CALIFORNIA INSTITUTE OF TECHNOLOGYControl and Dynamical SystemsCDS 101/110R. M. MurrayFall 2003Homework Set #6 Issued: 10 Nov 03Due: 17 Nov 03All students should complete the following problems:1. Plot the Nyquist and Bode plots for the following systems and compute the gain and phase marginof each. You should annotate your plots to show the gain and phase margin computations. For theNyquist plot, mark the branches corresponding to the following sections of the Nyquist “D” contour:negative imaginary axis, positive imaginary axis, semi-circle at infinity (the curved part of the “D”).(a) Disk drive read head positioning system, using lead compensator (we’ll learn about how to designthese in a week or two):P (s) =1s3+ 10s2+ 3s + 10C(s) = 1000s + 1s + 10(b) Second order system with PD compensator:P (s) =100(100s + 1)(s + 1)C(s) = s + 10Note: you may find it easier to sketch the Nyquist plot from the Bode plot (taking some liberties withthe scale) rather than relying on MATLAB.2. In this problem we will design a PI controller for a cruise control system, building on the exampleshown in class. Using the following transfer function to represent the vehicle and engine dynamics:P (s) =ra/m(s + a)(s + b/m)where r = 5 is the transmission gain (this was labelled k in previous sets), a = 0.2 is the engine lagcoefficient, m = 1000 kg is the mass of the car, and b = 50 N sec/m is the viscous damping coeffient.(a) Consider a proportional controller for the car, u = Kp(r − y). Assuming a unity gain feedbackcontroller, this givesC(s) = Kp.Set Kp= 100 and compute the steady state error, gain and phase margins, and poles/zeros forthe closed loop system. Remember that the gain and phase margins are computed based on theloop transfer function L(s) = P (s)C(s).(b) Consider a proportional + integral controller for the car,C(s) = Kp+Kis.Fill in the following table (make sure to show your work):KpKiStable? Gain Margin Phase Margin Steady State Error Bandwidth500 10050 100050 15 1For each entry in the table, plot the pole zero diagram (pzmap) for the closed loop system andthe step response.Only CDS 110a students need to complete the following additional problems:3. Continuing the previous problem, we will now insert a small amount of time delay into the feedbackpath of the system. A pure time delay of τ seconds satisfies the equationy(t) = u(t − τ )This system is a linear input/output system and it can be shown that its transfer function isG(s) = e−sτ.Unfortunately, MATLAB is not able to perfectly represent a time delay in this form, and so we have touse a Pade approximation, which gives a constant gain transfer function with phase that approximatesa time delay. Using a 2nd order Pade approximation, we can approximate our time delay asG(s) =1 − Tds/2 + (Tds)2/121 + Tds/2 + (Tds)2/12(this function can be computed using the pade function in MATLAB).Assume that there is a time delay of τ seconds, which we will insert between the output of the plantand the controller (as we did in Monday’s lecture).(a) For the case Kp= 50, Ki= 1, insert time delays of τ = 0.25 sec and τ = 0.75 seconds. Usinga Pade approximation, compute the resulting gain and phase margin for each case and computethe overshoot and settling time (2%) for the step responses.(b) Repeat part (a) using Kp= 20, Ki= 0.5 and time delays of 0.75 sec and 1.5 sec.(c) Optional: Plot the Nyquist plot for Kp= 20, Ki= 0.5, τ = 0.75 (with the exact time delay, notthe Pade approximation).4. Consider a simple DC motor with inertial J and damping b. The transfer function isP (s) =1Js2+ bs.For simplicity, choose J = 2, b = 1. In this problem you will design some simple controllers to achievea desired level of performance.(a) Design a proportional control law, C(s) = Kp, that gives stable performance and has a bandwidthof at least 1 rad/s and a phase margin of at least 30 degrees. Plot the step response for the closedloop system using your controller.(b) Design a proportional + derivative controller (PD) of the formC(s) = Kp+ Kdsthat gives closed loop bandwidth ω = 10 rad/sec and has phase margin of at least 30 degrees.Plot the step response for the closed loop system using your
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