solutions2HW7

# solutions2HW7 - 7.3 A certain nonlinear system is given by...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.3 A certain nonlinear system is given by the equations i1: —x1+x2 +x1(x12 HE) .h-‘z = —x1-ﬁx2 +x2(x12 +x22)l a) Determine the equilibrium points of the system. Solution: Set i=0.Then 2 2 0=—x1+x2+xl(xl +x2) 0=—x1—x2+x2(xlz+x§) x+x 2 l 2, xlxz—xglel+x1x2, xlz+x2=0 xl _ x 2 x1 x 2 Therefore, the origin of the system is ONLY THE EQUILIBRIUM POINT. b) Choose a candidate Lyapunov function and use it to determine the bounds of stability for the system. Sketch the region in the x; -— x2 plane in which stability is guaranteed by the Lyapunov function. Solution: . . l 1 . _ Use the candidate Lyapunov ﬁll‘lCtion V(xl,x2) = 3x12 +36 to test the stablhty of the system. Then V(xl,x2) : 5511:] + J‘szz _ 2 2 2 2 ——x1—-xl+x2 +.\c1(xl +x2) +1:2 —x1 —x2 +x2(xl +152) 2 2 2 2 2 2 2 2 :—.74:1 +x1x2 +x1(x] +Jc2)ex1:c2 wxz +x2(x1 +x2) 2 2 2 2 2 2 2 2 =—(x1 +x2)+xl (1] +x2)+x2(xl +172) =(—1+x]2+x§)(xlz+x§) Since the derivative Von ,xg) must be negative deﬁnite for the system to be Lyapunov stable, («1+x]2 + x§)(x12+ x5) < 0 x12 + 1-; <1 c) Draw a phase portrait for the system. Solution: Phase Portrait 7.5 An LTI system is described by the equations b a 0 x= x. 1 r1 Use Lyapunov’s direct method to determine the range of variable for which the system is asymptotically stable. Solution: Lyapunov Equation: ATP + PA = —Q P11 P should be symmetirc ; P =[ P12 1 p12]. Let Q=[ 0].Then P22 0 1 [61 1:“:Pn P12]+|:P11 P12]? 0]=[*1 0] 0 *1 P12 P22 P12 P22 1 ‘1 0 ‘1 2(aP11+P12): 4 Independent equations; (a — 1) pl: + p22 2 0 , solving (bottom to top) 4.022 = ’1 1 P22 =5 (> 0) __ _ 1 P12 201 _ I) l l p” _ 2a(a — 1) 2a In order for P to be POSITIVE DEFINITE, p“ and IF} should both be positive (principal minors test), i.e., l l 27a ———#:—m>0 2a(a—1) 2a 2a(a71) This implies (2 — a)2a(a -1)> 0 , which means either a < 0 or 1< a < 2 . And, P11: 27.: i > a(a—1) (CI-1f (2*GXR4)>1 a W is only POSITIVE for 1<a<2, or all a<0. But W is a a NEVER> 1 when l<a<2,andisALWAYS> 1 when a<0. a<0 Since or AND {a < 0} can’t happen simultaneously, the system is 1<a<2 ASYMPTOTICALLY STABLE for ALL a < 0 . [should be obvious from triangular form of A matrix] 7.8 Find a Lyapunov function that shows asymptotic stability for the system it = —(l + 0x ‘ Solution: Let V(x) = x2 be a candidate Lyapunov function. Then V(x) 2 71(x) : %x2 > 0 for all x ¢ 0 and for all 1‘20, and ﬁx) 2 2127': = 2x{—(l +I)x} = —2(l+1)x2 <Y2(x) : —x2 <0 for all xaeﬂ andforall 120‘ Therefore, the system is ASTMPTOTICALLY STABLE. 7.13 Determine whether or not the system shown in the following diagram is BIBO stable. Consider the input to be current u(r), and output to be voltage v(t) . P 7‘ 13 Repeat if R = 00 (i.e., remove resistor). Solution: i) 1. .. 1 _. _ EV(E)+CV(I)+EV(I)—u(f) 2? 6(3)— . l The system poles has negative real parts at —2—RC-. Therefore, the system is BIBO stable. ii) When R : co , dvn) 1 _ C d! + LJv(t)dI _ an) 1 .. 1 _ 55 Cu (I) + 7120‘) = 710‘) :3 C(5) =— L 2 l s + — LC 1 m. The poles are at if So, NOT BIBO stable. 7.15 Consider the system given by . 72 0 1 x : x + £1 0 0 1 y 2 [fl 21x a) Find all equilibrium solutions xe . Solution: 0 Axe = 0 : equilibrium points are x9 , such that xe 6 NM), or 2:8 =[1] or ANY SCALAR MULTIPLE. b) Determine which equilibria are asymptotically stable. Solution: Since one eigenvalue is zero, none of the equilibria are ASYMPTOTICALLY stable. c) Determine if the equilibrium solutions are Lyapunov stable. Solution: A has a non-positive real eigenvalue and non-repeated zero eigenvalue. So, the equilibria are Lyapunov stable. d) Determine if the system is BIBO stable. Solution: BIBO 3 +4 5(5 + 2) ' POLE on IMAGINARY AXIS (s = 0) => NOT BIBO stable. For stability, ﬁnd transfer function. [NUM,DEN]=ssth(A,B,C,D) :9 A Let zl :xl,and 22 = —x1 +172. Also let u{t‘)=0. Ifwe denote z=[z] zz]T and 2 : 1-12 , ﬁnd the equilibrium solutions ze , and sketch them on the z] —22 plane. Solution: 1 0 Let 2 =[_l 2].:3Px or x = P'lz. Then, 2 = Pi: = PA: = PAP—‘z A —2 0 A = PAP" =[ ] 2 0 So, Equilibrium ze Draw, on the same plane, a phase portrait. Solution: Phase Portrait To sketch phase portrait, ﬁnd eigenvectors: 1 For A: = -—2, el =[ ] (trajectories will move toward origin in direction of at) For 112 = 0, 92 = (trajectories will not move if they are on e2) 7.18 Given the system —25 —45 —55 2 it: —5 *i5 —l5 15+ .2 u [5 35 35 —.8 ’ y=[1 0 up: determine whether the system is Lyapunov stable, asymptotically stable, andfor BlBO stable. If it is not BIBO stable, give an example of a bounded input that will result in an unbounded output. Solution: eig(A) = {110, —le, —5} => Lyapunov STABLE, but NOT ASYMPTOTICALLY. 232 —55' + 50 (s + 5)(.¢2 +100] system is NOT BIBO STABLE. So any input of the form u(r) = Acos(10t+ d3) will excite the (D = [0 modes, causing an unbOunded output. Transfer function H(s)= has poles at —5 and ile, therefore the ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

solutions2HW7 - 7.3 A certain nonlinear system is given by...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online