Write MATLAB codes ONLY, algebraic calculations and graphs solutions are included.5.32 is 33 in the pdf
5047 (c) The equivalent plant with unity feedback isG0 = 200 s2(s2 + 12 + 40) + 200s : Thus the system is type 1 with Kv = 1: If the velocity feedback were zero, the system would be type 2 with Ka = 200 40 = 5: (d) The transfer function Y W1 = 100s2 s2(s2 + 12s+ 40) + 200(s+ 1) : The system is thus type 2 with Ka = 100: (e) The transfer function Y W2 = 100 s2(s2 + 12s+ 40) + 200(s+ 1) : The system here is type 0 with Kp = 1: (f) To get more damping in the closed-loop response, the controller needs to have a lead compensation. 33. Consider the plant transfer function G(s) = bs+ k s2[mMs2 + (M +m)bs+ (M +m)k] to be put in the unity feedback loop of Fig. 5.60. This is the transfer function relating the input force u(t) and the position y(t) of mass M in the non-collocated sensor and actuator problem. In this problem we will use root-locus techniques to design a controller D(s) so that the closed- loop step response has a rise time of less than 0.1 sec and an overshoot of less than 10%. You may use MATLAB for any of the following questions. (a) Approximate G(s) by assuming that m �= 0, and let M = 1, k = 1, b = 0:1, and D(s) = K. Can K be chosen to satisfy the performance speci
cations? Why or why not? (b) Repeat part (a) assuming D(s) = K(s+ z), and show that K and z can be chosen to meet the speci
cations. (c) Repeat part (b) but with a practical controller given by the transfer function D(s) = K p(s+ z) s+ p ; and pick p so that the values forK and z computed in part (b) remain more or less valid. (d) Now suppose that the small mass m is not negligible, but is given by m = M=10. Check to see if the controller you designed in part (c) still meets the given speci
cations. If not, adjust the controller pa- rameters so that the speci
cations are met. Solution: (a) The locus in this case is the imaginary axis and cannot meet the specs for any K: 5048 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD (b) The specs require that � > 0:6; !n > 18: Select z = 15 for a start. The locus will be a circle with radius 15: Because of the zero, the overshoot will be increased and Figure 3.32 indicates that wed better make the damping greater than 0.7. As a matter of fact, experimentation shows that we can lower the overshoot of less than 10% only by setting the zero at a low value and putting the poles on the real axis. The plot shows the result if D = 25(s+ 4): (c) In this case, we take D(s) = 20 s+ 4 :01s+ 1 : (d) With the resonance present, the only chance we have is to introduce a notch as well as a lead. The compensation resulting in the plots shown is D(s) = 11 s+ 4 (:01s+ 1) s2=9:25 + s=9:25 + 1 s2=3600 + s=30 + 1 : The design gain was obtained by a cycle of repeated loci, root location
nding, and step responses. Refer to the
le ch5p35.m for the design aid. Root Locus Real Axis Im ag A xi s -60 -50 -40 -30 -20 -10 0 -20 -10 0 10 20 0 0.2 0.4 0.6 0.8 0 0.5 1 1.5 Root Locus Real Axis Im ag A xi s -60 -50 -40 -30 -20 -10 0 -20 -10 0 10 20 0 0.2 0.4 0.6 0.8 0 0.5 1 1.5 Root Locus Real Axis Im ag A xi s -10 -8 -6 -4 -2 0 2 -4 -2 0 2 4 0 0.2 0.4 0.6 0.8 1 0 0.5 1 Root loci and step responses for Problem 33 34. Consider the type 1 system drawn in Fig. 5.64. We would like to design the compensation D(s) to meet the following requirements: (1) The steady- state value of y due to a constant unit disturbance w should be less than 4 5 , and (2) the damping ratio � = 0:7. Using root-locus techniques: (a) Show that proportional control alone is not adequate. 6156 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 58. For a unity feedback system with G(s) = 1 s( s20 + 1)( s2 1002 + 0:5 s 100 + 1) (2) (a) A lead compensator is introduced with � = 1=5 and a zero at 1=T = 20. How must the gain be changed to obtain crossover at !c = 31:6 rad/sec, and what is the resulting value of Kv? (b) With the lead compensator in place, what is the required value of K for a lag compensator that will readjust the gain to a Kv value of 100? (c) Place the pole of the lag compensator at 3.16 rad/sec, and determine the zero location that will maintain the crossover frequency at !c = 31:6 rad/sec. Plot the compensated frequency response on the same graph. (d) Determine the PM of the compensated design. Solution : (a) From a sketch of the asymptotes with the lead compensation (with K1 = 1) : D1(s) = K1 s 20 + 1 s 100 + 1 in place, we see that the slope is -1 from zero frequency to ! = 100 rad/sec. Therefore, to obtain crossover at !c = 31:6 rad/sec, the gain K1 = 31:6 is required. Therefore, Kv = 31:6 6157 (b) To increase