Signal Processing and Linear Systems-B.P.Lathi copy

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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= Also 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 t ransformation 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 (13.73c) (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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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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