# Design with Root Locus Lecture 9.

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Design with Root Locus Lecture 9

Objectives for desired response
Improving transient response Percent overshoot, damping ratio, settling time, peak time Improving steady-state error Steady state error

Compensator Allows us to meet transient and steady state error.
Composed of poles and zeros. Increased an order of the system. The system can be approx. to 2nd order using some techniques.

Improving transient response
Point A and B have the same damping ratio. Starting from point A, cannot reach a faster response at point B by adjusting K. We have pole at A on the root locus, but we want response like at B. Compensator is preferred.

Compensator configulations
Cascade Compensator Feedback Compensator The added compensator can change a pattern of root locus

compensator Method of implementing compensator:
1. Proportional control systems: feed the error forward to the plant. 2. Integral control systems: feed the integral of the error to the plant. 3. Derivative control systems: feed the derivative of the error to the plant. Es gibt eine Reihe von verschiedenen Kompensationsanlagen, die eingesetzt werden können, um zu beheben bestimmte System-Metriken, die außerhalb eines richtigen Betriebsbereich sind werden. Am häufigsten sind die Phaseneigenschaften in der Notwendigkeit der Kompensation, insbesondere wenn der Amplitudengang ist konstant bleiben.

Types of compensator Active compensator Passive compensator
PI, PD, PID use of active components, i.e., OP-AMP Require power source ss error converge to zero Expensive Passive compensator Lag, Lead use of passive components, i.e., R L C No need of power source ss error nearly reaches zero Less expensive

Placing a pole at the origin to increase system order; decreasing ss error as a result!! (a) Original system without compensation (b) Add a pole at the origin but angular contribution at point A is no longer 180

Also add a zero close to the pole at the origin. As angular contribution of the compensator zero and pole cancels out, point A is still on the root locus, and the system type has been increased.

Choose zero at -1 Damping ratio = in both uncompensated and PI cases

>> z=[1]; >> n=conv([1 3 2],[1 10]); >> sys=tf(z,n) >> sgrid(0.174,[2,10]) >> [k p]=rlocfind(sys)

As shown in the figure, the step response of the PI compensated system approaches unity in the steady-state, while the uncompensated system response approaches 1−0.108 = The simulation shows that it takes 18 seconds for the compensated system to reach and stay within ±2% of the final value of unity, while the uncompensated system takes about 6 seconds to settle to within ±2% of its final value of This is because there is no pole-zero cancelation and the pole not canceled is very close to the origin.

Finding an intersection between damping ratio line and root locus
Damping ratio line has an equation: where a = real part, b = imaginary part of the intersection point, Summation of angle from open-loop poles and zeros to the point is 180 degrees

Arctan formula

Magnitude of open loop system is 1
Use the formula to get the real and imaginary part of the intersection point and get Magnitude of open loop system is 1 No open loop zero

Draw root locus with compensator (system order is up by 1--from 3rd to 4th)
Needs complex poles corresponding to damping ratio of (K=158.2) From K, find the 3rd and 4th poles (at and ) Pole at can do phase cacellation with zero at -1 (3th order approx.) Compensated system and uncompensated system have similar transient response (closed loop poles and K are aprrox. The same)

PI Controller A compensator with a pole at the origin and a zero close to the pole is called an ideal integral compensator, or PI controller

Lag Compensator Ideal integral compensation: pole is in the origin, requires active network (costly). Real (passive) integral compensation: pole is close to origin (not in the origin), cheaper. (a) Type 1 uncompensated system (b) Type 1 compensated system Type is not increased. What about steady-state error

Example With damping ratio of 0.174, add lag
Compensator to improve steady-state error by a factor of 10

Step I: find an intersection of root locus and
damping ratio line ( j3.926 with K=164.56) Step II: find Kp = lim G(s) as s0 (Kp=8.228) Step III: steady-state error = 1/(1+Kp)= 0.108 Step IV: want to decrease error down to [Kp = (1 – )/ = ] Step V: require a ratio of compensator zero to pole as /8.228 = Step VI: choose a pole at 0.01, the corresponding Zero will be at *0.01 = 0.111

3rd order approx. for lag compensator
(= uncompensated system)  making Same transient response but 10 times Improvement in ss response!!!

If we choose a compensator pole at 0.001 (10 times
closer to the origin), we’ll get a compensator zero at (Kp=91.593) New compensator: 4th pole is at -0.01 (compared to ) producing a longer transient response.

SS response improvement conclusions
Can be done either by PI controller (pole at origin) or lag compensator (pole closed to origin). Improving ss error without affecting the transient response. Next step is to improve the transient response itself.

