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FAQ (Frequently Asked Questions)
Category:
CDS 101/110 Fall 2003
Identifiers: FN H0 H1 H2 H3 H4 H5 H6 H7 H8 L0.0 L1.1 L1.2 L10.1 L2.1 L2.2 L3.1 L3.2 L4.1 L4.2 L5.1 L5.2 L6.2 L8.1 L9.1 L9.2
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Why are the dimensions of A "n x n"?
Submitted by: waydo
Submitted on: October 1, 2003
Identifier:
L1.2
The dimension of the state vector is n, and so the dimension of its derivative xdot is n as well. Because we multiply A by a size n vector, x, to get another size n vector, xdot, A must be an n x n square matrix.
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How do we express control laws?
Submitted by: waydo
Submitted on: October 1, 2003
Identifier:
L1.2
, H1
...Is it an equation or more of a statement? What should we write on homework 1?
For example, for an elevator would we write
Ydot = k(Ydesired - Ycurrent)
or “If the elevator is below the desired floor go up, else if it is above go down?”
ANSWER:
Broadly speaking, a “control law” is anything that describes how to take control action, so either description could be a control law. In real life control laws take both forms depending on the system at hand. Note that the two alternatives expressed in the question really mean different things - the first says to move the elevator at a rate proportional to the error in location (proportional control), while the second says to simply move in the correct direction (assuming that the elevator moves up or down at a constant rate) (on-off control). Ignoring startup and slowdown transients, I would say that most elevators are better described by the second statement.
Something like a cruise control in a car, however, is probably better described by the first statement - no one would want to ride in a car that controlled its speed by using full throttle when going too slow and none when going too fast; a throttle level proportional to the error in speed would be more appropriate.
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Is the size of the Jordan block Ji equal to the algebraic, or geometric multiplicity of (lambda)i?
Submitted by: lars
Submitted on: October 1, 2003
Identifier:
L1.2
Note: This answer has changed since the first (incorrect) posting.
Actually, the answer is neither.
The essential fact is that each of the Jordan blocks corresponds to a linearly independent eigenvector of the matrix A (taking the Jordan form J = T-1AT). A given eigenvalue can have multiple linearly independent eigenvectors, therefore in this case it will have multiple Jordan blocks.
If the algebraic multiplicity (am)i equals the geometric multiplicity (gm)i for a given eigenvalue (lambda)i, then there is a linearly independent eigenvector for each of the (am)i repeated eigenvalues, and there are (am)i blocks, each of size 1.
In the case where (gm)i < (am)i for a given eigenvalue (lambda)i, the difference will be compensated by an increase in the size of the Jordan blocks.
For a given eigenvalue, there will be (gm)i Jordan blocks, and the number of "1"s on the superdiagonal will equal (am)i-(gm)i.
For some MATLAB fun regarding this topic, try investigating the Jordan form of the 3x3 identity matrix:
eig(eye(3))
[V,J]=jordan(eye(3))
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Can you explain the "(An+pn-1An-1+...+p0I)x=0, for all x" part again?
Submitted by: lars
Submitted on: October 1, 2003
Identifier:
L1.2
Sure. Professor Murray was trying to show that the quantity in parentheses above equals the zero matrix (i.e. the matrix with zeros in all entries) --- let's call this quantity X.
One way to show that X=0 (the matrix) is to show that if you multiply X by any vector, you get the zero vector.
Since you can write any vector in a linear space as a linear combination of the basis vectors (the eigenvectors, in this example), Prof. Murray checked the above condition by making sure that X v = 0 for all of the eigenvectors.
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You wrote rank(A) + nullity(A) = n for an nxm matrix. Is this right?
Submitted by: demetri
Submitted on: October 1, 2003
Identifier:
L1.2
This statement is actually not true except for a square matrix. In general, the result is:
rank(A) + nullity(A') = n
where the apostrophe indicates transposition. For a square matrix, the nullity of the transpose equals the nullity of the original matrix, and the equation from class holds.
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In the lecture, a one-to-one relation was called an injection. Is this common? Doesn't a one-to-one relation define a bijection?
Submitted by: mreiser
Submitted on: October 1, 2003
Identifier:
L1.2
The terminology used by Prof. Murray in the lecture is quite standard. But there is some confusion about this...from the excellent Mathworld web site, there is an entry pointing out that term 'A and B are in one-to-one correspondence' is used to mean that 'A and B are bijective.' This is less standard and is NOT the way we use the term in this and any future Controls class. The standard definitions can be found on the Mathworld web site, just keep in mind that these definitions apply to a general mapping, and for our purposes, the mapping we are concerend about are matrices, which map an element (a vector) of one vector space to another. Here are the links for the definitions: injection ,
surjection, and
bijection.
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Do you use "&sub" to mean a proper subset or a general subset, as in: {Au : u &isin &realm } &sub &realn
Submitted by: mreiser
Submitted on: October 1, 2003
Identifier:
L1.2
Recall that a proper subset is any subset that does not contain the entire set. In the context used in class we need this to be a general subset. In the definition of the range of a matrix:
{Au : u ∈ ℜm } ⊂ ℜn, we need to allow for the possibility that the range is in fact the entire vector space (which the matrix A maps vectors onto, this is ℜn in our definition). When this occurs, the map is surjective (onto). We could never have this property if we used proper subsets in the definition of a range.
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What is the normal form for a linear system with complex eigenvalues? Is it the Jordan Form?
Submitted by: demetri
Submitted on: October 1, 2003
Identifier:
L1.2
While normal form theory is not within the scope of this course, we will provide an informal answer to this question.
The answer depends on whether you are willing to accept complex-valued normal forms for real-valued ODEs. In this case, the Jordan form is indeed the normal form.
If one is after a "real" normal form, then there is the so-called "real" Jordan form. For this form, rather than treat a complex-conjugate pair as two independent eigenvalues, and writing a diagonal matrix for these values, we instead write a 2x2 "real Jordan block" for these eigenvalues. This matrix is precisely the prototypical matrix with a complex-conjugate pair of eigenvalues a +/- ib:
[a -b;b a]
Here we have used the MATLAB convention that a semicolon denotes the end of a row. This matrix is easily exponentiated, as follows:
expm([a -b;b a]) = exp(a)*[cos(b) -sin(b);sin(b) cos(b)]
The real part of the eigenvalue gives the exponential growth rate, and the imaginary part gives an oscillation frequency. The utility of the above formula is the primary motivation for using the "real Jordan form".
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Why do we need to learn all of these theorems?
Submitted by: murray
Submitted on: October 2, 2003
Identifier:
L1.2
Some of this will become more clear as the course goes on, but this is actually an important aspect of control. Because feedback systems can behave in counter-intuitive ways, the field has developed very systematic approaches to study the dynamics of feedback systems. The level of formalism in control is sometimes more abstract than it needs to be (and we are trying to change that), but some level of mathematical rigor is required to make sure that the control systems that we design really do behave correctly (according to a formal specification, not just our best guess).
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Where can I find out more about the Jordan canonical form?
Submitted by: murray
Submitted on: October 2, 2003
Identifier:
L1.2
Stephen Boyd at Stanford has written a nice set of lecture notes on this:
http://www.stanford.edu/class/ee263/jcf.pdf
Another good reference is the book by Gilbert Strang on Linear Algebra and Its Applications (International Thomson Publishing, 1988).
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