EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO

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EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total 100 1. You have 2 hours to complete this exam. INSTRUCTIONS 2. This is a closed book exam. You may use one 8.5 11 note sheet. 3. Calculators are allowed. 4. Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, Problem 1.2 Continued. NO credit will be given to solutions that do not meet this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. 1

Problem 1: (25 Points) 1. (13 points) Consider the feedback control system in Figure 1. Figure 1: Feedback control system. (a) (7 points) Determine the system type. (b) (3 points) Determine the position, velocity, and acceleration error constants. 2

(c) (3 points) For each of the following command inputs, determine the steady-state value of e(t): r 1 (t) = 10 u o (t) r 2 (t) = 10 t u o (t) r 3 (t) = 10 t2 2 u o(t). 3

2. (12 points) Figure 2 shows the block diagram of a nonunity feedback system. Figure 2: Nonunity feedback system. (a) (9 points) Determine the system type. (b) (3 points) Determine the value of the nonzero, finite, error constant. 4

Problem 2: (25 Points) Figure 3 shows the block diagram of a feedback system that regulates the angular displacement θ(t) of a servomotor. Figure 3: Servomotor system. 1. (7 points) In response to the input θ r (t) = πu o (t), the closed-loop system generates the zero-state where τ is a positive constant. θ(t) = C 0 + C 1 e t/τ cos(ω o t + θ), (a) (3 points) Without directly calculating θ(t), what must be the value of the constant C 0? Justify your answer with one or two sentences. (b) (4 points) Determine τ and ω o in terms of the controller gains K 1 and K 2. 5

2. (4 points) Given that the closed-loop system is BIBO stable, determine the velocity error constant in terms of the controller gains K 1 and K 2. 3. (4 points) Determine the sensitivity of the velocity error constant with respect to the controller gain K 1. 6

4. (10 points) Choose the controller gains K 1 and K 2 that achieve the following design specifications: Peak overshoot of 16.3% Error velocity constant of 20 sec 1. 7

Problem 3: (25 Points) The root locus for the closed-loop system in Figure 4 is shown in Figure 5 as the compensator gain K is varied from zero towards infinity. Figure 4: Closed-loop system with proportional control gain K. 1.5 Root Locus 1 0.5 Imaginary Axis 0 0.5 1 1.5 2.5 2 1.5 1 0.5 0 0.5 1 1.5 Real Axis Figure 5: Root locus as the proportional gain K is varied from zero towards infinity. 1. (7 points) Estimate the value of proportional gain K for which the closed-loop poles break-in to the real axis. 8

2. (18 points) Once again consider the control system in Figure 4, and whose root locus is shown again in Figure 6. Ignoring the effect of the zeros, an engineer chooses the value of K so that the location of the complex conjugate poles yields a settling time 9.2 sec. 1.5 Root Locus 1 0.5 Imaginary Axis 0 0.5 1 1.5 2.5 2 1.5 1 0.5 0 0.5 1 1.5 Real Axis Figure 6: Root locus as the proportional gain K is varied from zero towards infinity. (a) (8 points) What value of gain K did the engineer choose? (b) (5 points) For the chosen value of K, estimate the percent overshoot of the closed-loop unit-step response. (c) (5 points) For the chosen value of K, estimate the rise-time of the closed-loop unit-step response. 9

Problem 4: (25 Points) 1. (18 points) Consider the closed-loop system in Figure 7 where Gp(s) = s + 2 3 s 2 (s + 6) Figure 7: Closed-loop system with proportional control gain K. Neatly sketch by hand root locus in Figure 8 as the proportional gain is varied from zero towards infinity. Be sure to specify breakaway and breakin points, the gain at the breakaway and breakin points, asymptotes, arrival and departure angles, and imaginary-axis crossings, if any. Root Locus 3 2 Imaginary Axis 1 0 1 2 3 8 6 4 2 0 2 Real Axis Figure 8: Root locus as the proportional gain K is varied from zero towards infinity. 10

11

2. (7 points) Write an m-file that constructs the root locus for the system in part 1, draws lines of constant ζ = 0.9 and ω n = 3 rad/s on the plot, and allows the user to determine the gain for a desired pole position by selecting a point in the graphics window. 12