Solutions2HW8 - 8.1 —7 —2 6 l 1 S: = 2 —3 —2 x + l...

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Unformatted text preview: 8.1 —7 —2 6 l 1 S: = 2 —3 —2 x + l —l u b) —2 —2 l l 0 _—I‘1 2 y‘ 1 14x Solution: rank(ctrb(A, 3)) = 2 :> NOT controllable rank(obsv(A, 3)) 2 2 => NOT observable 2 —-5 M4 0] up] up ,1 Solution: rank(ctrb(A, 3)) = 2 => controllable rank(obsv(A, 3)) = 2 => observable _ 0 1 0 d) x z [-4 nix +i1l“ y = {1 113‘ Solution: rank(ctrb(A, 8)) = 2 => controllable rank(obsv(A,B)) = 2 => observable 8.8 For the electrical circuit shown below, find conditions on C1 and C; that will make the system uncontrollable. Consider vg to be the input, and v1 and v2 to be the state variables. P88 Solution: C, =C1 Using the KCL, VJ'V2+VI‘”g=Cfl Vz—V1+V2—vz=czm R R ‘ d: R R d: 2 1 1 i; l v+ 2 v 1 v v -———-v ———v ———-v —— — ‘ C,R ‘ CIR 2 C112 g l CZR ‘ C212 2 CZR g 2 I l _ l _ 2 + l ._ CIR CIR CIR v P_ CIR GER? 0,02}? " l 2 H 7 1 3 ' 1 1 7 2 CZR CIR CZR CZR CICZR (3331 For the system to be uncontrollable, matrix P must be rank degenerate; i.e., det( P) = 0 . By some calculus, we can find that the condition Cl : C2 leads to det[P) : 0 and the system uncontrollable. 3.10 For the system given in the block diagram below, find necessary and sufficient conditions for the values of milk“ and k; such that the system will be both controllable and observable. 108.10 Solution: According to diagram, )3] =-0Lx] +3! y=klxl +k2x2 STATE SPACE: i2 =—2Bx2 + 21: —o. 0 j; = [ 0 on y = [k] kzlx l 2 l...[ For controllability, we need or at 2B . For observability, we need a #213 AND kl #0 and k2 at 0. ...
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This note was uploaded on 01/14/2012 for the course EEL 5173 taught by Professor Staff during the Fall '11 term at University of Central Florida.

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