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

# Since all t he s tates a re decoupled because o f t

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Unformatted text preview: T he u pper l imit o n t he s ummation i n E q. ( 13.93a) is n onnegative. H ence k :::: 1, a nd t he s ummation is recognized as t he c onvolution s um IAI - AI =IA i -1 A-~ 1= A2 - ~A + ~ =(A - ~) (A - ~) =° 6 6 3 2 Hence, Al = l and A2 = ~ are t he eigenvalues of A and [see Eq. (13.95)] H ence + A k -1 u [k - x [k] = A kx[O] ' --v--&quot; 1] * Bf[k] y z ero-input ( 13.93b) , where [see Eq. (B.95b)] z ero-state a nd y[k] f3o] = [ f3I = C x+Df k -1 = C A kx[O] + L C A k -l-jBf{j] + D f [1 1 l] -1 ~ [ Wk] k = W [3 -2] -6 6 [ (3)-k] =[ ( 2)-k 3 (3)-k - 2(2)-k ] _ 6(3)-k (13.94a) j =o k1 = C Akx[O] + C A - u[k - 1] * Bf{k] + D f (13.94b) I n Sec. B.6-5, we showed t hat Ak = [ 3(3)-k - 2(2)-k] [ : = [ 3(3)-k _ 2 (2)-k A k = f30I + (31 A + (32 A 2 + . .. + (3n-1 A n -1 + 6 (2)-k a nd (13.95a) ( 3)-k _ ( 2)-k :] + [ -6(3)-k + 6 (2)-k] [:i + 6 (2)-k] _ 2(3)-k + 3 (2)-k ~] - 6(3)-k (13.97) We can now determine t he s tate vector x[k] from Eq. (13.93b). Since we are interested in t he o utput y[k], we shall use Eq. (13.94b) directly. Note t hat w here ( assuming n d istinct e igenvalues o f A ) C A k =[_15]A k =[2(3)-k_3(2)-k (30 - 4(3)-k + 9 (2)-k] (13.98) a nd t he zero-input response is C A kX[O], w ith (31 (13.95b) ( 3n-1 n An x[O] 1 a nd A1, A2, . .. , An a re t he n e igenvalues of A . We c an a lso d etermine A k f rom t he z -transform f ormula, w hich will b e d erived l ater i n E q. (13.102): (13.95c) = [ :] Hence, t he zero-input response is (13.99a) T he zero-state component is given by t he convolution sum of C A k -Iu[k - 1] a nd Bf(k]. Using t he shifting property of t he convolution sum [Eq. (9.46)], we can obtain t he zerostate component by finding t he convolution sum of C A kU[k] a nd Bf(k] and t hen replacing 13 82.8 S tate-Space A nalysis 13.6 o S tate-Space A nalysis o f D iscrete- Time S ystems 829 C omputer E xample C I3.S Using MATLAB find t he zero-state response of t he system in Example 13.12. A =[O 1 ;-1/6 5 /6]; B =[O; 1]; C =[-1 5]; D =O; [ num,den]=ss2tf{A,B,C,D); k =0:25; u =ones{ 1 : length{k) ); y =filter{ n um,den, u ); s tem(k,y) 0 J [k] 1 3.6-2 F ig. 1 3.12 System for Example 13.12. The Z - Transform Solution T he z -transform o f E q. (13.91) is given b y k w ith k - 1 in t he result. We use this procedure because the convolution sums are listed in Table 9.1 for functions of t he type f [k]u[k] r ather t han f [k]u[k - 1]. zX[z] - zx[O] = A X[z] + BF[z] ( zl - A)X[zJ) = zx[O] + BF[z] T herefore C A kU[k] * B f[k] = [ 2(3)-k _ 3 (2)-k - 4(3)-k + 9 (2)-k] * [ 0] u[k] a nd = _ 4(3)-k * u[k] + 9(2)-k * u[k] X[z] * B f[k] = - 4 [1 3 -Ck+1l] -1 _ ~ u[k] +9 [1 2- +] x[k] = Z -l[(1 - z-l A)-l]x[O] + Z -l[(zl - A)-lBF[zJJ , 1] = [12 + 6 (3)-k - 18(2)-k]u[k - 1] v &quot; , v z ero-input c omponent Now t he desired (zero-state) response is o btained by replacing k by k - 1. Hence * B f[k - ( 13.101a) u[k] = [12 + 6 (3- Ck +1l ) - 18(2- Ck +1l)]u[k] C A kU[k] (I - H ence Ck 1l -1 _ ~ ( zl - A)-lzx[O] = U sing Table 9.1 (Pair 2a), we...
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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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