经典教材Computer_Controlled Systems答案.pdf-资料库

vince426-1920759-TextBookSolutions.pdf-第1页.png

第1页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第2页.png

第2页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第3页.png

第3页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第4页.png

第4页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第5页.png

第5页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第6页.png

第6页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第7页.png

第7页 / 共116页

vince426-1920759-TextBookSolutions.pdf-第8页.png

第8页 / 共116页

Computer-ControlledSystems Third Edition SolutionsManual Karl J. Åström Björn Wittenmark Department of Automatic Control Lund Institute of Technology October 1997

Preface This Solutions Manual contains solutions to most of the problems in the third edition of Åström, K. J. and B. Wittenmark H1997I: Computer controlled Systems – Theory and Applications, Prentice Hall Inc., Englewood Cliffs, N. J. Many of the problems are intentionally made such that the students have to use a simulation program to verify the analytical solutions. This is important since it gives a feeling for the relation between the pulse transfer function and the time domain. In the book and the solutions we have used Matlab/Simulink. Information about macros used to generate the illustrations can be obtained by writing to us. It is also important that a course in digital control includes laboratory exercises. The contents in the laboratory experiments are of course dependent on the avail- able equipment. Examples of experiments are P P P Illustration of aliasing Comparison between continuous time and discrete time controllers State feedback control. Redesign of continuous time controllers as well as controllers based on discrete time synthesis Controllers based on input-output design Control of systems subject to stochastic disturbances. P P Finally we would like to thank collegues and students who have helped us to test the book and the solutions. Karl J. Åström Björn Wittenmark Department of Automatic Control Lund Institute of Technology Box 118 S-220 00 Lund, Sweden i

Solutions to Chapter 2 Problem 2.1 The system is described by −ax bu Sampling the system using H2.4I and H2.5I gives ˙x y cx xHkh hI e− ahxHkhI yHkhI cxHkhI ah 1 − e− uHkhI b a The pole of the sampled system is expH −ahI. The pole is real. For small values of h the pole is close to 1. If a 0 then the pole moves towards the origin when h increases. If a 0 then the pole moves along the positive real axis. Problem 2.2 a. Using the Laplace transform method we ﬁnd that 9>>;! 1 − 1 9>>; 8>>: cos h sin h sI − A 9>>; ds 8>>: sin s hZ hZ − sin h cos h eAsB ds L− 1 cos s 0 0 1 −1 s 1 s2 1 8>>: s 9>>; 8>>: 1 − cos h sin h F eAh L− G b. The system has the transfer function GHsI s 3 s2 3s 2 Using the Table 2.1 gives HHqI 2 c. One state space realization of the system is s 3 2 2h 2h h h − 1 − e− q − e− Hs 1IHs 2I 2 s 1 − 1 − e− q − e− 8>>>>>>>: 0 0 0 9>>>>>>>; x 8>>>>>>>: 1 9>>>>>>>; u 9; x 8: 0 0 1 9>>>>>>>; A3 0 8>>>>>>>: 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 1 s 2 1 ˙x y Now A2

then and hZ 0 G 9;T ds 0 0 1 0 h h2/2 h 1 8>>>>>>>: 1 9>>>>>>>; 8: h h2/2 h3/6 9; 9>>>>>>>; 8>>>>>>>: h 9>>>>>>>; x x h2/2 h3/6 Hq − 1I2 s2/2 9;T 8: 0 0 1 Hq − 1I3 x x eAh I Ah A2h2/2 ((( 8: 1 s hZ 0 eAsB ds HHqI CHqI − 1G I− 8>>>>>>>: x x h3Hq2 4q 1I h2Hq 1I/2 hHq − 1I 6Hq − 1I3 Problem 2.3 F eAh yHkI A 1 h ln F − 0.5yHk − 1I 6uHk − 1I 1y 6q− y − 0.5q− 1u qy − 0.5y 6u xHkh hI 0.5xHkhI 6uHkhI yHkhI xHkhI Hdiscrete-time systemI ˙xHtI axHtI buHtI yHtI xHtI Hcontinuous time systemI easb ds 6 0 a hR eah 0.5 eas b ds b a 8<: F hZ ah ln 0.5 8>>: −0.5 8><>: xHkh hI 9; xHkhI 8: 1 1 8>>: s 0.5 b 6a − 1 eah yHkhI : eas 0 0 6 eah − 1 h ln 2 − h b a 12 ln 2 0 1 −0.3 h 9>>; xHkhI 9>>; 0 −1 s 0.3 8>>: 0.5 0.7 9>>; uHkhI Eigenvalue to F detHsI − l 1 I −0.5 l 2 Both eigenvalues of F uous system exists. Hs 0.5IHs 0.3I 0 0 −0.3 on the negative real axis No corresponding contin- a. b. 2 F G F

