Variable Structure Control ~ Disturbance Rejection Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University
Outline Linear Tracking & Disturbance Rejection Variable Structure Servomechanism Nonlinear Tracking & Disturbance Rejection
Disturbance Rejection Setup w disturbance Controller u control Plant z performance variables y measurements state equations x Ax Ew Bu disturbance w Zw performance z Cx Fw Du measurements y Cx Fw Du x R, wr, ur, zr, y R n q m m r Eigenvalues of Z ordinarily on Im axis In effect, the error
Objectives Regulator Problem: Find a controller to achieve the following ) Regulation: z t 0 as t 2) (Internal) Stability: Achieve specified transient response Robust Regulator Problem: Find a solution to the Regulator Problem that satisfies 3) Robustness: ) and 2) should be maintained under specified small perturbations of plant and/or control parameters
Solution: Part - Regulation Consider the possibility of a control u t that produces a trajectory x t for some unspecified initial state x and any initial disturbance vector w, 0 0 so that the corresponding z t 0. Then, x, u, w must satisfy x Ax EwBu w Zw 0 Cx Fw Du Assume a solution of the form: x Xw, u Uw XZw AXw Ew BUw w 0 CXwFwDUw Thus, the hypothesized control u t exists if there are X, U that satisfy XZ AX BU E CX DU F Solutions typically are not unique
Solution: Part 2- Stability Define u, x u u u Uwu x AxBu x x x Xwx Now, if AB, is controllable, it easy to choose u Kx so that the closed loop x ABK x has desired transient characteristics. With K chosen, the control can be written as a function of the system states xw, u u u Uwu UwKx xx u UwKxXwK U KX KTOT w w
Solution: Part 3- Observation The control will be implemented using estimates of the composite state xw,. Consider the composite system d xa ExB u dt w 0 Z w 0 y CxFw If the composite system is observable, we can choose a matrix L, so that the following observer has the desired dynamics: d dt xˆa ExˆB u LCxˆ Fwˆ y wˆ 0 Z wˆ 0
Properties of the Loop K Disturbance Plant x ˆA ExˆB L C F u Ly wˆ 0 Z wˆ 0 u xˆ K U KX w ˆ U X Observer xˆ A EBxˆ L C F K U KX Ly wˆ 0 Z 0 wˆ u xˆ K U KX w ˆ Compensator contains copy of Z ~ internal model of disturbance Compensator Observer u Disturbance w Plant y
Example y, command s w, load torque dc motor s s controller s 3 y, motor speed
Example ŵ 3 y command observer ŵ 2 K 3 u w s 3 y K ŵ ˆx system disturbance observer ŵ 3 y command observer ŵ 2 K 3 compensator u w s 3 y G c 9420.3 s 9.8897 3 209 s s 49.8393
VS Servo Disturbance ˆx Plant U X Observer We make only one change u ˆ i si x 0 u Kxˆ u xˆ, ˆ, ˆ ˆ i x s x Gx u ˆ i si x 0
Closed Loop Dynamics Combine the estimator equation: d x ˆ A xˆ EwBuLC xx LF ww dt with the estimator equation ˆ ˆ d w ˆ Zw ˆ L 2 C x ˆ x L 2 F w ˆ w dt d xˆ x x ˆ Xw ˆ Ax ˆ Ew XZw ˆ L XL 2 C F BUwˆ B u dt wˆ w apply XZwˆ AXwˆ Ewˆ BUwˆ d dt xˆ x x ˆ Xwˆ Axˆ XwˆE w wˆl XL2C F B u wˆ w
Closed Loop Dynamics, 2 d dt xˆ xˆ A L ˆ XL2 C L XL2 F x x B x xˆ 0 ALC ELF xxˆ 0 u wwˆ0 L ˆ 2C Z L2F ww 0
Sliding Behavior Let xˆ xˆxˆ and define a new coordinates xˆ,, R, R nm 2 2 M ˆ ˆ Note: MB 0, KN 0, sliding 0 d dt x N B 2 x 2 KB K 2 MAN M L XL2 C M L XL2 F x xˆ 0 ALC ELF x xˆ wwˆ 0 L ˆ 2C Z L2F ww m
