Wife: "Hey Honey, have any plans for this weekend?" Husband: "I'm going to stress test my accelerometer and optical sensors with chemical input to see how much it takes for my vertical control systems fail." W: "So, you're going to get so drunk you fall over?" H: "Precisely! For science!"
Thank you for posting this amazing content. You are the best maths teacher on the internet, bar none. It would be amazing if one day you did a playlist on stochastic calculus and calculus of variation!
Hi Steve, when you linearize a system and use this linear model in real life and everything works fine, does this phisically mean that those terms you eliminated of the Taylor expansion (from the square term on) only account (at least locally) for microscop effects and this is why we can neglect them?
yes part of the Taylor expansion are the terms 1/1!*dt + 1/2!*dt^2 + 1/3!*dt^3. As most technical control systems work at 50Hz, 100Hz or faster dt is small already and dt^2 and dt^3 run down the decimal places really quick. 1 over factorial makes it even marginally smaller. It`s only when dt is really big that you might need more terms. But if it works well it just means it covers the dynamics very well. The inverted pendulum is relatively easy in simulation but might fail because of unmodeled dynamics like wind resistance, flex in the rod, friction in the bearing, and a small error in the angle measurement can throw it off too.
@@MrHaggyy those magic numbers you cite for controller update rates can be nuanced. The key is that your update rate is at least twice as fast as the response time of the closed system, and then faster than that by an amount that makes all of your unmodeled error sources small enough that they don't drive the system unstable. This is why having some intuition (at least) regarding your noise sources is important: running the controller faster can amplify errors, depending on their nature. Faster is often, but not always, better.
Not a control engineering nor an expert perspective, but my take is: linearization is always done around fixed points; the assumption is that the dynamics close to those points are practically linear. As long as your system operates in that zone, you can pretty much forget about the extra terms! Is not that they dont matter, is that in your particular area of operation they are not that relevant. However, this breaks down for systems with strong non-linearities (broadband turbulence, for instance) so watch out for those
@@jmchdjaimerporkpuedolol3681 yes that's true. The Taylor expansion tries to fit any dynamics with polynomials. A linearization would be Taylor from f(x, a) = f(a) + f'(a)(x-a), with a being your fixed point. If you want to model linear dynamics job is done great. But if the system you want to model is non-linear you get an ever-increasing error: so the real f(x, a) - Taylor from f(x, a) is no longer approximately zero. At this point you have two options: Either you approximate your system from multiple points and keep the linearization and superposition, which is great for the control algorithm choice. Or you add the quadratic, cubic ... etc terms to get a better model and deal with a nastier control algorithm. Basically, all systems I have dealt with evolve in time that's why I took dt in the first comment. In T f(x,a) f(a) + f'(a)(x-a) + f''(a)/2!*(x-a)^2 + f'''(a)/3!*(x-a)^3 if (x-a) is small (x-a)^2 is tiny and (x-a)^3 is basically zero. If you choose a bigger area (x-a) and so a bigger scope of dynamics you want to approximate the error gets bigger. And if (x-a) > 1 and f''(a), f'''(a) ... is not close to zero, meaning your system is not really linear, you get in trouble as the quadratic, cubic ... etc terms work against you.
Hello dear teacher, thank you for this fantastic series, I want to ask you if we don't know the value of theta and omega just the expression of thau, let f*cos (wt), how can we find the solution of the problem, I encountered this kind of problem in the Duffing oscillator?
thank you so much profesor. I have just confused 'tau' in the integral and torque symbol. can every one help me please? is there any relation between them?
The control bootcamp (which is awesome) is at ruclips.net/p/PLMrJAkhIeNNR20Mz-VpzgfQs5zrYi085m
Thanks!
Wife: "Hey Honey, have any plans for this weekend?"
Husband: "I'm going to stress test my accelerometer and optical sensors with chemical input to see how much it takes for my vertical control systems fail."
W: "So, you're going to get so drunk you fall over?"
H: "Precisely! For science!"
