EECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.

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1 Name: SID: EECS C128/ ME C134 Final Wed. Dec. 14, am Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth 1 points total. Problem Points Score Total 1 tan = 26.6 tan 1 1 = 4 tan = 18.4 tan 1 3 = 6 tan = 14 tan = 3 sin3 = 1 2 cos 6 = log 1 1 = db 2 log 1 2 = 6dB 2 log 1 2 = 3dB 2 1 log1 2 = 6dB 2 log 1 = 2db 6dB = 14dB 2 log 1 1 = 1 db 1/e.37 1/e /e In the real world, unethical actions by engineers can cost money, careers, and lives. The penalty for unethical actions on this exam will be a grade of F and a letter will be written for your file and to the Office of Student Conduct. 1

2 Problem 1 (16 pts) reference input r(t) + Σ error e(t) Controller D(s) control input u(t) plant G(s) output y(t) grid 8 pixels For the above system, the root locus is shown for 4 different controller/plant combinations, D 1 (s)g 1 (s),..., D 4 (s)g 4 (s). (Note: the root locus shows open-loop pole locations for D(s)G(s), and closed-loop poles for DG 1+DG ). Root Locus D1(s)G1(s) Root Locus D2(s)G2(s) Imaginary Axis (seconds 1 ) Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) Real Axis (seconds 1 ) Root locus D3(s)G3(s) Root Locus D4(s)G4(s) Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) [4 pts a) For each set of open-loop poles and zeros given above, choose the best corresponding open-loop Bode plot W,X,Y, or Z from the next page: (i) D 1 (s)g 1 (s): Bode Plot (ii) D 2 (s)g 2 (s): Bode plot (iii) D 3 (s)g 3 (s): Bode plot (iv) D 4 (s)g 4 (s): Bode Plot 2

3 Problem 1, cont. The open-loop Bode plots for 4 different controller/plant combinations, D 1 (s)g 1 (s),..., D 4 (s)g 4 (s) are shown below. Bode Diagram Bode Diagram Magnitude (db) Magnitude (db) Phase (deg) 9 9 Phase (deg) Frequency (rad/s) W Frequency (rad/s) X Bode Diagram Bode Diagram Magnitude (db) Magnitude (db) Phase (deg) 9 9 Phase (deg) Frequency (rad/s) Y Frequency (rad/s) Z [8 pts b) For each Bode plot, estimate the phase and gain margin: (i) Bode plot W: phase margin (degrees) at ω = Bode plot W: gain margin db at ω = (ii) Bode plot X: phase margin (degrees) at ω = Bode plot X: gain margin db at ω = (iii) Bode plot Y: phase margin (degrees) at ω = Bode plot Y: gain margin db at ω = (iv) Bode plot Z: phase margin (degrees) at ω = Bode plot Z: gain margin db at ω = 3

4 Problem 1, cont. s c) For each closed loop controller/plant with root locus as given in part a), choose the best corresponding closed-loop step response (A-D) (i) D 1 (s)g 1 (s): step response (ii) D 2 (s)g 2 (s): step response (iii) D 3 (s)g 3 (s): step response (iv) D 4 (s)g 4 (s): step response Step A Step B Amplitude Time (seconds) Amplitude Time (seconds) Step C Step D Amplitude Amplitude Time (seconds) Time (seconds) 4

5 Problem 2 (12 pts) reference input r(t) + Σ error e(t) Controller D(s) control input u(t) plant G(s) output y(t) grid 8 pixels 1 You are given the open loop plant G(s) =. The system is to be controlled using a lag s 2 +17s+6 controller. with D(s) = s+1 s+α. Given: the roots of s s s + 1 (s + 1.7)(s j)(s jj) [8 pts a) Sketch the positive root locus as α varies, noting asymptote intersection point and angle of departure. [4 pts b) (i) approximate asymptote intersection point s = (ii) approximate angle of departure for the poles: 1 Root Locus 8 6 Imaginary Axis (seconds 1 ) Real Axis (seconds 1 )

6 Problem 3 (8 pts) Consider the following circuit for a lag compensator: [4 pts a) Find the transfer function Y (s) U(s) [4 ptsb) Suppose the desired behaviour of this circuit is that the (asymptotic) phase response is 9 between 1rad/s and 1rad/s. At every other frequency the phase response should be greater than 9. If C = 1µF, what are the resistor values, R 1 and R 2? 6

7 Problem 4 (16 pts) You are given the following plant [ ẋ = Ax + Bu = x + [ 2 1 [ u(t), y = [4 1 x x(t = ) = 3 [2 pts a) Determine if the system is controllable and observable. such that with control u = K(r x), the con- [ k1 [4 pts b) Find feedback gains K = k 2 troller has closed loop poles at -2 and -4. k 1 = k 2 = [2 pts c) Draw a block diagram of the controlled system using integrators, summing junctions, and scaling functions. (Every signal should be a scalar, no vectors.) 7

