state transition matrix derivation?

In one of my modern control's class, the prof said we should be able to prove (as a possible midterm/final problem):

Phi = M L M^-1

Where Phi = state transition matrix

M = modal matrix

L = diagonal matrix of Eigen values

M^-1= inverse of M

I really have no idea where to start and am hoping someone here can help. midterms are coming up soon and I want to be prepared.

Thanks,

Bo

Reply to
Bo
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Is this not in your book? It's in mine from nearly 20 years ago: "Linear Systems" by Kailath. Look in your table of contents and your index for "Modal Decomposition". If you don't find it, see if your library has Kailath's book.

It may not be a bad idea to bring it up with your prof -- sensible teachers know that if there's one person asking a question, there's probably ten more who are too shy to speak up.

Reply to
Tim Wescott

I searched for modal matrix and the following link popped up near the top of list.

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what do you do with modal matrices after you pass the test?

Peter Nachtwey

Reply to
pnachtwey

You really should have covered this already in a linear algebra class. The chapter will be called eigensystems or some such. The idea is below.

xdot=A*x Ax=xdot A*X=s*X A*X=lambda*X A*v1=v1*lambda1 A*v2=v2*lambda2 A*v3=v3*lambda3 A*[v1 v2 v3]=[v1 v2 v3] *diag([lambda1,lambda2,lambda3])

dave y.

Reply to
dave y.

What is this used for besides passing a test? Some how I missed all of this when I was in college. I thought you might know since you seem to be mathematically inclined.. I remember you helped me show the IAE tuning coefficient were in need of correction.

I would be an annoying student now. I would always be asking why bother? Because it is on a test is not enough. The answer may help Bo too.

Peter Nachtwey

Reply to
Peter Nachtwey

Give a man a fish, feed him for a day. Teach a man to fish, and he'll ask if fish roe is on the exam.

Reply to
Peter K.

I had been thinking that if I ever get my seminar business off the ground I would invite you to a local one just to hear you heckle.

Now I'm not so sure I wouldn't tie you up and leave you in a corner half way through. :-).

The most important part of the modal decomposition for me was the insight in how one _could_ if one wanted to, separate the system such that there was one integrator per mode. This isn't something that's practically useful, but much like knowing that I-84 will take you farther than Boise is useful even if that's only as far as you ever go, it's nice to have that modal decomposition model in the back of your head when you do work in state space.

I also strongly suspect that if you ever get it piled higher and deeper you'll find some branch of mathematics where the modal decomposition will come in extremely handy. Fer instance, as I was writing the last sentence I realized that you could probably use modal decomposition to do a numerically robust "partial fraction" decomposition on a system expressed in state-space using modal decomposition, avoiding the inevitable problems with finding the roots of large polynomials in the process.

Reply to
Tim Wescott

Linear systems covers a huge swath of the field of engineering--many disciplines such as control systems, filter design, mechanical vibrations, etc. The question implicitly dealt with how one studies the inner works of such systems, by decomposing them into their essential characteristics (i.e. the eigenvectors/eigenvalues of the system). For example, in the field of vibrations, the eigenvectors represent the mode shapes of the vibration and the eigenvalues represent the resonant frequencies of those modes. To study these fields of engineering you need a language to describe how systems work, and that language is linear algebra.

dave y.

Reply to
dave y.

If you do get it off the ground, let me know when you are in the Los Angeles area.

Michael

Reply to
Herman Family

I'll surely be putting a shameless plug in here.

Reply to
Tim Wescott

That cracked me up...

I eventually figured it out-- by using eigen functions-- but I have not had linear algebra--so I am at a slight disadvantage....

Thanks,

Bo

Reply to
Bo

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