POLITONG - SHANGHAI BASIC AUTOMATIC CONTROL Exam grade March 8, 4 Academic Year 3/4 NAME (Pinyin/Italian)... STUDENT ID Ue only thee page (including the back) for anwer. Do not ue additional heet. Ue of any book, note, or other teaching material i not allowed. Write clearly and be explicit and concie in your anwer Subtitute [N] in the text with the number of letter of your firt (Pinyin/Italian) name EXERCISE Given the following dynamic ytem [N] x& x + u x& x y 5x 4x a) Claify it type. b) Compute it equilibrium tate when u [N] and determine their type and their tability. c) Compute the output at equilibrium. a) The ytem i continuou time, invariant, nonlinear, SISO, nd order. b) Setting the derivative to, we compute the unique equilibrium tate x x 4x y 5x + u x [ N]/ x ([ N]/ ) y 5x Which olve alo point c). /4 To determine the tability and type of the equilibrium, we have to linearize the ytem around the equilibrium condition f x x, u x 4 [ N] 4 x, u The matrix i triangular, o the eigenvalue can be immediately read on the main diagonal. They are - and -4. They are both real and negative, thu the equilibrium i aymptotically table and it i a nodal ink. March 8,4- POLITONG /4
EXERCISE Determine, a preciely a poible, the output y(t) of the feedback ytem in the figure when u (t)tep(t) and u (t)5tep(t). u + + u + + y [N] We firt compute the two tranfer function: Y ( ) G ( ) + U [ ] ( ) N + + [ N] + + Y( ) + G( ) U ( ) [ N] + + [ N] + + A expected, they have the ame firt order denominator. Then, we take into account the two input by applying to them the Laplace tranform: L[u (t)]/ and L[u (t)]5/. Thank to the uperimpoition principle, the overall output will be the um of the effect of the individual input. We can examine them eparately, or compute Y()G ()U ()+G ()U (). The numerator of Y() depend on [N], but the denominator i alway (+). So we can antitraform Y() by etting it equal to the um of two term A B Y ( ) + which mean that A ( + [ N] + ) + B mut be equal to the numerator of + [ N] + Y(). Thi provide two equation that allow to determine A and B and then a traightforward antitraformation. Auming, for intance, [N]3, [N]+7 and we obtain A, B4. The final reult i thu L - [A/]tep(t), or for t, and L - [B/(+7)]4 exp(-7t)tep(t), or 4e -7t for t and thu y(t) + 4e -4t for t. Such a concluion can be verified by uing the initial and final value theorem (the lat can be applied becaue the feedback ytem i aymptotically table, ince the pole of G and G i negative). Clearly, uing only the theorem intead of the analytical antitranformation, one obtain ome information on y(t), but not the complete behavior. March 8, 4 - POLITONG /4
EXERCISE 3 I the following control ytem aymptotically table for any value of K? Why? Determine at leat one value of K for which the ytem i aymptotically table and the bandpa i at leat [,[N]/]. U + - K ( + ) ( +.) Y You can ue the chart below, if ueful. db Auming K i a real number (a tf would have been written K()), the open-loop tf of the ytem i c K L( ) ( + ) ( +.) While the cloed loop tf i K ( + ) ( +.) F( ) K + ( + ) ( +.) We cannot analye the tability of F() by phae looking at the eigenvalue ince it ha a third order denominator, but we can apply Nyquit and Bode -9 theorem that allow u to -8 ay omething on the tability of F() by -7 conidering only L(). The firt quetion can be anwered by imply conidering the Nyquit plot. Since L() i 3 rd order without zeroe, the Nyquit plot (ee figure) will terminate ( ) with a phae -7, thu it will certainly cro the negative real axi and, for a ufficiently high gain K, it will pa below -. So there are (high) value of K for which F() become untable. Imaginary Axi 3 Nyquit Diagram To anwer the econd quetion, one may ue the Bode plot. The aymptotic value for K are hown, while the real one are in the following page. - - -3 - -5 5 5 5 3 35 4 Real Axi March 8, 4 - POLITONG 3/4
5 Bode Diagram Magnitude (db) -5 - -5 Phae (deg) -9-8 -7-3 Frequency (rad/ec) A a firt approximation, the Bode plot of F() i equal to (db) till c and coincide with L() for frequencie higher than c. So, looking at the aymptotic Bode plot, it i eay to undertand that, for K, the bandpa i about 3 and the ytem i aymptotically table (the condition of the Bode theorem apply and the critical phae i till higher than -8, which mean a poitive phae margin). To be afer, we can decreae K, thu decreaing the bandpa, but increaing tability. If [N]<, even K. will atify the requirement (the bandpa of F() would be _ and the phae margin cloe to 9 ). March 8, 4 - POLITONG 4/4
EXERCISE 4 Anwer the following quetion, uing only the available pace and WRITING CLEARLY, pleae. a) Two DISCRETE TIME ytem have the following tate tranition matrice. Which i more table? Why?.5 A [.8] [N]/ A For dicrete time ytem, we have to compute the magnitude ( abolute value real + imaginary ) of the eigenvalue. The farther they are from the tability limit i, the more table the ytem i. Thu, auming for intance [N]7, we obtain, -.5±.9j and o,.95 which mean the econd ytem i more table. b) Can a LTI continuou ytem with the following matrice be tabilized ( made aymptotically table) with a direct tate feedback u(t)kx(t)+v(t)?.5 A [ ] B [N] C We can compute the reachability matrix R[B AB], which in thi cae i quare. So, if det(r) the rank of Rorder of the ytem and thu the eigenvalue can be arbitrarily fixed with a direct tate feedback. By the way, the ytem wa already table. c) What i the gain margin of a feedback control ytem? The gain margin i an indicator of the robutne of the ytem to gain uncertainty. It i -/x A where xa L(j π ) and π i the frequency at which arg(l(j))-8. d) What are the main problem of uing an open-loop tructure to control a LTI ytem? With an open-loop controller, we cannot modify the tability of the ytem, we cannot reduce the effect of diturbance and, to make it feaible (at leat proper), we have to limit the bandpa. March 8, 4 - POLITONG 5/4