Linear System Theory and Design Chapter 9

Chapter 9 9.1 partial i plus/minus 1 = \mathcal{S} ^ 2 - i Nist05-2 Berasite Dis) are Nios are copreme, solutious H(z) emptyset (z) + beta ,z+N(z)=J^ + +2 pi+1=F(z) ting (9.8) and Example 91, we have, for any polynomial Qis j overline I(s) + 1/3 * (s ^ 2 + 2s + 2) = 2(s)(- j, z) B (s, t) = ii 3 (ir2)( epsilon^ 2 +isr2)+Qis)(s^ 2 -i) a solution, der auf S(s) = f_{c} of depres 0 we have gB(s) > alpha*yA(v) der suy G(s) = g_{p}*r*E_{t} * s' of degree! we have deg Bls>> dey Als), for any Q(x) = F_{2}*r 2,525 of degrees we have dey Bu deg Als) Proceeding forward, we conclude that there wheels no solution with deg B(x) \le deg A's) in the equation. "We give a different argument. Consider A(s) = A_{0} + A_{1}*J = A_{1} * s ^ 2 and B(r) = B_{w} + B_{1}*x + B_{2} * x ^ 2 Than A_{4} and Be muut meet -2 -2000. -200 A A 3. AL Then, for any is we hieve the slution [[d_{2}], [d_{2}], [d_{1}], [d_{1}], [d_{2}], [d_{2}], [d_{2}]] =[ matrix 0\\ -1 2 \\ -1 2 \\ -1 2 \\ -1 2 \\ -1 3 \\ -1 3 \\ -1 3 \\ -1 3 \\ -1\\ 1\\ 1\\ 1\\ 1 matrix ] (Theoresn 3.2), for any d. A_{2} = 0 df 2 Chen B perp ne C and dig B(s) 7 deg A (1). If alpha = 5/J chen tur 8,10 ad A_{1} = C theta_{2} = 0 3/x This we still have alpha*gG(s) > alpha*lgA(s) In conclution, we have no colations with deg B(5) A(s). zeta(s) = (- i + s)/(- q + 0.5 + s ^ 2) \approx (Nu)/(D(s)) C(s) = (beta(y))/(beta(s)) = (beta_{i}(s))/(beta_{k} + beta_{1}*s) F(s) = (s + 2)(s ^ 2 + 2s + 2) = 4 + 6c + 4s ^ 2 + s ^ 2 CAUA, [4641] [[- 4, 0, 1, 0], [- 1, 1, 0, 0], [0, - 4, 0, 1], [0, - 4, 0, 1], [0, - 1, 1, 0]] I MATLAB we type 14:1-46105-110030-40 130-1101; f = \mathcal{L}_{Psi} which yielch [A B A, B,] [-620110] Thus the compensator is C(s) = (B(s))/(A(s)) = ((8s) + 20)/(s - b) and hat g e (s)= PN(1) partial(s) F(s) equiv (p(s - 1)(10s + 2c))/(s ^ 2 + 4s + 4) To trach step reference imputs, we reppure hat g_{p}(c) = (p(- lambda * 0))/4 = - 5p = 1 p = - 1/5 = - 0.2 9.3 If F(y) becomes hat f(s) = (x - e, y)/(s ^ 2 - v, z) then hat j 1 ^ 2 = pC(z) hat g (z) 1+ zeta(z) hat g (z) = -p 1 + ((c*e_{2} + 2k)/(xi - k) * (j - e_{1})/(j - k))/(i + (c*e_{2} + 2k)/(f - k) * (j - e_{1})/(j - e_{1})) i+ ce 2 +2k f-6 j-e 1 f^ 2 -q.1 = (- (2z + w) * (z + 0, g))/(z ^ 2 + w * z ^ 2 + 6, 9z + 6.6) This system is still stable Becaus [A BA, B, AB0401 00 hat tilde g epsilon(c) = ((- w)(- c * 0.9))/6.6 = j.c 6.6 =0. #1 The output will track any step reference mput bout with 45 Do elros det phi*f(S) = S und consider A(y) * D(y) * 3 -> Bis )N(x)=F(yJ Because omega(s) hus degree I we med C(s) = (i * 3(s))/(Ar(s)) = hat a c +B 1 s+B 2 s^ 2 A 2 +A 1 s+A 2 s^ + From the coefficients of D(s) * s = s ^ 2 - 4s N(j) = j - j = - i + j' + j * j ^ 2, j * s ^ 2 * mu*alpha form Thow the fudback syetem has fwe poles and we can assign them all let F(s)=(s+1)( s + 1 + j(xs + 1 + j) * (s + 1) ^ 2 5 ^ 5 + 10s ^ 4 + 39s ^ 3 + 76s ^ 2 + 98s + 38 Then we have sqrt = = 78 76 34 10 13 ind the solution is CAS BA, BA n >= 1 [-41-36:10 2501134] Thus we have C(3) = (i * 195 ^ 2 + 2sqrt(0) * x - 38)/(x ^ 2 + 10s - 91) and the total compensator s (C(1))/(Phi(s)) = (ij * 9s ^ 2 + 255s - jt)/(J(s ^ 2 + (EI - 9))) The compranta has degree 3. The denyn in nchieved by introducing first the internal model 1/11/ F Next me goue in robust design without introducing first in internal model. consider with n = li*g_{j}*D*w_{j} = 2 , A(s) * D(s) + beta(s) * N(s) = F(s) If C(3) = B(1) / A * (s) is of degree, then