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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 steadystate 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 steadystate error. (c) Show that although a PI controller satis...es the condition derived in part (a), it will not yield a stable closedloop 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 secondorder 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 unityfeedback 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
2
0 To get zero steadystate 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 steadystate 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 steadystate 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 steadystate 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
 Chen

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