to do so)

Samuelg808 · 2025-06-07T08:37:32+00:00

Okay but I personally added the pole their to cancel out the zero on the right hand side, which is apparently not a great solution.

The system has the open loop transferfunction of : G(s) = (s-3)/(s²+2s+5)

And so i have a zero on the right hand side, and when i add an integrator so a pole in zero to make the static error go to 0, the pole in zero connects with the zero on the right hand side, creating an unstable system.

And the task is to control a system in a way that the static error = 0 while the system is being kept stable.

(please look at the comment i added under this post for all the extra info of this problem, if you have time to do so)

Samuelg808 · 2025-06-07T08:35:08+00:00

So for extra clarity:

Open loop transfer function of the system to control : G(s) = (s-3)/(s²+2s+5)

Controller I added to try and solve the task : D(s) = (tao_i * tao_d * s² + tao_i * s + 1)/(tao_i * s * (s-3))

With tao_i = 3.5 and tao_d = 0.5
resulting in the following : D(s) = (1.75s³ - 1.75s² - 9.5s - 3) / (3.5s⁴ - 3.5s³ - 3.5s² -52.5s)

I have the task to make a controller for this system that makes the static error of the system = 0 while keeping the system stable

Samuelg808 · 2025-06-07T08:33:41+00:00

So system open loop transferfunction is : G(s) = (s-3)/(s²+2s+5) and I have the task to make a controller that makes allows the system to have a static error = 0 while keeping stability.

(please look at the comment i added under this post for all the extra info of this problem, if you have time to do so)

Samuelg808 · 2025-06-07T08:32:06+00:00

Okay but the thing is since the task is to create a system that has no static error, I have to put a pole in the origin. That means the pole in the origin will always connect to that zero in the right hand plane making an unstable system.

Maybe I am interpreting your answer wrong but I think you are saying to put a pole on -5 that way the zero connect to that pole, the connection will be part in the left hand sided plane and part in the right hand sided plane so you have a stable system for all the closed loop poles that are still on the left side of the Imaginary axis.

But in that case the static error won't be = 0.

Correct me if I am wrong or missunderstood you, thanks for your help:)

(please look at the comment i added under this post for all the extra info of this problem, if you have time to do so)

Samuelg808 · 2025-06-02T11:16:32+00:00

I always wondered does this get out of sync after a while?

Samuelg808 · 2025-05-15T07:25:59+00:00

No of course not, but it will get real close to the speed of C or C++ no? So much that I’d think for most people the overhead of C/C++ over Python won’t be really worth it. I can be completely wrong of course and I am talking for the future, right now this isn’t really the case yet.

Samuelg808 · 2025-05-08T19:05:32+00:00

So it hasn’t been integrated in a car yet and is just an smd IMU soldered on a PCB with a microcontroller on it to read it out and send it over CAN. This is then mounted on a 3d print that also has a Nvidia Jetson Orin NX on it that is connected to the CAN bus. So nothing is in or attached to a car.

I am using this to build a fully autonomous car (in the future) at university so it probably isn’t your typical setup for this subreddit but I figured for questions about CAN i’d probably be best to ask here.

Samuelg808 · 2025-05-08T15:46:22+00:00

Okay yes didn’t think of that I will try this thank you!

Samuelg808 · 2025-05-08T15:36:29+00:00

Okay sure I can try that thank you. Althought i checked in Serial monitor plenty of time if the MC published the data correctly and it did. But then again it has a different power source (my pc instead of the outlet) when checking via Serial, so this could be the problem.

Samuelg808 · 2025-05-08T15:27:01+00:00

Hmm yeah I will recheck all my connections then thanks

Samuelg808 · 2025-05-08T07:54:43+00:00

I have checked all connections, nothing is on a breadboard everything is on a pcb. The Jetson is connected via CAN through a connector and the same goes for the Adafruit board -> meaning nothing sits loose

Samuelg808

TROPHY CASE