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Unformatted text preview: W n t o t he various i nputs m ust also b e t he eigenvalues of m atrix
A . O n t he o ther h and, t he s ystem is also specified by Eq. (13.68d). T his m eans
t hat t he poles o f t he t ransfer functions must b e t he eigenvalues of A. T herefore,
t he eigenvalues o f m atrix A r emain unchanged for t he l inear t ransformation of
variables represented by Eq. (13.67), a nd t he eigenvalues o f m atrix A a nd m atrix
A(A = P AP-I) a re identical, implying t hat t he c haracteristic equations of A a nd
A are also identical. T his r esult also c an b e proved a lternately as follows.
Consider t he m atrix P (sl - A)p-I. We have P (sl - A)P- I = P slP- 1 - PAP-I = s PIP- 1 - A= sl - A 814 13 S tate-Space Analysis 13.4 Linear Transformation O f S tate Vector 815 Taking t he d eterminants o f b oth sides, we o btain (s - ad 0 0 o I sI- AI = T he d eterminants IFI a nd I F-II a re reciprocals of each other. Hence This is t he desired result. We have shown t hat t he c haracteristic equations of A
a nd A are identical. Hence t he eigenvalues o f A a nd A a re identical.
In Example 13.9, m atrix A is given as =0 o (13.71) l si - AI = l si - AI o (s - a2) o 0 o o or
Hence, t he eigenvalues of A are aI, a2, . .. , a n. T he nonzero (diagonal) elements
of a diagonal m atrix a re therefore its eigenvalues AI, A2, . .. , An. We shall denote
t he diagonal m atrix by a special symbol, A: T he c haracteristic equation is l si - AI = I: o
A= o [-2 0]
3 -1 [S + 2 0] =
-3 s s2 +1 + 3s + 2 = 0 Diagonalization o f Matrix A 0 We shall assume t hat AI, A2, . .. , An, the eigenvalues of A , a re distinct (no repeated
roots). Let us transform t he s tate vector x into t he new s tate vector z, using t he
Z =Px a2 (13.73a) 0 (13.73b)
We desire t he t ransformation t o be such t hat P A P-l is a diagonal matrix A given
by Eq. (13.72), or
(13.74a) A P=PA 0 z = Az +:EU
A = P AP-I Hence (13.74b) or
0 An Then, after t he development of Eq. (13.68c), we have For several reasons, i t is desirable to make m atrix A diagonal. I f A is n ot
diagonal, we can transform t he s tate variables such t hat t he r esulting m atrix A is
diagonaLt O ne c an show t hat for any diagonal m atrix A , t he d iagonal elements
o f t his m atrix m ust necessarily be AI, A2, . . , , An ( the eigenvalues) of the matrix.
Consider t he diagonal m atrix A:
al 0 x =Ax+Bf T his r esult verifies t hat t he c haracteristic equations of A a nd A a re identical. 13.4-1 0 L et us now consider t he t ransformation of t he s tate vector A such t hat t he resulting
m atrix A is a diagonal matrix A.
Consider t he system a nd
l si _ AI = (13.72) I - 1 = s2 + 3s + 2 = 0
s +3 0 0 A=
. ............. We know A a nd A . Equation (13.74b) therefore can be solved t o determine P . o • 0 0 an T he c haracteristic equation is given by
t In t his discussion we a ssume d istinct eigenvalues. I f t he eigenvalues a re n ot d istinct, we c an
r educe t he m atrix t o a modified diagonalized (Jordan) form. E xample 1 3.10
F ind t he diagonalized form o f t he s tate e q...
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