线性控制课件module1008415english_new.pptVIP

- Next step: Determine the angle of emergence from the complex poles. Set a trial point st on a small circle around the s=s-2+2j. For argument equation: = Therefore, the locus leaves the complex poles at angle of ±12o. Sample Problem 10.1 Sample Problem 10.2 - Determine the locations of poles on locus for K=1 Rule 9: On the root from origin to zero, the imaginary axis crossing point K=10. Locate the two roots by trials. Rule 10: Determine the other two root. * Linear Control Systems Engineering 线性控制系统工程 Module 10 Rules for Plotting the Root Locus （绘制根轨迹的规则） 根轨迹示例 j 0 j 0 j 0 j 0 j 0 j 0 0 j 0 j 0 j j 0 0 j 同学们，头昏了吧？ 根轨迹示例2 j 0 j 0 j 0 0 j j 0 j 0 j 0 j 0 0 j j 0 0 j j 0 n=1;d=conv([1 2 0],[1 2 2]);rlocus(n,d) n=[1 2];d=conv([1 2 5],[[1 6 10]);rlocus(n,d) Example: open loop transfer function Draw the root locus and (1) calculate K for closed pole at s=-1(2)determine the locations of all other roots at this value of gain. Rule #1: Start and finish points Plot open loop n poles and m zeros. The locus starts at a pole for K=0 and finishes at a zero or infinity when K= ∞; The number of segments going to infinity is n-m. - The example: zero s=-5, m=1 poles s=0, s=-2±2j, n=3 segment: 2 segment go to infinity and 1 segment go to the zero Rule #2: Root locus on the real axis Segments of real axis to the left of an odd sum of real poles and real zeros are segments of the root locus. (The complex poles or zeros have no effect). Note: This rule is defined by argument equation. - The example: There is a locus segment between s=-0 and s=-5. Rule #3: The angle between adjacent asymptotes Rule 3: (1)The loci are symmetrical(对称) about real axis since roots are always in conjugate pair. (2)The angle between adjacent asymptotes is 360o/(n