MAE 143B  Homework 9


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1 MAE 43B  Homework a) G(s) = (s+)(s+) Polar plot; red for negative ; no encirclements of, a.s. under unit feedback h) G(s) = s+ s(s+) Polar plot; red for negative ; no encirclements of, a.s. under unit feedback; semicircle is traversed
2 a) G(s) = 2 (s+)(s+2) Polar plot; red for negative f) G(s) = s s Polar plot; semicircle traversed at radius ρ ; same contour for positive/negative. 2
3 7.7 a) G(s) = s + s + d) G(s) = s + s 2 h) G(s) = s + (s + ) As seen in the polar plot below, no point on the negative real axis is encircled. Given that the openloop transfer function has no poles in the righthalf plane, this guarantees asymptotic stability for any K >. We achieve both asymptotic tracking of a constant reference and rejection of a constant torque disturbance because our controller has a pole at the origin. Gm = Inf db (at Inf rad/s), Pm = 88.9 deg (at.33 rad/s) Polar plot; red for negative ; semicircle traversed Bode and polar plots for 7. 3
4 7. Note to graders: These solutions explore all possible combinations of positive and negative gains. It is sufficient to give one combination of K p and K i that yields asymptotic stability. As we have seen in a previous assignment, we can design PI controllers in two steps by first choosing the zero of our controller by fixing the ratio between the two gains K p and K i and then choosing K p using the methods from class. This follows from writing our controller transfer function as K(s) = K p(s + K i /K p ). s First note that by the zero at the origin, tracking and disturbance rejection are as in 7. provided K i. calling the polar plot in 7., notice that we have no chance of stabilizing the system with a zero in the righthalf plane and positive K p, which would cause the circle traversed at radius ρ to cover the entire lefthalf plane and thus result in an encirclement of any given point on the negative real axis. Given that our openloop transfer function does not have poles in the righthalf plane, this means that we cannot stabilize the system with a controller zero in the righthalf plane and K p >, eliminating all combinations of K p > with K i <. For an example with the zero placed at s = K i /K p =, see the following plots. 4 Gm = Inf, Pm = 34 deg (at 6. rad/s) Polar plot; red for negative ; semicircle traversed Bode and polar plots for 7., zero at s = K i /K p =. We can further see from the above plots how K p < does indeed allow for stabilization with the controller zero in the righthalf plane, provided K p is sufficiently small. This becomes apparent by noting that negative values of K p amount to rotating the entire polar plot contour by π around the origin. K p sufficiently small in this case is required to avoid encirclement of the point /K p by the small circle next to the origin. That is, we can stabilize the system by K i > and K p < provided K p is sufficiently small. This leaves us with placing the zero in the lefthalf plane. However, again using the insights from our polar plot to 7., we can place this lefthalf plane zero anywhere and choose any positive gain K p to asymptotically stabilize the system. Given that the zero specifies the ratio of our control gains, this implies that any combination of K i > and K p > yields asymptotic stability (from 7. we know that K p = works, too). For an example with the zero placed at s = K i /K p =, see the following plots. 4
5 4 Gm = Inf, Pm = 9 deg (at 6. rad/s) Polar plot; red for negative ; semicircle traversed Bode and polar plots for 7., zero at s = K i /K p =. By the same argument as above (rotation of the polar plot contour by π around the origin), no combination of K i < and K p < can be used to asymptotically stabilize the system.
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