Reaching Behavior Assume the switching control is designed to stabilizes the switching manifold K x x AxB s, s Kx 0 for the perturbation system Two important results follow trajectories are steered in finite time to a domain D that contains the subspace K xˆ 0 D shrinks exponentially to the subspace K xˆ 0
Reaching, 2 Theorem : the domain For a fixed >0, there exists a finite time T such that xˆ is confined to T k xˆ, i,, m, t T. i Moreover, 0, T. this means that sliding does not occur on s ˆ asymptotically as xˆ t x t, w t w t xˆ t xˆ t 0 but this manifold is reached * let denote an ideal sliding solution. can be viewed as non-ideal sliding in that it can be shown that there exists a constant c such that for all t T ˆ ˆ * x t x t c The performance variables can be expressed as ˆ ˆ ˆ ˆ z C xx F ww D u as t, we have xx 0, ww 0, uu 0 z0 eq
Example: Drum Level Control Drum Throttle Valve Feedwater Valves Riser Downcomer
Drum Level, 2 N number of riser sections Ldo,L downcomer length and riser section length (total riser length/n) Ado,A downcomer, riser cross section areas wi mass flow rate at ith node Pi pressure at ith node Ti temperature at ith node si aggregate entropy at ith node vi specific volume at ith node wr,wdc,ws mass flow rates, riser, downcomer and turbine, respectively vdf,vdg drum specific volume, liquid and gas, respectively Pd drum pressure Td drum temperature V total drum volume Vw volume of water in drum xd net drum quality, xd=vw/v ws0 throttle flow at rated conditions Pd0 drum pressure at rated conditions At normalized throttle valve position, at rated conditions At=
Drum Level, 3 u = q, u 2 = e, u 3 = A t d av = f dt ( av,s,s 2,s 3,P av,p d ) ds = f dt 2 ( av,s,p av )+g 2 (P av,s )u +g 22 ( av,p d )u 2 ds 2 = f dt 3 ( av,s,s 2,P av )+g 3 (P av,s 2 )u ds 3 = f dt 4 ( av,s 2,s 3,P av )+g 4 (P av,s 3 )u ds 4 = f dt 5 ( av,s 3,s 4,P av )+g 5 (P av,s 4 )u dp av = f dt 6 ( av,s,s 2,s 3,s 4,P av,p d )+g 6 ( av,s,s 2,s 3,s 4,P av )u dp d = f dt 7 ( av,s,s 2,s 3,s 4,P av,p d,v w )+g 7 ( av,s,s 2,s 3,s 4,P av,p d,v w )u +g 72 (P d,v w )u 2 -g 73 (P d,v w )u 3 dv w = f dt 8 ( av,s,s 2,s 3,s 4,P av,p d,v w )+g 8 ( av,s,s 2,s 3,s 4,P av,p d,v w )u +g 82 (P d,v w )u 2 -g 83 (P d,v w )u 3 y = P d, y 2 =, = h 2 (V w ), y 3 = s = h 3 (P d )+d 3 (P d )u 3
Linearized Dynamics, Poles 0.4 0.3 0.2 Poles as a function of load level, 5%-00% imaginary 0. 0-0. -0.2-0.3-0.4 - -0.8-0.6-0.4-0.2 0 0.2 real
Linearized Dynamics, Zeros 0.8 0.6 0.4 0.2 0-0.2-0.4-0.6 e -0.8-3 -2.5-2 -.5 - -0.5 0 0.5
Transmission Zeros 0.6, A P, e t d 0.4 0.2 imaginary 0-0.2-0.4-0.6 -.2 - -0.8-0.6-0.4-0.2 0 real
Transmission zeros 0.8 0.6 0.4 Q,, A P,, e t d s imaginary 0.2 0-0.2-0.4-0.6-0.8 -.4 -.2 - -0.8-0.6-0.4-0.2 0 real
Drum Level, Normal Form u u 2 u 3 Furnace Valve Actuator Valve Actuator q e A t Drum & Circulation Loop P d, s s y y 2 y 3 v v 2 v 3 s s s z z 2 4 z 6 s s P d, s s y y 2 y 3 = z = z 3 = z 5
Drum Level, Switching Controller s 0.5z z s 0.5z z z * * T sgn,,2,3 z P P / P * d d d 2 2 2 2 s * 3 0.000z3 0.02 z 3 z3 s s/ s dt u x U s i s Qs z i i i 0.007 0.09436 0.93505 0 0 0.003 0.06793 0.46, Q 0 0. 0 0 0 0.976 0 0 0.00
