Lecture Outline Regeltechniek Previous lecture: Nyquist plot and stability criterion. Lecture 11 – State-space models and state feedback control (We are finished with frequency domain methods.) Robert Babuˇ ska Today: Delft Center for Systems and Control • State-space models, representation. Faculty of Mechanical Engineering Delft University of Technology • State feedback, pole placement. The Netherlands e-mail: r.babuska@dcsc.tudelft.nl www.dcsc.tudelft.nl/ ˜ babuska tel: 015-27 85117 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 1 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 2 The big picture Transfer Function vs. State Space Methods Type�of�model Transfer function State-space First�principles Simulation,�prediction, Nonlinear�differential�eq. Physical�world�(process) better�understanding graphical tools computational Linearization provide insight numerical, less insight Data Basis�for control - Linearized differential�eq. oriented�models interactive / iterative “numbers in – numbers out” Analysis,�control�design Transfer�function State-space�model SISO systems SISO and MIMO systems Design Influence�a�process, Controller modify�behavior Implementation Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 3 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 4
Linear State-Space Model State-Space Model: Block Diagram x ( t ) = Ax ( t ) + Bu ( t ) ˙ x ( t ) = Ax ( t ) + Bu ( t ) ˙ y ( t ) = Cx ( t ) + Du ( t ) y ( t ) = Cx ( t ) + Du ( t ) A . . . state matrix B . . . input matrix . ( ) ( ) u t y t ( ) ( ) C . . . output matrix x t x t D . . . direct transmission matrix Interpretation: Derivative of each state is given by a linear combination of states plus a linear combination of inputs. Similarly for the output . . . Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 5 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 6 DC Motor: State-Space Model DC Motor: State-Space Model � T � L di ( t ) dθ ( t ) dt + Ri ( t ) = V ( t ) − K t electrical part state: x ( t ) = , input: u ( t ) = V ( t ) , output: i ( t ) ω ( t ) θ ( t ) dt y ( t ) = θ ( t ) J d 2 θ ( t ) dt 2 + b dθ ( t ) = K t i ( t ) mechanical part dt L − K t 1 − R x 1 ˙ 0 x 1 L L Introduce velocity: ω ( t ) = ˙ θ ( t ) K t − b = + u ˙ 0 0 x 2 x 2 J J x 2 ˙ 0 1 0 x 3 0 Rewrite the above equations as a set of three 1st order DE: i ( t ) = − R Li ( t ) − K t L ω ( t ) + 1 ˙ LV ( t ) x 1 � � ω ( t ) = K t J i ( t ) − b y = 0 0 1 x 2 ˙ Jω ( t ) x 3 ˙ θ ( t ) = ω ( t ) Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 7 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 8
Compare to the Input Output Model State-Space Models → Transfer Function G ( s ) = θ ( s ) K t x ( t ) = Ax ( t ) + Bu ( t ) ˙ V ( s ) = s [( Ls + R )( Js + b ) + K 2 t ] y ( t ) = Cx ( t ) + Du ( t ) Use Laplace: � � LJs 3 + ( RJ + Lb ) s 2 + ( Rb + K 2 sX ( s ) = AX ( s ) + BU ( s ) θ ( s ) t ) s = K t V ( s ) Y ( s ) = CX ( s ) + DU ( s ) Express X ( s ) : Corresponds to the following differential equation: ( sI − A ) X ( s ) = BU ( s ) ... θ ( t ) + ( RJ + Lb )¨ θ ( t ) + ( Rb + K 2 t ) ˙ X ( s ) = ( sI − A ) − 1 BU ( s ) LJ θ ( t ) = K t V ( t ) � � C ( sI − A ) − 1 B + D Y ( s ) = CX ( s ) + DU ( s ) = U ( s ) Input-output models do not use internal variables, instead use higher derivatives of input and output. Transfer function: G ( s ) = C ( sI − A ) − 1 B + D Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 9 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 10 Poles of State-Space Models State Feedback Control – Main Idea Transfer function: G ( s ) = C ( sI − A ) − 1 B + D • Assume a state-space model with all states measured: x ( t ) = Ax ( t ) + Bu ( t ) ˙ y ( t ) = x ( t ) ( sI − A ) − 1 = adj( sI − A ) det( sI − A ) • Controller = linear combination of states: u ( t ) = − Kx ( t ) = − k 1 x 1 ( t ) − k 2 x 2 ( t ) · · · − k n x n ( t ) Poles = roots of the characteristic equation of A : • Goal: obtain desired dynamics (e.g., fast, well damped) det( sI − A ) = 0 Therefore the poles are the eigenvalues of state matrix A . • Design parameters: location of closed-loop poles Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 11 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 12
