(电气工程与自动化专业英语)Chapter 12MIMO Systems

1、Add Your Company Slogan( (电气工程与自动电气工程与自动化专业英化专业英语语)Chapter )Chapter 12MIMO Systems12MIMO Systems6 A helicopter has three primary flight control modes with cross-coupling, including: Collective control: all the rotor blades are required to be simultaneously tilted. Cyclic control: all the rotor blade

2、s are required to be tilted in some specific sequential cycle. Tail Rotor control: the tail rotor blades are controlled by the foot pedals. 27 Each of the three controls has one primary effect on helicopter flight behavior. However, changes to any of the control inputs cause additional effects that

3、the control system (usually a pilot) must counteract. For example, the primary effect of the collective control mode is to increase upward force. However, this control input also result in the main rotor torque increasing, making the helicopter body rotate around the main rotor axis. While its body

4、rotation can be prevented by adjusting the tail rotor tilt, a desired change in one plant output (helicopter altitude) requires at least two of the control inputs (the collective and the tail rotor), resulting in the cross-coupling.28 Although it is often possible to apply techniques such as PID con

5、troller design and root locus design in these situations, the results are seldom satisfactory. Because SISO design techniques do not account for plant cross-coupling, the resulting controllers might be unable to satisfy the design requirements. The pole placement and optimal state-space design techn

6、iques are inherently capable of MIMO controller design due to the fact that they can effectively deal with the cross-coupling in the plant. 29 Note that in the helicopter example, the controller of the helicopter system works as a regulator. That is, all the states, inputs, and outputs are defined a

7、s deviations from their equilibrium values. Consequently the feedforward gain N is not necessary, and all inputs to the observer-controller should be the required inputs subtracting the equilibrium values, and all the outputs are the obtained output adding the equilibrium values. Also notice that th

8、e inputs and outputs should be adjusted accordingly because of the nonlinearity of the helicopter model.2Text A: Pole placement for an aircraft glide slop control system 12.1 Concepts of MIMO systems2 12.2 Control design for an aircraft glide slop systemOutline111 The plant had two inputs, (aircraft

9、 attack angle and engine throttle setting), and three outputs (the measured aircraft speed V, the measured flight path angle , and the measured distance of the aircraft above or below the glide slope herr). For simplicity, we assume that the aircraft has no side-to-side motion and no roll orientatio

10、n. We also consider that the other flight aspects are handled adequately by other control systems. We ignore the effects of wind and the actions which are required to initially position the aircraft on the glides lope and to land the aircraft when it reaches the runway. 312 Even though the simplific

11、ations above are made, the glide slop control problem has significant cross-coupling between the inputs and outputs. Breaking the model into separate SISO systems would not be a satisfactory approach. We will use the pole placement method to design an observer-controller in order to achieve a satisf

12、actory control performance.313x : the forward direction in aircraft body coordinates. : the angle between the aircraft velocity vector V (m/s) and the body x-axis. : the angle between the horizontal and the velocity vector V. The engine thrust T acts in the x direction. The drag force D acts in then

13、egative direction of the velocity. The lift force L acts in an upward direction perpendicular to the velocity vector.314 The goal of the autopilot control system is to drive to zero and simultaneously maintain the commanded velocity Vc (m/s).Note that: The aircraft attack angle is manipulated by the

14、 elevator which is the moveable surface along the back edge of the horizontal tail. The engine throttle commands is represented by the engine thrust T in which the response time is ignored.errh315 Based on modelling of the aircraft glide slop control described above, the simplified system is describ

15、ed by the nonlinear equations shown in Eq.(12.1) ,cossin,sincossincostanerrgsmVDVTmgmVLVTmghV(12.1) 316 The lift and drag force terms L and D in Eq. (12.1) are further derived as in equation (12.2), in which is the air density and S is the aircraft reference area. 002221,21,2LLLDDLLDCCCCCKCLVCV SDVC

16、V S(12.2)317 The parameters in Eqs. (12.1) and (12.2) are defined in Table 12.1, in which the typical values represent a nominal flight condition during approach for aircraft landing.3183Note that： (12.1) and (12.2) should be linearized in order to design the controllers using the linear model appro