Kv to be 100, we need an additional gain of 3.16 from the lag compensator at very low frequencies to yield Kv = 100: (c) For a low frequency gain increase of 3.16, and the pole at 3.16 rad/sec, the zero needs to be at 10 in order to maintain the crossover at !c = 31:6 rad/sec. So the lag compensator is D2(s) = 3:16 s 10 + 1 s 3:16 + 1 and D1(s)D2(s) = 100 s 20 + 1 s 100 + 1 s 10 + 1 s 3:16 + 1 The Bode plots of the system before and after adding the lag com- pensation are 100 101 102 103 10-2 100 102 ω (rad/sec) M ag ni tu de Bode Diagrams 100 101 102 103 -400 -300 -200 -100 0 ω (rad/sec) P ha se (d eg ) 100 101 102 103 10-2 100 102 ω (rad/sec) M ag ni tu de Bode Diagrams Lead and Lag Lead only 100 101 102 103 -400 -300 -200 -100 0 ω (rad/sec) P ha se (d eg ) 6158 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (d) By using the margin routine from Matlab, we see that PM = 49� (!c = 34:5 deg/sec) 59. Golden Nugget Airlines had great success with their free bar near the tail of the airplane. (See Problem 5.39) However, when they purchased a much larger airplane to handle the passenger demand, they discovered that there was some exibility in the fuselage that caused a lot of unpleasant yawing motion at the rear of the airplane when in turbulence and was causing the revelers to spill their drinks. The approximate transfer function for the dutch roll mode (See Section 10.3.1) is r(s) �r(s) = 8:75(4s2 + 0:4s+ 1) (s=0:01 + 1)(s2 + 0:24s+ 1) where r is the airplanes yaw rate and �r is the rudder angle. In performing a Finite Element Analysis (FEA) of the fuselage structure and adding those dynamics to the dutch roll motion, they found that the transfer function needed additional terms that reected the fuselage lateral bending that occurred due to excitation from the rudder and turbulence. The revised transfer function is r(s) �r(s) = 8:75(4s2 + 0:4s+ 1) (s=0:01 + 1)(s2 + 0:24s+ 1) � 1 ( s 2 !2b + 2� s!b + 1) where !b is the frequency of the bending mode (= 10 rad/sec) and � is the bending mode damping ratio (= 0:02). Most swept wing airplanes have a yaw damperwhich essentially feeds back yaw rate measured by a rate gyro to the rudder with a simple proportional control law. For the new Golden Nugget airplane, the proportional feedback gain, K = 1; where �r(s) = �Kr(s): (3) (a) Make a Bode plot of the open-loop system, determine the PM and GM for the nominal design, and plot the step response and Bode magnitude of the closed-loop system. What is the frequency of the lightly damped mode that is causing the di¢ culty? (b) Investigate remedies to quiet down the oscillations, but maintain the same low frequency gain in order not to a¤ect the quality of the dutch roll damping provided by the yaw rate feedback. Speci
cally, investigate one at a time: i. increasing the damping of the bending mode from � = 0:02 to � = 0:04: (Would require adding energy absorbing material in the fuselage structure) ii. increasing the frequency of the bending mode from !b = 10 rad/sec to !b = 20 rad/sec. (Would require stronger and heavier structural elements) 6101 (d) The characteristic equation for PM of 45� : 1 + 1:1 s(s+ 1) h� s 5 �2 + 0:4 � s 5 � + 1 i = 0 =) s4 + 3s3 + 27s2 + 25s+ 27:88 = 0 =) s = �1:03� j4:78; �0:47� j0:97 32. For the system depicted in Fig. 6.94(a), the transfer-function blocks are de
ned by G(s) = 1 (s+ 2)2(s+ 4) and H(s) = 1 s+ 1 : (a) Using rlocus and rloc
nd, determine the value of K at the stability boundary. (b) Using rlocus and rloc
nd, determine the value of K that will produce roots with damping corresponding to � = 0:707. (c) What is the gain margin of the system if the gain is set to the value determined in part (b)? Answer this question without using any frequency response methods. (d) Create the Bode plots for the system, and determine the gain margin that results for PM = 65�. What damping ratio would you expect for this PM? 6102 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.94: Block diagram for Problem 32: (a) unity feedback; (b) H(s) in feedback (e) Sketch a root locus for the system shown in Fig. 6.94(b).. How does it di¤er from the one in part (a)? (f) For the systems in Figs. 6.94(a) and (b), how does the transfer func- tion Y2(s)=R(s) di¤er from Y1(s)=R(s)? Would you expect the step response to r(t) be di¤erent for the two cases? Solution : (a) The root locus crosses j! axis at s0 = j2. K = 1 jH(s0)G(s0)j js0=j2 = jj2 + 1j jj2 + 4j jj2 + 2j2 =) K = 80 6103 (b) � = 0:707 =) 0:707 = sin � =) � = 45� From the root locus given, s1 = �0:91 + j0:91 K = 1 jH(s1)G(s1)j js1=�0:91+j0:91 = j0:01 + j0:91j j3:09 + j0:91j j1:09 + j0:91j2 =) K = 5:9 (c) GM = Ka Kb = 80 5:9 =