EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =

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1 1. Pole Placement Given the following open-loop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the state-variable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback vector, to yield a 5% overshoot and a settling time of.8 s. What is the closed-loop representation of the system in phase-variable form? Answer: (2 pts) 1 ẋ = 1 x + r k 1 (1 + k 2 ) (7 + k 3 ) 1 y = 3 1 x We write the open-loop plant as G(s) = Y X X U. The inverse Laplace transform of Y X gives the output equation, y = Cx, and the inverse Laplace transform of X gives the state equation, U ẋ = Ax + Bu. Plugging in the feedback we get ẋ = Ax + B( Kx + r) = (A BK)x + r. Where should the second-order poles be to satisfy the design requirements? Answer: (1 pt) s = 5. ± j5.24 Using the design specifications, ζ = ln(os) =.69, π 2 + ln 2 (OS) ω n = 4 ζt s = 7.26, so the desired second-order poles are at s = ζω n ± jω n 1 ζ 2 = 5. ± j5.24. (c) Calculate the desired closed-loop characteristic equation by selecting the third closed-loop pole to cancel the closed-loop zero. Is this a good design choice? Explain. Answer: (3 pts) s s s = Note that this feedback does not change the open-loop zero. Since we can cancel the zero with a stable pole, the closed-loop system behaves like a second-order system after cancellation, and the design requirements can be satisfied exactly. The desired closed-loop characteristic equation is (d) Find the control gains k 1, k 2, k 3. (s 2 + 2ζω n + ω 2 n)(s + 3) = s s s =. 1 of 5

2 Answer: (2 pts) k 1 = 157.5, k 2 = 72.5, k 3 = 6 The determinant of Is (A BK), where A and B are from part, gives the characteristic equation: s 3 + (7 + k 3 )s 2 + (1 + k 2 )s + k 1 =. Matching coefficients with the desired characteristic equation, we find k 1 = 157.5, k 2 = 72.5, k 3 = 6. (e) Simulate the step response. Did you meet the design specifications? Answer: (1 pt) The overshoot is 5 % and the settling time is.83 s, slightly greater than the desired settling time. 2. More pole placement Repeat the above for the following open-loop plant: G(s) = 2(s + 15)(s + 2) (s + 3)(s + 5)(s + 7). For part (c), choose the third pole to cancel one of the zeros. Which one should you choose? For part (e), you should find that you did not meet all the design specifications. What went wrong and where should you have placed the third pole instead? What is the closed-loop representation of the system in phase-variable form? Answer: (2 pts) 1 ẋ = 1 x + r (k ) (71 + k 2 ) (15 + k 3 ) 1 y = x Where should the second-order poles be to satisfy the design requirements? 2 of 5

3 (c) Answer: (1 pt) same as 1: s = 5. ± j5.24 Calculate the desired closed-loop characteristic equation by selecting the third closed-loop pole to cancel the closed-loop zero. Which zero should you choose? Explain. Answer: (3 pts) s s s We place the third pole at 15. We choose 15 instead of 2 to reduce overshoot. As the other zero becomes larger negative, the closed-loop transfer function becomes closer to a second-order approximation. (d) Find the control gains k 1, k 2, k 3. Answer: (2 pts) k 1 = 682.4, k 2 = 131.5, k 3 = 1. The determinant of Is (A BK), where A and B are from part, gives the characteristic equation: s 3 + (15 + k 3 )s 2 + (71 + k 2 )s + (k ) =. Matching coefficients with the desired characteristic equation, we find k 1 = 682.4, k 2 = 131.5, k 3 = 1. (e) Simulate the step response. You should find that you did not meet all the design specifications. What went wrong and where should you have placed the third pole instead? Answer: (2 pts) The step response is plotted below. The overshoot is 5.5% and settling time is.78 s. The specifications were not met since the system is not close enough to second-order As we move the third pole in the desired characteristic equation closer to the origin, the overshoot decreases and settling time increases. At 12, the overshoot is 4.8% and settling time is 7.9 s, so it is possible to satisfy the design specifications. The step response is plotted below. 3 of 5

4 Controllability and observability For the following systems, compute by hand the observability matrix and determine whether the system is observable ẋ = x + u y = x Answer: (3 pts) Controllable and observable C M = [B AB A 2 B] = rank(c M ) = 3 = order of plant so system is controllable. C O M = CA = CA rank(o M ) = 3 = order of plant so system is observable ẋ = x + u y = 3 5 x 4 of 5

5 Answer: (3 pts) Observable, not controllable C M = [B AB A 2 B] = rank(c M ) = 2 < order of plant so system is not controllable. C 3 5 O M = CA = CA rank(o M ) = 3 = order of plant so system is observable. 5 of 5