Drum Level - Conventional, PID with Steam/Water FF load change 80-75%.0 Drum Pressure 0.77 Steam Flow 0.76 0.99 0.75 0.98 0 50 00 50 200 time (s) 0.74 0 50 00 50 200 time (s) 0.4 Drum Level 0.85 Feedwater Flow 0.2 0.8 0 0.75-0.2 0 50 00 50 200 time (s) 0.7 0 50 00 50 200 time (s)
Drum Level - Conventional, PID load change 5-20% 0.84 Drum Pressure 0.205 Steam Flow 0.82 0.2 0.8 0.95 0.78 0 50 00 50 200 time (s) 0.9 0 50 00 50 200 time (s) 2 Drum Level 0.3 Feedwater Flow 0.2 0 0. - 0 50 00 50 200 time (s) 0 0 50 00 50 200 time (s)
Drum Level - VS Control, load change 80-75% Q,, A P,, e t d s Drum Pressure 0.995 0.99 0.985 0 00 200 300 400 time (s) Steam Flow 0.85 0.8 0.75 0 00 200 300 400 time (s) Drum Level 0.2 0-0.2 0 00 200 300 400 time (s) Feedwater Flow 0.8 0.75 0.7 0.65 0 00 200 300 400 time (s)
Drum Level VS Control 5-20% Drum Pressure 0.83 0.82 0.8 0.8 0 50 00 50 200 Steam Flow 0.25 0.2 0.5 0. 0 50 00 50 200 time (s) time (s) Drum Level 0.4 0.2 0-0.2 0 50 00 50 200 Feedwater Flow 0.4 0.3 0.2 0. 0 50 00 50 200 time (s) time (s)
Regulation of Nonlinear Systems The nonlinear regulator problem Exponential stabilizability and detectability Existence and construction of controllers
The (local) Nonlinear Regulator Problem x f x, w, u state w w z y h x, w g x, w disturbance error measurement Assume an equilibrium point at the origin, i.e. 0 f 0,0,0,0 0,0h 0,0,0 g 0,0 Assume disturbance dynamics are neutrally stable
The Nonlinear Regulator Problem, 2 Consider either a state feedback controller: u k x, w x f x, w : f x, w, k x, w, x xr or an output feedback controller cl cl cl cl cl cl f x, w, v, g x, w r u v, y, v v, y, vr xcl fcl xcl, w: vg, xw, Local Regulator Problem : Determine a feedback control law such that ww xw t Regulation z t ) Stability, for each 0, a neighborhood of the origin, the closed loop has an exponentially stable trajectory, 2), 0, as neighborhood of the origin. t for each w 0 W and x 0 X, a cl n x x v cl
Exponential Stabilizability and Detectability Definition (Exponential Stabilizability) : The system x f x, u with f 0,0 0 is exponentially stabilizable if there exists a feedback control u k x with k 0 0 such that x f x, k x is exponentially stable. Definition (Exponential Detectability) : The system x f x, u, y g x with f 0,0 0 and g 0 0 is exponentially detectable around the origin if there exists a system, y, R where 0,0 0, x, g x f x,0 n and 0 is an exponentially stable equilibrium point of, 0.
Remarks
Regulation of Feedback Linearizable Systems - 2 w,, x f x, w G x u state w z h x w y g x w disturbance error measurement Assume an equilibrium point at the origin, i.e. 0 f 0,0,0 0,0 h 0,0,0 g 0,0 Assume disturbance dynamics are neutrally stable
Regulation of Feedback Linearizable Systems - 2 F 2 0 2,, u,, 2 A02 E0 x w x w u z C choose: u x, w x, w v take v K x, w so that Re A E K 0 lim z t 0 t 2 0 0
Example: SISO Linear Systems x AxEwbu w Zw z cx fw z caxcew fzw cbu c f cb Z w u 0 A E x A Ex z c f cab 0 Z w u r A Ex r z c f ca bu r 0 2 Z w r A Ex ca b, c f 0 Z w r
Example Continued x ca ce A A Z AZ Z w r r r2 r2 r ca r b x u r ca b w x c f w x 2 ca ce fz w r r r2 r2 r ca ce A A Z AZ Z v ca ce A A Z Z r r2 r3 r2 r r fz x w