机电控制技术考试试卷及答案解析

Mechatronic Control SystemsSpring 2013Dr. Bin YaoFINAL EXAMApril 30, 2013INSTRUCTIONS:1.This is a Closed book exam. You are allowed one help sheet of hand-writtensummary.2.Your exams must be stapled.3.Circle your final answers.4.Be neat and clear.PROBLEM 1 (20Points)Consider the following feedback system:where()3()(1)sP ss s+=-.You are required to design a controller to meet the following performance specifications: (P1).Zero steady-state error for ramp type reference input ()r t and constant disturbance ()d t(P2).The resulting closed-loop system should not have excessive transient responses for step reference input ()R s, i.e., your design should avoid either excessive large overshoot or large undershoot in the step responses.To solve this problem, you are required to follow the following procedure:a)Determine the correct controller structure that is needed to meet the performancerequirement P1. To receive full credit, you need to justify your answer as well.b)Determine the suitable desired pole locations of the closed-loop system so that theperformance requirement P2 can be satisfied. Again, to receive full credit, you need to justify your answer as well.c)Determine the unknown controller parameters to meet the above performancerequirements.Solutions:由上面两条定理可以得到结论：1、a). As the plant has an integrator, to satisfy (P1), the controller only needs one integrator, i.e.,()()()(),with order()order()1()CC CCN sC s N sD ssD s=≤+(1) With the controller (1), the closed-loop output is given by()()()()()()223()3()()()()1()3()1()3()C CC C C Cs N s s sD sY s R s d ss s D s s N s s s D s s N s++=+-++-++(2)Y(s)-()R s()d s()C s()P sController PlantThus, for ramp type reference input (i.e., 2()/c R s r s =) and constant disturbance (i.e., ()/d d s A s =), the system output tracking error is()()()()()()()()()()2222()()()1()3() ()()1()3()1()3()1()3() 1()3()C C C C C C C c C dC C E s R s Y s s sD s s sD s R s d s s s D s s N s s s D s s N s s D s r s D s A s s D s s N s =--+=--++-++--+=-++ (3)So as long as the CL system is stable (i.e., the denominator in (3) has all roots in LHP), the condition for applying FVT is satisfied. By FVT, you can easily show that the steady-state error in (3) is zero.PROBLEM 1 (conts)b). As the plant has an unstable pole at 1, to avoid excessive overshoot due to this unstable pole, the CL bandwidth should be higher than the break frequency of this unstable pole, which can be roughly met by imposing the following conditions on dominant CL poles:{}Re 1CL p > (4) As the plant has a stable zero at -3 which tends to increase the overshoot significantly when itis slower comparing to the CL bandwidth, to avoid excessive overshoot, the following condition on dominant CL poles should be imposed normally:{}Re 3CL p < (5) Thus we can place the dominant CL poles around 2 to make a compromise between theconflicting requirements of (4) and (5). Note that as this zero is stable, you can also cancel this zero in the controller design (by placing one CL pole at -3) to remove its effect on the CL response with respect to the reference input as well. In that case, its effect still appears in the CLTF from the disturbance input to the output.c). With a second-order controller of the form (1),22101()()c c c c b s b s b C s s s a ++=+ (6) we have four controller parameters free to choose and the resulting CL system has four poles. Thus we can arbitrarily place all four CL poles with the controller form of (6). For simplicity, let all CL poles at -2, which leads to the following desired CL characteristic polynomial (CLCP):4432()(2)8243216CLd A s s s s s s =+=++++ (7)From (2), the actual CLCP with the controller (6) is()()()()()22121043212112010()1()(3)1333CL c c c c c c c c c c c c A s s s s a b s b s b s s a b s a b b s b b s b =-+++++=+-++-+++++ (8) Comparing (7) and (8), we obtain12111220110018107/36 2.97324217/36633280/98.931616/3 5.3c c c c c c c c c c c c a b a a b b b b b b b b -+===⎧⎧⎪⎪-++===⎪⎪⇒⎨⎨+===⎪⎪⎪⎪===⎩⎩ (9)Thus,268.9 5.3()( 2.97)s s C s s s ++=+ (10)PROBLEM 2 (20 Points)Consider the following two-DOF feedback system:Fig.2.1where1()P s s=and the system has the following characteristics:(C1) The input disturbance ()i d t has significant energy in the frequency band [0, 1] rad/s. (C2) The measurement noise ()n t has significant energy in the frequency band [5, 100]rad/s. (C3) The reference signal ()r t has significant energy in the frequency band [0, 10] rad/s. You are required to synthesize proper controller transfer functions ()F s and ()C s to meet the following specific design goals while taking into account the above system characteristics: (P1) Zero steady-state errors for ramp type output disturbances ()o d t(P2) The response of the closed-loop system for step reference input ()r t has no oscillations. (P3) The closed-loop system should follow the reference signal well in the frequency bandspecified in (C3).To solve this problem, you may want to follow the following procedure:a) Determine the correct structure of feedback controller ()C s that is needed to meet the steady-state performance requirement P1.b) Determine the suitable desired pole locations of the closed-loop system that take into account the system characteristics (C1)-(C3). To receive full credit, you need to justify your answer as well.c) Determine the parameters of the feedback controller ()C s to place the closed-loop poles at the desired locations.d) Determine a suitable filter transfer function ()F s so that (P2) and (P3) are satisfied.Solutions:a). As the plant has an integrator, to satisfy (P1), the controller only needs one integrator, i.e.,()()()(),with order ()order ()1()C C C C N s C s N sD s sD s =≤+ (1)b). (C1) demands that the CL bandwidth should be at least higher than 1 rad/s to have certain attenuation to the input disturbance in the frequency band of [0, 1] rad/s. (C2) implies that the CL bandwidth should not be set too high to amplify the effect of noise in the frequency band of [5, 100] rad/s. Thus a good compromise for the CL bandwidth to meet both requirementsshould be around 2 to 3 rad/s. So assume that we would like to place dominant CL poles at -3 in the following.c). With a first-order controller of the form (C1)10()c c b s b C s s+= (2)there will be two controller parameters free to choose and the resulting CL system will be of order 2. Thus we can arbitrarily place the two CL poles. With 3CLd p =-, the desired CLCP is22()(3)69CLd A s s s s =+=++ (3)The actual CLCP with the controller (2) is210()CL c c A s s b s b =++(4) Comparing (3) and (4), we obtain10669()9c c b s C s b s =⎧+⇒=⎨=⎩ (5) d). With the controller (5), the CLTF from ()R s to ()Y s is()12()69()()3CL Y s s G s R s s +==+ (6)Thus, to be able to track reference signal ()r t in the frequency band of [0, 10] rad/s, a feedforward TF ()F s is needed so that the resulting CLTF from ()R s to ()Y s has abandwidth far more than 10 rad/s. As such, we needs to cancel the slow CL poles at -3 in (6). Furthermore, to avoid overshoot, the stable zero in (6) should be cancelled as well. With all these in mind, we can choose()()()23()1()()691()1CL f f s Y s F s G s s s R s s ττ+=⇒==+++ (7)where 1/100.1f τ<<= is the small time constant of the additional filter needed to make()F s proper.Problem 3 (40 Points)Consider the control of an inertia load such as the rigid ECP emulator introduced in the lectures and the homework. In the presence of disturbance forces such as the Coulomb friction force, the inertia load dynamics can be described by:()[]12010(),41110x y x x u d t x x y y x⎡⎤⎡⎤⎡⎤⎡⎤=++==⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦⎣⎦=where ()d t represents the disturbance force. Assume that the disturbance force is constant butunknown (i.e., ()d t =unknown constant), and only the output y is measured.a) Design a minimum-order observer to estimate the unmeasured state (i.e., the velocity) and the unknown constant disturbance force. The observer gain should be chosen to place all observer poles at -5.b) Consider the following output feedback control law with disturbance estimate:ˆˆ()u Kxd t r =--+ where ˆ()xt and ˆ()d t represent the plant state and disturbance estimates from part a), and r represents the filtered reference input. Determine the feedback gain K so that allun-cancelled poles of the closed-loop transfer function from the filtered reference inputr to the output, i.e., ()()()CL