# Ch05Solns - 5000 Solutions Manual Feedback Control of...

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5000 Solutions Manual . Feedback Control of Dynamic Systems . . Gene F. Franklin . J. David Powell . Abbas Emami-Naeini . . . . Assisted by: H.K. Aghajan H. Al-Rahmani P. Coulot P. Dankoski S. Everett R. Fuller T. Iwata V. Jones F. Safai L. Kobayashi H-T. Lee E. Thuriyasena M. Matsuoka

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Chapter 5 The Root-Locus Design Method Problems and solutions for Section 5.1 1. Set up the following characteristic equations in the form suited to Evans’s root-locus method. Give L ( s ) ,a ( s ) , and b ( s ) and the parameter, K, in terms of the original parameters in each case. Be sure to select K so that a ( s ) and b ( s ) are monic in each case and the degree of b ( s ) is not greater than that of a ( s ) . (a) s +(1 )=0 versus parameter τ (b) s 2 + cs + c +1=0 versus parameter c (c) ( s + c ) 3 + A ( Ts +1)=0 i. versus parameter A , ii. versus parameter T , iii. versus the parameter c , if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all (d) 1+[ k p + k I s + k D s τs +1 ] G ( s . Assume that G ( s )= A c ( s ) d ( s ) where c ( s ) and d ( s ) are monic polynomials with the degree of d ( s ) greater than that of c ( s ) . i. versus k p ii. versus k I iii. versus k D iv. versus τ 5001
5002 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Solution: (a) K =1 ; a = s ; b (b) K = c ; a = s 2 +1; b = s +1 (c) Part (c) i. K = AT ; a =( s + c ) 3 ; b = s /T ii. K = AT ; a s + c ) 3 + A ; b = s iii. The parameter c enters the equation in a nonlinear way and a standard root locus does not apply. However, using a polynomial solver, the roots can be plotted versus c. (d) Part (d) i. K = k p ; a = s ( s ) d ( s )+ k I ( s ) c ( s k D τ s 2 Ac ( s ); b = s ( s ) c ( s ) ii. K = Ak I ; a = s ( s ) d ( s Ak p s ( s k D τ s 2 Ac ( s ); b = s ( s ) c ( s ) iii. K = Ak D τ ; a = s ( s ) d ( s Ak p s ( s ) c ( s Ak I ( s + 1 ) c ( s ); b = s 2 c ( s ) iv. K ; a = s 2 d ( s k p As 2 c ( s k I Asc ( s ); b = sd ( s k p sAc ( s k I Ac ( s k D s 2 Ac ( s )

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5003 Problems and solutions for Section 5.2 2. Roughly sketch the root loci for the pole-zero maps as shown in Fig. 5.48. Show your estimates of the center and angles of the asymptotes, a rough evaluation of arrival and departure angles for complex poles and zeros, and the loci for positive values of the parameter K . Each pole-zero map is from a characteristic equation of the form 1+ K b ( s ) a ( s ) =0 , where the roots of the numerator b ( s ) are shown as small circles o and the roots of the denominator a ( s ) are shown as × 0 s on the s -plane. Note that in Fig. 5.48(c), there are two poles at the origin. Solution: Root Locus Real Axis Imag Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Root Locus Real Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Root Locus Real Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Root Locus Real Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Root Locus Real Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Root Locus Real Axis -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 Figure 5.48: Pole-zero maps (a) a ( s )= s 2 + s ; b ( s s +1 Breakin(s) -3.43; Breakaway(s) -0.586 (b) a ( s s 2 +0 . 2 s +1; b ( s s
5004 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Angle of departure: 135.7 Breakin(s) -4.97 (c) a ( s )= s 2 ; b ( s )=( s +1) Breakin(s) -2 (d) a ( s s 2 +5 s +6; b ( s s 2 + s Breakin(s) -2.37 Breakaway(s) -0.634 (e) a ( s s 3 +3 s 2 +4 s 8 Center of asymptotes -1 Angles of asymptotes ± 60 , 180 Angle of departure: -56.3 (f) a ( s s 3 s 2 + s 5; b ( s s +1 Center of asymptotes -.667 Angles of asymptotes ± 60 , 180

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Ch05Solns - 5000 Solutions Manual Feedback Control of...

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