ch5soln - Chapter 5 The Root-Locus Design Method Problems...

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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 )aremon icineachcaseandthedegreeo f b ( s ) is not greater than that of a ( s ) . (a) s +(1 / τ ) = 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 τ s +1 ] G ( s )=0 . Assume that G ( s )= A c ( s ) d ( s ) where c ( s d ( s )are mon icpo lynom ia lsw iththedegreeo f d ( s )greater than that of c ( s ). i. versus k p ii. versus k I iii. versus k D iv. versus τ 229

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230 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Solution: (a) K =1 / τ ; 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 τ ; a = s ( s / τ ) d ( s )+ k I ( s / τ ) c ( s k D τ s 2 Ac ( s ); b = s ( s / τ ) c ( s ) ii. K = Ak I ; a = s ( s / τ ) d ( s Ak p s ( s / τ k D τ s 2 Ac ( s ); b = s ( s / τ ) c ( s ) iii. K = Ak D τ ; a = s ( s / τ ) d ( s Ak p s ( s / τ ) c ( s Ak I ( s + 1 / τ ) c ( s ); b = s 2 c ( s ) iv. K / τ ; 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 ) Problems and solutions for Section 5.2 2. Roughly sketch the root loci for the pole-zero maps as shown in Fig. 5.62. 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 , wheretherootsofthenumerator b ( s ) are shown as small circles o and the roots of the denominator a ( s )areshownas & 0 s on the s -plane. Note that in Fig. 5.62(c), there are two poles at the origin. Solution: (a) a ( s )= s 2 + s ; b ( s s Breakin(s) -3.43; Breakaway(s) -0.586 (b) a ( s s 2 +0 . 2 s b ( s s
231 Root Locus Real Axis I m a g A x i s -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 5.62: Pole-zero maps 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 Angle of departure: -90 Breakin(s) -2.06 Breakaway(s) 0 . 503

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232 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 3. For the characteristic equation 1+ K s ( s +1)( s +5) =0: (a) Draw the real-axis segments of the corresponding root locus.
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ch5soln - Chapter 5 The Root-Locus Design Method Problems...

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