sc46 - Control Systems Design Q Sem Ac Yr / Mock Exam originally given November 5 9 Notes: Please be reminded that only an A4 paper with formulas may be used during the exam no other material is to be allowed (in particular certainly no programmable calculators) The actual distribution of points for the single questions of all the exercises is reported right before the questions You may decide what exercise to start with depending on that Design Problem Consider the following dynamical model: ẋ x + ẋ x y x x 4 pts a) Consider zero initial conditions Assume that the model has been excited by a step input u(t): compute the steady-state error in the output y(t) 5 pts b) Consider a direct output feedback controller with the following form: u(t) ky(t) + r(t) k > (k positive) Find the range of values for k with which the closed-loop system is stable If you think there is no such k then explain why Additionally can you choose the controller gain k to make the closed-loop system asymptotically stable? 5 pts c) Assume that the state may be measured precisely Design a state feedback controller u(t) x F F (t) so that the trajectories of the x (t) closed-loop system dynamics are bounded above on each of their spatial component in time by the function e 4t In other words you have to select proper distinct eigenvalues for the closed-loop system so that this happens u Solution: The steady state error for a step input is equal to infinity Let us compute the
following quantity: C(sI a) B + D s s s s + Thus the steady-state error can be computed as follows: s s which is not finite As a confirmation the following is the step response of the above model plotted in MATLAB: x 5 Step Response 8 Amplitude 6 4 5 5 Time (sec) Let us consider an output feedback controller of the following kind: u(t) ky(t) + r(t) k > This results in the following closed-loop dynamics: ẋ x A + ( k) x + r ẋ x x x + r k x y x x We then obtain: det(si A) s + k k > which has roots in s ±j k Thus regardless of the choice of the gain k the closed-loop dynamics will be stable but never asymptotically stable
Let us proceed with the state feedback controller Considering a control feedback with the indicated structure the closed-loop dynamics become: ẋ x x F F ẋ x x x F F x The associated characteristic polynomial is det(si ) s + F F F s + F The requirement on the convergence speed can be obtained by placing poles at the left of the s 4 vertical threshold For instance let us select points 4 4 which obtains: This can be obtained by choosing (s + 4) s + 8s + 6 F 6 F 8 Actually we could be even more precise: the roots of the characteristic polynomial are in s F ± F 4F If F > F /4 the roots are complex conjugate thus they re well positioned if F 4 On the contrary if F F /4 the roots are real thus the condition is verified if (consider worst case) F + F 4F 4 which is verified if F F 4 5 5 5 F 5 5 5 8 6 4 4 6 8 F 3
Two-Tanks Process Consider the system in figure known as a two-tanks process It deals with two tanks the first positioned on top of the second which are communicating through a small hole at the bottom of the first Additionally the first tank is added fluid (with rate u) through a hose operated by a pump whereas the second is releasing fluid through two holes The rate of one of the two bottom holes can be controlled by an outflow rate of v We are interested in describing the dynamics of the fluid level in the two tanks which will be denoted with the variables x and x A non-linear continuous-time state-space description of the process can be derived from Bernoulli s energy equations and reduces to: ẋ α x + β u ẋ α x α x β v where α α are characteristic parameters of the two fluids and β β are parameters that describe the top hose and the bottom outflow The quantities u v are interpreted as the two control inputs pts a) Determine the equilibrium point as a function of the inputs and of the parameters 4 pts b) Assume the following parameters: α α 5 β β 5 Linearize the model around the corresponding equilibrium point and the input (ū v) ( ) 5 pts c) Assume again the following parameters: α α 5 β β 5 Your friend from TU Eindhoven suggests you to linearize the model around the equilibrium point arising from the input choice (ū v) ( )? Can you comment on his/her suggestion? 4 pts d) Consider the linearized model from question two Assume that you can select exclusively a single actuator to control your system and that this 4
can be either the hose u at the top or the aperture v at the bottom For each choice let the other input be a fixed constant reference Reasoning in terms of controllability which of the two inputs would you choose? Please provide an explanation for this 4 pts e) Consider the linearized model from question two Assume that you can measure the fluid level only in one of the two tanks that is either x or x Which one would you prefer to measure? Please provide an explanation for this pts f) Consider the linearized open-loop model obtained from question two Before you test your model in the lab on a real setup you come up with the following discretization of the state-space model using sampling time h 4463 seconds: Φ 8 89 894 Γ 4 45 4 After a little experimentation you find out that