A friend of mine's daughter waited till the last minute to ask for help from me for her graduate ODE class. Happens. But I haven't taken ODE for decades and when I did, well, let's say the results were less than impressive. My friend's daughter's professor seemed, from the murky lecture notes I got, to not be a great communicator, so I needed to get a quick warm-up on this topic. Thank goodness I ran into your clear-as-a-bell lecture series. College professors, beware! There is some serious online talent who will teach your kids if you're too lazy to write coherent notes and give coherent lectures. In short, bravo! You were a life-saver!!!
I just wanna call this professor and let him know he is awesome. I got what exactly I needed. Learn something and see how it can be applied in real world. You are just amazing.
I joined my master's course after 7 years being away from college. and your videos are my saviour for my control systems class! Thank you so much for such brilliant inductive way of introducing and explaining concepts!
Thank you Sir....I have seen the whole playlist and it cleared a lot of my concepts about control theory. Your videos are just great and your way of teaching complex things in simple manner is appreciable. Thanks Again.
i dont mean to be so off topic but does someone know a trick to get back into an instagram account? I was dumb forgot my account password. I would appreciate any tips you can offer me!
@Jayce Marcelo Thanks for your reply. I got to the site through google and I'm in the hacking process now. Seems to take quite some time so I will get back to you later with my results.
Love this series. Are we able to linearize around non-fixed points? For many real world mechanical problems, we're not trying to control to an equilibrium position, so I guess this must already be a solved problem.
Awesome, glad you like it! Yes, we can definitely linearize about non-fixed points. The vector field will flow through this point though, so it is not an equilibrium, and there will be a "constant" vector field plus the Ax term
Does this imply that an undampened pendulum cannot be linearized around the fixed points? The eigenvalues of the Jacobian would be +/- i, with no real part.
You have some great instructional video for dynamical systems, stability, and control (linear and nonlinear). This series of videos would be very helpful to new graduate level control engineering students, and I’d recommend they watch them all.
Hi, I want to say that your videos are just AMAZING!!! no words to say, this series also amazing like your other series. but why you have stopped posting more such videos on this bootcamp playlist? I am eagerly waiting for more videos on this bootcamp, apecially MPC controller and controlling of Non-linear systems. Please keep this playlist running and post more such contents please....
Hello Steve. Thanks for the lecture. I cannot help but think that we cheated a bit because we have some previous knowledge of the physics and we know about those two fixed points. But I’m wondering about the sort of systems where we “honest to goodness” don’t know if there are fixed points and where do they could be although we expect them to exist because we have this intuition it can become stable even if we can tell nothing else. Thanks for supplement the lecture with Matlab, it makes it somewhat easier to grasp from an engineering perspective.
Really good question. You are right, usually we start with some knowledge of the system. When we truly have no knowledge, we can sometimes use system identification and then analyze the resulting system. We have a method (sparse identification of nonlinear dynamics; SINDy) that will give equations of motion from data, and then these can be analyzed. But it is extra tricky when the system is naturally unstable. (we have a video on the SINDy method too)
I have never been the brightest student, so even I like and find fascinating control it was alwasy dificult for me. Thank you this was really easy to understand.
just finished about 16 lectures from ur ME564 course and wanted to start getting into the control playlist .. i gotta say having examples with numbers after evey part of the lecture + the matlab examples was a luxury :D
Thank you so much for the series. It really gives me a chance to pass the test with only one day studying. The cool part is, during the semester I wasn't interested at all in this subject (hence the one day study) but seeing your videos, and understanding these concepts so easily, now I find this area really fascinating, and will learn more about it. So thanks again for the solid explainers and for showing me how interesting this subject really is.
In the first part, around 12 min mark, when deriving the linearization around x^bar, where do you use that x^bar has to be a fixed point? It seems like it allowed you to cancel f(x^bar)=0, but other than that, this same process could be applied at any x, it seems. Thanks, great video!
Is there a reason why we linearize around fixed points, apart for the sake of simplicity ? Couldn't we, almost as easily, use an affine approximation around any point, even if its not an equilibrium, and then design a controller that stabilizes the system around that point ?