8 Problem 4, cont You are given the following plant [ 1 ẋ = A 1 x + B 1 u = x + [ 1 [ u(t), y = [4 1 x x(t = ) = 3 [2 pts d) Determine if the system {A 1, B 1, C 1 } is controllable and observable. [4 pts e) Find feedback gains K = [k 1 k 2 such that with control u = K(r x), the controller has closed loop poles at -2 and -4. k 1 = k 2 = [2 pts f) Draw a block diagram of the controlled system using integrators, summing junctions, and scaling functions. (Every signal should be a scalar, no vectors.) 8

9 Problem (11 pts) [3 pts a) Given the following system: ẋ = Ax + Bu y = Cx The state is transformed by a non-singular P such that x = Px. Thus x = Ā x+ Bu and y = C x. Find Ā B C in terms of A, B, C, P: A =: B =: C =: [4 pts b) You are given the following system: [ [ x + 1 u(t), y = [3 1 x Find the transformation P and Ā such that Ā = P 1 AP is in modal canonical (diagonal) form. [ [ P = Ā = [4 pts d) Find eāt and e At : [ eāt = [ e At = 9

10 Problem 6 (16 pts) The simplified dynamics of a magnetically suspended steel ball are given by: mÿ = mg c u2 y 2 y is the position of the ball; u is the current through the coil (in amps); c is a constant that describes the magnetic force between the coil and the ball. The system is linearized at equilibrium position y with equilibrium input u e : y = y + δy The linearized state space equations are: [ [ [ẋ1 1 x1 = ẋ 2 = 2g y [ 1 2 x 2 [ x1 δy = [ 1 [ x 1 x 2 x 2 [ cg y m [ 2 u = u e + δu u e = y mg c δu δu [1 pts (a) What are the units of c? Assume that all other quantities are SI standard (kilograms, meters, amps, etc). [4 pts (b) We want to build a regulator to keep the ball at y. We will design a state feedback scheme, δu = Kx, so that the poles of the linearized system are at s = 2, 12. Find K. K = [ [3 pts (c) Assume that you can directly access x 1 and x 2. You build your regulator as described above, and it successfully levitates the ball. You decide to try levitating four steel balls at the same time. Now m is four times bigger; everything else stays the same. Is your linearized system still stable? Will the steel balls be stable at y? Why or why not? 1

11 [4 pts (d) Return to the one-ball problem. Assuming that the only accessible output of the plant is y, you will need an observer in order to implement state feedback. Draw a block diagram of your regulator system. Use one block labelled Plant, with input u and output y; one block labelled Observer, with output ˆx 1 and ˆx 2 (you decide what the input should be); static gains; and addition junctions. Every signal should be scalar (no vectors). Label as many signals as you can. (Note that you re not being asked to design the observer gain). [2 pts (e) What are some sensible values for the poles of the observer? [2 pts (f) Does using an observer introduce any new problems if you try to levitate four balls, as in (c)? 11

12 Problem 7. (13 pts) You are given a continuous time plant described by the following state equation. ẋ = Ax + Bu The system is driven with a D/A converter such that u(t) = u[n for nt < t < nt + T. (That is, the input is held constant, by a zero-order hold equivalent.) Every T seconds the state of the system is measured with an A/D converter, that is x[n = x(nt). Recall that the solution for the continuous time system is given by: t x(t) = e A(t to) x(t o ) + e A(t τ) Bu(τ)dτ. t o (1) [3 pts a) For the zero input response, (x(t = ) = x o, u(t) = ) Find:(in terms of A and x o ) x[ =: x[1 =: x[n =: [3 pts b) For the zero state response, (x(t = ) = ) Find:(in terms of A, B, u) x[ =: x[1 =: x[n =: [3 pts c) Consider the CT system ẋ = x+u. With u(t) as shown, sketch x(t) for < t < 6sec with initial condition x() = u(t) s d) Let T = 1sec. For the CT system ẋ = x + u, x o =, with zero-order hold on input, determine the value of x at following steps (Answers may be left in terms of e.) Consider u[ = 1, u[1 = 1 etc. x[ =: x[1 =: x[2 =: x[3 =: 12

13 Problem 8 (8 pts) You are given the following plant [ 2 1 ẋ = Ax + Bu = 1 x + [ 1 [ 1 u(t), y = [1 4 x and x(t = ) = The LQR method is used to find the linear control u = Kx which minimizes the cost J = (xt Qx + u T Ru)dt, where Q and R are positive semi-definite. Four responses of the closed-loop system A,B,C,D are shown below for different choices of Q, R. Match the plots with Q R weights below. 1 State Response Using LQR Controller x1 x2 1 State Response Using LQR Controller x1 x2 x x u Control Inputs Using LQR Controller A State Response Using LQR Controller u1 x1 x2 u Control Inputs Using LQR Controller B State Response Using LQR Controller u1 x1 x2 x x Control Inputs Using LQR Controller u Control Inputs Using LQR Controller u1 u 1 u C D [4 pts For each plot of x(t) and u(t) choose the appropriate Q, R pair, writing in the appropriate letter. Q = [ 1 1 R = [1 Q = [ 1 1 R = [1 Q = Plot: [ 1 1 Plot: R = [1 Q = Plot: [ 1 1 Plot: R = [1 13

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