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Unformatted text preview: EE3530, Spring 2008 Homework # 7 Problems Due date: Monday, April 7th, 2008 4019 12. Consider the system shown in Fig. 4.36 which represents control of the angle of a pendulum which has no damping. Figure 4.36: Control system for Problem 12 (a) What condition must D(s) satisfy so that the system can track a ramp reference input with constant steady-state error? (b) For a transfer function D(s) that stabilizes the system and satis...es the condition in part (a), ...nd the class of disturbances w(t) that the system can reject with zero steady-state error. (c) Show that although a PI controller satis...es the condition derived in part (a), it will not yield a stable closed-loop system. Will a PID controller work; that is, satisfy part (a) and stabilize the system? If so, what constraints must kp , kI , and kD satisfy? (d) Discuss qualitatively and brie the eects of small variations on the y controller parameters kp , kI , and kD on the system' step response s CHAPTER 4. BASIC PROPERTIES OF FEEDBACK rise time and overshoot. 4022 13. A unity Solution: system has the overall transfer function feedback
(a) Y (s) !2 n : = T1 = 2 (s) R(s)Y = (W +s + 2 Y ) s + ! 2 ! n KY n D(R ) 2
s s2 + D + K W + DR Give the system type and corresponding error constant for tracking polyY( )= 2 nomial reference inputs in terms s of and ! n . s2 Y = s2 D 1 R+ 2 W +D+K s +D+K D + s2 + D + K 2 R(s) Y (s)s= + 2 2 n s s !+ D + K = Solution:
E(s) =E(s) R(s) R(s) = s2 + 2 ! n s + ! 2 2 n
s +K R(s) +D+K = 1 4 2 2 s+( )s + : : : !n !2 n s2 1 !n !2 n Kv = : Therefore, the two can be set independently and Ka = 2 1 4 2 except that K = 1 for = 0:5: EE3530, Spring 2008 Homework # 7 Problems Due date: Monday, April 7th, 2008 4023 14. Consider the second-order system G(s) = 1 : +2 s+1 s2 We would like to add a transfer function of the form D(s) = K(s+a)=(s+b) in series with G(s) in a unity-feedback structure. (a) Ignoring stability for the moment, what are the constraints on K; a, and b so that system type 1? (b) What are the constraints placed on K, a, and b so that the system is stable and type 1? (c) What are the constraints on a and b so that the system is type 1 and remains stable for every positive value for K? Solution: K 0 (s + a) : (a) Rewrite D(s) as D(s) = s ( + 1) b E(s) = R(s) Y (s) = R(s) K 0 (s + a) !2 n R(s) s s2 + 2 ! n s + ! 2 n ( + 1) b s ( + 1)(s2 + 2 ! n s + ! 2 ) K 0 (s + a)! 2 n n = b R(s) s ( + 1)(s2 + 2 ! n s + ! 2 ) n b ess s ( + 1)(s2 + 2 ! n s + ! 2 ) K 0 (s + a)! 2 1 n n = lim s b s s!0 s2 ( + 1)(s2 + 2 ! n s + ! 2 ) n b
0 To get zero steady-state error we must have b = 1, K a = 1, and !n 2 ! 2 K 0 = 2 ! n . This means a = and K 0 = . n EE3530, Spring 2008 Homework # 7 Problems Due date: Monday, April 7th, 2008 4025 16. Consider the system shown in Fig. 4.37(a). (a) What is the system type? Compute the steady-state tracking error due to a ramp input r(t) = ro t1(t). (b) For the modi...ed system shown in Fig. 47(b), give the value of Hf so the system is type 2 for reference inputs and compute the Ka in this case. (c) Is the resulting type 2 property of this system robust with respect to changes in Hf ? i.e., will the system remain type 2 if Hf changes slightly? Figure 4.37: Control system for Problem 16 Solution: (a) System is Type 1. E(s) = [1 = = T (s)]R(s) 1 R(s) 1 + G(s) s( s + 1) ro s( s + 1) + A s2
3 The steady-state tracking error using the FVT (assuming stability) is ro ess = lim sE(s) = : EE3530, Spring 2008 Homework # 7 Problems Due date: Monday, April 7th, 2008 4027 17. A controller for a satellite attitude control with transfer function G = 1=s2 has been designed with a unity feedback structure and has the transfer 10(s + 2) function D(s) = s+5 (a) Find the system type for reference tracking and the corresponding error constant for this system. (b) If a disturbance torque adds to the control so that the input to the process is u + w; what is the system type and corresponding error constant with respect to disturbance rejection? Solution: (a) Kp = lim D(s)G(s) = 1
s!0 ess = 1 = 0: 1 + Kp Kv = lim sD(s)G(s) = 1
s!0 ess = 1 = 0: Kv Ka = lim s2 D(s)G(s) = 4
s!0 ess = 1 = 0:25: Ka (b) For the disturbance input, the error is E(s) W (s) = = G 1 + GD s2 (s s+5 + 5) + 10(s + 2) 1 : There1 + Kp The steady-state error to a step is thus ess = 0:25 = fore, Kp = 3 4 ...
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This note was uploaded on 02/03/2012 for the course EE 3530 taught by Professor Chen during the Fall '07 term at LSU.
- Fall '07