Uppsala University Department of Information Technology Systems and Control Professor Torsten Söderström Final exam: Automatic Control II (Reglerteknik II, TT495) Date: October 22, 2 Responsible examiner: Torsten Söderström Preliminary grades: 3 23 32p, 4 33 42p, 5 43 5p. Instructions The solutions to the problems can be given in Swedish or in English, except for Problem 2 that must be solved in English. Problem is an alternative to the homework assignment. (In case you choose to hand in a solution to Problem you will be accounted for the best performance of the homework assignments and Problem.) Solve each problem on a separate page. Write your code 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, copies of OH transparencies. Note that the following is not allowed: Exempelsamling med lösningar. Good luck!
Problem Consider the system Ü(Ø + ) Ý(Ø) 5 2 3 2 2 Ü(Ø) + 2 3 Ü(Ø) Ù(Ø) (a) Is the system controllable? (b) Is the system observable? (c) Is the system controllable from the input Ù? Problem 2 Consider the system ( ) controlled with the regulator À ( ) according to the block diagram. ¹ Σ ¹ À ¹ In this case, the system is a simple DC motor. Set ( ) ( + ) À ( ) à (a) For which values of à is the closed-loop system stable? (b) Assume the system is computer-controlled, so that the closed loop system actually can be represented as in the block diagram below. Let the sampling interval be equal to. ¹ Σ ¹ À ¹ ZOH ¹ 2
Determine the discrete-time transfer function (Þ) that corresponds to sampling ( ). Set «. 4 points Hint. It may be an advantage to first split ( ) as a sum of two first order systems: ( ) ( + ). (c) Assume that the discrete-time regulator is a proportional regulator, so that that À (Õ) Ã. Show that the closed-loop system can be unstable for some values of Ã. 3 points Problem 3 Consider a random process in discrete-time Ý(Ø) with covariance function Ö() 5 ( ) (a) Assume that we want to make prediction steps ahead of the process. A very simple estimator of the Ý(Ø + ) from data available at time Ø would be to take ˆÝ(Ø + ) Ý(Ø) that is to take the last available measurement. How large will the prediction error variance Î [Ý(Ø + ) ˆÝ(Ø + )] 2 become? (b) A somewhat more sophisticated predictor would be to multiply the last available measurement with a constant, say ˆÝ(Ø + ) «Ý(Ø) Determine how the prediction error variance depends on the parameter «, and how it can be minimized with respect to «. 3 points (c) Determine the spectrum of the process. 4 points (d) Use the spectrum to derive a time-domain model of the process on state-space form Ü(Ø + ) Ü(Ø) + Ú(Ø) Ý(Ø) Ü(Ø) + (Ø) and specify also the variances of Ú(Ø) and (Ø). 3 points Problem 4 Consider a first order system given on polynomial form ( + Õ )Ý(Ø) Õ Ù(Ø) + ( + Õ )(Ø) where (Ø) is white noise, with zero mean and variance 2. Assume that holds. 3
(a) Show that the system can be written in standard state-space form as follows Ü(Ø + ) Ü(Ø) + Ù(Ø) + Ú(Ø) Ý(Ø) Ü(Ø) How are Ú and related? (b) Assume both states can be measured. Determine the optimal feedback Ù(Ø) ÄÜ(Ø) that minimizes È Ø Ý2 (Ø). 4 points (c) Determine the Kalman filter (the predictor form) giving ˆÜ(ØØ ). 5 points Problem 5 Consider a first order process Ü(Ø) observed with some noise (Ø). This may be modelled as Ü Ý Ü + Ú Ü + where Ú and are white noise processes, with intensities Ö and Ö 2, respectively. Also, assume that. (a) Determine the spectrum and the variance of the signal Ü(Ø). (b) As the signal Ü(Ø) is not measured directly, consider estimating it with an observer ˆÜ ˆÜ + à (Ý ˆÜ) Show that this can be written as ˆÜ(Ø) (Ô)Ý(Ø) and determine the transfer function operator (Ô), and sketch the character of its Bode plot, that is sketch how () varies with. Let à be a fixed parameter. (c) Examine the estimation error Ü Ü the error can be written in the form ˆÜ as a function of Ú and. Show that Ü(Ø) À (Ô)Ú(Ø) + À 2 (Ô)(Ø) and determine the two transfer function operators À (Ô) and À 2 (Ô). (d) Determine the variances of the two error terms in part (c), that is the variances of À (Ô)Ú(Ø) and of À 2 (Ô)(Ø). 4
