Download presentation

Presentation is loading. Please wait.

Published byVictor Manly Modified about 1 year ago

1
1 Modern Control Theory Lecture 5 By Kirsten Mølgaard Nielsen Based on notes by Jesper Sandberg Thomsen

2
2 Outline Root Locus Design Method Root locus (review) The phase condition Test points, and sketching a root locus Dynamic Compensation Lead compensation Lag compensation Notch compensation Time Delay (In AnaDig 3) Padé approximation Direct application The Discrete Root Locus

3
3 Root Locus Closed loop transfer function Characteristic equation, roots are poles in T(s)

4
4 Root Locus Root locus Determination of the closed loop pole locations under varying K. For example, K could be the control gain. The characteristic equation can be written in various ways

5
5 Root Locus Definition 1 The root locus is the values of s for which 1+KL(s)=0 is satisfied as K varies from 0 to infinity (pos.). Definition 2 The root locus of L(s) is the points in the s-plane where the phase of L(s) is 180°. The angle to the test point from zero number i is i. The angle to the test point from pole number i is i. Therefore, In def. 2, notice,

6
6 Root Locus s0s0 p1p1 What is meant by a test point ? s 0 - p 1

7
7 Root Locus s0s0 p p*

8
8 Root Locus Rules for sketching a root locus Rule 1: The n branches of the locus start at the poles L(s) and m of these branches end on the zeros of L(s). Rule 2: The loci on the real axis are to the left of an odd number of (real axis) poles plus zeros. end start

9
9 Root Locus Rule 3: For large s, the loci are asymptotic to lines at angles l radiating out from the center point s = . 1 = 60 deg. 2 = 180 deg. 3 = 300 deg.

10
10 Root Locus Rule 4: The angles of departure (arrival) of a branch of the locus from a pole (zero) of multiplicity q. 1,dep = 60 deg. 2,dep = 180 deg. 3,dep = 300 deg. 3 poles: q = 3

11
11 Root Locus Rule 5: The locus crosses the j axis at points found by Routh’s criterion. Rule 6: Identification of multiple roots on the locus. crossings point with multiple poles

12
12 Dynamic Compensation Some facts We are able to determine the roots of the characteristic equation (closed loop poles) for a varying parameter K. The location of the roots determine the dynamic characteristics (performance) of the closed loop system. It might not be possible to achieve the desired performance with D(s) = K. Controller design using root locus Lead compensation (similar to PD control) overshoot, rise time requirements Lag compensation (similar to PI control) steady state requirements Notch compensation

13
13 Dynamic Compensation PD Lead PI Lag

14
14 Lead Compensation PD P Lower damping ratio with PD !

15
15 Lead Compensation PD control Pure differentiation Output noise has a great effect on u(t) Solution: Insert a pole at a higher frequency controller D(s) plant G(s) r(t)u(t)y(t)e(t) + - noise + +

16
16 Lead Compensation

17
17 Lead Compensation Selecting p and z Usually, trial and error The placement of the zero is determined by dynamic requirements (rise time, overshoot). The exact placement of the pole is determined by conflicting interests. suppression of output noise effectiveness of the zero In general, the zero z is placed around desired closed loop n the pole p is placed between 5 and 20 times of z

18
18 Lead Compensation Design approach (A) Initial design z= n, p=5z Noise: additional requirement, max(p)=20z Iterations First design: z= n, p=5z (neglecting the add. req.) Second design: z= n, p=20z Third design: new z, p=20z

19
19 Lead Compensation Possible pole location Matlab – design 1 sysG = tf([1],[1 1 0]) sysD = tf([1 7],[1 5*7]) rlocus(sysD*sysG) First design: z=wn =7 p=5z=35

20
20 Lead Compensation No possible pole location ! Matlab – design 2 sysG = tf([1],[1 1 0]) sysD = tf([1 7],[1 20*7]) rlocus(sysD*sysG) Second design: z=wn, p=20z

21
21 Lead Compensation Possible location Matlab – design 3 sysG = tf([1],[1 1 0]) sysD = tf([1 4],[1 20*4]) rlocus(sysD*sysG) New zero location

22
22 Lead Compensation Design approach (B) Pole placement from the requirements Noise: Additional requirement, max(p)=20. So, p=20 to minimize the effect of the pole.

