2.004 Dynamics and Control II, Solution of Problem Set 6

Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Solution of Problem Set 6 Problem 1: Note that for part a and c we have a zero-pole cancellation in LHP or origin. However, the MATLAB does not recognize this cancellation and we have to manually deal with it. The poles/zeros are extracted by zpkdata command and s-plane plot is obtained by pzplot command. (a) dy du s+3 + 3y = + 3u ⇒ G = =1 dt dt s+3 (b) d2 y dy du 2s + 1 (s + 0.5) + 7 + 12y = 2 + u ⇒ G = 2 =2 2 dt dt dt s + 7s + 12 (s + 3)(s + 4) (c) d2 y dy du d3 y + 5 +7 = ⇒ 3 2 dt dt dt dt s+0 1 1 1 √ G= 3 = 2 = = 2 2 s + 5s + 7s + 0 s + 5s + 7 (s + 2.5) + 0.75 (s + 2.5)2 + ( 23 )2 1 MATLAB Command − line : >> [Z,P,K]=zpkdata(tf([1 3],[1 3])) % Part a Z = [-3] P = [-3] K = 1 >> [Z,P,K]=zpkdata(tf([2 1],[1 7 12])), P=P{:} % Part b Z = [-0.5000] P = [2x1 double] K = 2 P = -4 -3 >> [Z,P,K]=zpkdata(tf([1 0],[1 5 7 0])), P=P{:} Z = 2 % Part c [0] P = [3x1 double] K = 1 P = 0 -2.5000 + 0.8660i -2.5000 - 0.8660i Problem 2: (a) Having a steady-state value means that the system is stable. Consequently, no poles can be in RHP (right half plane). This condition is satisfied by all options. Furthermore, the steady-state unit step response is equal to G(0). If there is a zero at the origin then G(0) = 0 ⇒ yss = 0, which eliminates option c. Besides, if there is a pole at the origin then G(0) = 10 = ∞ ⇒ yss = ∞ ( yss is unbounded), which eliminates option b and d. Hence, the only feasible option is option a. (b) G(s) = K yss = G(0) ⇒ 2 = K (s − 1) (s − 1) =K (s − (−2))(s − (−5)) (s + 2)(s + 5) (0 − 1) 1−s ⇒ K = −20 ⇒ G(s) = 20 (0 + 2)(0 + 5) (s + 2)(s + 5) (c) Modal functions correspond to e−2t and e−5t : G(s) = 20( 1 2 − ) ⇒ g(t) = 20(e−2t − 2e−5t ) s+2 s+5 Problem 3: (a) Since the system, G(s), has a zero at s = z, we can say that G(s) = G1 (s)(s − z) such that G1 (s) has no pole at s=z . The forced response can be computed from A A = G1 (s)(s − z) s−z = AG1 (s). This means that Y (s) has Y (s) = G(s)U(s) = G(s) s−z no pole at s = z as well. Consequently particular solution (corresponding to a pole at s = z) is zero. 3 (b) G(s) = G1 (s)(s−z1 )(s−z2 ) = G1 (s)(s−(−σ+jω))(s−(−σ+jω)) = G1 (s)((s+σ)2 +ω 2)) Again G1 (s) has no pole corresponding to complex conjugate pair of −σ ± jω. u(t) = Ae−σt sin (ωt + φ) = Ae−σt (sin(ωt) cos(φ) + cos(ωt) sin(φ)) U(s) = A ω cos(φ) + (s + σ) sin(φ) (s + σ)2 + ω 2 Y (s) = G(s)U(s) = AG1 (s)(ω cos(φ) + (s + σ) sin(φ)) Consequently, Y (s) has no pole at complex conjugate pair of −σ ± jω as well. Hence the forced response is zero due to zero-pole cancellation. Problem 4: (a) u(t) = e−bt ⇒ U(s) = s+1b a a 1 Y (s) = H(s)U(s) = (s+a)(s+b) = a−b ( s+b − a −bt −at y(t) = a−b (e − e ) (b) H(s)|s=−a = a0 = ∞ and Y (s)|s=−a = a −bt yp (t) = a−b e and lim yp (t) = ∞ . a 0×(b−a) 1 ) s+a = ∞ . The particular solution is equal to b→a (c) The overall system response is yet bounded when b = a. This can be computed by below limit from L’Hospital’s rule for 00 limit: a lim y(t) = lim (e−bt − e−at ) = a lim b→a b→a a − b b→a ∂(e−bt −e−at ) ∂b ∂(a−b) ∂b −te−bt = ate−bt b→a −1 = a lim Problem 5:. (a) Gp (s) = K 1 (s + 2)(s + 4) Unity Gain ⇒ G(0) = 1 ⇒ K = 8 ⇒ Gp (s) = 8 8 = 2 (s + 2)(s + 4) s + 6s + 8 (b) Gcl (s) = Kp Gp (s) 8Kp 8Kp = 2 = 2 1 + Kp Gp (s) s + 6s + 8 + 8Kp s + 6s + 8(1 + Kp ) (c-d) The closed-loop poles for the values of Kp = 0, 0.025, 0.05, 0.075, 1, 1, 25, 1, 2, 3, . . . , 10 are plotted via attached code. Alternatively we could use rlocus command or rltool toolbox. 4 (e) Critically Damped ⇒ ς = 1 ⇒ 6 = 2ωn ⇒ ωn = 3 1 ωn2 = 8(1 + Kp ) = 9 ⇒ Kp = = 0.125 8 MATLAB m − file : clc, close all, clear P_cl=[]; plot([-2 -4],[0 0],’k+’),hold on for Kp=[0:.025:.125 1:10] [Z,P,K]=zpkdata(tf(8*Kp,[1 6 (1+Kp)*8])); P_cl(end+1,:)=P{:}; end plot(real(P_cl(:,1)),imag(P_cl(:,1)),’r-*’) 5 plot(real(P_cl(:,2)),imag(P_cl(:,2)),’b-*’) grid on, box on title (’Root Locus’,’fontsize’,16) xlabel (’Real Axis’,’fontsize’,16) ylabel (’Imaginary Axis’,’fontsize’,16) axis([-4.5 0.5 -10 10]) ------------------ Computed Poles by Varying Kp ------------------­ >> P_cl P_cl = -4.0000 -3.8944 -3.7746 -3.6325 -3.4472 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 + + + + + + + + + + + 0.0000i 2.6458i 3.8730i 4.7958i 5.5678i 6.2450i 6.8557i 7.4162i 7.9373i 8.4261i 8.8882i -2.0000 -2.1056 -2.2254 -2.3675 -2.5528 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 - 0.0000i 2.6458i 3.8730i 4.7958i 5.5678i 6.2450i 6.8557i 7.4162i 7.9373i 8.4261i 8.8882i 6