better then netflix :D

@Eigensteve

4 жыл бұрын

Nice!

I had to watch it a few times to finally get the joke. Yay! Thank you professor!

Thank you so much for your wonderful explanation, professor!

Thank you very much for your great and intuitive explanations!

Cheap vodka, I'm dying over here... XD prof. letting us know he had a totally normal undergrad experience.

Thank you, Professor!

@Eigensteve

4 жыл бұрын

Glad you like it!

Oh, thank you sir!

Thank you so much for these videos, they are very useful. I have been asked to show that controllability and observability are not affected by replacing A with (A+αI ). How can be proved such statement? replacing A with A+αI would change its eigenvalues, so the PBH test can't be exploited. And I can't think on how to prove that the rank of the Controllability Matrix will not change either. Do you have any suggestion?

Hi Steve. If you maximize the projection of B along the least stable directions of the state space, does that result in optimal control (assuming all eigenvalues of A have multiplicity 1 for now)? Or is it possible to set this up as a form of control optimization? Edit: By optimization here I think I mean minimizing control input energy, but I'm not sure if it's necessary to specify what type of optimization I mean. It seems like it could be generally true.

Thank you. Amazing work and incredibly helpful. Cheers!! Great Idea to record using a mirror, I think. The wedding ring gave it away else you'd be some kind of savant dyslexic.

Great video, but not going to lie, that marker was killing me lol!

A question about point 2: The purpose of adding B is to supplement the left null space of (A-\lambda I) so that the final matrix is full-rank. Then I don't quite understand the saying of point 2 that "B needs to have some components in each eigenvector direction" (Eigenvectors lie in the null space of (A-\lambda I), is there any relation between left null space and null space)?

Csn you suggest some linear algebra tutorial or book specifically for controls problem? Just to map linear algebra concepts to control problems.

@Eigensteve

4 жыл бұрын

Just added a link to our data science book in the comments.

A not-so-straightforward topic tackled really well by you! Thumbs up! The importance of repeated eigenvalues, a concept I wasn't taught in my grad studies. I just wanted to clarify one little thing. At 12:04, you said that two eigenvectors that're really really close, what does it signify in the physical system? Mathematically I understood PBH test but I was just thinking of a simplest system where A = Identity matrix, B=[1;1], then with a single control input variable u, I can influence system in both of the eigenvectors direction set by A. The two equations are essentially the same and decoupled with some input u available as well. Intuitively, I think it should be controllable but it isn't according to PBH criteria because the matrix of PBH test is rank-deficient. I could not fit the takeaways of this test here intuitively when my matrix A has repeated eigenvalues and B has components in all the eigenvector directions set by A, then essentially, all the eigenvector directions could be reached by my control input u or not? What am I missing here? Could you please explain a little? Thanks!

@Eigensteve

4 жыл бұрын

Thanks! Also, great question. When I talk about two eigenvectors being very close, this happens a lot in fluid systems, and it can cause them to go unstable. This is called "non-normality", and I will be making a video on this soon. Your example of A=identity and B=[1;1] is a great illustration of what can go wrong when you have a repeated eigenvalue in A. You are right that B can essentially "access" both states through the A matrix, but because the A matrix is symmetric (both eigenvalues are 1), you can't independently change the first and second state. Whatever you do to the first state you are also doing to the second state, and vice versa. So you can't move the state to any arbitrary value, say x=[1;5], because the two states have exactly the same dynamics. So controllability is not the same as "being able to stabilize" the system. It really states that you can move the state to any arbitrary value with control.

@adeeljamal6846

4 жыл бұрын

Will be waiting for that video. Hmm.... this makes a lot of sense. Many thanks for clarifying! I have already taken advanced control classes in my grad studies but the mathematical kinda concepts are now making more sense after taking your lectures. Grateful to you!

Nice video, Professor Steve. Explanation is clear and easy to understand. You mentioned that you do not use PBH in practice though. Does that imply that you use something else? If so, what?

@Eigensteve

4 жыл бұрын

In practice I look at the controllability matrix and the controllability Gramian. They are usually easier to work with.

@DobbinsMath

4 жыл бұрын

​@@Eigensteve Hi Steve. Great Video! To follow up on this question, how do we address stabilizability and detectability in practice? I have seen 2 methods based on the 2 forms of PBH but if Gramian is preferable for controllability, is there a way to use the gramian as a stabilizability check in code?

