Root Locus Techniques


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1 4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since T s = 8 seconds, we search along  2, the real part of poles with this settling time, for 80o. We find the point to be j6.9 with 76.39K = , or K = Secondorder approximation is OK since third pole is much more than 5 times further from the imaginary axis than the dominant secondorder pair. b. Program: numg= ; deng=poly([ ]); 'G(s)' G=tf(numg,deng) rlocus(g) axis([2,0,0,0]); title(['root Locus']) grid on [K,p]=rlocfind(G) K=K/76.39 Computer response: ans = G(s) Transfer function: s^ s^ s Select a point in the graphics window selected_point = i
2 Solutions to Problems Closedloop poles will be in the righthalfplane for Therefore, stable for K < 4/9; unstable for K > 4/9. (2)(2) 4 K > = (gain at the origin). (3)(3) Imag Axis Real Axis Breakaway: σ = for K =.78. System is never unstable. System is marginally stable for K =.
3 274 Chapter 8: Root Locus Techniques. System : (a) a. Breakaway: σ =.4 for K = 0.03; Breakin: σ = .4 for K = b. Imaginary axis crossing at j.4 for K =. Thus stable for K >. c. At breakin point, poles are multiple. Thus, K = d. Searching along 35 0 line for 80 0, K = 5 at System 2: (b) a. Breakin: σ = .4 for K = 28.4.
4 Solutions to Problems 275 b. Imaginary axis crossing at j.4 for K = Thus stable for K > c. At breakin point, poles are multiple. Thus, K = d. Searching along 35 0 line for 80 0, K = 4 at a. Root locus crosses the imaginary axis at the origin for K = 6. Thus the system is stable for K > 6. b. Root locus crosses the imaginary axis at j0.65 for K = Thus, the system is stable for K < Imag Axis Real Axis
5 278 Chapter 8: Root Locus Techniques 7. a. b. σ a = ( )  () 3 = ; Angle = (2k+)π 3 = π 3, π, 5π 3 c. Root locus crosses imaginary axis at j4.28 with K = d. K = Assume that root locus is epsilon away from the asymptotes. Thus, σ a = Angle = (2k+)π ( )  (α) 2 ; = π 2, 3π 2. Hence α = 7. Checking assumption at ± j00 yields 80o with K = a. Breakaway: for K = Breakin:.37 for K = 3.93 b. Imaginary axis crossing: ±j0.7 for K = 0.33 c. System stable for K < 0.33
6 Solutions to Problems 279 d. Searching 20 o find point on root locus at ο = ± j0.433 for K = a Imag Axis Real Axis b. 0 < K < c. K = ± j3.04 d. At the breakin point, s = , K = a. Asymptotes: σ int = ( )  (0) 4 = ; Angle = (2k+)π 4 = π 4, 3π 4, 5π 4, 7π 4 b. Breakaway: .38 for K = and for K =
7 Solutions to Problems 283 ( ) (2) σ a = = (2k +)π θ a = = π 4 2 2, 3π 2 c. At the jω axis crossing, K = 5.6. Thus for stability, 0 < K < 5.6. d. Breakaway points at σ = K = and σ = K = e. For 25% overshoot, Eq. (4.39) yields ζ = Searching along this damping ratio line, we find the 80 0 point at j.496 where K = f ± j7.425 g. Secondorder approximation not valid because of the existence of closedloop zeros in the rhp. h. Program: numg=35.98*[ 2 2]; deng=poly([ ]); G=tf(numg,deng); T=feedback(G,) step(t) Computer response: Transfer function: s^ s s^4 + 7 s^ s^ s + 32 Simulation shows over 30% overshoot and nonminimumphase behavior. Secondorder approximation not valid. 26. a. Draw root locus and minimum damping ratio line.
8 284 Chapter 8: Root Locus Techniques Minimum damping ratio o Minimum damping ratio is ζ = cos ( ) = cos o = Coordinates at tangent point of ζ = line with the root locus is approximately + j The gain at this point is b. Percent overshoot for ζ = is.09%. c. T s = 4 = 4 seconds; T π p = = 4.57 seconds d. Secondorder approximation is not valid because of the two zeros and no polezero cancellation. 27. The root locus intersects the 0.55 damping ratio line at j0.959 with K = A justification of a secondorder approximation is not required. The problem stated the requirements in terms of damping ratio and not percent overshoot, settling time, or peak time. A secondorder approximation is required to draw the equivalency between percent overshoot, settling time, and peak time and damping ratio and natural frequency. 28. Since the problem stated the settling time at large values of K, assume that the root locus is approximately close to the vertical asymptotes. Hence, σ int =  + α =  4. Since T 2 T s is given s as 4 seconds, σ int =  and α = 9. The root locus is shown below.
9 Solutions to Design Problems 303 c. Using G e (s) in part b, we find T s G e = = + G e M c s 2 + s+ s 2 + 2s+ 2 M c s 2 + s 3 + 2s 2 + 2s+ M c s 2 + s+. Dividing numerator and denominator by s 3 + 2s 2 + 2s+, T s = s 3 + 2s 2 + 2s+ s 2 + 2s+ 2 M c s 2. Thus, the system s 3 + 2s s+ has the same root locus as a system with G(s)H(s) = s2 + 2s+ 2 M c s 2 s 3 2s 2 = s2 + 2s+ 2 M c s s+ s+ s 2. + s+ Sketching the root locus, jω o x x oo σ o x SOLUTIONS TO DESIGN PROBLEMS 46. a. For a settling time of 0. seconds, the real part of the dominant pole is = Searching along the σ =  40 line for 80 o, we find the point 40 + j57.25 with 20,000K = x 0 9, or K = 02,300. b. Since, for the dominant pole, tan  ( %OS = e ζπ / ζ 2 x00 =.4%. ) = o, ζ = cos ( o ) = Thus, c. Searching the imaginary axis for 80 o, we find ω = rad/s for 20,000K =.43 x 0 0. Hence, K = 75,000. Therefore, for stability, K < 75, G(s) = 6.73K (s+0) 3 (s 2 +.s )
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