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FAQ (Frequently Asked Questions)
Category:
CDS 101/110 Fall 2004
Identifiers: H0 H1 H2 H3 H4 H5 H6 H7 H8 L0.0 L1.1 L1.2 L2.1 L2.2 L2.3 L3.1 L3.2 L4.1 L4.2 L5.1 L5.2 L6.1 L7.1 L9.1
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In lecture you said we want to place the eigenvalues of (A+BK), but MATLAB's "place" places eigenvalues of A-BK. What's the deal?
Submitted by: waydo
Submitted on: October 27, 2004
Identifier:
L5.1
, L5.2
Whether our closed loop dynamics matrix is A-BK or A+BK just depends on the sign convention we use when applying the feedback. Sometimes you will see the signal inverted as it is fed back (so u = -Kx), in which case the closed loop dynamics are given by A-BK, and sometimes it will not be inverted (so u = Kx), in which case we use A+BK. The two formulations are exactly equivalent, and the sign convention just determines the sign of the elements of K.
In the book and in MATLAB we use the A-BK sign convention, so we will be more careful to do this in lecture in the future.
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How come you can always find u(t) such that the integrals come out to be what is needed to drive the state where you want it to go if the reachability condition holds?
Submitted by: waydo
Submitted on: October 27, 2004
Identifier:
L5.2
The answer to this question is somewhat involved and beyond the scope of this course. For one take on it, check out Dullerud and Paganini, "A Course in Robust Control Theory: A Convex Approach" (Chapter 2).
If we consider the discrete-time case, however, the intuition is much easier. Let
x(k+1) = A x(k) + B u(k)
We can drive x to zero (or to any arbitrary other value) in n steps or less (where as usual n is the dimension of the system) if the reachability condition
rank( [B AB A^2B ... A^(n-1)B] ) = n
holds, just like in the continuous case. The difference is that it in general takes a finite number of steps to do this, rather than an arbitrarily short period of time as in the continuous case.
To see this, look at x as k increases:
x(k+2) = A x(k+1) + Bu(k+1)
= A^2 x(k) + ABu(k) + Bu(k+1)
...
...
x(k+n) = A^n x(k) + A^(n-1)Bu(k) + ... + ABu(k+n-2) + Bu(k+n-1)
Now all the u's can be chosen independently, so if the columns of
[B AB A^2B ... A^(n-1)B]
are linearly independent (i.e. the rank of this matrix is n) we can pick the u's to "cancel out" any A^n x(k), so we can drive the state to zero.
It isn't too hard to find a counterexample that shows that, depending on A, B, and x, we may not be able to drive the system to zero in less than n steps.
Also, the Cayley-Hamilton theorem tells us that A^n, A^(n+1), etc are linearly dependent on A, A^2, ..., A^(n-1), so we can't do any better (i.e. we can't increase the rank of the reachability matrix if it is deficient) by considering driving the system to zero in more than n steps. Thus, if the system is reachable, we can certainly drive the state anywhere we want it to go in n steps or less, and if it is not, we can pick an initial condition and final condition such that we can't drive the initial condition to the final condition no matter how many steps we get to take.
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What is a "pole"?
Submitted by: asa
Submitted on: October 27, 2004
Identifier:
L5.2
A "pole" is the location of a singularity of a complex function. More generally, we talk about finding the "poles" and "zeros" of a function. If we imagine the function as the ratio of two functions, the poles are the zeros of the denominator, and the zeros are the zeros of the numerator. We will cover poles and zeros in much more depth when we talk about transfer functions in the next chapter.
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Prof. Mabuchi mentioned that if the observability matrix is full rank, we may still have to wait some time to determine a derivative. Is this true?
Submitted by: waydo
Submitted on: October 29, 2004
Identifier:
L5.2
The question continues: If you know the dynamics, and they are linear, surely you can solve for the derivative given only one datum and initial conditions. For example, if you tell me initial conditions and position x, then immediately I can tell you x-dot.
There are a couple of issues here. First is that the observability matrix is telling us if, given measurements, we can back out what the initial condition was. The idea is that we don't know exactly where we are, only what our observations tell us, which may not give us the entire state. For example, we may measure our position but not our rate, so we have to use our measurements of position over time to determine the rate.
The second issue is one of noise. In an ideal world, the rank condition tells us that we can back out the state of the system from measurements not instantaneously, but arbitrarily fast. However, in the real world our measurements are noisy, and if we make our estimator converge very quickly our estimate will amplify the noise present in the measurements. The noisier the system, the longer it takes us to get a good idea of what the state is based on measurements.
In CDS110b we discuss how to strike an optimal balance between converging our estimator quickly and filtering out noisy data, a process known as "Kalman filtering."
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