23
23 Lead Compensation

24
24 Lead Compensation Pole location Matlab – design B sysG = tf([1],[1 1 0]) sysD = tf([1 5.4],[1 20]) rlocus(sysD*sysG)

25
25 Lead Compensation Exercise Design a lead compensation D(s) to the plant G(s) so that the dominant poles are located at s = -2 ± 2j controller D(s) plant G(s) r(t)u(t)y(t)e(t) + - You can use, triangle a/sin(A) = b/sin(B)

26
26 Lead Compensation Answer we have three poles 0,0,20 and one zero 11 22 Notice, triangle a/sin(A) = b/sin(B)

27
27 Dynamic Compensation PD Lead PI Lag

28
28 Lag Compensation Lag compensation Steady state characteristics (requirements) Position error constant K p Velocity error constant K v Let us continue using the example system and the designed controller (lead compensation)

29
29 Lag Compensation Calculation of error constants Suppose we want K v ≥ 100 We need additional gain at low frequencies !

30
30 Lag Compensation Selecting p and z To minimize the effect on the dominant dynamics z and p must chosen as low as possible (i.e. at low frequencies) To minimize the settling time z and p must chosen as high as possible (i.e. at high frequencies) Thus, the lag pole-zero location must be chosen at as high a frequency as possible without causing any major shifts in the dominant pole locations

31
31 Lag Compensation Design approach Increase the error constant by increasing the low frequency gain Choose z and p somewhat below n Example system: K = 1 (the proportional part has already been chosen) z/p = 3, with z = 0.03, and p = 0.01.

32
32 Lag Compensation Notice, a slightly different pole location Lead-lag Controller

33
33 Lag Compensation z p

34
34 Lag Compensation Step

35
35 Lag Compensation Ramp

36
36 Notch compensation Example system We have successfully designed a lead-lag controller Suppose the real system has a rather undampen oscillation about 50 rad/sec. Include this oscillation in the model Can we use the original controller ?

37
37 Notch compensation Step response using the lead- lag controller

38
38 Notch compensation Aim: Remove or dampen the oscillations Possibilities Gain stabilization Reduce the gain at high frequencies Thus, insert poles above the bandwidth but below the oscillation frequency – might not be feasible Phase stabilization (notch compensation) A zero near the oscillation frequency A zero increases the phase, and ≈ PM/100 Possible transfer function If n < 1, complex zeros and double pole at 0

39
39 Notch compensation

40
40 Notch compensation Cancellation Root Locus The lead-lag controller with notch compensation 0 = 50 = 0.01

41
41 Notch compensation Too much overshoot !! Also, exact cancellation might cause problems due to modelling errors Thus, we need different parameters

42
42 Notch compensation Modified parameters Notch compensation Dynamic characteristics Increase n to obtain less overshoot Make sure that the roots move into the LHP In general, obtain a satisfactory dynamic behavior After some trial and error

43
43 Notch compensation The overshoot has been lowered to an acceptable level ! The oscillations have almost been removed ! So, we have reached a final design !

44
44 Notch compensation Root Locus The lead-lag controller with notch compensation 0 = 60 = 0.3

45
45 Time Delay Notice Time delay always reduces the stability of a system ! Important to be able to analyze its effect In the s-domain a time delay is given by e - s Most applications contain delays (sampled systems) Root locus analysis The original method does only handle polynomials Solutions Approximation (Padé) of e - s Modifying the root locus method (direct application)

46
46 Time Delay

47
47 Time Delay Root locus for some example system (heat exchanger) No delay Padé (1,1) Padé (2,2) Exact (p,p) Approx. in Matlab : [num,den]=pade(T,p)

48
48 Time Delay However, Matlab does not support this approach...

49
49 The Discrete Root Locus Discrete (z-domain) Closed loop transfer function Characteristic equation Thus, same sketching techniques as in the s-domain, the unit circle is the stable region. However, different interpretation !!!

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google