# SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

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1 FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman

2 Problem 1: Frequency-domain analysis and control design (15 pt) Given is a plant consisting of 3 first-order time delay elements connected in series. The time constants are T 1 = 0.01, T 2 =0.5, T 3 =0.1. The overall plant gain is K=40. a) Determine the transfer function of the plant. How many stable poles has the system? Explain your answer. Show the Bode diagram of the system. Are there sufficient amplitude and phase margins in the system? Explain your answer. (4 pt) Transfer function of the plant The system has 3 stable poles: 100, 2, 10 since all roots of characteristic polynomial (denominator of G) are in the left-hand side of s-plane. Bode diagram with the amplitude and phase margins: Both, the available amplitude margin (4.52 db) and phase margin (9.63 deg) are not sufficient: a low amplitude increase (about 1.7 times) and low phase shift (9.63 deg) would lead to a closed-loop instability. b) Design the PID controller which compensates for two corner frequencies of the plant. Write down the transfer function of the open-loop and closed-loop systems. Use MATLAB to simulate the step response of the closed-loop system. Copy the simulated response to paper. (4 pt) Both PID-control time constants compensate for two largest time constants of the plant (T 2 =0.5, T 3 =0.1). The PID control gain accounts for the plant gain and smallest time constant:... with K r = 1 / (2 K T 1 ). The open- and closed-loop transfer functions are:,... Step response of the closed-loop system:

3 c) Using the Lead-Lag compensators design the control system with the following specification: phase margin 30 deg and step response steady-state error 5 %. (4 pt) The DC gain of the closed-loop system: is : {matlab -> dcgain(g/(1+g))}. Thus the step response steady-state error is already 5 %. Design a Lead compensator only for improving the phase margin which is 9.63 deg. The required phase enhancement is m deg. Assume e.g. m =25 deg, then 1/ 2.5 this from the characteristic Lead diagram, then = 0.4. At the cross-over frequency m = c = , rad/s. The Lead-compensator transfer function:,. Then, reduce the Lead gain K p so that the open loop comes back to the original cross-over frequency c = 26.9 rad/s, -> K p = The total compensator transfer function is: d) Use Simulink to simulate and to compare the step response of the closed-loop control systems from problems 1b) and 1c). Copy the step response and block diagram from Simulink to paper. Label all blocks, signals, and axes. (3 pt) Closed-loop step responses and block diagram (Simulink) of both controllers PID and Lead: Problem 2: State-space analysis and control design (15 pt) Given is a linear time-invariant system in the state-space form , a) Determine poles of the system and draw the pole diagram in complex plane. What can be said about system stability? Which type has the system? Explain your answers. (3 pt) The poles are the eigenvalues of system matrix {matlab -> A = [ ; 4 0 0; 0 1 0]; eig(a)}. -> 1 = 0, 2 = -2, 3 = -10. The system is stable as there are no poles in the righthand side of s-plane. There is one pole in origin (0,0) of the complex s-plane -> the system has an integral behavior -> the system type (number of free integrators) is 1.

4 b) Verify whether the system is controllable and observable by using Hautus criteria. (4 pt) Hautus criterion for controllability: rank i I A, B = n, and Hautus criterion for ii A observability: rank = n, where n is the system order and i is the system pole (each C one to be evaluated separately), {matlab -> A = [ ; 4 0 0; 0 1 0]; B = [8 0 0] ; C = [0 0 1]; I = eye(3,3);} -> for all three poles 1 = 0, 2 = -2, 3 = -10 evaluate the rank of matrices above -> for all poles the rank is 3 -> the system is fully controllable and observable. c) Design the state-feedback control so that the system matrix of the closed-loop control system has the following form Design the pre-filter to ensure the steady-state accuracy. Use MATLAB to simulate and to show the step response of closed-loop control systems with state feedback and pre-filter. What is the drawback of a static pre-filter? Explain your answer. (4 pt) According to Kalman canonical decomposition, the system matrix of the closed-loop control system A c is in the modal canonical form so that the real poles (eigenvalues) can be immediately read out from the main diagonal: c, 1 = -30, c, 2 = -25, c, 3 = -20. Use pole placement command place in Matlab to determine the state feedback gains; determine the new system matrix with K feedback, determine the state-space model with the new system matrix; determine the prefilter by the explicit formula:{matlab -> K = place(a,b,[-20,-25,- 30]; Ac = A-B*K; Gc = ss(ac,b,c,d); V = (C*(B*K-A)^(-1)*B)^(-1);}. The step response of the closed-loop control systems with state feedback and pre-filter: {matlab -> step(v*gc)}: The drawback of a static pre-filter V is that its computation relies on the plant parameters, i.e. matrices A, B, and C, and there is no comparison between the reference and real (output) value in the control loop. Thus, the model uncertainties will lead to that no steady-state accuracy (zero control error) can be guaranteed. d) Assuming x 3 is the single measurable state in the system, design the Luenberger state observer. The observer poles should be approximately 5 times faster than poles of the closed-loop control system from problem 2c). Use Simulink to simulate and to show the response of observer-based state-feedback control with pre-filter from problem 2c). Use sinusoidal with angular frequency 2 rad/s as a reference value. Copy the system response and block diagram from Simulink to paper. (4 pt)

