# Question I didn't get on my exam on controllability of a system...

Discussion in 'Homework Help' started by blazedaces, Dec 9, 2008.

1. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
So there was a question on my final I just took that had a bunch of parts to it. I could do all but the last part. Here is basically what the question said without giving actual numbers:

You have a system described by the following state space:

dx(t)/dt = Ax(t)+Br(t)
y(t) = Cx(t) + Dr(t)

If it helps at all, D was equal to 0. The rest were given in my problem of course, but they are not relevant to answering my question.

It then asks the following question (and I'll highlight the last one, the one I didn't know how to do)

a) Is the system BIBO stable?
b) Is the system completely observable?
c) Is the system completely controllable?
d)
based on part c above: i) If the system is NOT CC, what needs to be done to matrices B and/or C to make it so
ii) If the system IS CC, what gain values for the K = [k1 k2] using the control law r(t) = K * x(t) would yield a double eigenvalue of -2?

Now, I know what eigenvalues are, and I know how to do everything up to that point, and I would have known how to do part i) if the system was NOT CC, but it was. So what in the world does it mean to say that these gain values would "yield" a double eigenvalue of -2?

In this case, just so you're made aware, the transfer function I solved for was not a matrix, it was something like

So basically, while on one hand I'm saying I don't know how to do the problem, my bigger question is what does the question mean? If I knew abstractly what it meant, I could have easily come up with an equation and solved for k1 and k2.

Thanks guys,

-blazed

2. ### steveb Senior Member

Jul 3, 2008
2,432
469
I can look at this in more detail when I get home later, but some initial comments. The eigenvalues are basically the poles I think. The given system has eigenvalues (two poles) at -4 and 1, so one of these is in the RHP and indicates an unstable system. If you use full control on the system (i.e. two control loops, one on each state, with gains K1 and K2) you can move those poles. With a particular choice of K1 and K2 you can move them both to -2 which would result in a stable system.

3. ### steveb Senior Member

Jul 3, 2008
2,432
469
OK, so I just got home from work and looked at this more carefully. I think I addressed your main question above in my previous answer, so I won't add anymore unless you need more clarification.

I do agree that the wording of the question is a little difficult, but I think my interpretation is valid.

One further comment. You probably already know this, but in case others might be interested; the benefit of adding feedback to move both poles to the same value is that it creates a critically damped system which is often desirable since it creates the fastest settling time with no overshoot. The choice of the value (-2) just sets the bandwidth response (i.e. the speed) of the system.

4. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
Thank you very much good sir. So I found this, which even has an example EXACTLY like the one above: http://en.wikipedia.org/wiki/Full_state_feedback

I for some reason or other forgot that the characteristic equation of the system was also given by the determinant of the (SI - A) matrix. I really hope that the method I used, which was multiplying by B, and then multiplying C by the result corresponds to the correct transfer function. In my head, it should, since that technically IS the transfer function, AND because B and C are just constants and thus in a non-vector form would correspond to only changing the constant in the numerator.

I think I just did extra, non-necessary work, but luckily I had the time.

-blazed