c. yHkI 0.5yHk − 1I 6uHk − 1I yHk 1I HHqI −0.5yHkI 6uHtI q 0.5 6 one pole on the negative real axis. No equivalent continuous system exists. Problem 2.4 Harmonic oscillator Hcf. A.3 and 3.2aI. xHk 1I F xHkI G uHkI yHkI C xHkI a. Pulse transfer operator 1 h p 2 9>>; HhI −1 0 G HhI y − sin h cos h sin h 8>>: cos h sin h 8>>: 1 − cos h 9>>; 9>>; G 8>>: 1 9>>; 8>>: 0 1 9; x 8: 1 0 9>>;− 9;8>>: q −1 18>>: 1 8: 1 0 9>>; q 1 9>>;8>>: 1 9;8>>: q 1 q2 1 1 1 1 z2 1 z − 1 p Hk − 1Iq Hk − 1I q Hk − 1I q Hk − 1I 1 − cos 9>>; 1 9;8>>: s −1 2 9>>;− 18>>: 0 8: 1 0 z z − 1 1G I− −1 q 9>>; 1 1 q k 1 s 1 s2 1 1 HHqI CHqI − q2 1 8: 1 0 YHzI HHzIUHzI z 1 z2 1 yHkI sin p 2 where q Hk − 1I is a step at k 1. GHsI Jsee Probl. 2.2K sHs2 1I 1 s − s s2 1 1 YHsI yHtI 1 − cos t yHkhI 1 − cos b. The same way as a. p 2 k 4 k Notice that the step responses of the continuous time and the zero-order hold sampled systems are the same in the sampling points. yHtI 1 − cos t, and yHkhI 1 − cos p Problem 2.5 Do a partial fraction decomposition and sample each term using Table 2.1: GHsI 1 s2Hs 2IHs 3I 1 6 1 s2 − HHqI 1 12 q 1 Hq − 1I2 − 5 36 1 q − 1 5 36 1 8 1 1 4 s 1 − e− q − e− 1 s 2 − 1 27 2 2 − 1 1 s 3 9 1 − e− q − e− 3 3 3 F F F

Problem 2.6 Integrating the system equation gives ˙x Ax Bu khhZ eAsBd Hs − khIuHkhIds xHkh hI eAhxHkhI kh eAhxHkhI BuHkhI Problem 2.7 The representation H2.7I is xHkh hI F xHkhI G uHkhI yHkhI C xHkhI and the controllable form realization is zHkh hI ˜F zHkhI ˜G uHkhI where 8>>: 1 h 8: 1 0 0 1 9>>; 9; F C G ˜C From Section 2.5 we get yHkhI ˜C zHkhI 8>>: h2/2 9>>; 8: h2/2 h2/2 h ˜F 9; 8>>: 2 −1 1 0 9>>; 9>>; 8>>: 1 0 ˜G 1 ˜F TF T − ˜G TG ˜C CT − This gives the following relations 1 or or ˜F T TF ˜C T C t11 2t11 − t21 t21 t11 ht11 t12 2t12 − t22 ht21 t22 t12 t11 ht12 1 h2 2 t21 ht22 0 h2 2 t21 1 t11 h2 h2 2 2 t12 h2 t22 0 h2 2 2 H1I H2I H3I H4I Equations H1I–H4I now give t11 t21 1/h2 t12 −t22 1/H2hI or T 1 2h2 4 9>>; 8>>: 2 h 2 −h

8>>: z − 0.2 0 z − 0.5 9>>;8>>: 2 1 9>>; Problem 2.8 The pulse transfer function is given by 1G I− 8: 1 0 9; HHzI CHzI − zHz − 0.5I zHz − 0.5I 2Hz − 0.1I 2z − 0.2 zHz − 0.5I 8>>: f 9>>; x 8>>: −a ˙x b c −d Problem 2.9 The system is described by 9>>; u Ax Bu g The eigenvalues of A is obtained from Hl aIHl dI which gives l r − bc l 2 Ha dIl ad − bc 0 a d 2 Ha − dI2 4bc 4 − The condition that aA bA c, and d are nonnegative implies that the eigenvalues, l 1 and l 2, are real. There are multiple eigenvalues if both a d and bc 0. Using the result in Appendix B we ﬁnd that eAh a 0I a 1 Ah el 1h a 0 a 1l 1h el 2h a 0 a 1l 2h and This gives To compute G we notice that a 0 and a 1 depend on h. Introduce l 2el 1h a 0 l 1el 2h − l 2 l 1 − − el 2h a 1 el 1h l 2Ih Hl 1 − hZ hZ a 0HsIds 0 b 0 b 1 sa 1HsIds l 1 l 2 1 l 1 − l 2 1 l 1 − l 2 8>>:a 0 − a 1ah a 1ch F a 1bh a 1dh a 0 − el 2h − 1 − l 2 l 1 el 1h − 1 1 l 1 el 1h − 1 − 1 l 2 el 2h 9>>; 0 then G b 0 B b 1 AB − 1 5 F –

Problem 2.10 a. Using the result from Problem 2.9 gives l 1 l 2 −0.0197 −0.0129 el 1h 0.7895 el 2h 0.8566 Further and a 0 0.9839 9>>; 0.176 0.857 a 1 0.8223 8>>: 0.790 8>>: 0.281 0.0296 0 9>>; F a 0I 12a 1A b 0 11.9412 b 1 63.3824 G Hb 0 I b 1 AI B b. The pulse transfer operator is now given by 8: 0 1 9;8>>: q − 0.790 −0.176 9>>;− 18>>: 0.281 0.0297 9>>; 0 q − 0.857 HHqI CHqI − 1G I− 0.030q 0.026 − 1.65q 0.68 q2 which agrees with the pulse transfer operator given in the problem formula- tion. Problem 2.11 The motor has the transfer function GHsI 1/HsHs 1II. A state space represen- tation is given in Example A.2, where 1 A 9>>; 1 0 8>>: −1 0 8>>: e− hZ h 1 9>>; F expHAhI L− 0 hZ h 1 1 − e− eAsBds G 0 0 0 B 8>>: 1 9>>; HsI − AI− 8>>: e− s s 1 − e− 9; 8: 0 1 C L− 1 1 sHs 1I 0 1 s 1 9>>;! 8>>: s 9>>; 9>>; ds 8>>: 1 − e− h e− h h − 1 This gives the sampled representation given in Example A.2. 6 F

资料库

经典教材Computer_Controlled Systems答案.pdf

相关推荐

课程资源

热门标签

最新资料