MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.+-R ( s )C ( s )P c o n t r o l l e rp l a n tK1s + 3 s + 5 s + 232� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 261 Reading: • Nise: Chapter 6 • Nise: Chapter 8 1 Determining Stability Bounds in Closed-Loop Systems Consider the closed-loop third-order system with proportional controller gain K with open-loop transfer function K Gf (s) = s3 + 3s2 + 5s + 2 shown below. The closed loop transfer function is: N(s) K Gcl(s) = = D(s) + N(s) s3 + 3s2 + 5s + (2 + K) Let’s examine the closed-loop stability by using the pzmap() function in MATLAB: sys = tf(1,[1 3 5 2]); pzmap(sys); hold on; for K = 2:2:30 sys = tf(K,[1 3 5 2+K]); pzmap(sys); end; which superimposes the closed-loop pole/zero plots for K = 0 . . . 30 on a single plot: 1copyright cD.Rowell 2008 26–1Pole−Zero MapReal AxisImaginary Axis−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1−3−2−10123K=0K=30K=30K=30K=0.K=0K increasingK increasingK increasingFrom the plot we note the following: • This system always has two complex conjugate poles and a single real pole. • When K = 0 the poles are the open-loop poles. • As K increases, the real pole moves deeper into the l.h. plane, and the complex con-jugate poles approach and cross the imaginary (jω) axis, and the system becomes unstable. • Close examination of the plot shows that the system becomes unstable at a value of K between K = 12 and K = 14. We now look at three methods for determining the stability limit of the prop ortional gain K for this system. Example 1 Use the Routh-Hurwitz method to find the range of proportional controller gain K for which the above system will be stable. The first two rows of the Routh array are taken directly from D(s): s3 1 5 0 s2 3 2 + K 0 26–2and the next two rows are computed as above 1 ����an an−2 ����1 ����1 5 ����1 b1 = −an−1 an−1 an−3 = − = − (K − 13)3 3 2 + K 3 1 ���an an−4 ���1 ���1 0 ���b2 = = 0−an−1 � an−1 an−5 � = −3 � 3 0 � Similarly, the s0 row is computed 1 ����an−1 an−3 ����3 ����3 24 ����c1 = = 2 + K−b1 b1 b2 = −K − 13 −(K − 13)/3 2 + K 1 ����an−1 an−3 ����3 ����3 0 ����c2 = = 0−b1 b1 b3 = −K − 13 −(K − 13)/3 0 and the complete Routh array is s3 1 5 0 s2 3 2+K 0 s1 -(K-13)/3 0 s0 (2+K) 0 We now examine the first column to determine the range of proportional gain for which this system will be stable. In order for there to be no sign changes we require −2 < K < 13 We conclude that if K < −2 there will be one (therefore real) unstable pole, while if K > 13 there will be two unstable poles. When K = 13 the denominator is D(s) = s 3 + 3s 2 + 5s + 15 = (s + 3)(s + j2.236)(s − j2.236) so that the closed-loop system has a pair of poles on the imaginary axis. The system will be marginally stable (a pure oscillator at a frequency of ω = 2.236 r/s). Example 2 Use the stability criterion for third-order systems developed in Example 3 of Lecture 25 to determine the stability bounds for the above system. In Lecture 25 we showed that for a third-order system with characteristic equa- tion: D(s) = a3s 3 + a2s 2 + a1s + a0 = 0 the system is stable only if a1a2 > a0a3 26–3In this case D(s) = s 3 + 3s 2 + 5s + (2 + K) and therefore for stability we require 15 > 2 + K or K < 13. Example 3 Use the characteristic equation directly to find the closed-loop stability limits for the above system. There are three closed-poles. We conjecture that at the stabil-ity boundary (marginal stability) there will be a pair of p oles on the imaginary axis at s = ±jω, and a single real pole at s = −a. The closed-loop characteristic polynomial will therefore be D(s) = (s + a)(s 2 + ω2) = s 3 + as 2 + ω2 s + aω2 Comparing coefficients with the actual closed-loop characteristic polynomial D(s) = s 3 + 3s 2 + 5s + (2 + K) we determine a = 3 ω2 = 5 ω = √5 → aω2 = K + 2 K = 13 → 2 Root Locus Methods We have seen that the closed-loop poles change as controller parameters vary. A root-locus is is an s-plane plot of the paths that the closed-loop poles take as a controller parameter varies. Let’s start with some simple examples. Example 4 Consider the first order plant under prop ortional control, as shown below: 26–4xsjws - p l a n eK = 0- aK i n c r e a s i n gt h e r o o t l o c u si n d i c a t e s t h e p a t ho f t h e p o l e a s Kv a r i e s .+-R ( s ) C ( s )P - D c o n t r o l l e rp l a n tK + K s1s + apd+-R ( s ) C ( s )P c o n t r o l l e rp l a n tK1s + aThe closed-loop transfer function is K Gcl(s) = s + (a + K) with a single pole pc] = −(a + K). The root-locus is simply the path of this pole as K varies from K = 0 to K = Clearly as K 0, the closed-loop pole ∞. →approaches the open-loop pole (s = −a), and as K → ∞, the closed-loop pole p → −∞. This is all the information we need to construct the root-locus for this system. Example 5 Construct the root-locus plot for the first-order system under P-D control with Gc(s) = Kp + Kds: 26–5xsjws - p l a n eK = 0- aK i n c r e a s i n g- boxsjws - p l a n eK = 0- aK i n c r e a s i n g- bo( a ) b > a ( b ) a > bIf we write � Kp �Gc(s) = Kd s + Kd we have a open-loop pole at s = −a and an open-loop zero at s = −Kp/Kd = −b. The closed-loop transfer function is Kd(s + b)Gcl(s) = (Kd + 1)s + (a + Kdb) with a single pole a + bKd pcl .= − 1 + Kd We now construct the root-locus as …
View Full Document