Ok so I actually worked it out on a piece of paper, by linearizing around theta = pi/3 in a damped pendulum. As it turns out, the dynamics is stable. Crazy, right ? Well it is stable but the equilibrium point is not theta/3, which would be what we want. We linearize around points of equilibrium because if we are not around a point of equilibrium the approximation does not hold in the long run, and is thus not interesting to study stability. The general workflow is to first shift the equilibrium (in that case by adding a constant term to the control -f(x bar)), then stabilize around this new equilibrium. edit : see www.cds.caltech.edu/~murray/courses/cds101/fa02/faq/02-10-09_linearization.html
Great series, I'm learning a lot. Not sure if this comment is getting lost, may be posting multiple times, so apologies up front. On the dynamics of the pendulum, you have the friction/dissipation term as $-\delta \dot{\theta}$. Since friction works to oppose the acceleration regardless of which way it is swinging, shouldn't the term use the absolute value of theta-dot?: $-\delta \lvert \dot{\theta} vert$? I am new to this, pulling on old physics education, so I may be misunderstanding something. Or perhaps it doesn't affect the main thrust of the lesson, and doesn't materially affect the Jacobian at the fixed point? Thanks if you are able to address this question.
This is awesome, but I'm in an application where I need to accurately control a pendulum throughout its entire range of motion, not just around the fixed points. Is there a solution for linearizing the dynamics around an arbitrary point?
Why does a hyperbolic point have neighboring stable and unstable manifolds, but not a center manifold? Don't you need an "inflection point" between them, which would have eigenvalues of zero and be a center manifold?
Thanks for this amazing series! Question: if the pendulum had no dampening, could we still linearize the system around the fixed points? The eigenvalues of the Jacobian would be +/- i, which the Hartman-Grobman theorem (IIUC) says would make it unsuitable for simple linearization.
I am using this method for the cart / inverted pendulum problem in video 12. I can linearize the A matrix with the Jacobian by setting u = 0, but then what do I do with the B matrix which also has nonlinear terms in it? Can I also take the Jacobian of the B matrix and evaluate that around the stability points? Or does even solving the A matrix require knowing what u(t) is prior, since the variables vary with u? Link to a text file with my code: drive.google.com/file/d/11aVEOvqi4T7-qTYThAk-zAFcJJEKwMWa/view?usp=sharing EDIT: I should clarify, my code is actually for a rod with uniform mass on a cart, so not exactly the same as the inverted pendulum.
I found this video here, which goes through how to linearize a non-linear system dx = Ax +Bu, answers my question (beginning around 9:30) m.ruclips.net/video/Bxo0_A5Dp3w/видео.html
Coming back to this video, I have a question. Usually, the theoretical justification for representing a NL system close to an equilibrium with its linearized version is the Hartman-Grobman theorem, which states that for non-hyperbolic equilibria the phase portraits of the NL and linearized systems are topologically equivalent. However, for an undamped pendulum the Jacobian matrix at the stable equilibrium point has two purely imaginary eigenvalues, which actually are hyperbolic (they have zero real part). So how come the linearization does not fail? What piece of information am I missing here? Thanks to anyone who is willing to answer!
Great video. I just want to get something straight. Maybe someone can help me out with it. I want to relate the eigenvalues (EVs) with the eigenvectors (EVecs). I **UNDERSTAND** what the EVs tell us. I **THINK** an EVec represents the angle of the pendulum and the rate of change of that angle. In the stable fixed point case when we have complex EVs, I can see what they tell me about the EVecs. Specifically, EVs tell me that the EVecs will oscillate with some damping. More specifically, the angle and the rate of change of angle will oscillate but the oscillations will die out. The correponding EVecs are complex as well which I am finding it hard to put a meaning to. I though the first coordinate meant angle and the second coordinate meant the rate of change of that angle. So does this mean that if a perturbation makes the pendulum move in a direction which is exactly the same as an EVec, only then the pendulum will behave the way EVs tell us? If it is true, what does this mean when the second coordinate of an EVec is negative or complex?
Man oh Man, amazing content! But if I may give one humble opinion, it would be a lot better if you could just present your lectures the usual way rs, Just a good old, run of the mill black board without the fancy refletive of shooting it. But that might be just me, keep up the great work though.
15:30 sir your cleaning seems to have an adaptive control with awesome actuators, no marks left :). apart from joke this video and this whole lecture are unmatched.
The A matrix for down position of the inverted pendulum without friction has eigenvalues +-i which are purely imaginary. Is it correct to point out that this violates the Hartman-Grobman condition and hence linearization might not work in this case?
Yes! In general, we can linearize about any point, but linearizing about a fixed point gives the advantage that the first term in the Taylor series i.e. f(x-bar) = 0 and so we do not get a constant in the linear equation.
