Download presentation

Presentation is loading. Please wait.

Published byMaribel Sleeper Modified about 1 year ago

1
Design with Root Locus Lecture 9

2
Objectives for desired response 1. Improving transient response Percent overshoot, damping ratio, settling time, peak time 2. Improving steady-state error Steady state error

3
Gain adjustment Higher gain, smaller steady stead error, larger percent overshoot Reducing gain, smaller percent overshoot, higher steady state error

4
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 2 nd order using some techniques.

5
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.

6
Cascade Compensator Feedback Compensator The added compensator can change a pattern of root locus Compensator configulations

7
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.

8
Types of compensator 1.Active compensator –PI, PD, PID use of active components, i.e., OP-AMP –Require power source –ss error converge to zero –Expensive 2.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

9
Improving steady-state error (PI) 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

10
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. Improving Steady-State Error (PI)

11
Improving Steady-State Error (PI):Example Damping ratio = in both uncompensated and PI cases Choose zero at -1

12
>> z=[1]; >> n=conv([1 3 2],[1 10]); >> sys=tf(z,n) >> sgrid(0.174,[2,10]) >> [k p]=rlocfind(sys) Improving Steady-State Error (PI)

13

14
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. Improving Steady-State Error (PI)

15

16
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

17
Arctan formula

18
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

19
Draw root locus with compensator (system order is up by 1--from 3 rd to 4 th ) Needs complex poles corresponding to damping ratio of (K=158.2) From K, find the 3 rd and 4 th 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)

20
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

21
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. Type is not increased. What about steady- state error ( a) Type 1 uncompensated system (b) Type 1 compensated system

22

23
Example With damping ratio of 0.174, add lag Compensator to improve steady-state error by a factor of 10

24
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)= 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

25
3 rd order approx. for lag compensator (= uncompensated system) making Same transient response but 10 times Improvement in ss response!!!

26

27
If we choose a compensator pole at (10 times closer to the origin), we’ll get a compensator zero at (Kp=91.593) New compensator: 4 th pole is at (compared to ) producing a longer transient response.

28
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.

29
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

30
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).

31
PD controller: Improving transient response Transfer function of the PD controller Gc(s) = K 2 s + K1 = K 2 (s+K1/K2) = K(s+zc) introduces a zero at −z c 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.

32
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

33
(a) Uncompensated system, (b) compensator zero at -2 (d) compensator zero at -3, (d) compensator zero at -4 Indicate settling time Indicate peak time

34
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

35

36
Example design a PD controller to yield 16% overshoot with a threefold reduction in settling time

37
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 3 rd pole (at -7.59) which is far away from the dominant poles 2 nd order approx. works!!!

38
More details in step II and III Characteristic equation:

39
Step IV: evaluate a desired settling time: Step V: get corresponding real and imagine number of the dominant poles ( and )

40
Location of poles as desired is at ±j6.192

41
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

42
Compensator: (s+3.006) Might not have a pole-zero cancellation for compensated system

43

44
PD Compensator

45
Lead Compensation Zeta2-zeta1=angular contribution 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.

46
Lead Compensation Zeta2-zeta1=angular contribution 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, 2 − 1 − 3 − 4 + 5 = (2k+1)180 where 2 − 1 = c is the angular contribution of the lead compensator.

47
The angular contribution c 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.

48
Example Design three lead compensators for the system that has 30% OS and will reduce settling time down by a factor of 2.

49
Step I: %OS = 30% equaivalent to damping ratio = 0.358, Ѳ = 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:

50
Step VI: let’s put a zero at -5 and find the net angle to the test point ( ) Step VII: need a pole at the location giving 7.31 degree to the test point.

51

52
Note: check if the 2 nd order approx. is valid for justify our estimates of percent overshoot and settling time –Search for 3 rd and 4 th 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!!!

53

54

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google