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 + GsHs () () = ss () + ++ 1 15 0 |( ) | Sj G M w w = 1 = 0.0289 S H M = - + 1 = - 50 0 ) | H M w w = 1 = 1.02 4.2 Ys Rs = M ( s )= P 11 D D P 1 = KK 31 2 1 ; + D = 1 5 1 1 1 2 12 2 -- + - + - + F H G I K J s ; D 1 = 1 M ( s 5 5 5 1 2 K K K + + S K M 1 = · M K K M 1 1 = K K s K K 2 2 2 55 5 5 +- + + ) | K M 1 0 w w = = 5 1 K + = 0.5 4.3 For G ( s ) = 20/( s + 1), and R ( s ) = 1/ s , yt | open-loop = 20 (1 – e t ) | closed-loop = 20 21 1 21 - - e t For G' ( s ) = 20/( s + 0.4), and R ( s ) = 1/ s , | open-loop = 50 (1 – e 0.4 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 11 0 + 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, Ke K i Af f - = L di dt RR i f sf ++ This gives Is Es f f = K sL R R K K A sA ++ + t f | closed-loop = L RR KK = 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( + ) Ks 1 Ws Ys s +1 s = 04 11 10 1 . sK
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26 CONTROL SYSTEMS: PRINCIPLES AND DESIGN For W ( s )= 1 / s , y ss = lim ( ) s sY s 0 = 04 10 1 . K + = 0.01 This gives K = 3.9 4.7 Case I: Ys Ws () = K t t + ++ 1 1 for W ( s )= 1/ s , y ss = 0 Case II: = K 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+ Ts I Plant 4.8 A unity-feedback configuration of the given system is shown below. + q r q Ds e 200 0.02 ¥ (+1 ) s (+2 ) s Gs = Es s r q = 1 1 + DsGs () () ; 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. 4.9 (i) s 2 + 1 = 0; oscillatory (ii) s 2 + 2s + 1 = 0; stable (iii) s 3 + s + 2 = 0; unstable
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SOLUTION MANUAL 27 The derivative term improves the relative stability; and the integral term has the opposite effect.
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Chapter4 - CHAPTER 4 BASIC PRINCIPLES OF FEEDBACK CONTROL...

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