17、aches. the equilibrium condition occurs when all the derivatives in Eq. (12.1) are zero. A standard engineering approach is to use calculus techniques to linearize the nonlinear plant model about the equilibrium point. An alternative is to execute linmod() at the MATLAB command to derive the lineari

18、zed model from the Simulink diagram.19 The resulting model is represented as in Eq. (12.3) where and 3TerrhVxTTU0.0180019.796604.83745.25740060.00292510.006276500.578623.01490092.081701781.912000exxeue(12.3) 20 Now we can begin the MIMO observer-controller design for the linear plant model in Eq. (1

19、2.3) using the pole placement approach.3 the controller are given as a settling time of 2 seconds and a damping ratio of 0.8. The outputs of the design process will be the controller gain matrix K and the observer matrix L, which will be used in the controller structure .213The first step is to crea

20、te a state-space plant model. a = -0.018001 -9.7966 0; 0.0029251 -0.0062765 0; 0 81.912 0; b = -4.8374 5.2573e-6; 0.57862 3.0149e-9; 0 0; c = eye(3); d = 0; ssplant = ss(a, b, c, d); The plants controllability and observa-bility must be checked.This plant model passes both of the tests. 3 This plant

21、 is a third-order system, three closed-loop pole locations should be selected. Function select_poles()fulfill this task. n = 3; t_settle = 2; damp_ratio = 0.8; p = select_poles(n, t_settle, damp_ratio);The controller gain is computed with a call to function place(). K = place(ssplant.a, ssplant.b, p

22、)Calling place() again determines the observer gain matrix L. obs_pole_mult = 3; q = obs_pole_mult * p; L = place(ssplant.a, ssplant.c, q)3 The state- space observer-controller can now be formed based on the controller gain K and observer gains L by the following commands. ssobsctrl=ss(ssplant.a-L*s

23、splant.c,L ssplant.b-L*ssplant.d,-K,0);We need one more step to form a modified plant model that passes the plant inputs as additional outputs. We do so because the observer-controller requires the plant inputs in addition to the plant outputs. r = size(b, 2); % Number of inputs n = size(a, 1); % Nu

24、mber of states ssplant_aug = ss(ssplant.a, ssplant.b, ssplant.c; zeros(r, n), ssplant.d; eye(r);3 Form a closed-loop system consisting of the augmented plant model and the observer-controller as below: sscl = feedback(ssplant_aug, ssobsctrl, +1); Plot the pole locations along with the design specifi

25、cations by the command below. plot_poles(sscl, t_settle, damp_ratio);The resulting plot is shown in Figure 12.2. 12.3 Control design of the controllerText B: Optimal control for an inverted pendulum system2 251Outline12.3 Control design of the controller26 An inverted pendulum on a cart has one inpu

26、t, the cart driving force F, and two outputs, the cart position x and the pendulum angle .1Example 12.3 l = 1 meter mp = 2.5 kg mc = 5 kgThe input is the cart driving FThis is a fourth-order plant with four system states, including ：carts horizontal position x (meters), pendulum angle (radians) with

27、 respect to (w.r.t.) the vertical, cart velocity (m/s), andpendulum angular velocity (rad/s).12.3 Control design of the controller27 The goal of controller design is to provide a cart position settling time of 1.5 seconds with satisfactory damping characteristics, while keep the pendulum staying in

28、its equilibrium point.Example 12.31 model the plant： One approach is to use kinematics and dynamics of the rigid body, resulting in a nonlinear equation system for the plant modelling. An alternative is to use software tools to perform this analysis for your problem in hand. 12.3 Control design of t

29、he controller28 we use SimMechanics in Simulink Toolbox for modelling the inverted pendulum, as shown in Figure 12.4Example 12.3112.3 Control design of the controller29 The linmod() command extracts the linear model from the Simulink diagram. The following commands extract the model matrices and cre

