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Unformatted text preview: tateSpace Analysis o f Discrete Time S ystems o
o 1 o o o o 1 o 825 o o
o
+ x ndk
x n[k + 1J + 1J ~ o o o o  ao  al  a2 x [k+lJ . o 1  a n2  anl A f[kJ xn[kJ 1 ''..''
x[kJ ~ B (13.89a)
a nd F ig. 1 3.11 C ontroller canonical realization o f a n n thorder d iscretetime s ystem. . T he i nput f[kJ a nd t he o utput y[kJ of this system are related by t he difference
e quation (13.89b) C I n general,
( En + a n_IE n 1 + ... + a lE + ao)y[kJ =
(bmEm + b m_IE m  1 + ... + b lE + bo)f[kJ x [k (13.86b) T he controller canonical realization of this equation is illustrated in Fig. 13.11.
S ignals appearing a t t he o utputs of n delay elements are denoted by x dk], x2[k],
. .. , xn[kJ. T he i nput of t he first delay is xn[k + IJ. We c an now write n equations,
o ne a t t he i nput of each delay: + IJ
x2[k + IJ
x dk + 1J = AX[kJ + Bf[kJ (13.90a) CX[kJ + Df[kJ (13.90b) Y[kJ = Here we have represented a discretetime system with s tate equations in controller
canonical form. There are several other possible representations, as discussed in
Sec. 13.2. We may, for example, realize t he s ystem by using a series, parallel, or
observer canonical form. In all cases, t he o utput o f e ach delay element qualifies as
a s tate variable. We t hen w rite t he e quation a t t he i nput of each delay element.
T he n e quations t hus o btained are t he n s tate equations. = x2[kJ 13.61 = x3[kJ Consider t he s tate e quation (13.87) + IJ = xn[kJ
xn[k + IJ =  aoxdkJ  Solution in S tateSpace x [k x ndk a1X2[kJ  ...  anIxn[kJ + 1J = AX[kJ + Bf[kJ (13.91) From this equation it follows t hat + f[kJ a nd a nd x[kJ = A x[k  IJ + Bf[k  IJ (13.88) + Bf[k  2J
3J + Bf[k  3J (13.92a) x [k  IJ = A x[k  2J Equations (13.87) are n firstorder difference equations in n variables x I(k), x 2(k),
. .. , x n(k). These variables should immediately be recognized as s tate variables,
since the specification of t he initial values o f these variables in Fig. 13.11 will
uniquely determine t he response y[kJ for a given f[kJ. T hus, Eqs. (13.87) represent t he s tate equations, a nd Eq. (13.88) is t he o utput equation. In m atrix form we
c an w rite these equations as (13.92b) x [k  2J = A x[k  (13.92c) x[IJ = Ax[OJ + Bf[OJ S ubstituting E q. (13.92b) in Eq. (13.92a), we o btain 13 8 26 S tateSpace A nalysis 13.6 S tateSpace A nalysis o f D iscreteTime S ystems 8 27 •
x [k] = A 2 x[k  E xample 1 3.12
Give a statespace description of t he s ystem in Fig. 13.12. F ind t he o utput Y[k] i f t he
i nput I [k] u[k] a nd t he initial conditions are X I [0] 2 a nd X2[O] 3.
T he s tate equations are [see Eq. (13.89)] 2] + ABf{k  2] + Bf{k  1] = S ubstituting E q. (13.92c) i n t his e quation, we o btain
x [k] = A 3 x [k  = 3] + A 2 Bf{k  3] + A Bf{k  2] + Bf{k  1] k Xdk+1]]
[0
[ x2[k + 1] =  i C ontinuing i n t his way, we o btain : ] [ Xd ]]
6
x2[k] = + [0] I (13.96a) 1 and y [k]=[1
k 1
= A kx[O] + L A k 1 j Bf{j] ( 13.93a) j =O 5] XI[k]]
[ x2[k] (13.96b) To find t he solution [Eq. (13.94)], we m ust first determine A k . T he characteristic equation
of A is...
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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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