EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions"

Transcription

1 EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller to drive the step response error to zero. (a) Draw the root locus for the uncompensated system. Answer: (pt) 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (b) (c) Determine the gain K and location of the poles at the operating point. Answer: (pt) K = 242, poles are.49,.4926 ± j Using sisotool, we have a damping ratio of. when K 242. The poles are.49,.4926± j G = tf(,poly([ 3 8])); 2 K = 242; 3 poles = pole(feedback(k*g,)) Add an ideal integral compensator with a pole at the origin and a zero at.. Draw the compensated root locus. Answer: (2pt) G c = K(s+.) s(s+)(s+3)(s+8) of 6

2 EE C28 / ME C34 Fall 24 HW 6.2 Solutions 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (d) (e) Again, determine the gain K, and the locations of the poles at the operating point. Answer: (pt) K 234, poles:.9387,.4848 ± j4.828,.98 Use MATLAB to simulate the step responses. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) Step Response.6.4 Uncompensated compensated.2 Amplitude Time (seconds) Indeed the compensated system has zero steady-state error. The response reaches. at t 85 s. 2. Lag Compensator For the system G = K (s+)(s+3)(s+8) in negative feedback, we are going to design a lag compensator reduce the steady-state error by a factor of, operating with a damping ratio of. (a) Using the gain K from Problem 2(b), determine K p and the steady-state error for a step input for both the uncompensated system and the desired lag-compensated system. 2 of 6

3 EE C28 / ME C34 Fall 24 HW 6.2 Solutions Answer: (pt) K p =.8, e step ( ) =.9, K des p = 9.83, e des step( ) = K p = lim s (s + )(s + 3)(s + 8) =.8, e step ( ) = =.9, + K p e des step( ) = e step( ) K des p = =.9, e des = step ( ) (b) (c) Using these values and a pole placed at., determine the location of the required zero. Answer: (pt) s =. K des p = lim s s + z c s + p c G(s) = z c p c K p. If we place a pole at s =., then p c =., so z c = Kdes p =.. Thus the zero must be placed at s =.. p c K p Draw the compensated root locus. Answer: (2pt) 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (d) (e) Again, determine the new gain K, and the new locations of the poles at the operating point Answer: (pt) K 234, poles:.9388,.485 ± j4.83,.9 Use MATLAB to simulate the step responses. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) 3 of 6

4 EE C28 / ME C34 Fall 24 HW 6.2 Solutions Step Response.6.4 Uncompensated compensated.2 Amplitude Time (seconds) The steady-state error of the uncompensated and compensated systems are.92 and.93, respectively, a reduction by a factor of 9.7 ( ). 3. Lead Compensator For the system G = (s+3)(s+4) in negative feedback, we are going to design a lead compensator to achieve a settling time of less than.7 s and a percent overshoot of less than.5%. (a) Draw the uncompensated root locus. Answer: (pt) Imaginary Axis (seconds ) Real Axis (seconds ) (b) A lead compensator moves the closed-loop poles to a new desired location. Where should the closed-loop poles be located to satisfy the design specifications? Answer: (2pt) s = 5.7 ± j4.27 for T s =.7 s and %OS =.5% Using a second-order approximation, the poles are located at s = ζω n ± jω n ζ 2. The settling time can be approximated by T s = 4 ζω n = 4 Re(s), 4 of 6

5 EE C28 / ME C34 Fall 24 HW 6.2 Solutions so we need Re(s) > 4.7 = 5.7. For at most.5% overshoot, we need damping ratio ζ > ln(%os/) = π 2 + ln 2 (%OS/) ln(.5) π 2 + ln 2 (.5) =.8 Using the representation of s = ζω n ±jω n ζ 2 in the complex plane,to ( satisfy ) the overshoot condition, the poles must lie below the line which makes angle θ = cos ζωn = cos (ζ) = 36.8 with the negative real axis. The poles should be in the region bounded by the solid lines in the graph below, with the poles at 5.7 ± j4.27 corresponding to the lower bounds. ω n Im(s) Re(s) (c) (d) In what general region should the compensator pole and zero be placed to shift the root locus in that direction? Why? Answer: (pt) The compensator zero needs to be put to the left of the current poles to pull the root locus to the left. The compensator pole should be placed even further off to the left to give the zero ample time to bring the root locus to the left before its effect is canceled by the compensator pole. For this example, we choose to place a zero at 25 and a pole at 6. Find the value of K that gives us an overshoot of.4% Answer: (pt) K = 9.2 For an overshoot of.4% we need ζ = ln(.4) =.85. π 2 + ln 2 (.4) Using sisotool, ζ =.85 when K = 9.2. (e) Draw the compensated root locus. Answer: (2pt) 5 of 6

6 EE C28 / ME C34 Fall 24 HW 6.2 Solutions 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (f) Use MATLAB to simulate the compensated step response and find the overshoot and settling time. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) T s =.82s, %OS =.44 Step Response Amplitude Time (seconds) The overshoot is.44% and setting time is.82s. We did not meet the design specifications, though we were close. We used a second-order approximation for the design step. Even though the pole at 6 is far away from the dominant poles of the system, in this case the approximation was not close enough. 6 of 6