He goes over how to calculate the fixed point in this video. Here f(x) is the following vector ( x2, -sin(x1) - delta*x2)^T Since f(x) = 0, x2 = 0 and sin(x1) = 0 sin(x1 ) is 0 when x1 = 0, n*pi which are fixed points.
As I expected, the guy forgot about 1/2 in the third term of the Taylor expansion (see 9:33). Of course, higher orders are later negclected, but still it does not look fine.
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This is an "honest-to-goodness" fantastic series. Thanks !
I want to like this comment but the like count is too nice. Also, is anyone else reminded of Mr. Mackey mkay?
A friend of mine's daughter waited till the last minute to ask for help from me for her graduate ODE class. Happens. But I haven't taken ODE for decades and when I did, well, let's say the results were less than impressive.
My friend's daughter's professor seemed, from the murky lecture notes I got, to not be a great communicator, so I needed to get a quick warm-up on this topic. Thank goodness I ran into your clear-as-a-bell lecture series. College professors, beware! There is some serious online talent who will teach your kids if you're too lazy to write coherent notes and give coherent lectures.
In short, bravo! You were a life-saver!!!
I just wanna call this professor and let him know he is awesome. I got what exactly I needed. Learn something and see how it can be applied in real world. You are just amazing.
Bro you are honestly the best. You make everything easier to understand.
That is really nice of you!
Thanks prof. for sharing this piece of art with public
I joined my master's course after 7 years being away from college. and your videos are my saviour for my control systems class! Thank you so much for such brilliant inductive way of introducing and explaining concepts!
Thank you Sir, this is another life-saving Series. Now I feel much better and more clear about what I'm dealing with in Modern Control
You're welcome!
This is so good. Tanks a lot for this lectures.
I did not know I needed this series
Thank you Sir....I have seen the whole playlist and it cleared a lot of my concepts about control theory. Your videos are just great and your way of teaching complex things in simple manner is appreciable. Thanks Again.
Thank you for posting this excellent series!
Glad you enjoy it!
@@Eigensteve if i could have lecturers like You :0. Thanks for video, it was very helpful :]
This is an absolutely fantastic video series. Thanks for making these!
i dont mean to be so off topic but does someone know a trick to get back into an instagram account?
I was dumb forgot my account password. I would appreciate any tips you can offer me!
@Messiah Justin Instablaster ;)
@Jayce Marcelo Thanks for your reply. I got to the site through google and I'm in the hacking process now.
Seems to take quite some time so I will get back to you later with my results.
@Jayce Marcelo it did the trick and I finally got access to my account again. I'm so happy!
Thank you so much, you really help me out!
@Messiah Justin glad I could help :)
Love this series. Are we able to linearize around non-fixed points? For many real world mechanical problems, we're not trying to control to an equilibrium position, so I guess this must already be a solved problem.
Awesome, glad you like it! Yes, we can definitely linearize about non-fixed points. The vector field will flow through this point though, so it is not an equilibrium, and there will be a "constant" vector field plus the Ax term
Thank you Professor! 13:35. Hartman-Grobman theorem
Does this imply that an undampened pendulum cannot be linearized around the fixed points? The eigenvalues of the Jacobian would be +/- i, with no real part.
You have some great instructional video for dynamical systems, stability, and control (linear and nonlinear). This series of videos would be very helpful to new graduate level control engineering students, and I’d recommend they watch them all.
Hi, I want to say that your videos are just AMAZING!!! no words to say, this series also amazing like your other series. but why you have stopped posting more such videos on this bootcamp playlist? I am eagerly waiting for more videos on this bootcamp, apecially MPC controller and controlling of Non-linear systems. Please keep this playlist running and post more such contents please....
Your manner of teaching is very amazing!
Hello Steve. Thanks for the lecture. I cannot help but think that we cheated a bit because we have some previous knowledge of the physics and we know about those two fixed points. But I’m wondering about the sort of systems where we “honest to goodness” don’t know if there are fixed points and where do they could be although we expect them to exist because we have this intuition it can become stable even if we can tell nothing else. Thanks for supplement the lecture with Matlab, it makes it somewhat easier to grasp from an engineering perspective.
