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时间:2010-06-01 00:54来源:蓝天飞行翻译 作者:admin
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        [<P(t)lk = Q(kt)
(5.191)
(5.192)
(5.193)
(5.194)
Characteristic equation.   Given a square matrix A, the equation
                                    A(1) = lAl - Al - 0                            (5.195)
is called the characteristic equation of matrix A. Here,  I-I denotes the deternunant
of the argument (square) matrix.
    Eigenva/ues and eigenvectors.   The roots ofcharacteristic Eq. (5.195) are
called the eigenvalues of the matrix A and are usually denoted by Ar, / : 1, . . . , n,
where n is the number of rows or columns of the matrix A. As an example,let
so that
A=[2 ;]
AI-A = [A_41  A,-ll]
(5.196)
(5.197)
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW       511
A(A) : IAI - Al = A2 - 2A - 3 = 0
(5.198)
Solving, we get Ai - 3 and A2 - -1.
     An important property of the eigenvalues is that they remain invariant under any
linear tra~sformation. A direct consequence of this property is that the closed-loop
characteristic equation also remains invariant under any linear transformation.
To make this p9int clear, suppose we are given a linear, time-invariant system
x - Ax + Bu and we transform this system using a linear transformation x - Pz
so that the transformed system is z = P-lAPz + P-IBu. Then, the eigenvalues
of matrix A and those of P-lAP are identical. Interested readers may verify this
statement by working out the details.
     If A, is an eigenvalue of the square matrix A, then any vector x that satisfies the
equation
Ax - A/x
(5.199)
is called the eigenvector corresponding to the eigenvalue A.,, In other words, every
eigenvalue will have an associated eigenvector. Returning to the above example,
                               [A   1] [f:] = c-i) [f:]
or
                                                 2xi +X2 = O
                                                        4xi +2x2 -. O
(5.200)
(5.201)
(5.202)
Note that the two equations are identical, stating that the eigenvector is not unique
and depends on our choice ofone ofthe two variables. Let X2 -.  1 SO that xi  = -1l2.
The eigenvector corresponding to A = -1 is [-1l2 1]T. Here, the superscript "T"
denotes the matrix transpose. It can be shown that the eigenvector obtained by
choosing any other value for X2 would be a scalar multiple of this eigenvector.
Similarly, the eigenvector corresponding to A2 = 3 is [1 2) r .
    To understand the physical meaning of eigenvalues and eigenvectors, consider
a system with two first-order, coupled linear differential equations
 x] - XI +x2
X2 -. 4xi +X2
(5.203)
(5.204)
Assume that the solution is given by xi(t) = uieAt and X2(t) = u2eA: Substitut.ing,
we get
AuleAr = uieAr + u2eAt
                                               Au2eAr = 4uieAr + u2eA'
Because eAt > O for all t > O, we can write
                                       Aui - ui +L12
                                         Au2 - 4ui +U2
(5.205)
(5.206)
 
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