Uppsala University Department of Information Technology Systems and Control Professor Torsten Söderström Final exam: Computer-controlled systems (Datorbaserad styrning, RT450, TS250) Date: December 9, 2008 Responsible examiner: Torsten Söderström Preliminary grades: 3 23 32p, 4 33 42p, 5 43 50p. Instructions The solutions to the problems can be given in Swedish or in English. Problem 6 is an alternative to the homework assignment. (In case you choose to hand in a solution to Problem 6 you will be accounted for the best performance of the homework assignments and Problem 6.) Solve each problem on a separate page. Write your name on every page. Provide motivations for your solutions. Vague or lacking motivations may lead to a reduced number of points. Aiding material: Textbooks in automatic control (such as Reglerteori flervariabla och olinjära metoder, Reglerteknik Grundläggande teori, and others), mathematical handbooks, collection of formulas (formelsamlingar), textbooks in mathematics, calculators. Note that the following are not allowed: Exempelsamling med lösningar, copies of OH transparencies. Good luck!
Problem Consider an oscillative system with transfer function ( ) 2 Ó 2 + 2 Ó + 2 Ó Design a feedback regulator using Internal Model Control and tuning. Will the regulator be integrating? 5 points Problem 2 Consider a two-input, two-output system with the transfer function ¼ ½ ( ) +2 +5 0 + 5 +3 (a) Determine RGA((0)). (b) Which input-output pairing should be preferred? 4 points Problem 3 Civ intends to use Kalman filter theory as a methodology to filter a noisy measurement signal. Consider the setup displayed in the figure below. Here Þ(Ø) is the useful signal, Ò(Ø) the measurement noise and Ý(Ø) the noisy measurements. The filter to be designed is denoted Ä, which output ˆÞ(Ø) should be a good estimate of the useful signal Þ(Ø). It is desired that Ä( ) should be a low-pass filter with the characteristics Ä(0) Ä( ) decreasing at least as for large Ú ¹ Þ ¹ Ý ¹ ( ) Ò + Ä( ) ˆÞ (a) Civ first tried to model the signal Þ(Ø) as a first order filtered signal with the transfer function ( ) + and with a white input signal Ú(Ø) having intensity Ö Ú. The measurement noise is also assumed to be white noise, and to have intensity Ö. 2
Show that the setup can in state space form be written as Ü Ü + Ú Ý ( )Ü + Ò What values can the parameter take? Writing the general observer in the form ˆÜ ˆÜ + [Ý ( )ˆÜ] derive the corresponding filter transfer function Ä( ). (b) What constraints on the model ( ) are needed to meet the specified properties of the filter Ä( )? (c) Using the constraints on ( ) derived in part (b), derive the Kalman filter. How is it influenced by the choice of the parameter? 3 points (d) Show that the way the noise intensities Ö Ú and Ö influence in the Kalman filter is only through the ratio Ö Ú Ö, and write out this dependence explicitly. (e) Assume instead that the signal model is taken as the second order system ( ) ( 0 0) ( + ) Will the constraint Ä(0) be satisfied? 3 points Problem 4 Consider À 2 control of the system with the weightings ( ) + Ï Ë ( ) «(«0) Ï Ì ( ) Ï Ù ( ) (a) Write the extended model in state-space form. (b) Find the solution of the associated Riccati equation. (c) Determine the optimal À 2 regulator. 4 points 5 points 3
Problem 5 Consider the feedback system in the figure below. Relä Ö 0 Σ ( ) Σ À( ) where À( ) ( ) ( + )( + 2) Use the circle criterion to design the feedback parameters and such that the closed loop system is guaranteed to be asymptotically stable. 6 points Problem 6 Consider a system with the transfer function ( ) 0 + where is a real-valued number. (a) When designing a controller for the system, Civerth chose not to consider the crosscouplings in the system, that is he used the simplified model 0 ( ) 0 + 0 Determine the relative model uncertainty. (b) Civerth made a choice to design two decoupling PI-regulators, which leads to the feedback Ì Ã + Ì Ý ( ) 0 Ì 0 Ã 2 + 2 Ì 2 Determine the regulator parameters, based on the simplified open loop model, so that the closed loop system becomes ( ) Á. + (c) Use some suitable robustness criterion to decide for which values of one can guarantee that the closed loop system is stable, when the feedback in (b) is applied. 3 points 4