due to some defect in the actuator circuitry your system response suffers from τ 896 msec of delay How would you modify your state-space system description (based on the original Φ and Γ matrices) to have an accurate representation of the delayed system? Solution: Select a nominal value for the inputs u ū and v v Obtain from the two dynamical relations the following equilibria ( x x ): x β α ū x β α ū β α v The choice of parameters and inputs yields the equilibrium point ( ) The linearization is: ( ) ẋ α x x u ẋ α x α + x x 5 v (x x )() ( ) / x u + / /4 5 v x The choice (ū v) ( ) gives the equilibrium ( ) However for x the second ODE has no unique solution This is related to the fact that the Jacobian of the model is unbounded for that value The choice of input u is related to a control matrix B This lead to the following controllability matrix: / / which has full rank The choice of input v is related to a control matrix B 5 This lead 5
to the following controllability matrix: / /8 which does not have full rank Clearly the choice of the top controller is preferable The measure of the fluid level in the first tank is associated with a matrix C This lead to the following controllability matrix: / which does not have full rank The measure of the fluid level in the second tank is associated with a matrix C This lead to the following controllability matrix: / / which has full rank Clearly the latter choice is preferable In order to incorporate the extra time delay in the discrete-time state-space system description we need to introduce extra states that will represent delayed versions of the control input signal Notice that the observed time-delay τ is exactly twice as large as the sampling time h thus we can write the delayed system as x(k + ) Φx(k) + Γu(k d) u where d and u This simplifies our problem since we only have v to delay the input signal by two sampling periods Since we have two control inputs we will need to introduce FOUR extra states as follows: where u(k) x(k + ) u(k ) u(k) u(k) v(k) Φ Γ I } {{ } Φ delayed x(k) u(k ) u(k ) + u(k) I }{{} Γ delayed 6
3 Position control of an electric motor The figure describes an electric motor: a torque applied to the motor shaft is controlled by a voltage applied to the motor windings The variables are: v the applied voltage; i the current on the windings; 3 τ the motor torque and τ d a disturbance torque coming from the environment; 4 θ the shaft angle and ω θ its velocity The model for the motor follows: v Ri + L di dt τ γi J dω dt τ τ d Bω where R is the resistance L is the inductance J a moment of inertia and B a mechanical damping Assume for simplicity that the parameters are normalized: R L γ J B The objective of the process is to track a reference θ r by acting on the input voltage v 4 pts a) Write out a third order state-space model for the process under study with variables (θ ω i) 3 pts b) Let us neglect the effect of the inductance L on the dynamics: simplify the model above into a new second order model by assuming L 7
4 pts c) Study the controllability of the two models obtained from questions one and two 4 pts d) For both models devise a state-feedback controller to track the reference signal: v K(θ r θ) where K > is a parameter to be determined Write out the closed-loop dynamics 6 pts e) Consider the closed-loop second order (simplified) model from question two Study the asymptotic stability of this model for varying values of the parameter K > If asymptotically stable find explicitly a Lyapunov function for the closed-loop model (You can determine it by solving the Lyapunov equation with a proper choice of a positive definite matrix) 4 pts f) Now consider the closed-loop third order (original) model from question one Study the asymptotic stability of this model for varying values of the parameter K > (Since finding roots of the characteristic polynomial may not be straightforward select a few small integer values for the parameter K: K 3 ) What do you conclude about the asymptotic stability of the closed-loop third-order model? As a consequence do you have any comment on the simplification procedure that assumed L? Solution: Third order model: θ ω i θ ω i + v + τ d Second order model: θ ω θ ω + v + τ d Both models are controllable since the Kalman rank condition for them is: rank rank 3 State feedback for second order model: θ ω v K(θ r θ) K θ K ω θ ω K + The roots of the characteristic polynomial are: s ± 4K + Kθ r θ r + τ d 8
which have negative real part for any K > A Lyapunov function can be found by selecting V (x) x T θ p p P x x P ω p p 3 and by selecting the parameters so that where Q which yields We obtain: P A + A T P Q Kp p p Kp 3 p p Kp 3 (p p 3 ) p K + + K p K p 3 K + This choice makes P positive definite as well State feedback for third order model: v K(θ r θ) K θ ω i K Characteristic polynomial is Try K to obtain: which has roots θ ω i θ ω i + s 3 + s + s + K (s + )(s + ) K + Kθ r θ r + τ d s s 3 ±j The last two roots indicate lack of asymptotic stability 9
3 Root Locus Imaginary Axis System: untitled Gain: Pole: Damping: Overshoot (%): Frequency (rad/sec): System: untitled Gain: 99 Pole: 54 998i Damping: 54 Overshoot (%): 998 Frequency (rad/sec): 998 3 5 4 3 Real Axis