ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II


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1 ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and Books All questions are equally weighted All necessary formulas are provided at the last page Student No. :.. Name : SURNAME :.. Signature :.. 3 Σ
2 PROBLEM The transfer function of a plant is given as: G p (s) s + 3s a) Determine the stability of the uncontrolled plant. b) Design a closedloop control system with unity feedback using a Pcontroller with a gain of K>0. Draw the block diagram of the closedloop system and determine the conditions on K to have a stable closedloop system. c) What is the steadystate error of the closedloop system in (b) to a unit step input? (Use K 5 for the Pcontroller.) d) Modify the closedloop system in (b) so that you have a PIcontroller illustrated below. Redraw the block diagram of the closedloop system. What is the steadystate error of the new closedloop system to a unit step input? PIcontroller + s 0 e) Comment on the steadystate errors you found in (c) and (d). Why are they different? Hint: Think about the type numbers of the closedloop systems in (c) and (d). f) Bonus: Why did we use different gains in (c) and (d)? Solution: a) The characteristic polynomial of the uncontrolled system is equal to characteristic polynomial of the plant. D(s) s + 3s. Since there is a sign change in the coefficients of the characteristic polynomial, D(s) cannot pass the Hurwitz test which results in same conclusion with Routh criterion for second order systems, so the uncontrolled system is unstable. b) The block diagram of the closedloop system with a PIcontroller having a gain of K is: R(s) E(s) K R(s) s + 3s K + s + 3s When simplified the transfer equation becomes: R(s) K s + 3s + (K ) In order to have a stable system all the coefficients of the characteristic polynomial should have the same sign due to the Hurwitz test. Since the coefficients of s and s are positive K should also be positive. This imposes that: K > 0; K > ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page
3 c) Taking K 5 for the above system gives us the following openloop transfer function (OLTF): G OL G H 5 G OL s + 3s If we put GG OOOO in the standard form, it becomes: G OL 5 s 3 s + K OL 5 Since there is no free s in the denominator the type number of the system is 0. The steadystate error (for unit step input) (e ss ) becomes. +K OL d) The block diagram of the new system with the PIcontroller is: PIcontroller R(s) E(s) The OLTF of the new closedloop control system is: 0(s + ) G OL s(s + 3s ) If we put it into the standard form: G OL 0 (s + ) s( s 3 s + ) K OL 0 Since there is free s in the denominator of the OLTF, the type number of the system is. The steadystate error (for unit step input) becomes 0 in that case. e) The steadystate errors in (c) and (d) are different due to the addition of a PIcontroller in (d). When we add a PIcontroller in the closedloop control system, we create add a free s in the denominator of the OLTF, hence increase the type number of the system by. f) We used different gains in (b) and (d) because the gain we used in (b) results in an unstable system in (d). Check the stability of the closedloop system in (d) for K5 to clarify the explanation. ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page
4 PROBLEM Below is the model of a first order system with a commercially available PID controller in unity feedback configuration. Switches I, II and III can be independently arranged for different controller configurations where the transfer function of the system for all the switches are on is: G s Cs s K d + sk p + K i R s s + Kd + s+ Kp + Ki Note that switches have the effect of multiplication of the corresponding gain (Kp, Ki and Kd) by (when closed/on) and by 0 (when open/off). a) For the Kp values below together with Ki and Kd0, determine whether the system exhibits an oscillatory behavior to a step input or not? If oscillatory, is the output converging or not? i) Kp 5 ii) Kp 3 iii) Kp b) When switch I is on, switch II is off, and switch III is off, how does the speed of the step response get affected when Kp is increased? c) For positive Kp, Ki and Kd, list down all possible states of the switches so that the step response have i. zero initial jump and ii. nonzero initial jump at t0 +? Justify your answer. d) For only proportional (with positive Kp) and only integral (with positive Ki), what are the corresponding type number of the systems? Solution: a) For Ki and Kd0 G s Cs skp + Ki skp + R s as + s b+ K + K s + s + K + so the poles are ( p) i ( p) p p + K ± + K 3 p, 6± For Kp 5 the poles are p,,, so the response in not oscillatory Kp 3 the poles are p, ± i, so the response is oscillatory but convergent. Kp the poles are p, ± i, so the response is sustained oscillations. ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page 3
5 b) When switch I is on, switch II is off, and switch III is off, the transfer function becomes G s Cs Kp R s as b K a T. b+ K p + ( + p ), so the only pole is As Kp is increased, response becomes faster. b+ K p p and the time constant is a + x s Kd + sk 0 p + Ki s Kd + skp + Ki K y 0 lim s lim x d 0 x0 s s s a+ Kd + s b+ Kp + Ki s s a+ K d + s b+ Kp + K + i d c) so if switch III is on, no matter in which state other switches are, there will be a nonzero initial jump. If switch III is off, no matter in which state other switches are, there will be no initial jump. d) Cs K i G 0 s Kp skd R s + + s s + For KdKi0 (Pcontrol) G ( s) For KpKd0 (Icontrol) G ( s) 0 0 Cs Kp, N0, so the type number is 0. R s s + Ki ( + ) Cs, N, so the type number is. R s s s ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page
6 PROBLEM 3 A dynamic system with the following feedback architecture will be controlled by selecting appropriate positive values for amplifier gain and sensor gain Kh. E(s) a) What will be the steady state error of the closed loop system to unit impulse input? b) Determine the gains and Kh that will make the closed loop system have the following unit step response.. Step Response. Time (seconds):.8 Amplitude:.6 Amplitude Time (seconds):. Amplitude: Time (seconds):.0 Amplitude: Time (seconds) Solution: a) Steady state error of the closed loop system to unit impulse input can be found by stability analysis of the closed loop system. Transfer function of the inner loop can be found as G i (s) s+ + K h s+ s+ s++k h s+ s+ K h +. The feedforward and openloop transfer functions are G(s) G i (s) G(s)H(s) s s +( K h +)s since H(s). Closed loop transfer function is R(s) s + K h + s + s + K h + s s + K h + s s + K h + s+ s + K h + s s +( K h +)s+. ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page 5
7 The closed loop system will be stable for all positive values of and Kh therefore the steady state error of the closed loop system to unit impulse input is zero. Alternatively; Since E(s) e ss lim se(s) lim s +G(s)H(s) R(s) where R(s). + s + ( K h + )s lim s s + ( K h + )s + s + ( K h + )s b) The step response shown has the following transient response specifications: t r. s t p.8 s M p 6% t s.0 s (% criterion). These can be satisfied with the system parameters given as ω n rad s ζ 0.5 using for example t s ζω n and M p e lim s 3 + ( K h + )s s + ( K h + )s + 0. ζ π ζ Closed loop transfer function can be shown to be equivalent to an ideal second order system R(s) s + ( K h + )s + ω n 8. K h + ζω n K h 3 8. s + K h + s + ω n s + ζω n s + ω n. ME30/6S/MT/Balkan,Azgın,Turgut,Yazıcıoğlu Page 6
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