ch4soln - Chapter 4 Basic Properties of Feedback Problems...

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Chapter 4 Basic Properties of Feedback Problems and Solutions for Section 4.1 1. Consider a system with the con & guration of Fig.4.2(b) where D c is the constant gain of the controller and G is that of the process. The nominal values of these gains are D c =5and G = 7. Suppose a constant dis- turbance w is added to the control input u before the signal goes to the process. (a) Compute the gain from w to y in terms of D c and G . (b) Suppose the system designer knows that an increase by a factor of 6 in the loop gain D c G can be tolerated before the system goes out of speci & cation. Where should the designer place the extra gain if the objective is to minimize the system error r y due to the disturbance? For example, either D c or G could be increased by a factor of 6, or D c could be doubled and G tripled, and so on. Which choice is the best? Solution: (a) Need y/w so set r =0: Y W = G 1+ GD c = 7 1+35 (b) Take r =0:= e = y So to minimize e due to w ,increase D c to 30 (since it is given that beyond this dynamic response goes out of speci & cations) 169
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170 CHAPTER 4. BASIC PROPERTIES OF FEEDBACK 2. Bode de & ned the sensitivity function relating a transfer function G to one of its parameters k as the ratio of percent change in k to percent change in G .W ed e & ne the reciprocal of Bode&s function as S G k = dG/G dk/k = d ln G d ln k = k G dG dk . Thus, when the parameter k changes by a certain percentage, S tells us what percent change to expect in G . In control systems design we are almost always interested in the sensitivity at zero frequency, or when s = 0. The purpose of this exercise is to examine the e f ect of feedback on sensitivity. In particular, we would like to compare the topologies shown in Fig. 4.36 for connecting three ampli & er stages with a gain of K into a single ampli & er with a gain of 10. (a) For each topology in Fig.4.36, compute β i so that if K = 10, Y = 10 R . (b) For each topology, compute S G k when G = Y/R . [Use the respective β i values found in part (a).] Which case is the least sensitive? (c) Compute the sensitivities of the systems in Fig. 4.36(b, c) to β 2 and β 3 . Using your results, comment on the relative need for precision in sensors and actuators. Solution: (a) For K =10and y = 10 r, we have: Case a: y r = β 1 K 3 = β 1 =0 . 01 Case b: y r =( K 1+ β 2 K ) 3 = β 2 . 364 Case c: y r = K 3 β 3 K 3 = β 3 . 099 (b) Sensitivity S G K ,G = y r Case a: dG dK = 3 β 1 K 2 S G K = K G dG dK = K β 1 K 3 ( 3 β 1 K 3 )=3 Similarly: Case b: S G K . 646
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171 Figure 4.36: Three ampli & er topologlies for Problem 4.2 Case c: S G K =0 . 03 Casecistheleastsens it ive . (c) Sensitivities w.r.t. feedback gains: Case b: S G β 2 = 2 . 354 Case c: S G β 3 = 0 . 99 The results indicate that the closed-loop system is more sensitive to errors in the feedback path than in the forward path (sensors need to have relatively higher precision than actuators).
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ch4soln - Chapter 4 Basic Properties of Feedback Problems...

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