16.30/31, Fall 2010 Recitation # 1
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1 6./, Fall Recitation # September, In this recitation we consiere the following problem. Given a plant with open-loop transfer function.569s +.5 G p (s) = s +.7s +.97, esign a feeback control system such that the close-loop ominant poles have unampe natural frequency ω n = ra/s an amping ratio ζ =. (This problem was mentione uring lecture, see the Topic notes, as a stability augmentation system for a B77.) r + - e G c (s) u G p (s) y Figure : The stanar block iagram for a unit-feeback loop. Since we are working with the root locus metho, it is convenient to rewrite the transfer function in the form z Zeroes(G (s z) p) s G p (s) = K p p Poles(G (s p) =.569 p) s +.7s Note that the root-locus gain of the plant s transfer function K p =.569 is negative. This is a fact that is easy to overlook, an may generate confusion if not recognize an hanle properly. More on this later. The pole-zero map for the system is shown in Figure, where: z =.969 is the zero (root of the numerator) of G p. p, =.75 ±.8888 are the poles (roots of the enominator) of G p. p, =.8 ±.j are the esire close-loop poles. (We know that the magnitue of the esire poles is equal to ω n =, an the real part is ζω n =.8.) Q. Can we achieve the control objective using proportional feeback? A first question one may want to answer is whether a simple proportional feeback (i.e., G c (s) = K, for some K) woul o the trick. You may convince yourself that this is not likely to be the case, e.g., by sketching the root locus. However, it is possible to check rigorously whether a point is on the root locus (i.e., whether the root locus goes through a given point), using the angle conition: (s z) (s p) = l8, l =, ±, ±,.... z Zeroes(G) p Poles(G) (In particular, the point will be on the positive-gain root locus for l o, an on the negativegain root locus for l even.)
2 Im(s) Re(s) Figure : The open loop poles (blue crosses) are in an unesirable locations: the response is too slow (ω n ra/s) an too lightly ampe (ζ.). The objective is to esign a control system such that the close loop system has its ominant poles in the location marke with re squares. Recall that, e.g., (s z) can be thought of as a vector going from z to s. The angle (s z) can be compute as, the arctangent of Im(s z)/re(s z). (You may want to use the atan function on your calculator or in Matlab to make sure you get the quarant right.) In our case, let us check where the esire poles are on the root locus. Due to symmetry, it is sufficient to check for one of them, i.e., for p =.8 +.j. We get p z =.5 +.j, hence (p z ) = atan(.,.5) =.6. p p = j, hence (p p ) = atan(.5,.95) =.. p p = j, hence (p p ) = atan(.888,.95) =.5. Hence, (p z ) (p p ) (p p ) =.8 = l8, l =, ±, ±,... an the point p is not on the root locus (for proportional feeback). Q. Can we achieve the control objective using a ynamic compensator? To make a long story short: yes. There are many ways one can achieve the objective. Let us consier one, we will iscuss others shortly. Let us use a ynamic compensator of the form G c (s) = K p s z c. s p c Let us choose p c = z, i.e., let us cancel the open-loop zero with a compensator pole. We still have two egrees of freeom remaining: the location of the zero z c an the gain K c. The angle conition in this case woul be (recall that the open-loop zero is cancele with the compensator pole): (p z c ) (p p ) (p p ) = l8, l =, ±, ±,...,
3 Root Locus Step Response Imaginary Axis Amplitue Real Axis Time (sec) Figure : Root locus an close-loop step response using the lag compensator propose in the text. that is (recall that we have alreay compute the angles of p p, ), (p z c ) = l8, l =, ±, ±,... In other wors, if we choose z c in such a way that (p z c ) = 66.9, we satisfy the angle conition, with l =. Since l is o, the esire poles will be on the positive-gain root locus for the propose feeback system. In orer to fin z c, just set (p z c ) = atan(.,.8 z c ) = 66.9, i.e.,. z c =.8 =.8. tan 66.9 Notice that given the location of the compensator pole an zero, this is a lag compensator (the pole is closer to the origin). The magnitue ratio between the pole an the zero is within the stanar guielines of a factor between 5 an. The last step in the esign of the compensator is the choice of the gain. The magnitue conition can be written as: z Zeroes(G) s z =, s p K z Poles(G) where G = G c G p an K = K c K p. Doing the calculations (remember we can ignore the cancele pole/zero), we get: K =.859, an ultimately K c = K / K p =.7. Since K p < as alreay mentione, we can conclue K c =.7, an s +.8 G c =.7. s For verification purposes, we can compute the characteristic equation of the close-loop system: s +.7s (s +.8) s +.6s G c G p = =,......
4 Root Locus Step Response Imaginary Axis Amplitue Real Axis Time (sec) Figure : Root locus an close-loop step response using the simple compensator with one pole iscusse uring the recitation. which shows that the close-loop poleas are inee at the esire locations. A similar lag-compensation approach woul have worke, without the nee for pole-zero cancellation, but the algebra woul have been a bit more complicate. The root locus an the (close-loop) step response using the propose compensator are shown in Figure. Other approaches Simple lag One approach propose in class was consier a ynamic compensator with only one pole p c. The angle conition can be written as (p z ) (p p ) (p p ) (p p c ) =.8 (p p c ) = l8, l =, ±, ±,..., i.e., we can choose (p p c ) = 55.6, satisfying the angle conition with l =. Since l is o, the esire close loop poles will be on the positive-gain root locus. Proceeing as in the previous case, we can compute p c =.8./ tan 55.6 =.7. The magnitue conition woul yiel K c = Inee, two of the close-loop poles will be at the esire location. However, a new pole has been introuce in the system, an will actually play a ominant role in the response, which is not esirable see Figure. Also note that the steay-state error is quite large.
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