(d) Determine for precisely what values of the closed loop system is indeed stable. 3 points 5
Uppsala University Department of Information Technology Systems and Control Prof Torsten Söderström Computer-controlled systems, December 9, 2008 Answers and brief solutions Problem The transfer function É( ) becomes É( ) and the regulator will be ( + ) 2 ( ) ( + ) 2 2 + 2 Ó + 2 Ó 2 Ó Ý ( ) [ É( )( )] É( ) 2 + 2 Ó + 2 Ó ( + ) ( +) 2 2 2 Ó 2 + 2 Ó + 2 Ó 2 2 + 2 2 Ó As Ý (0) ½, the regulator is integrating. Problem 2 (a) In this case Hence RGA((0)) (0) 2 0 5 53 2 0 53 5 5 53 56 2 0 2 6 56 2 5 2 7 2 56 7 2 5 (b) Avoid pairing leading to negative diagonal elements in RGA. Hence combine Ù with Ý 2, and Ù 2 with Ý. Problem 3 (a) The state-space model means that the transfer function from Ú to Þ is ( Á ) Æ ( ) + ( ) Any non-zero value of can be used. 6
The observer leads to so ˆÞ ( )ˆÜ ( ) Ô + + Ý Ä( ) + + (b) The condition Ä(0) leads to that Civ must make the choice (c,d) The Riccati equation becomes in this case simply 0 0 È + È Ì + ÆÊ Ú Æ Ì È Ì Ê È 0 0 + 0 + 2 Ö Ú Ô 2 ( ) 2 Ö with the solution Ô 2 Ô ÖÚ Ö The corresponding filter gain is (Ô)( Ö ) Ö ÖÚ Ö In the filter the parameter does not depend at all on. Ö ÖÚ Ö (e) Represent the system as Ü 0 0 Ý 0 Ü + 0 Ü + Ú Then it holds Ä 0 Ô + + 2 Ô 2 Ô Ô(Ô + + ) + 2 2 Ô + 2 Ô 2 + ( + )Ô + 2 which shows in particular that Ä(0), no matter of the values of and 2. 7
Problem 4 (a) Write the system to be controlled as Ü Ü + Ù Ý Ü Set Choose the second state as Þ Ù Þ 2 Ü Þ 3 «Ô (Ü + Û) Ü 2 Þ 3 This gives together the extended model 0 Ü Ü + «0 0 Check the matrix Æ Ý 0 Ü + Û ¼ Þ 0 0 ½ ¼ 0 Ü + 0 0 0 0 «0 0 «0 Ù + «½ Ù 0 Û 0 0 0 As it does not have any eigenvalue in the right half plane, the model is in innovations form. (b) The Riccati equation 0 Ì Ë + Ë Å Ì Å Ë( Ì ) Ì Ë gives in partitioned form with 0 0 0 0 «0 0 Ë 2 2 22 2 2 22 2 2 22 0 + 2 0 2 22 «0 2 2 22 2 0 Evaluating the matrix equation componentwize, + 0 0 0 2 + 2«2 + 2 2 0 2 + «22 2 2 0 2 2 2 8
The second equation gives 22 2 «+ 2 from which we conclude that 2 0. Hence, the third equation gives The first equation implies 2 0 2 + 2 2 2«3 2 2 4 + 2«3 + 2 2 The positive sign must be chosen, as 0 must hold. Hence, ¼ Ë 2 + ( 4 + 2«3 + 2 ) 2 2 + 2«+ «2 (c) For the optimal regulator it holds and then Problem 5 Ä ( Ì ) Ì Ë 2 2 2 Ý ( ) Ä Á + ( Ì ) Ì Ë + Æ Æ + + 2 0 2 «+ ««( + + ) «2( + ) ( + + ) «2 ( + ) 2 ( + + 2 2 0 + + 0 «2 ) «( + ) 2 + 2 2 + Ô 4 + 2«3 + 2 «( + ) + Ô 2 + 2«+ The relay gives the bounds 0 2 ½. The circle in the circle criterion becomes the full left half plan. The linear part must hence lie fully in the right half plan. The linear part has transfer function 0 ( ) ( ) [ + À( )] 9 + ( + )( + 2) ½
The requirement Re 0 () 0 is equivalent to Problem 6 Re + ( + )( + 2) 0 µ Re ( + )( + )( + 2) 0 µ Re [ (2 2 ) + 3 2 + (2 2 ) 3) ] 0 µ (2 2 ) + 3 2 0 µ 2 + (3 ) 2 0 3 0 µ (a) Introduce ( ) [Á + ( )] ( ), which gives 0 ( ) ( ) ( ) Á 0 (b) The closed loop system will be ( ) Ã( + Ì ) Ì (0 + ) + Ã( + Ì ) Á + Á precisely when Ì Ì 2 0 Ã Ã 2 0. (c) The robustness criteria for the uncertainty in part (a) is Ì ½, that is the largest singular value of Ì () () must be less than one for all frequencies. Here Ì ( ) Á. Using + ( ) as in part (b), Ì ( ) ( ) + 0 0 The singular values of this matrix happens to be equal, and () 2 () Hence, robust stability is guaranteed if which finally leads to the condition Ô 2 + Ô 2 + 0
(d) The loop gain will be Ä( ) ( ) Ý ( ) The sensitivity function becomes Ë( ) [Á + Ä( )] + ( + ) 2 2 + One can then see that the system is stable precisely when ( + ) 2 2 has all zeros strictly in the left half plane, which is the case when.