Really good question. You are right, usually we start with some knowledge of the system. When we truly have no knowledge, we can sometimes use system identification and then analyze the resulting system. We have a method (sparse identification of nonlinear dynamics; SINDy) that will give equations of motion from data, and then these can be analyzed. But it is extra tricky when the system is naturally unstable. (we have a video on the SINDy method too)
I have never been the brightest student, so even I like and find fascinating control it was alwasy dificult for me. Thank you this was really easy to understand.
just finished about 16 lectures from ur ME564 course and wanted to start getting into the control playlist .. i gotta say having examples with numbers after evey part of the lecture + the matlab examples was a luxury :D
Wonderful .. Highest quality of presentation
3blue1brown has a really good video about Jacobian and how it represents linearization. Definitely worth checking out.
@@vikeshbubbles205 ruclips.net/p/PL2QY0xcsWhz6ghPUixeNQm47xapSKw4SM
@@vikeshbubbles205 ruclips.net/video/wCZ1VEmVjVo/видео.html
@@vikeshbubbles205 ruclips.net/video/bohL918kXQk/видео.html&ab_channel=KhanAcademy
Jacibain is a covector (if $f$ just a function of several vars), means vector-row. So, \partial f_i are in the rows
Multiplying by delta(x) - top tip! Makes life helluva easier
This makes me want to do another masters
Thank you so much for sharing your knowledge. This is truly remarkable and amazing content
Another legendary video. I have an 'honest-to-goodness' question. What if there is no fixed point?! Is it possible to analyze the system?
Thank you so much for the series. It really gives me a chance to pass the test with only one day studying. The cool part is, during the semester I wasn't interested at all in this subject (hence the one day study) but seeing your videos, and understanding these concepts so easily, now I find this area really fascinating, and will learn more about it. So thanks again for the solid explainers and for showing me how interesting this subject really is.
Thanks a lot, I've just started the series and im in love with them. Really good work!
In the first part, around 12 min mark, when deriving the linearization around x^bar, where do you use that x^bar has to be a fixed point? It seems like it allowed you to cancel f(x^bar)=0, but other than that, this same process could be applied at any x, it seems. Thanks, great video!
just a one word 'Awesome'
very nice, can anyone upload some more practice problems where we need to determine System Matrix(A) for given Differential equations ?
Is there a reason why we linearize around fixed points, apart for the sake of simplicity ? Couldn't we, almost as easily, use an affine approximation around any point, even if its not an equilibrium, and then design a controller that stabilizes the system around that point ?
Ok so I actually worked it out on a piece of paper, by linearizing around theta = pi/3 in a damped pendulum. As it turns out, the dynamics is stable. Crazy, right ? Well it is stable but the equilibrium point is not theta/3, which would be what we want. We linearize around points of equilibrium because if we are not around a point of equilibrium the approximation does not hold in the long run, and is thus not interesting to study stability.
The general workflow is to first shift the equilibrium (in that case by adding a constant term to the control -f(x bar)), then stabilize around this new equilibrium.
edit : see www.cds.caltech.edu/~murray/courses/cds101/fa02/faq/02-10-09_linearization.html
Fantastic class except for the shrill noises ...
Excellent explanation. When you are obtaining the fixed points don't you have to consider the control signal/force?
Amazing as usual
Great series, I'm learning a lot. Not sure if this comment is getting lost, may be posting multiple times, so apologies up front.
On the dynamics of the pendulum, you have the friction/dissipation term as $-\delta \dot{\theta}$. Since friction works to oppose the acceleration regardless of which way it is swinging, shouldn't the term use the absolute value of theta-dot?: $-\delta \lvert \dot{\theta}
vert$? I am new to this, pulling on old physics education, so I may be misunderstanding something. Or perhaps it doesn't affect the main thrust of the lesson, and doesn't materially affect the Jacobian at the fixed point? Thanks if you are able to address this question.
Why there isnt a factorial term in the taylor series? Great lecture, btw.
Yeah, he missed it in the third term (2!), but it didn't really matter since we neglected that term anyway.
This is awesome, but I'm in an application where I need to accurately control a pendulum throughout its entire range of motion, not just around the fixed points. Is there a solution for linearizing the dynamics around an arbitrary point?
Seriously, you are awesome!!. Thank you very much for all this information.
Why does a hyperbolic point have neighboring stable and unstable manifolds, but not a center manifold? Don't you need an "inflection point" between them, which would have eigenvalues of zero and be a center manifold?
Thanks for this amazing series! Question: if the pendulum had no dampening, could we still linearize the system around the fixed points? The eigenvalues of the Jacobian would be +/- i, which the Hartman-Grobman theorem (IIUC) says would make it unsuitable for simple linearization.