Improving Transient Response
Objective is to Decrease settling time Get a response with a desired %OS (damping ratio) Techniques can be used: PD controller (ideal derivative compensation) Lead compensator

PD controller: Improving transient response
System above controlled by a pure gain (P controller) in the forward path has its root locus going through point A for some value of gain K. • Our goal is to speed up the response at A to that at B, while keeping the percent overshoot unchanged. • The above root locus with a P controller cannot go through point B (sum of angles from the open-loop finite poles and zeros to point B is not an odd multiple of 180◦). A solution is to add a (nonzero) zero to the forward path (e.g., PD controller).

PD controller: Improving transient response
• Transfer function of the PD controller Gc(s) = K2 s + K1 = K2(s+K1/K2) = K(s+zc) introduces a zero at −zc Into the forward path. • Effect of the added zero: The added zero will contribute to make the sum of angles from the open-loop finite poles and zeros to the desired point (point B) be an odd multiple of 180◦. Note: an added zero has the effect of pushing the root locus to the left while an added pole has the effect of pushing it to the right. • The new root locus can meet the specific transient response (with shorter settling time) by going through point B for some value of gain K.

Ideal Derivative Compensator
So called PD controller Compensator adds a zero to the system at –Zc to keep a damping ratio constant with a faster response

(a) Uncompensated system, (b) compensator zero at -2 (d) compensator zero at -3, (d) compensator zero at -4 Indicate peak time Indicate settling time

Settling time & peak time: (b)<(c)<(d)<(a)
%OS: (b)=(c)=(d)=(a) ss error: compensated systems has lower value than uncompensated one cause improvement in transient response always yields an improvement in ss error

Example design a PD controller to yield 16% overshoot with a threefold reduction in settling time

Step I: calculate a corresponding damping ration (16% overshoot = 0
Step I: calculate a corresponding damping ration (16% overshoot = damping ratio) Step II: search along the damping ratio line for an odd multiple of 180 (at ±j2.064) and corresponding K (43.35) Step III: find the 3rd pole (at -7.59) which is far away from the dominant poles  2nd order approx. works!!!

More details in step II and III
Characteristic equation:

Step IV: evaluate a desired settling time:
Step V: get corresponding real and imagine number of the dominant poles ( and )

Location of poles as desired is at -3.613±j6.192

Step VI: summation of angles at the desired pole location, -275
Step VI: summation of angles at the desired pole location, , is not an odd multiple of 180 (not on the root locus) need to add a zero to make the sum of 180. Step VII: the angular contribution for the point to be on root locus is =95.6  put a zero to create the desired angle

Compensator: (s+3.006) Might not have a pole-zero cancellation for compensated system

PD Compensator

• A PD controller can be approximated with a lead compensator, which is implemented with a passive network. • If the lead compensator pole is farther from the imaginary axis than the compensator zero, the angular contribution of the compensator is still positive and thus approximates an equivalent single zero. • The advantages of a passive lead compensator over an active PD controller are that (1) no additional power supplies are required and (2) noise due to differentiation is reduced.

• The concept behind lead compensation: the difference between 180◦ and the sum of the angles from the uncompensated system’s poles and zeros to the design point (desired pole location) must be the angular contribution required of the compensator. That is, q2 −q1 −q3 −q4 +q5 = (2k+1)180◦ where q2 −q1 = qc is the angular contribution of the lead compensator.

• The angular contribution qc can be determined from the rays originating from the desired closed-loop pole and terminating at the compensator pole and zero. These rays can be rotated about the desired closed-loop pole and thus different pairs of compensator pole and zero can be used to meet the transient response requirement. • Different possible lead compensators: differences are in the values of the static error constants, the static gain, the difficulty in justifying a second-order approximation when the design is complete, and the ensuing transient response. • For design we arbitrarily select either a lead compensator pole and zero and find the angular contribution at the design point of this pole or zero with the uncompensated system’s open-loop poles and zeros. The difference between this angle and 180◦ is the required contribution of the remaining compensator pole or zero.

Example Design three lead compensators for the system
that has 30% OS and will reduce settling time down by a factor of 2.

Step I: %OS = 30% equaivalent to damping ratio = 0.358, Ѳ= 69.02
Step II: Search along the line to find a point that gives 180 degree (-1.007±j2.627) Step III: Find a corresponding K ( ) Step IV: calculate settling time of uncompensated system Step V: twofold reduction in settling time (Ts=3.972/2 = 1.986), correspoding real and imaginary parts are:

Step VI: let’s put a zero at -5 and find the net angle to the test point (-172.69)
Step VII: need a pole at the location giving 7.31 degree to the test point.

Note: check if the 2nd order approx
Note: check if the 2nd order approx. is valid for justify our estimates of percent overshoot and settling time Search for 3rd and 4th closed-loop poles (-43.8, ) -43.8 is more than 20 times the real part of the dominant pole is close to the zero at -5 The approx. is then valid!!!