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# Chapter4 - CHAPTER 4 BASIC PRINCIPLES OF FEEDBACK CONTROL...

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CHAPTER 4 BASIC PRINCIPLES OF FEEDBACK CONTROL 4.1 S G M = 1 1 + G s H s ( ) ( ) = s s s s ( ) ( ) + + + 1 1 50 | ( )| S j G M w w = 1 = 0.0289 S H M = - + G s H s G s H s ( ) ( ) ( ) ( ) 1 = - + + 50 1 50 s s ( ) | ( )| S j H M w w = 1 = 1.02 4.2 Y s R s ( ) ( ) = M ( s ) = P 1 1 D D P 1 = K K s s 3 1 2 1 ( ) ; + D = 1 5 1 1 1 2 1 2 3 1 2 - - + - + - + F H G I K J s s K K s K K s s ( ) ( ) ; D 1 = 1 M ( s ) = 5 1 5 5 5 1 2 1 1 K s s K K ( ) + + + + S K M 1 = · M K K M 1 1 = s s K K s s s K K 2 1 1 2 2 1 1 1 5 5 5 1 5 5 5 ( ) ( ) + + + - + + + + | ( )| S j K M 1 0 w w = = 5 5 5 1 K + = 0.5 4.3 For G ( s ) = 20/( s + 1), and R ( s ) = 1/ s , y t ( )| open-loop = 20 (1 – e t ) y t ( )| closed-loop = 20 21 1 21 ( ) - - e t For G' ( s ) = 20/( s + 0.4), and R ( s ) = 1/ s , y t ( )| open-loop = 50 (1 – e 0.4 t ) y t ( )| closed-loop = 20 20 4 1 20 4 . ( ) . - - e t The transient response of the closed-loop system is less sensitive to variations in plant parameters

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SOLUTION MANUAL 25 4.4 For G ( s ) = 10/ ( ), t s + 1 and R ( s ) = 1/ s , e ss | open-loop = 0 e ss | closed-loop = 1 1 10 + K p = 0.0099 For G' ( s ) = 11/ ( ) t s + 1 , and R ( s ) = 1/ s , e ss | open-loop = – 0.1 e ss | closed-loop = 0.009 The steady-state response of the closed-loop system is less sensitive to variations in plant parameters. 4.5 t f | open-loop = L R = 2 50 = 0.04 sec For the feedback system, K e Ki A f f ( ) - = L di dt R R i f s f + + ( ) This gives I s E s f f ( ) ( ) = K sL R R K K A s A + + + t f | closed-loop = L R R K K s A + + = 2 51 90 + K = 0.004 This gives K = 4.99 4.6 The given block diagram is equivalent to a single-loop block diagram given below, + 0.4 10( + ) K s 1 W s ( ) Y s ( ) s + 1 s + 1 Y s W s ( ) ( ) = 0 4 11 10 1 . s K + +
26 CONTROL SYSTEMS: PRINCIPLES AND DESIGN For W ( s ) = 1/ s , y ss = lim ( ) s sY s 0 = 0 4 10 1 . K + = 0.01 This gives K = 3.9 4.7 Case I: Y s W s ( ) ( ) = s s s s K ( ) ( ) t t + + + 1 1 for W ( s ) = 1/ s , y ss = 0 Case II: Y s W s ( ) ( ) = K s s K ( ) t + + 1 For W ( s ) = 1/ s , y ss = 1 The control scheme shown in the following figure will eliminate the error in Case II. + + K c 1 1 + T s I W s ( ) Y s ( ) Plant 4.8 A unity-feedback configuration of the given system is shown below. + q r q D s ( ) e 200 0.02 ¥ ( + 1) s ( + 2) s G s ( ) = E s s r ( ) ( ) q = 1 1 + D s G s ( ) ( ) ; e ss = lim ( ) s sE s 0 ; q r s ( ) = 1 s (i) e ss = 1/3 (ii) e ss = 0 (iii) e ss = 1/3 The integral term improves the steady-state performance and the derivative term has no effect on steady-state error.

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