Y s G s R s =, are at -1. c) Draw the equivalent block diagram of the closed-loop system with the above controller and estimator using transfer functions. To receive full credit, you need to obtain the explicit expressions of all relevant transfer functions.d) Obtain the closed-loop transfer function from the filter reference input r to the output,()()()CL Y s G s R s = and verify that all its un-cancelled poles are at -1 as required. e) Obtain the closed-loop transfer function from the disturbance input ()d t to the output,()()()dCL Y s G s d s =. Use this transfer function to show that constant disturbances will not cause any steady-state error in the output as expected.Solutions:a). 降阶观测器极点配置的方法For constant disturbance ()d t, the augmented system model is[]01004111,0000100a ay y y x d y y u x y d dt d d d y x ⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=--+==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦= (1)which is in the standard form for designing minimum-order observer with 2e x y x d d ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦and[]1111141110,10,,,0,0000e e ee e a A A A b B --⎡⎤⎡⎤⎡⎤======⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦(2) The observer gain matrix 12e e e l L l ⎡⎤=⎢⎥⎣⎦can then be determined by placing the eigenvalues of112110e ee e e e l A L A l --⎡⎤-=⎢⎥-⎣⎦(3)at -5, i.e.,()()212212211151025e e e e s l s l s l s s s ls ++-⎡⎤=+++=+=++⎢⎥⎣⎦(4) ⇒129,25e e l l ==(5)The minimum-order observer is thus given by[]111111ˆˆ()()()101691ˆ 25022509ˆˆˆ25p ee e e p ee e e e e e e e p e p e p xA L A x A L A L A L a yB L b u x y u xx L y x y =-+-+-+---⎡⎤⎡⎤⎡⎤=++⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦⎡⎤=+=+⎢⎥⎣⎦(6)b).状态反馈极点配置的求法 ：根据独立性原则，状态观测器和状态反馈互不影响，也就是K 和L 互不影响The closed-loop poles due to the state feedback gain K are determined by()[]()1212221101041411 14p p ssI A B K sI k k k s k s k s k -⎛⎫⎡⎤⎡⎤⎡⎤--=--= ⎪⎢⎥⎢⎥⎢⎥+++--⎣⎦⎣⎦⎣⎦⎝⎭=++++ (7)Thus, to have the un-cancelled CL poles at -1,()()2221121413,1s k s k s k k ++++=+⇒=-=(8)c). With the minimum-order observer (6), the control law is given by[][][]21ˆˆˆ()31ˆˆ 311ˆ 3111e e e p y u Kx d t r x r x y x r y xr ⎡⎤=--+=---+⎢⎥⎣⎦=-+=--+ (9)Substituting (9) into (6),101691ˆˆ25022501101001ˆ 2502250p p p x x y u x y r --⎡⎤⎡⎤⎡⎤=++⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦--⎡⎤⎡⎤⎡⎤=++⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦(10)Thus,()11101001ˆ()()()2502250010011()()2511225011p X s sI Y s R s s Y s R s s s s -⎛-⎫⎧-⎫⎡⎤⎡⎤⎡⎤=-+⎨⎬ ⎪⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦⎝⎭⎩⎭⎧-⎫⎡⎤⎡⎤⎡⎤=+⎨⎬⎢⎥⎢⎥⎢⎥-+-+⎣⎦⎣⎦⎣⎦⎩⎭(11)and the control law (9) in s-domain is given by[]()()()()()22ˆ()31()11()()3252525 31()1()11113116255 ()()1111pU s Y s X s R s s s Y s R s s s s s s s s Y s R s s s s s =--+--⎛⎫⎛⎫=-++- ⎪ ⎪++⎝⎭⎝⎭+++=-+++ (12)The equivalent block diagram of the above CL system can then be drawn below ：图中从u+d 到Y 的输出是由原系统中的A 、B 、C 矩阵求出来的！-)(s R ()d s ()()2511s s s ++214s s ++()Y s ()223116255s s s +++or-)(s R ()d s ()225311625s s s +++214s s ++()Y s ()231162511s s s s +++d). From Fig.1, the CLTF from ()R s to ()Y s is()()()()()2224322222()5311625()()3116251246602551151CL Y s s s s G s R s s s s s s s s s s s +++==+++++++==+++ (13)which has all uncancelled poles at -1.e). From Fig.1, the CLTF from ()d s to ()Y s is()()()22()11()()15dCL Y s s s G s d s s s +==++ (14)which has a s in the numerator. As such, (0)0dCL G =, indicating that the constant disturbances will not cause steady-state error.Problem 4 (20 Points)Consider the same second-order system as in Problem 3 but with an input disturbance, i.e.,()[]011()10111x x u d t y x⎡⎤⎡⎤=++⎢⎥⎢⎥⎣⎦⎣⎦= where ()d t represents the disturbance force. Assume that the disturbance force is constant butunknown (i.e., ()d t =unknown constant), and only the output y is measured. Design an output feedback controller using the technique of state-estimator with disturbance estimation andcompensation (i.e., ˆˆ()u Kxd t r =--+) to achieve the following performance requirement: a) Stable closed-loop system.b) Zero state-steady error for any constant input disturbance ()d t .c) All the un-cancelled poles of the closed-loop transfer function from the filtered reference input r to the