Chapter #4 EEE Automatic Control

Transcription

1 Spring 008 EEE 00 Chapter #4 EEE 00 Automatic Control Root Locu Chapter 4 /4

2 Spring 008 EEE 00 Introduction Repone depend on ytem and controller parameter > Cloed loop pole location depend on ytem and controller parameter. Ta: Find the controller gain( uch a we have a atifactory performance > Find the controller gain( uch a we have a atifactory pole location. To do that we need a good undertating of how pole move in the -plane. We have already een three cae: OL +, a OL +, ζω n + ωn OL ( + a( + ζω + ω n n Example (H: OL + CL + + o pole location: : cnt; for 0:0.0:0; r(cntroot([ +]; cntcnt+; end plot(real(r,imag(r,'.' Chapter 4 /4

3 Spring 008 EEE Propertie: Start from the OL pole location. End at. Example (H: OL CL o pole location: : clear, clc cnt; for 0:0.0:0; r(:,cntroot([ 3 +]; cntcnt+; end plot(real(r,imag(r Chapter 4 3/4

4 Spring 008 EEE Propertie: Start from the OL pole location. Collide at a point. Move on a traight line perpendicular to the real axi. Symmetrical with repect to the real axi End at. 5 ±. Example (H: OL + + CL o pole location: : Chapter 4 4/4

5 Spring 008 EEE 00 clear, clc cnt; for 0:0.0:0; r(:,cntroot([ +]; cntcnt+; end plot(real(r,imag(r,'*' Propertie: Start from the OL pole location. Start already with complex pole o no colliion. Move on a traight line perpendicular to the real axi. Symmetrical with repect to the real axi End at ±. Chapter 4 5/4

6 Spring 008 EEE 00 Example (H: OL CL o pole location: : clear, clc cnt; for 0:0.0:0; r(:3,cntroot([ 6 6+]; cntcnt+; end plot(real(r,imag(r,'.' Propertie: Start from the OL pole location. Two pole collide and then they follow aymptotically a line which i not perpendicular to the real axi. Chapter 4 6/4

7 Spring 008 EEE 00 Symmetrical with repect to the real axi The other pole goe to. Example (H: ( + + ( + + CL ( + + OL o 3 + pole location: 3 3 ( ( ( clear, clc cnt; for 0:0.0:0; r(:3,cntroot([ 6+ +* 6+*]; cntcnt+; end plot(real(r,imag(r,'.' Chapter 4 7/4

8 Spring 008 EEE Propertie: Start from the OL pole location. Two pole collide and then they follow aymptotically a line which i not perpendicular to the real axi. They do not diverge to infinity but they converge to the two complex zero. Symmetrical with repect to the real axi The other pole goe to. Hence it can be een that the root location depend on the order and the number of zero of the ytem. The procedure of finding the cloed loop pole location for variou value of the proportional gain,, i nown a root locu method. Chapter 4 8/4

9 Spring 008 EEE 00 We can find that location numerically or graphically (W. R. Evan. Matlab provide the command rlocu(num_ol,den_ol but it i better to have a deeper undertanding of the graphical method. Thi method can alo be ued for other parameter (apart from the proportional controller but i rather tricy. In EEE05 we will uperficially cover thi analyi method. In EEE30 you will cover thi topic extenively with Prof. J.W. Finch. To include the effect of the feedbac TF we name a OLTF the ( H( : We alway tart from the location of the OL pole (i.e. pole of (H(: CL + H N D N + D N D H H D N + N N D H H D NDH D + N H N H CE : D DH + N NH 0. For 0 (firt value: CE : D D H 0, o pole of (H(. We ue the ymbol x to denote OL pole. We ue the ymbol o to denote OL zero. We ue the ymbol to denote CL pole. Chapter 4 9/4

10 Spring 008 EEE 00 No need for CL zero a they do not move. Angle and magnitude condition ( ( 0 ( ( : ( ( ( ( ( + + H H CE H R C Thi implie that ( ( H and ( (, 80 ( ( arg + ± n H n 0,,... For example: ( ( ( ( ( ( 3 p p p z H Chapter 4 0/4

11 Spring 008 EEE 00 A 3 θ 3 B p 3 A φ θ z p A θ p arg( θ θ θ + φ ( H ( B A A A 3 3 Hence every point that belong onto the Root Locu (RL atifie thee two criteria. For example for the ytem that we tudied before CL the point point.5 ±. 5 j doe. OL ±. 5 j doe not belong while the Chapter 4 /4

12 Spring 008 EEE 00 The OL cae i zero: OL o I have two pole at - and - and no x ±. 5 j A A θ θ p p x ±. 5 j.5 A θ x x.5 tan x.5 x ( 80 θ θ 80 atan Chapter 4 /4

13 Spring 008 EEE 00 A (.5 + x ± xj A.5 θ x x.5.5 tan θ x x A ( θ atan.5 + ( x For the point.8 ±. 5 j :.5 θ 80 atan o.5 θ atan o So θ θ o that point doe not belong to the root locu. Chapter 4 3/4

