EE3530 Spring 2008 Homework 8 Problems

EE3530 Spring 2008 Homework 8 Problems - 232 (a) (b) (c) =...

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Unformatted text preview: 232 (a) (b) (c) = 4; i= CHAPTER ! o = 1:77 ROOT-LOCUS DESIGN METHOD 45; 135; 5. THE 6=3 = 2; i = 60; 180 i EE3530, Spring 2008 30 (b) K== 4:67; (c) o Homework # 8 Problems Breakaway(s) = 60; = 4; i = 0:503 180; ! o = 5:98 > none R oot Locus 90; ! o 10 Due date: Monday, April 21st, 2008 3. (d) the characteristic! o > none For = 5; i = 90; equation 5 Root Locus Imag Axis 6 4 1+ 0 K =0: 6 s(s + 1)(s + 5) 4 Root Locus (a) Draw the real-axis segments of the corresponding root locus. 2 2 -5 Imag Axis 0 0 (b) Sketch the asymptotes of the locus for K ! 1. -10 (c) For what value of K are the roots on the imaginary axis? -15 -10 -5 0 5 R eal A xis -2 -4 Imag Axis -2 -4 (d) Verify your sketch with a MATLAB-6plot. -6 -10 -5 0 -10 Real Axis Solution for Problem 5.3 -5 Real Axis 0 Imag Axis (c) Ko = (a) L(s) = 30 -2 -4 0 Imag Axis 4. Real poles and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to Solution: give the asymptotes, arrival and departure angles at any complex zero or Root Locus Root Locus 6 6 pole, and the frequency of any imaginary-axis crossing. After completing (a) The real axis segments are 0 > > 4 1; > 4 each hand sketch verify your results using MATLAB.5Turn in your hand sketches2= 6=3 MATLAB results on the same scales. and the = 2; 2 (b) = 60; 180 i 1 s(s + 1)(s + 5)(s + 10) 0 -2 (s + 2) (b) L(s) = -6 -10 s(s + 1)(s + 5)(s + 10) -5 0 10 R oot Locus -4 -6 -10 -5 Real Axis 0 Real Axis (s + 2)(s + 6) Solution for Problem 5.4 s(s + 1)(s + 5)(s + 10) 5. Complex poles and zeros Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. (c) L(s) = 5 Imag Axis 0 -5 -10 -15 -10 -5 0 5 R eal A xis Solution for Problem 5.3 1 s2 + 3s + 10 4. Real poles and zeros. Sketch the root locus with respect to K for the 1 (b) L(s) =1 + 2 equation s(s KL(s) = 0 and the following choices for L(s). Be sure to + 3s + 10) (a) L(s) = give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. (a) L(s) = (b) L(s) = (c) L(s) = 1 s(s + 1)(s + 5)(s + 10) (s + 2) s(s + 1)(s + 5)(s + 10) (s + 2)(s + 6) s(s + 1)(s + 5)(s + 10) 1 234 EE3530, Spring 2008 Homework # 8 Problems CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Due date: Monday, April 21st, 2008 (c) L(s) = (s2 + 2s + 8) s(s2 + 2s + 10) (s2 + 2s + 12) (d) L(s) = s(s2 + 2s + 10) (e) L(s) = (s2 + 1) s(s2 + 4) (s2 + 4) 235 (f) L(s) = 2 + 1) s(s Solution: 6. Multiple poles at the origin Sketch the root locus with respect to K for the equation 1 +locus plots and the following choices for L(s). solutionto All the root KL(s) = 0 are displayed at the end of the Be sure set give the asymptotes, arrival and departure angles at any complex zero or for this problem. pole, and the frequency of any imaginary-axis crossing. After completing each hand i = 90; d = 90 ! o > MATLAB. Turn in your hand (a) = 3;sketch verify your results usingnone sketches and the MATLAB results on the same scales. (b) = 3; i i = 60; 180; 180; d d d d d = 28:3 ! o = 3:16 a a 1 (a) L(s) = 2 8) (d) = 2; s = +180; i (s (c) = 2; = = = 161:6; 18:4; 180; a a = 200:7; ! o 16:8; ! o > none > none = (e)(b) = 0; = = 1180; L(s) i s3 (s + 8) = = 180; ! o > none > none (f) 1 (c) L(s) = 4 s (s + 8) = 0; i = 180; = 0; = 0; ! o (d) L(s) = 10 Imaginary Axis 5 0 (e) L(s) = -5 -10 -2 (s + 3) s3 (s + 4) -1.5 (s + 1) -1 Real Axis s3 (s + Root Locus 4) Imaginary Axis Root (s + 3) Locus s2 (s + 8) plot a Root Locus 5 plot b 0 -5 2 (f) L(s) = 4 Imaginary Axis -0.5 0 -6 -4 -2 0 Real Axis Root Locus 2 4 6 (g) L(s) = 3 1 Imaginary Axis (s + 1)2 s3 (s + 10)2 2 Solution: plot c 4 3 2 1 plot d All 0 the root locus plots are displayed at the end of the solution set 0 -1.5 -0.5 0 -2 -1.5 -1 -0.5 for -2 this problem. -1 Real Axis Real Axis Root Locus 3 Imaginary Axis 0 Root Locus i 2 Imaginary Axis (a) (b) (c) (d) = = 1 2:67; 2; 1:6; -1 = 60; plot e 3 180; w0 > none 2 1 plot f i = i 45; 36; 0 Real Axis 135; w0 108; w0 1 > none 2 2 -1 0 Real Axis 1 2 = 0 -2 = = > none 0 -2 = 2:5; 90; w0 > none Figure 5.64: Control system for problem 12 -40 EE3530, Spring 2008 -10 Homework # 8 Problems -5 0 Real Axis -5 -4 -2 0 Real Axis 2 Due date: Monday, April 21st, 2008 Solution for Problem 5.7 S(0:1S+0:9) (c) L(s)= 0:1S^3+0:9S+1:5 = a(s) b(s) 8. Right half plane poles and zeros Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to Root Locus Root Locus give the asymptotes, arrival and departure angles at any complex zero or 20 plot b pole, and 20 plot a the frequency of any imaginary-axis crossing. After completing 10 each hand sketch verify your results using MATLAB. Turn in your hand 10 sketches and the MATLAB results on the same scales. Imag Axis 0 Imag Axis 0 (a) (b) (c) (d) 238 s+2 1 -10 L(s) = ; The model for a -10 case of magnetic levitation 2 s + 10 s 1 -20 with lead compensation. -20 -30 s + 2 -20 -10 1 0 10 -25 -20 -15 -10 -5 0 5 Real Axis Real Axis L(s) = ; The magnetic levitation system with in2 s(s + 10) (s 1) tegral control and lead compensation. s 1 Root Locus L(s) = plot2c s 1 s2 + 2s + 1 L(s)0.5= : What is the largest value that can s(s + 20)2 (s2 2s + 2) 0 be obtained for the damping ratio of the stable complex roots on this CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD -0.5 locus? Imag Axis (s + 2) ; -10 -5 0 s(s 1)(s + 6)2 Real Axis 1 Solution for problem 5.11 (f) L(s) = (s 1)[(s + 2)2 + 3] Solution: 12. For the system shown in Fig. 5.64, determine the characteristic equation and sketch the root locus of it with respect to positive values of the241 pa(a) rameter4; Give L(s), w0 > none and be sure to show with arrows the = c. i = 90; a(s); and b(s) (e) L(s) = -15 -1 the locus. (b) direction in iwhich60; 180; w0 on > none = 4; = c increases (c) (e) (f) = 1; i = 180; w0 > none 3:24; 15:37; 1:732; = d (d) (a) Solution: = = 12; i = 1; i s2 + 9 a(s) = =L(s) =i = 60; 180; w0 3; s3 + 144s b(s) 60; 180; w0 = = 92:4 > none d = 60; 180; w0 = 40:9 1 plot c 0.5 Imag Axis Root Locus 10 10 Figure 5.64: Control system for problem 12 plot a 5 Imag Axis plot b 5 S(0:1S+0:9) (c) L(s)= 0:1S^3+0:9S+1:5 = 0 a(s) b(s) Imag Axis 0 0 -5 Root Locus -5 20 -0.5 plot b -5 0 10 Real Axis 0 20 plot a -10 -10 plot d Imag Axis 10 0 -5 0 Real Axis -10 -10 Imag Axis -1 -1 0 1 Real Axis 3 2 10 -10 4 plot e plot f 15 ...
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