(e) Determine the observer gain à that minimizes the total error variance Î (Ã) [À (Ô)Ú(Ø)] 2 + [À 2 (Ô)(Ø)] 2 (f) Determine the Kalman filter related to the state space model set up in this problem. What is the variance Ü 2 (Ø) when the Kalman filter is used? 5
Uppsala University Department of Information Technology Systems and Control Prof Torsten Söderström Automatic control II, October 22, 2 Answers and brief solutions Problem (a) As is nonsingular, the controllability matrix has full rank. The system is controllable. (b) As is nonsingular, the observability matrix has full rank. The system is observable. (c) As the element and the system is in diagonal form, the state Ü is not controllable. The system is not controllable using Ù. Problem 2 (a) The closed loop characteristic equation becomes + ( )À ( ) µ 2 + + à (b) The poles will lie in the left half plane for all positive values of Ã. ( ) ( + ) + The sampled-data system can be written as follows. Using «, (Þ) Þ «Þ «Þ «(Þ )( «) (Þ )(Þ «) (c) The closed loop characteristic equation becomes leading to or yet + (Þ)À (Þ) (Þ )(Þ «) + à [Þ( + «) + ( ««)] Þ 2 + Þ ( «+ Ã( + «)) ßÞ Ð A necessary condition for stability is 2 2 + «+ Ã( ««) ßÞ Ð 2 Both these two inequalities are violated if à is large enough. 6 Þ( + «) + ( ««) (Þ )(Þ «)
Problem 3 (a) In this case (b) In this case Î [Ý(Ø + ) ˆÝ(Ø + )] 2 [Ý(Ø + ) Ý(Ø)] 2 Ý 2 (Ø + ) + Ý 2 (Ø) 2Ý(Ø + )Ý(Ø) Ö() + Ö() 2Ö() ( ) Î («) [Ý(Ø + ) ˆÝ(Ø + )] 2 [Ý(Ø + ) «Ý(Ø)] 2 Ö() + «2 Ö() 2«Ö() Minimizing Î («) with respect to «gives that the best value of «is «Ö()Ö(), and min Î («) Ö() Ö 2 ()Ö() 5 5 2 5( 2 ) «which is always smaller than ( ). (c) The spectrum becomes directly from the definition () ½ Ö() ½ ½ Ö() + Ö() ½ ½ 5( ) + ½ 5( ) 5 Ö() 5 + 5 5 5 + ( )( ) 5( 2 ) + 2 2 cos() 2 + + (d) The process is apparently an AR() process, and can be represented as a first order system with Ê 5( 2 ) Ê 2 7
Problem 4 (a) The transfer function operators from Ù(Ø) and Ú(Ø) to Ý(Ø) appear from the calculations Ý(Ø) Õ + Õ(Õ + ) Õ Õ(Õ + ) Õ Õ Õ + (Õ + ) Ù(Ø) + Õ + Õ(Õ + ) Ú(Ø) Ù(Ø) + Ù(Ø) + Ù(Ø) + Ú(Ø) This describes precisely the given system if Ú(Ø) (Ø + ). Ú(Ø) (b) In this case É Ì É 2. The Riccati equation becomes Ë Ì Ë + Ì Ì Ë Ë Ì Ì Ë Set Ë 2 2 22 Spelling out the Riccati equation elementwise leads to 2 2 22 + The feedback gain becomes Ú(Ø) Ä ( Ì Ë) Ì Ë 2 (c) Set È Ô Ô 2 Ô 2 Ô 22 The Riccati equation is in this case È È Ì + Æ 2 Æ Ì È Ì È Ì È Ì È È + 2 Ô È 8
2 + È Ô Ô 2 Ô Ô Ô 2 2 + 2 + Ô 22 Ô 2 2 Ô Ô 22 Ô 2 2 Ô Hence È 2 + where is the remaining unknown to be determined. It holds leading to Ô 22 Ô 2 2 Ô µ 2 2 ( 2 ) 2 2 + 2 4 2 + 2 2 4 2 µ with the two solutions 2 2 ( 2 ) and hence must be chosen. Therefore, È 2 2 + + 2 2 2 The Kalman gain becomes Ã È Ì È Ì 2 2 ( ) The Kalman filter will be ˆÜ(Ø + Ø) ( Ã)ˆÜ(ØØ ) + Ù(Ø) + ÃÝ(Ø) ˆÜ(ØØ ) + Ù(Ø) + Ý(Ø) It even follows that ˆÜ 2 (Ø + Ø). As this applies at all times, we also have ˆÜ (Ø + Ø) ˆÜ (ØØ ) + Ù(Ø) + ( )Ý(Ø) 9
Problem 5 (a) Ü(Ø) Ô + Ú(Ø) µ Ü() Ö 2 + 2 The variance of Ü(Ø) can be obtained by integrating the spectrum, but even easier by solving a Lyapunov equation, (b) One gets directly 2È + Ö µ È Ü 2 (Ø) Ö 2 (Ô) Ã Ô + + à which is a first order, lowpass, filter. The static gain is Ã( + Ã). (c) Set Ü Ü ˆÜ. Then Ü ( Ü + Ú) ( ˆÜ + ÃÝ ÃˆÜ) ( Ã) Ü + Ú Ã It follows that À (Ô) Ô + + à À 2(Ô) Ã Ô + + à (d) Using part (a), [À (Ô)Ú(Ø)] 2 Ö 2( + Ã) [À 2(Ô)Ú(Ø)] 2 à 2 Ö 2 2( + Ã) (e) Î (Ã) Ö + à 2 Ö 2 2( + Ã) Î ¼ (Ã) µ ( + Ã) 2ÃÖ 2 (Ö + à 2 Õ Ö 2 ) Õ µ à 2 Ö 2 + 2ÃÖ 2 Ö µ à 2 + Ö Ö 2 à + 2 + Ö Ö 2 The positive sign is chosen as + à must hold to guarantee stability. (f) The Riccati equation gives in this case È È + Ö È 2 Ö 2 È 2 + 2Ö 2 È Ö Ö 2 Õ È Ö 2 + 2 Ö2 2 + Ö Ö Õ 2 à ÈÖ 2 + 2 + Ö Ö 2 which is the same result as in part (e).