output, i.e., ()()()CL Y s G s R s =, are at -2.d) All other assignable closed-loop poles should be placed at -10.Solutions 1:As shown in Problem 3, the given system is not observable but detectable, and is not controllable but stabilizable.By introducing the coordinate transformation of11221211,,or 11c c x z z x T z T x z z =+⎡⎤==⎢⎥=--⎣⎦(a1)The system matrices in the new coordinate z are[]11101,,20010c c c c A T AT B T B C CT --⎡⎤⎡⎤======⎢⎥⎢⎥-⎣⎦⎣⎦(a2) which isolates the uncontrollable and unobservable mode 21λ=- represented by thecoordinate 2z . Though this mode cannot be moved with any state feedback and observable design, it is stable and does not contribute to the overall TF from the input to the output. Thus we can ignore this mode and only consider the controllable and the observable mode in synthesizing the output feedback controller. Thus the given system is reduced to()111()2z z u d t y z =++=(a3)For constant disturbance ()d t , the augmented system model is[]1111,00020a aaa a a B A aC z x x u x d y x ⎡⎤⎡⎤⎡⎤=+=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦= (a4)A full-order observer can then used to estimate the augmented states in (a4). The observer gain matrix []12TL l l =should be chosen such that the eigenvalues of[]112212*********a a l l A LC l l -⎡⎤⎡⎤⎡⎤-=-=⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦(a5)are at -10, -10, i.e.,()()212122121212102s l s l s l s l s -+-⎡⎤=+-+=+⎢⎥⎣⎦ (a6)⇒1210.5,50l l == (a7)With the above observer gain, a full-order observer can be constructed as()201110.5ˆˆˆ1000050a a a a a a x A LC x B u Ly x u y -⎡⎤⎡⎤⎡⎤=-++=++⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦(a8) To have the CL poles by the state feedback at -2, from (a3), the state feedback gain z K for 1z should be chosen as （忽略Z1，因为不可观）21)(+=+-=--s K s BpK Ap sI3z K =(a9)With the observer (a8) and the above gain z K in (a9), the following stabilizing output feedback control law can be used:[]1ˆˆˆ()31z a u K z d t r x r =--+=-+ (a10)Solutions 2:For constant disturbance ()d t , the augmented system model is[]1201111011,0000110aaaa a a B A aC x x x x u x x d d y x ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=+==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦= (1)As shown in Problem 3, the above system is not observable but detectable, which means thatwe cannot move the unobservable mode 21λ=- with any observable designs but we still can design a stable state observer. So when a full-order observer is used to estimate theaugmented states in (1), we can arbitrarily place the other two observer CL poles while the third one should be at -1. Thus, the observer gain matrix []123TL l l l =should be chosensuch that the eigenvalues of[]11122233301111101110110000a a l l l A LC l l l l l l --⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥-=-=--⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦(3) are at -10, -10, and -1, i.e.,()()()()112222213331111112101s l l l s l s s l l s l s s l l s +-+-⎡⎤⎢⎥⎡⎤-++-=+++-+=++⎣⎦⎢⎥⎢⎥⎣⎦(4) ⇒21321,50l l l +== (5)The fact that there are infinite number of solutions to the observer gains is due to theappearance of unobservable mode. With the observer gains satisfying (5), a stable full-order observer can be constructed as()111111111ˆˆˆ2021112150500050a a a a a a l l l x A LC x B u Ly l l x u l y --⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=-++=-+-++-⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦(6) where 1l can be any value.Again, as shown in Problem 3, the given system is not controllable but stabilizable with the uncontrollable mode given by 21λ=-. Thus, we cannot use state feedback to move theuncontrollable mode 21λ=-. The state feedback gain K should thus be chosen such that the eigenvalues of()[]()121212212121210111101 1(1)(1)p p s k k sI A B K sI k k k s k s k k s k k s s k k +-⎛⎫⎡⎤⎡⎤⎡⎤--=--= ⎪⎢⎥⎢⎥⎢⎥-++⎣⎦⎣⎦⎣⎦⎝⎭=++++-=+++- (7)at -1 and -2 as required, which leads to123k k +=(8)Again, the non-unique solution to K is due to the appearance of uncontrollable mode.With the observer (6) and the gain K in (8), the following stabilizing output feedback control law can be used:[]22ˆˆˆ()31a u Kx d t r k k x r =--+=--+ (9)where 2k can be any value.。