5 Determine the observer feedback gain by pole placement using place command in Matlab: {matlab -> L = place(a',c', 5*[-20,-25,-30]) ;}. Determine the state-space model of Luenberger observer {matlab -> Alc = A-L*C; Blc = [B L]; Clc = C; Dlc = D;}. The observer-based state-feedback control with pre-filter (in Simulink) and the controlled response of x 3 output value to the sinusoidal reference with 2 rad/s angular frequency: Problem 3: Digital control systems (15 pt) Given are the continuous-time transfer functions of the plant G and PD controller 1, For the digital control system with sampling and without quantization of the signals, assume the sampling time T s =0.01 sec. a) Transform the plant transfer function from the continuous-time (s) domain into the discrete-time (z) domain. Determine the difference equation (discrete time-series equation) with the input u(k) and output y(k), where k is a discrete time instant. (4 pt) According to the Tustin s method with the correspondence between s -1 and z: 1/s -> Ts/2 (z+1)/(z-1), the plant transfer function in the discrete-time domain is: This results in the input-output equation: U(z)(2.5e-5 + 5e-5z e-5z -2 )=Y(z)(1-2z -1 +z -2 ). After applying the time shift transformation X(z)z -n -> x(k-n) the resulted output equation is: y(k) = 2y(k-1) - y(k-2) + 2.5e-5 u(k) + 5e-5 u(k-1) + 2.5e-5 u(k-2). b) How the difference equation from problem 3a) can be represented by using the forward and backward shift operators? Explain you answer. Draw the corresponding block diagram of the signal flow from u(k) to y(k). (4 pt) The forward shift operator (in z-domain) means multiplication with z and backward shift operator means division by z. In the discrete time domain the backward shift operator is a time-delay (or memory operator) for which the output value is the input value delayed by one discrete time instant. The block diagram using the backward shift operator in the time domain, e.g. using Simulink modeling is:

6 c) Determine the difference equation (discrete time-series equation) of the PD controller. Draw the corresponding block diagram with the signal flow from e(k) to u(k). (4 pt) Similar as in solution 3b), the input to output equation of the PD control is: 2 z z Uz ( ) Kp Kd Ez ( ) Ez ( ) Ts z1 z1 1 1 U( z)(1 z ) Ez ( )( z ) uk ( ) 8400 ek ( ) 7600 ek ( 1) uk ( 1). The block diagram (in Simulink notation) is d) Use Simulink to simulate and to show the step response of the digital control system consisting of the discrete-time plant G and PD controller from problems 3b) and 3c). Copy the step response and block diagram from Simulink to paper. (3 pt) The block diagram in Simulink and the controlled step response are:

7 Problem 4: Miscellaneous system modeling and analysis (15 pt) In Figure below is shown an electromagnet with movable armature. The controllable input value is the terminal voltage V. The measurable output value is the armature displacement Y. Figure: Electromagnet with movable armature The linearized equations of magnetic circuits and mechanical mass-spring-damper part are:,,. The system parameters are the coil inductance L, connection resistance R, electro-mechanical coupling constant C, armature mass M, viscous damping D, and return spring stiffness K. The current in electromagnetic circuits is I. The induced magnetic force is F. a) Define the dynamic states of the system and write down the overall state-space model. (4 pt) The dynamic states of the system are the current, velocity, and position: x = [I, dy/dt, Y]. The resulted state-space model is I R/ L C/ L 0 I 1/ L dx d Ax + Bu Y C/ M D/ M K / M Y 0 V dt dt, Y Y y Cx Y I Y Y T b) Draw the block diagram of the system. Label all signals and parameters in the block diagram, including summations nodes and signal flow direction. (4 pt) The block diagram of the system, e.g. using Simulink modeling: c) Write down the transfer function of the system. Determine the parameter conditions for which the system is stable. (4 pt) To construct the system transfer function from V(s) to Y(s) consider (with help of solution 4b) two sub-transfer functions: from I to Y and from V to I. The latter is first without feedback by

8 C! G 1 (s)=y(s)/i(s) = C/(Ms 2 +Ds+K) and G 2 (s)=1/(ls+r). Then the overall transfer function is the feedforward of G 1 (s)g 2 (s) with the feedback through Cs: G(s) = G 1 G 2 /(1+ CsG 1 G 2 ) -> C Gs () LMs ( DL MR) s ( DR KL C ) s KR To derive the conditions for system to be stable apply the Routh criterion by evaluating the coefficients of characteristic polynomial a 3 =LM, a 2 =DL+MR, a 1 =DR+KL+C 2, a 0 =KR. The first column of the Routh array should have no sign changes: a 3 > 0, a 2 > 0, b 1 > 0, c 1 > 0. The b 1, c 1 coefficients should be computed according to the Routh array. d) For the given set of parameters, L=0.001, R=0.1, C=1, M=0.2, D=0.01, K=1000, show the Bode diagram and step response of the plant. Is the closed-loop system with negative feedback of the armature displacement stable? Explain your answer. (3 pt) The Bode diagram and step response of the system with assumed numerical values are: In order to see whether the closed-loop system with a negative feedback of the armature displacement is stable, use the root locus {matlab -> rlocus(g)}. One can see that the closed-loop is stable. However when increasing the feedback gain, the system becomes unstable up from a certain gain value. Evaluation scale (max 60 points): points < 5 F 5 <= points <= 11 E 12 <= points <= 23 D 24 <= points <= 35 C 36 <= points <= 47 B 48 <= points A

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