Chapter6 - CHAPTER 6 6.1 THE PERFORMANCE OF FEEDBACK...

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CHAPTER 6 THE PERFORMANCE OF FEEDBACK SYSTEMS 6.1 (a) Refer Section 6.3 for the derivation (b) The characteristic equation is s 2 + 10 s + 100 = s 2 + 2 zw n s + w n 2 = 0 (i) w n = 10 rad/sec; z = 0.5; w d = w n 1 2 - z = 8.66 rad/sec (ii) t r = pz w - - cos 1 d = 0.242 sec; t p = p w d = 0.363 sec M p = e -- pz z /1 2 = 16.32%; t s = 4 zw n = 0.8 sec (iii) K p = lim s 0 G ( s ) = ¥ ; K v = lim s 0 sG ( s ) = 10; K a = lim s 0 s 2 G ( s ) = 0 (iv) e ss = 1 1 + K p = 0; e ss = 1 K v = 0.1; e ss = 1 K a = ¥ 6.2 Characteristic equation is s 2 + 10 s + 10 K A = s 2 + 2 zw n s + w n 2 = 0 (a) K A = 2.5 gives z = 1 which meets the response requirements. (b) K p = ¥ , K v = 2.5 e ss = 5 1 1 + + KK p v = 0.4 6.3 (a) Using Routh criterion, it can be checked that the close-loop system is stable. (i) e ss = 10 K v = 10 ¥ = 0 ; (ii) e ss = 10 0 2 a v + . = 10 0 01 ¥ + .2 . = 2 (b) The closed-loop system is unstable. 6.4 The closed-loop system is stable. This can easily be verified by Routh criterion. G ( s )= 5 1 1 0 1 ( . )( )( .2 ) ss s ++ +
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SOLUTION MANUAL 43 e ss | r = 10 = 10 1 + K p = 10 15 + Ys Ws () = 01 1 10 1 1 5 . ( ) (. 2 ) ) s sss + +++ + For W ( s )=4 / s , y ss = 4/6 e ss | total = 7/3 6.5 (a) The characteristic equation of the system is 900 s 3 + 420 s 2 + 43 s + 1 + 0.8 K A = 0 Using Routh criterion, we find that the system is stable for K A < 23.84. (b) Rearrangement of the block diagram of Fig. P6.5 is shown below. + q r q 50 1 0.016 30 1 s+ 10 1 31 K A G ( s )= 0 016 50 3 1 30 1 . ( ) · ++ K ss A H ( s 1 10 1 s + ; the dc gain of H ( s ) is unity. K p = lim s 0 G ( s ) = 0.8 K A ; e ss = 10 1 + K p = 1 This gives K A = 11.25 6.6 (a) The system is stable for K c < 9. K p = lim s 0 D ( s ) G ( s ) = K c ; e ss = 1 1 + K c e = 0.1 (10%) is the minimum possible value for steady-state error. Therefore e ss less than 2% is not possible with propor- tional compensator.
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44 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) Replace K c by D ( s ) = 3 + K s I . The closed-loop system is stable for 0 < K 1 < 3. Any value in this range satisfies the static accuracy requirements. 6.7 (a) K v = 1000 10 K c = 1000 z = 10 1000 2 100 + · K D = 0.5 These equations give K c = 10 and K D = 0.09. (b) The closed-loop poles have real part = – zw n = – 50. The zero is present at s = – K c / K D = – 111.11. The zero will result in pronounced early peak. z is not an accurate estimate of M p .
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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Chapter6 - CHAPTER 6 6.1 THE PERFORMANCE OF FEEDBACK...

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