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

# 1375a a nd 1375c a nd f our unknowns t his o

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Unformatted text preview: tate-Space Analysis o f Discrete- Time S ystems o o 1 o o o o 1 o 825 o o o + x n-dk x n[k + 1J + 1J ~ o o o o - ao - al - a2 x [k+lJ . o 1 - a n-2 - an-l 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 th-order d iscrete-time 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 discrete-time 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 13.6-1 = x3[kJ Consider t he s tate e quation (13.87) + IJ = xn[kJ xn[k + IJ = - aoxdkJ - Solution in S tate-Space x [k x n-dk a1X2[kJ - ... - an-Ixn[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 first-order 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 tate-Space A nalysis 13.6 S tate-Space A nalysis o f D iscrete-Time S ystems 8 27 • x [k] = A 2 x[k - E xample 1 3.12 Give a state-space 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.

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