30、ate a state-space plant model.Example 12.31 a,b,c,d =linmod(Pendulum); ssplant = ss(a,b,c,d); The resulting linear model is shown in Eq. (12.4).00100100000010001019.590000.266601003.2560000.17780001xxuyx(12.4)12.3 Control design of the controller30 We now begin the control design using optimal contr

31、ol techniques for this system. The controller gain K is determined by the LQR method, and the observer will function as a steady-state Kalman filter. The inverted pendulum on cart is a fourth-order system with one input, thus the weighting matrix Q is 4 4 in dimension and R is a scalar. we can fix R

32、 = 1 and vary the diagonal elements of Q during the design process.212.3 Control design of the controller31 As a first attempt, create a linear plant model, set Q to be the identity matrix, and design a controller gain with lqr()by the following commands.2 Q = eye(4); R = 1; K = lqr(ssplant, Q, R)Th

33、is sequence of commands produces the controller gain K = -165.9 -1 -38.2 -4.5resulting in a closed-loop system by the command below. cl_sys = feedback(ssplant, -K*inv(ssplant.c), +1);12.3 Control design of the controller32 Now plot the closed-loop pole locations, as shown in Figure 12.52 t_settle =

34、1.5; damp_ratio = 0.7; plot_poles(cl_sys, t_settle, damp_ratio);12.3 Control design of the controller33 The figure shows two of the poles are much slower than the design specifications. Now we try increasing the element of Q corresponding to the x state by a factor of 10. Q = diag(1 10 1 1); Repeat

35、the steps to compute the controller gain. After a few more iterations, the following Q matrix is found to produce a response the design requirement. Q = diag(1 1e6 1 1);212.3 Control design of the controller34 This system has the gain vector K = -2012 -1000 -520 -629. The close-loop poles, along wit

36、h the settling time of 1.5 second and the damping ratio of 0.7, is obtained as in Figure 12.62 The faster poles correspond to motion of the pendulum relative to the cart. The slower poles relate to moving the cart to the commanded position.Text B: Optimal control for an inverted pendulum systemOutli

37、ne 12.3 Control design of the controller2 13536 Assume that the cart position x and pendulum angle are measured by sensors that provide quantized samples. It will be necessary to esti-mate the cart velocity and pendulum angular rate from the measurem-ents, resulting in a linear system model as in Eq

38、. (12.5), 10010000010100019.590000.266601003.2560000.1778xxuyx(12.5)outputs : inputs : 12.4 Control design of the observer 1 Now consider that the system contains nonlinearities, such as friction and random disturbances Here we assume that the all effects (e.g., friction, random disturbances etc.) c

39、ould be synthesized as random force (i.e., white noises). And this random force, together with F, is applied to the cart instead of pendulum. This simplification allow us to set matrix G = B as in Eq. (12.5), and the matrix H = 0. Let the standard deviation of this random force be 0.1 Newtons, resul

40、ting in the process noise covariance QN = 0.12.3712.4 Control design of the observer 1 Assume that LSB for the cart position is 0.01 meters and the quantization of is 0.0044 radians (0.25). This results in meas-urement noise covariance :220.01 /12000.0044 /12 With the assumptions above, the steady-s

41、tate Kalman observer gain is computed as follows: a = ssplant.a; b = ssplant.b; c = 1 0 0 0; 0 1 0 0; d = 0; 0; g = b; h = d;3812.4 Control design of the observer obs_plant = ss(a, b g, c, d h); QN = 0.12; RN = 0.012/12 0; 0 0.00442/12; kest, L = kalman(obs_plant, QN, RN);1The resulting observer gai

42、n matrix is14.2822.70-5.044.03- 6.27-38.800.78-63. 8L3912.4 Control design of the observer We can now form a state-space observer-controller by the command below.1 ssobsctrl = ss(a-L*c, L b-L*d, -K, 0);And augment the plant model to pass its inputs as additional outputs. r = size(b, 2); % Number of inputs n = size(a, 1); % Number of states ssplant_aug = ss(a, b, c; zeros(r, n), d; eye(r);4012.4 Control design of the observer 2 ssplant2 = ss(a, b, c, d); sys_r_to_x = ssplant2(2,1);Compute