MAE 143B  Homework 8 Solutions


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1 MAE 43B  Homework 8 Solutions P6.4 b) With this system, the root locus simply starts at the pole and ends at the zero. Sketches by hand and matlab are in Figure. In matlab, use zpk to build the system with desired poles and zeros and gain, then use rlocus to plot its root locus. For example, sys = zpk([],[],); rlocus(sys).6 Root Locus.4 Imaginary Axis (seconds  ) Real Axis (seconds  ) Figure : Root locus for P6.4(b): by hand (left) and matlab (right). d) The portion from  to of the real axis belongs to the root locus. Since we have 3 poles and zeros, there are asymptotes. The intersection of the asymptotes is c = (j j ) 3 and the angles that they make with the real axis are =, φ = π and φ = π + π = 3π. Root locus sketches by hand and matlab are in Figure.
2 8 Root Locus 6 4 Imaginary Axis (seconds  ) Real Axis (seconds  ) Figure : Root locus for P6.4(d): by hand (left) and matlab (right). e) The portion from  to of the real axis belongs to the root locus. There are no zeros and 4 poles, which means there are 4 asymptotes. Their intersection is c = and the angles they make with the real axis are ( + j j) 4 =, φ = π 4, φ = 3π 4, φ 3 = 5π 4, φ 4 = 7π 4. Root locus sketches by hand and matlab are in Figure 3..5 Root Locus.5 Imaginary Axis (seconds  ) Real Axis (seconds  ) Figure 3: Root locus for P6.4(e): by hand (left) and matlab (right).
3 P6.5 b) Using sisotool in matlab, one possible controller is C(s) =.8 s + 6 s 5 which produces the root locus in Figure 4. Note that the gain can be chosen so that the closedloop system is stable. The controller is not asymptotically stable because it has a pole at 5. Figure 4: Root locus for P6.5(b). d) Referring to Part (d) of P6.4, we can see that the system is already stable. Therefore any proportional control of gain K > will suffice. e) A possible controller has a gain of.8 and complex zeros at.6 ± j.3 and a real zero at so that a pole/zero cancelation happens, and complex poles at 6.5 ± j and a real pole at 9. The resulting root locus is in Figure 5. Again, the gain can be select differently, making sure that the closedloop poles are in the left half plane. P6. Let the states w w w 3 r = ṙ, ω 3
4 Figure 5: Root locus for P6.5(e). and the output y = r = w. Rearrange the differential equations to get the following statespace form. The block diagram is in Figure 6. ẇ = w = f (w, u) ẇ = w ω GM = f (w, u) w u m ẇ 3 = w ω = f 3 (w, u) w y = w = h(w, u) Figure 6: Block diagram for P6.. 4
5 P6. To find the equilibrium points, from the statespace found in P6., set ẇ =. ẇ = ṙ = ẇ = w ω GM w ẇ 3 = u = = : true if w = r = R, ω = Ω and Ω R 3 = GM And so, if w = r = R, ω = Ω and Ω R 3 = GM, then u = ṙ =, r = R, ω = Ω is the equilibrium point of the nonlinear system. Define the new states as follows. x x = r R ṙ, x 3 Rω RΩ and the new output ỹ = r R. Then the equilibrium above becomes u x x = x 3 Linearization of the system around the above equilibrium point is ẋ = 3Ω Ω x + u = Ax + Bu Ω ỹ = [ ] x = Cx + Du. Using symbolic matlab, the transfer function can be calculated as m G(s) = C(sI A) B + D = For the period T = h = 396s, the angular velocity is Ω = π T =.59 4 rad/s. With m = 6kg, the transfer function G(s) becomes G(s) = that has no zeros and poles at and ±j Using sisotool in matlab, a possible controller can be C(s) =. 7 s(s ), Ω ms(s + Ω ). (s +.)(s +.4) (s +.4), which produces the root locus in Figure 7. Note that with the above poles and zeros of the controller, a range of gain K could be selected such that the closedloop poles are stable. 5
6 Figure 7: Root locus for P6.. 6
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