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Name:_________________. In order to have transient die out, all closed-loop poles of the system must __ in the LHP, or stable __ Use z, s, w n , and w d to fill in the spaces below. Settling time is inversely proportional to ____ s _____

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  1. Name:_________________ In order to have transient die out, all closed-loop poles of the system must __in the LHP, or stable__ Use z, s, wn , and wd to fill in the spaces below. Settling time is inversely proportional to ____s_____ Rise time is inversely proportional to ____wn_____ Percentage overshoot is most directly determined by ___z____ Oscillation frequency determined by ____wd_____

  2. Prototype 2nd order system: target

  3. Settling time:

  4. Effects of additional zeros Suppose we originally have: i.e. step response Now introduce a zero at s = -z The new step response:

  5. Effects: • Increased speed, • Larger overshoot, • Might increase ts

  6. When z < 0, the zero s = -z is > 0, is in the right half plane. Such a zero is called a nonminimum phase zero. A system with nonminimum phase zeros is called a nonminimum phase system. Nonminimum phase zero should be avoided in design. i.e. Do not introduce such a zero in your controller.

  7. Effects of additional pole Suppose, instead of a zero, we introduce a pole at s = -p, i.e.

  8. L.P.F. has smoothing effect, or averaging effect Effects: • Slower, • Reduced overshoot, • May increase or decrease ts

  9. Matlab program template % enter plant transfer function Gp(s) nump = …. ; denp=…. ; % enter desired closed loop step response specification: % you may allow both uppper and lower limits … … % convert from specs to zeta, omegan, sigma, omegad … … % plot allowable region for pole location … … % obtain controller transfer function % for now make C(s) = 1 numc=1; denc=1; % obtain closed loop transfer function from Gp(s) and C(s) … … numcl=…; dencl=…; … … % obtain closed-loop step response … … % compute actual step response specs, using your program from last week … … % are they good? % compute the actual closed-loop poles, place “x” at those locations … … % are they in the allowable region?

  10. Root locus • A technique enabling you to see how close-loop poles would vary if a parameter in the system is varied • Can be used to design controllers or tuning controller parameters so as to move the dominant poles into the desired region

  11. Recall: step response specs are directly related to pole locations • Let p=-s+jwd • ts proportional to 1/s • Mp determined by exp(-ps/wd) • tr proportional to 1/|p| • It would be really nice if we can • Predict how the poles move when we tweak a system parameter • Systematically drive the poles to the desired region corresponding to desired step response specs

  12. Root Locus k s(s+a) y e r Example: + - Two parameters: k and a. would like to know how they affect poles

  13. The root locus technique • Obtain closed-loop TF and char eq d(s) = 0 • Rearrange terms in d(s) by collecting those proportional to parameter of interest, and those not; then divide eq by terms not proportional to para. to get this is called the root locus equation • Roots of n1(s) are called open-loop zeros, mark them with “o” in s-plane; roots of d1(s) are called open-loop poles, mark them with “x” in s-plane

  14. The “o” and “x” marks falling on the real axis divide the real axis into several segments. If a segment has an odd total number of “o” and/or “x” marks to its right, then n1(s)/d1(s) evaluated on this segment will be negative real, and there is possible k to make the root locus equation hold. So this segment is part of the root locus. High light it. If a segment has an even total number of marks, then it’s not part of root locus. For the high lighted segments, mark out going arrows near a pole, and incoming arrow near a zero.

  15. Let n=#poles=order of system, m=#zeros. One root locus branch comes out of each pole, so there are a total of n branches. M branches goes to the m finite zeros, leaving n-m branches going to infinity along some asymptotes. The asymptotes have angles (–p +2lp)/(n-m). The asymptotes intersect on the real axis at:

  16. Imaginary axis crossing • Go back to original char eq d(s)=0 • Use Routh criteria special case 1 • Find k value to make a whole row = 0 • The roots of the auxiliary equation are on jw axis, give oscillation frequency, are the jw axis crossing points of the root locus • When two branches meet and split, you have breakaway points. They are double roots. d(s)=0 and d’(s) =0 also. Use this to solve for s and k. • Use matlab command to get additional details of root locus • Let num = n1(s)’s coeff vector • Let den = d1(s)’s coeef vector • rlocus(num,den) draws locus for the root locus equation • Should be able to do first 7 steps by hand.

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