Can't we say that if the determinant is 0 then the equation dimension is reduced .That means eigen values are reducing the dimension of equation(A- lamda*I) corresponding to their eigen vector directions ?

Amazing Video! But just one question. Shouldn't we look at left eigenvectors of A?

@zacharymcnulty3490

Жыл бұрын

Seems like it yeah. Because you’re interested in when Col(A- lambda I) is not all of Rn, in which case we need the columns of B to reach everything in the perpendicular complement. The perpendicular complement is exactly Null((A- lambda I)^T) = Null(A^T - lambda I). So it’s really the left eigenvectors we are interested in. Part 3 (advanced) also doesn’t seem clear to me. Suppose for example A is the identity. Then for lambda = 1 we just have A-lambda I is the zero matrix so there’s no way a single vector can make up the missing rank. If the eigenvalues are distinct so all the eigen spaces are one dimensional then this would seem more likely

Hi Steve , in the 1st point of test , we need to test only for A-(lambda)I if lambda is not its eigenvalue , but if in this case B is pointing in a direction different than A-(lambda)I, rank may be greater than n , we didn't account for that case . Please correct me if I am wrong.

@ALDead666

4 жыл бұрын

@kshitij bithel If lambda is not an eigenvalue, then concatenating any B with A - lambda I will not change the rank of [A - lambda*I B]. This is because the columns of A - lambda I already form a full basis of the n-dimensional space. Hence, it is impossible to find a B in a direction that is not already accounted for by the column space of A - lambda * I, and thus the rank of [A - lambda*I B] cannot be greater than n.

@wizardOfRobots

4 жыл бұрын

like A P says....in short - the greatest possible rank of a matrix is the number of rows(size of the column)

good lecture! thank steve! I'm taking modern control , but we have learned pbh test. I have a question. When applying pbh test, Is the B of (A-(lambda)I B) laplace transformed B? for example, If the B has time varying component, can original B be substituted to (A-(lambda)I B) ? or should we transform B to laplace transformed B? sorry for my bad english.

@hariprasadhparthasarathy6360

8 ай бұрын

this B is coming from the statespace description.

Really great videos Can someone tell me about Fact 2: How does a component in an eigenvector direction look like, what is meant by component? If I take for example a b that's just a vector it just points in one direction, right?

@micknamens8659

3 жыл бұрын

My understanding is that B should not be in the direction (i.e. should not be a multiple of) of any of the eigenvectors to get full rank of (A-lambda*I B)

Regarding the PBH test, I have a bit of confusion over repeated eigenvalues. Lets say A matrix is an identity matrix (order 2). Hence the eigenvalues are repeated. The eigenspace is whole of R^2. I believe if I have an actuation matrix B with only one column (lets say [1,3]^T), I can still control the system. In that case, the algebraic multiplicty of the repeated eigenvalues are not the same as the no. of columns of B. Can you please clarify this case in comparison to your statement at 11.51? Thanks so much.

@prabinrath7105

3 жыл бұрын

In that case, C becomes [1 3; 1 3] which has rank 1 and hence not controllable

At 3:45, Are we assuming that the rank of A is n??

@animeshsinghal3405

2 жыл бұрын

I realised that this scenario is taken care of when eigen_value=0

Hi Steve it is me here again. My state matrix (A) is 10x10 and my control matrix (B) is 10x6. So when I compute the controllability matrix it has rank 10 . So I belive that my system is controllable, but when I do a numeric simulation in MatLab I got two state (position and velocity) that I can not control. That means, my system is stable but not controllable ? . if I try to increase the number of columns in matrix B It could help my in someway ? Thanks for attention !

@Eigensteve

4 жыл бұрын

It sounds like maybe your system is technically controllable, but not very controllable in practice. I discuss this a bit on my lecture on the controllability Gramian. So you can look at the eigenvalues of the controllability matrix to get a feeling for how controllable a given state is. Adding more columns to B should help.

@devonho5399

4 жыл бұрын

@@Eigensteve Good english there

will remember the 'cheap vodka' test lmao

Point 3 is false if A has eigenvalues with multiplicity larger than one! Example: A = 0 in IR^(2x2).