The best 16 minutes of the day! Thank you!
Thank you for the compact description.
Thank you for posting this amazing content. You are the best maths teacher on the internet, bar none. It would be amazing if one day you did a playlist on stochastic calculus and calculus of variation!
You are very welcome, and thanks for the suggestion -- I'm already thinking about these!
What a force of nature.
amazing, new achievement unlocked
9:31
Per your request.
Link needed to control boot camp.
Awesome, thanks for the reminder and the timestamp!!
Dear sir, please make a video on Boussinesq approximation.
Would love to see that too
Don't forget to add the links ☺
Thanks for the reminder!
thank you professor
Thank you for the video !
👍👍👍👍👍👍👍👍
Hi Steve, when you linearize a system and use this linear model in real life and everything works fine, does this phisically mean that those terms you eliminated of
the Taylor expansion (from the square term on) only account (at least locally) for microscop effects and this is why we can neglect them?
yes part of the Taylor expansion are the terms 1/1!*dt + 1/2!*dt^2 + 1/3!*dt^3. As most technical control systems work at 50Hz, 100Hz or faster dt is small already and dt^2 and dt^3 run down the decimal places really quick. 1 over factorial makes it even marginally smaller. It`s only when dt is really big that you might need more terms.
But if it works well it just means it covers the dynamics very well. The inverted pendulum is relatively easy in simulation but might fail because of unmodeled dynamics like wind resistance, flex in the rod, friction in the bearing, and a small error in the angle measurement can throw it off too.
@@MrHaggyy those magic numbers you cite for controller update rates can be nuanced. The key is that your update rate is at least twice as fast as the response time of the closed system, and then faster than that by an amount that makes all of your unmodeled error sources small enough that they don't drive the system unstable.
This is why having some intuition (at least) regarding your noise sources is important: running the controller faster can amplify errors, depending on their nature. Faster is often, but not always, better.
Not a control engineering nor an expert perspective, but my take is: linearization is always done around fixed points; the assumption is that the dynamics close to those points are practically linear. As long as your system operates in that zone, you can pretty much forget about the extra terms! Is not that they dont matter, is that in your particular area of operation they are not that relevant.
However, this breaks down for systems with strong non-linearities (broadband turbulence, for instance) so watch out for those
@@jmchdjaimerporkpuedolol3681 yes that's true. The Taylor expansion tries to fit any dynamics with polynomials. A linearization would be Taylor from f(x, a) = f(a) + f'(a)(x-a), with a being your fixed point. If you want to model linear dynamics job is done great. But if the system you want to model is non-linear you get an ever-increasing error:
so the real f(x, a) - Taylor from f(x, a) is no longer approximately zero.
At this point you have two options:
Either you approximate your system from multiple points and keep the linearization and superposition, which is great for the control algorithm choice.
Or you add the quadratic, cubic ... etc terms to get a better model and deal with a nastier control algorithm.
Basically, all systems I have dealt with evolve in time that's why I took dt in the first comment. In T f(x,a) f(a) + f'(a)(x-a) + f''(a)/2!*(x-a)^2 + f'''(a)/3!*(x-a)^3 if (x-a) is small (x-a)^2 is tiny and (x-a)^3 is basically zero. If you choose a bigger area (x-a) and so a bigger scope of dynamics you want to approximate the error gets bigger.
And if (x-a) > 1 and f''(a), f'''(a) ... is not close to zero, meaning your system is not really linear, you get in trouble as the quadratic, cubic ... etc terms work against you.
Fantastic video, even by your high standards! Now, about that link? 😁
Thanks for the reminder!
Hello dear teacher, thank you for this fantastic series, I want to ask you if we don't know the value of theta and omega just the expression of thau, let f*cos (wt), how can we find the solution of the problem, I encountered this kind of problem in the Duffing oscillator?
thank you so much profesor.
I have just confused 'tau' in the integral and torque symbol. can every one help me please? is there any relation between them?
I must miss thing. Should A matrix be [[0 1],[-1,0]]