the compensator in mngue now of we increase the degree of C(S) le 2, 26m Clss is not unque coned we may be able to find a Ciss duet contams the facts 45. Ler C(s) = (B_{0}*r*B_{1}*s + B_{2} * s ^ 2)/(A_{0}*r*A_{1}*s + A_{2} * s ^ 2) and select F(s) = (s + 2)(s ^ 7 + 2s + 2)(s + 3) = i2r * 22s + 14s ^ 2 + 7s ^ 2 + s ^ 4 We form [A x otimes x ; theta x ; theta x , theta x , theta x ] [[phi, phi, 0, 0, 0, 0], [- 1, phi, 0, 0, 0, 0], [0, - phi, 0, 0, 0, 0], [0, - phi, 0, 0, 0, 0], [0, 0, - phi, 0, 0, 0], [0, 0, - 1, 0, 0, 0], [0, 0, 0, 1, 0, 0]] [12 22 18 7/1 Solutions are not unique, ause they have one free paremeter. The general solution is for any a [ A_{0}*B_{p}*A_{1}*B_{1}*A_{2}*B_{2} \ = [- 30 / (- 18 50/25)] + alpha[1 - 4/- 1 * w, 1] If 23. then [A_{c}*B_{c}*A_{1}*B_{1}*A_{3}*B_{A}] +[0-12;-21 50;1-2 beta] C(s) = (29s ^ 2 + 50s + 2)/(s ^ 2 - 21s) = (29s ^ 2 + gps - r * 2)/(S(s - 1)) This proper compensator has de facin Is and can achive robust Tractory. It has regese 2 intead of 3 an in the previous design. 9.4 hat f(s) = (j - i)/(5(S - 2)) = -i ne j 0+2S* S^ 2 C(s) = (B_{0} + B_{1}*s)/(A_{0} + A_{1}*s) F(s) = (s + 2)(1 ^ 2 + 2s + 2) =g+(5+45^ 2 +s^ 2 matrix ( partial x , partial x , partial x , partial x matrix ][ matrix 0&-2&t\\ -1&1&0&-\\ -1&0&-1&0&-\\ 0&-1&0 matrix ]&+[ matrix 0&0&0&+&1\\ 0&-1&0 matrix ] is [44:122] thur and hat zeta c ^ hat zeta = hat zeta (z) hat zeta (z) 1+C(11) hat zeta ^ hat zeta (z) + hat overline zeta * 2 overline zeta * overline zeta - overline zeta 1+ 2 overline zeta * overline zeta - overline zeta 1+ 2 overline zeta * overline zeta * z- overline zeta overline zeta(z - 2) = p * ((12s - 6)(x - i))/(s ^ 3 + 4s ^ 2 + 6s + 4) Because hat J c (c)=p- ( - i_{c} * j(c) omega =p=l, Zhare 10 no need to introdure fedforward. Jam (p = 1) to astuce tracking. The reason is the mistence of the facter 516 4.5 of fls) becomes hat hat f epsilon (3) = C(s) hat hat g (s) 1+C(s) hat g ^ hat g (s) hat f (z)= z - c, y f( overline z - overline z ,t) = ((125 + 4)(5 - c, Phi))/(5(5 - 2, 1)(s - 16) + (225 - 4)(s + 0, Phi)) = (215-4)(y-0.9) y ^ 3 + 1.4s ^ 2 + 9.85 +j.6 This system is BIBO stable. Bersamas tilde J_{epsilon}(epsilon) = (4 * 0, q)/(x, epsilon) = 1 . the system still track suy etep reference imput. Thus the design is robuat. 9.6 hat g(s) = 1/(s - 1) = (N(s))/(D(s)) Jar C(s) = overline A (s) A(s) In noder to schisme the ) mult contam 21174) Let A(z) = vec A s (s^ prime (s^ prime (s^ - ,z^ prime )) B(x) = delta c + overline 6 ,s+ delta c s^ 2 = delta d s^ J A(s) * D(s) + B(s) * N(s) = F(x) Select F(r)=(x^ 3 +2z+5^ - ) (s ^ (- w) * s + 5) =s^ prime prime +6s^ prime prime +i hat r s^ prime prime + 3ca + 25 tilde A_{c}*s(s ^ 2 + 4)(s - i) +[0 c +19s+0 z s^ 2 +8,s^ 3 ]*1 =(5) tilde A_{k}(s ^ 4 - s ^ 3 + 4s ^ 3 - 4s) +(8 k + theta j z+B k s^ 3 +B j z^ 3 ) =F (x) Egnating the coefficients pietols [ matrix tilde g k & tilde g k & tilde g k & tilde g k matrix ] [[0, - phi, - i, i, 0, 0], [1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]] = [35 20 18 61 th solution so 1125 34 14 71 Thus the componcatne C(s) = (7s ^ 3 + 14s ^ 2 + 34.5 + 25)/(s(s ^ 2 + 4)) This compensator will place che poles in the assryened positions, track robustly sny clép reference imput and weet wy disturbance of form a em (2t+B). 97 2/(zeta(x + t)) The model is is not in the forerent path from te e derefore, for any tits 2. ecayor Indeed, we have hat hat g f^ prime (z) = 3/(j(z + 1))/(i + 1/(i(z + 1))) = 3/(i ^ 4 + z + 3) WE)