State Feedback Control – Remarks State Feedback Control Scheme d • Note that the controller is a static system: u ( t ) = − Kx ( t ) = − k 1 x 1 ( t ) − k 2 x 2 ( t ) · · · − k n x n ( t ) u . y=x (as opposed to e.g. a lead-lag compensator or PID). -K x = Ax + Bu • Any desired location of closed-loop poles can be obtained (which is not the case with e.g. root-locus) • Often not all the states are measured (need an observer). Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 13 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 14 State Feedback Control – Problem One Possible Solution Given the state-space model: x ( t ) = Ax ( t ) + Bu ( t ) ˙ u ( t ) = − Kx ( t ) x ( t ) = Ax ( t ) + Bu ( t ) ˙ Construct the closed-loop system: We want to design the state-feedback gain K (a vector) x ( t ) = Ax ( t ) − BKx ( t ) ˙ u ( t ) = − Kx ( t ) x ( t ) = ( A − BK ) ˙ x ( t ) � �� � A cl Central question: Compute the closed-loop characteristic polynomial: How can we compute K , such that the closed-loop det( sI − A cl ) = det( sI − A + BK ) = s n + a 1 s n − 1 + · · · + a n − 1 s + a n poles are at a desired location? Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 15 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 16
✂ ✁ Solution: Finding K Desired Dynamics (Closed-Loop Poles) Closed-loop characteristic polynomial: Control goals are typically stated in terms of: det( sI − A + BK ) = s n + a 1 s n − 1 + · · · + a n − 1 s + a n • desired frequency ω n and damping ζ The coefficients a i are linear functions of the gains k i . for dominant second-order dynamics Define desired characteristic polynomial • time domain characteristics: (i.e., desired dynamics in terms of poles): rise time, settling time, overshoot d ( s ) = s n + d 1 s n − 1 + · · · + d n − 1 s + d n ω 2 α ∗ ω n n G d ( s ) = · · · · s 2 + 2 ζω n s + ω 2 s + α ∗ ω n n � �� � � �� � faster dynamics dominant dynamics Compute K by equating coefficients a i and d i Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 17 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 18 Performance Specifications for Closed-Loop Tracking a Reference Input (Servo) 1 8 . Goal: respond to a reference signal in a specified way. ω ≥ n t r t t M M ζ ≥ ζ r s p p Replace u ( t ) = − Kx ( t ) by: u ( t ) = − Kx ( t ) + K ff r ( t ) 4 6 . σ ≥ t s d r K ff θ Im Im Im u . y=x σ -K x = Ax + Bu ω n Re Re Re Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 19 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 20
Computing the Feedforward Gain Computing the Feedforward Gain y u . b ( s ) x r H cl ( s ) = C ( sI − A + BK ) − 1 BK ff = K ff K ff x = Ax + Bu C d ( s ) - K Given a desired DC gain of the closed loop: H cl (0) = DC des Transfer function from r to y : b ( s ) H cl ( s ) = C ( sI − A + BK ) − 1 BK ff = K ff Compute: d ( s ) K ff = d (0) b (0) · DC des with: b ( s ) the process open-loop TF numerator d ( s ) is the desired closed loop TF denominator Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 21 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 22 Example: Disk Drive Arm Control Example: Disk Drive Arm Control y ( t ) = c ¨ Ju ( t ) 1. Form a state-space model A , B , C , D 2. Analyze open-loop dynamics (poles) y ( t ) 3. Define desired closed loop dynamics d ( s ) 4. Compute state-feedback gain vector K 5. Compute feedforward gain vector K ff u ( t ) Specifications: settling time < 10 ms, overshoot < 2% Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 23 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 24
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