14 Spring 008 EEE 00 For the point.5 ±. 5 j :.5 θ 80 atan o.5 θ atan o So θ θ 80 o that point belong to the root locu. By uing the magnitude condition we can find the gain: A A ( x.5 + (.5.58 ( x.5 + ( A A A A.5 Analytically:.8 ±. 5 j + H j + 0 > WRON ince i real..5 ±. 5 j + H Chapter 4 4/4

15 Spring 008 EEE 00 x.5 th80-atand(.5/(x- thatand(.5/(-x th+th Aqrt(.5^+(x-^ Aqrt(.5^+(-x^ A*A -x+.5*j; K-^-3*- Simple root locu Aume that the OLTF i ( ( + ( + pole with angle: arg (, arg( +, arg( +.. There are three real OL Firt of all the number of branche equal the number of pole. The root locu i alway ymmetrical with repect to the real axi and the branche tart from pole and end up to zero or infinity (infinite zero. In thi pecific cae I have three brache, which will diverge to imaginary zero at infinity. If >0 then arg ( 0, arg ( + 0, arg ( + 0 and therefore that area doe not belong to the root locu a arg ( arg( + arg( + 0 If <0 and >- then arg ( 80, arg ( + 0, therefore that area belong to the root locu a. ( 0 arg + and Chapter 4 5/4

16 Spring 008 EEE 00 ( arg( + arg( arg. If <- and >- then arg ( 80, arg ( + 80, therefore that area doe not belong to the root locu a ( arg( + arg( arg. If <-3 then arg ( 80, ( 0 arg + and arg ( + 80, arg ( + 80 that area belong to the root locu a and therefore ( arg( + arg( ( arg + with n. So the root locu on the real axi i: 0 We have een that when we numerically calculate the root locu. Chapter 4 6/4

17 Spring 008 EEE 00 At ome point the pole from - and - will collide. To find thi brea-out point we rearrange the CE a: B CE : f ( 0 B( + A( 0 and then we find the t A ( ( derivative with repect to and we equate the reult with zero: d d 0. Hence in thi cae: ( + ( ( + ( + d d hence ( + ( + and Since the nd 0.4 point only belong to the root locu the other one (-.5 i ignored: 0 The ame procedure can be ued to find brea-in point. To find the aymptote that the locu of thee two pole will follow: The angle of the aymptote i: ( n + ± 80, where n p i the number of n p n z finite pole and n z i the number of finite zero. Hence in thi cae Chapter 4 7/4

18 Spring 008 EEE 00 n 3, n 0 and hence their angle i 80/360. The point of interection p z with the real axi i: n p p n z p zi > : n n 3 i i i z Aymptote 0.5 The point of interection of the root locu with the imaginary axi can be calculated by uing the CE at jω. So in thi cae jω 3 3 ( + ( jω 3ω + jω + 0 So 3 0 ω + ω 0 ω and 3ω Chapter 4 8/4

19 Spring 008 EEE A imilar reult will be obtained if we ue Matlab: >> num; den[ 3 0]; rlocu(num,den 4 Root Locu 3 Imaginary Axi Real Axi Chapter 4 9/4

20 Spring 008 EEE 00 Example: Find the root locu of: ( ( +, H(: I have two complex pole and real zero.. Hence I have two branche that are ymmetrical with repect to the real axi. 3. Place the OL pole and zero:.4 j.4 j 4. Find the loci on the real axi: a. For >- I have no pole or zero o even number and hence that area doe not belong to the locu. b. From to - I have zero, i.e. odd number and hence thi belong to the locu: Chapter 4 0/4

21 Spring 008 EEE 00.4 j.4 j 5. Determine brea in and out point: ( ( CE : ( d d The acceptable value i Determine behaviour of complex pole: Thee pole will move toward the real axi by increaing. After the colliion at one will move toward to the zero at -and the other will diverge to. The angle of departure of the complex pole i calculated by 80 -angle with repect to other pole +angle with repect to other + zero: 80 θ φ : Chapter 4 /4

22 Spring 008 EEE 00 θ.4 j φ θ 90.4 j So θ 80 θ + φ The angle of departure for --.4j i -45 ince the locu i ymmetric..4 j.4 j Chapter 4 /4

23 Spring 008 EEE 00 Now we imply have to draw the locu from the complex pole to the real axi:.4 j.4 j With Matlab: >> num[ ]; den[ 3]; rlocu(num,den Root Locu.5 Imaginary Axi Real Axi Chapter 4 3/4

24 Spring 008 EEE 00 enerally the rule are:. The branche are alway ymmetric with repect to the real axi.. The number of the branche equal the number of the OL pole. 3. Every branch tart from an OL pole (0 and end at an OL zero (. If we have more pole than zero then we aume that there are imaginary zero at. 4. There are np n z aymptote where they interect on the real axi at: n p p n i i i n p z n z z i 5. The angle between them i π /( np n z aymptote with the real axi i π /( n p n z.the angle of the t 6. The branch exit on a point on the real axi if there are odd number of real pole and zero on it right. Complex pole do not contribute. d 7. The brea in and out point are found by olving 0. d 8. The point where we have interection with the imaginary axi are found by replacing with j ω at the CE. 9. The departure angle from a complex pole i 80 θ + φ. 0. The arrival angle at a complex zero i 80 + θ φ enerally the root locu of more complicated ytem will be very difficult to be derived and for that reaon we ue Matlab. Chapter 4 4/4