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Unformatted text preview: o btain C A kU[k] + ( zl  A)lBF[z]
zlA)lX[O] + ( zl  A)lBF[z] = (13.101b) ' z erostate c omponent A c omparison o f E q. ( 13.101b) w ith E q. ( 13.93b) shows t hat
(13.99b) (13.102) I t follows t hat y[k] = [ 8(3)k + 21(2)ku[k] + [12 + 6(3)k  18(2)k]u[k  1] (13.100a) T he o utput e quation is given b y Y[z] y[k] = [12  2(3)k o + 3(2)k]u[k] (13.100b) A =[O 1 ;1/6 5 /6]; B =[O; 1]; C =[1 5]; D =O;
x O=[2;3];
k =0:25;
u =ones(I,26);
[ y,x]=dlsim(A,B,C,D,u,xO);
s tem(k,y)
0 CX[z] C [(I  zl A)lx[O] + ( zl  A)lBF[zJJ + DF[z] = C (I  zl A)lx[O] + [ C(zl  A)lB + D]F[z] •
C omputer E xample C 13.7
Solve Example 13.12 using MATLAB. + DF[z] =
= T his is t he desired answer. We can simplify this answer by observing t hat 12 + 6 (3)k IB(2)k = 0 for k = O. Hence, u[k  1] may b e replaced by u[k] i n Eq. (13.99b), and = ? (I  zl A )lx[O! + (13.103a) H[z]F[z]
' v' z eroinput r esponse z erostate r esponse w here H[z] = C (zl  A  l)B + D (13.103b) N ote t hat H[z] is t he t ransfer f unction m atrix o f t he s ystem, a nd H;j[z], t he i jth
e lement ofH[z], is t he t ransfer f unction r elating t he o utput Yi(k) t o t he i nput I j(k).
I f we define h[k] a s
h[k] = Z l[H[zJJ 13 S tateSpace Analysis 8 30 t hen h[kJ represents t he u nit i mpulse function response m atrix o f t he s ystem. Thus,
hiAkJ, t he i jth element of h (k), r epresents t he z erostate response Yi(k) when t he
i nput f j(k) = h[kJ a nd all o ther i nputs a re zero.
• E xample 1 3.13
Using the ztransform, find the response y[k] for t he system in Example 13.12.
According to Eq. (13.103a) Y [z]=[1 1  2.
6z 6z z (6z5) = ~ [ 1 5]
[ z 6 z 2 5z+1 13z 2  3z
= 5
z2  SZ
1 [2] _~ 5][~ 1
+S+ 1 ] + [1 z~
6 3 1 [ 0] =z 1 6Z2~Z:Z+1l [:] +[1 831 response a nd t he t ransfer f unction. Statevariable description c an also b e e xtended
t o t imevarying p arameter s ystems a nd n onlinear systems. A n e xternal s ystem
d escription m ay n ot d escribe a system completely.
T he s tate e quations o f a s ystem c an b e w ritten d irectly from t he knowledge
of t he s ystem s tructure, from t he s ystem equations, o r from t he block d iagram
r epresentation o f t he s ystem. S tate e quations consist of a s et o f n f irstorder differential equations a nd c an b e solved by timedomain o r f requencydomain (transform)
methods. Because a s et o f s tate variables is n ot unique, we c an h ave a variety of
s tatespace d escriptions of t he s ame s ystem. I t is possible t o t ransform one given
s et o f s tate v ariables into a nother by a linear transformation. Using such a t ransformation, we c an see clearly which of t he s ystem s tates a re controllable a nd which
are observable. References 6 z 2 5z+1 (5z  l)z
5
( z  1)(z 2  SZ + 1. 21z
12z
12z
6z
18z
   + z ! +z 1 +   +~ z !
    z 1 z  z~ Therefore + 2 1(2)k +,12 + 6 (3)k  '...,'
z eroinput r esponse 18(2)k]u[k] z erostate r esponse , • Linear Transformation, Controllability, and Observability
T he p rocedure for linear transformation is p arallel t o t hat i n t he c ontinuoustime c ase (Sec....
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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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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