I am using this method for the cart / inverted pendulum problem in video 12. I can linearize the A matrix with the Jacobian by setting u = 0, but then what do I do with the B matrix which also has nonlinear terms in it? Can I also take the Jacobian of the B matrix and evaluate that around the stability points? Or does even solving the A matrix require knowing what u(t) is prior, since the variables vary with u? Link to a text file with my code: drive.google.com/file/d/11aVEOvqi4T7-qTYThAk-zAFcJJEKwMWa/view?usp=sharing
EDIT: I should clarify, my code is actually for a rod with uniform mass on a cart, so not exactly the same as the inverted pendulum.
I found this video here, which goes through how to linearize a non-linear system dx = Ax +Bu, answers my question (beginning around 9:30)
m.ruclips.net/video/Bxo0_A5Dp3w/видео.html
This is really cool video. Thank a lot.
I have a Question at 29:47 min: What is the stable and the unstable direction? Are those "directions" represented physically somehow?
Fantastic video!! What were to happen if the equilibrium points were 0 for both? Would there be any reason to linearize the problem?
“Honest to goodness” love this term😂
Coming back to this video, I have a question.
Usually, the theoretical justification for representing a NL system close to an equilibrium with its linearized version is the Hartman-Grobman theorem, which states that for non-hyperbolic equilibria the phase portraits of the NL and linearized systems are topologically equivalent. However, for an undamped pendulum the Jacobian matrix at the stable equilibrium point has two purely imaginary eigenvalues, which actually are hyperbolic (they have zero real part).
So how come the linearization does not fail? What piece of information am I missing here?
Thanks to anyone who is willing to answer!
I have many questions. How would I contact you?
Great video. I just want to get something straight. Maybe someone can help me out with it.
I want to relate the eigenvalues (EVs) with the eigenvectors (EVecs).
I **UNDERSTAND** what the EVs tell us.
I **THINK** an EVec represents the angle of the pendulum and the rate of change of that angle.
In the stable fixed point case when we have complex EVs, I can see what they tell me about the EVecs. Specifically, EVs tell me that the EVecs will oscillate with some damping. More specifically, the angle and the rate of change of angle will oscillate but the oscillations will die out.
The correponding EVecs are complex as well which I am finding it hard to put a meaning to. I though the first coordinate meant angle and the second coordinate meant the rate of change of that angle. So does this mean that if a perturbation makes the pendulum move in a direction which is exactly the same as an EVec, only then the pendulum will behave the way EVs tell us? If it is true, what does this mean when the second coordinate of an EVec is negative or complex?
Man oh Man, amazing content! But if I may give one humble opinion, it would be a lot better if you could just present your lectures the usual way rs, Just a good old, run of the mill black board without the fancy refletive of shooting it. But that might be just me, keep up the great work though.
You are amazing! Thanks!
Thank you too!
15:30 sir your cleaning seems to have an adaptive control with awesome actuators, no marks left :).
apart from joke
this video and this whole lecture are unmatched.
I'm confused between this and the extended the kalman filter. In EKF we linearize about the mean and not a stationery point. Why is that any thoughts?
I'm just curious, how did he write oppositly on board?
The A matrix for down position of the inverted pendulum without friction has eigenvalues +-i which are purely imaginary. Is it correct to point out that this violates the Hartman-Grobman condition and hence linearization might not work in this case?
Sir is this book available in paperback format in India..?? Hard cover book is very very expensive & as a student I cannot afford..
Thanks
you are amazing ❤
What if you want to control to a non stable point?
The pleasure is mine:)
Thanks.
Is the approximation at the fixed point is a Taylor expansion?
Yes! In general, we can linearize about any point, but linearizing about a fixed point gives the advantage that the first term in the Taylor series i.e. f(x-bar) = 0 and so we do not get a constant in the linear equation.
Great teacher thank you
👌
where is 2nd part?
Did we already cover how to find the fixed point? It is not clear to me how to solve f(x) = 0 here
He goes over how to calculate the fixed point in this video.
Here f(x) is the following vector
( x2, -sin(x1) - delta*x2)^T
Since f(x) = 0, x2 = 0 and sin(x1) = 0
sin(x1 ) is 0 when x1 = 0, n*pi which are fixed points.
Since you missed the 1/2 in the Taylor series you must mean it when you say you don’t use nonlinear approximations often.
=) loved it.
is he writing in reverse?
@@mouseShyu He is writing on the mirror!
As I expected, the guy forgot about 1/2 in the third term of the Taylor expansion (see 9:33). Of course, higher orders are later negclected, but still it does not look fine.
Ugh still above my head