1. Prototype 2nd order system:
target

2. Settling time:

3. 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. 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

10. 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

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

20. 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

21. 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.

22. Let n=#poles=order of system, m=#zeros
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:

23. 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.