Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0

Transcription

1 Objective Root Locu Diagram Upon completion of thi chapter you will be able to: Plot the Root Locu for a given Tranfer Function by varying gain of the ytem, Analye the tability of the ytem from the root locu plot. Determine all parameter related to Root Locu Plot. Plot Complimentary Root Locu for negative value of Gain Plot Root Contour by varying multiple parameter. Introduction The tranient repone of a cloed loop ytem i dependent upon the location of cloed loop pole. If the ytem ha a variable gain then location of cloed loop pole i dependent on the value of gain. It i therefore neceary that we mut know the how the location of cloed loop varie with change in the value of loop gain i varied. In control ytem deign it may help to adjut the gain to move the cloed loop pole at deired location. So, root locu plot give the location of cloed loop pole a ytem gain K i varied. Root loci: The portion of root locu when k aume poitive value: that i Complementary Root loci: The portion of root loci k aume negative value, that i k Root contour: Root loci of when more than one ytem parameter varie. k Criterion for Root Loci Angle condition: It i ued for checking whether any point i lying on root locu or not & alo the validity of root locu hape for cloed loop pole. For negative feedback ytem, the cloed loop pole are root of Characteritic Equation 1G H G H 1 G H 18 tan q 1 18 Kreatryx. All Right Reerved. 1

2 Angle condition may be tarted a, for a point to lie on root locu, the angle evaluated at that point mut be an odd multiple of 18 Magnitude condition: It i ued for finding the magnitude of ytem gain k at any point on the root locu. 1G H G H 1 G H 1 Solved Example Problem: Determine whether the point 3 4 j & 3 j lie on the root locu of 1 k GH 4 1 Solution: K K G H j 1 4 j G H Not an odd multiple of 18 k k GH j 1 j GH =Odd multiple of 18. Kreatryx. All Right Reerved.

3 Advantage of Root Locu The root locu i a powerful technique a it bring into focu the complete dynamic repone of the ytem and further, being a graphical technique, an approximate root locu can be made quickly and deigner can be eaily viualize the effect of varie ytem parameter an root location. The root locu alo provide a meaure of enitively of root to the vacation in parameter being conidered. It may further be pointed out here that root locu technique i applicable to ingle a well a multiple loop ytem. Contruction Rule Of Root Locu Rule 1: The root locu i ymmetrical about real axi (G()H() = -1) Rule : Each branch of Root Locu originate at an open loop pole and terminate at an open loop zero or infinity. Let P = number of open loop pole Z = number of open loop zeroe And if P > Z Then, No. of branche of root locu = P No. of branche terminating at zeroe = Z No. of branche terminating at infinity = (P Z) Rule 3: A point on real axi i aid to be on root locu if to the right ide of their point the um of number of open loop pole & zeroe i odd. Ex: G k P 3 ; Z ; P Z 1 NRL Not Root Locu; RL Root Locu Kreatryx. All Right Reerved. 3

4 Rule 4: Angle of aymptote The P Z branche will terminate at infinity along certain traight line known a aymptote of root locu. Therefore, number of aymptote = (P - Z) Angle of aymptote i given by: q 118 P Z ; q,1,,3... Suppoe if P Z =, then angle of aymptote would be: and Solved Example Problem: Find the angle of aymptote of a ytem whoe characteritic equation i 4 1 k 1 Solution: The characteritic equation can be written in tandard form a hown below: K G H K P 4 ; Z 1 ; P Z Kreatryx. All Right Reerved. 4

5 Rule 5: Centroid It i the interection point of aymptote on the real axi, it may or may not be a part of root locu. centroid Realpart of openloop pole Realpart of openloop zeroe P Z Solved Example Problem: Find centroid of the ytem whoe characteritic equation i given a: 3 5 K 6 k => 3 Solution: 3 5 k k 6 k 1 k P = 3, Z = Pole =, -, -3 ; zeroe = centroid, k 1 Rule 6: Break Point They are thoe point where multiple root of the characteritic equation occur. Procedure to find Break Point: 1. Contruct 1 + G()H() =. Write k in term of 3. Find dk d 4. The root of dk give the location of break-away point. d 5. To tet the validity of breakaway point ubtitute in tep. If k >, it mean a valid breakaway point. Kreatryx. All Right Reerved. 5

6 Breakaway Point If the break point lie between two ucceive pole then it i termed a a Breakaway Point. Prediction 1) The branche of root locu either approach or leave the breakaway point at an angle of 18 n Where n = no. of branche approaching or leaving breakaway point. ) The complex conjugate path for the branche of root locu approaching or leaving or breakaway point i a circle. 3) Whenever there are adjacently placed pole on the real axi with the ection of real axi between there a part of RL, then there exit a breakaway point between the adjacently placed pole. Solved Example Problem: Plot Root Locu for the ytem hown below: Solution: Applying Contruction Rule of the Root Loci C k R k G k Rule : P =, Z =, P Z = Rule 3: Section of real axi lying on Root Locu Kreatryx. All Right Reerved. 6

7 1 1 Rule 4: 18 9 ; Rule 5: centroid 1 Rule 6: Breakaway point k => k dk 1 d From the third prediction about Breakaway point we know that there mut be a breakaway point between = and =-. Hence = -1 i a valid breakaway point. Since centroid and breakaway point coincide the branche of root locu will leave breakaway point along the aymptote. Break in point The point where two branche of root locu converge are called a break in point. A Break-in Point lie between two ucceive zeroe. To differentiate between break in & break away point. dk d For break in point dk For break-away point d Kreatryx. All Right Reerved. 7

8 Otherwie by obervation a breakaway point lie between two pole and a break-in point lie between two zeroe. Prediction about Break-in Point: 1. Whenever there are zeroe on real axi & the portion of real axi between zeroe lie on root locu then there i a break in point between zeroe.. Whenever there exit a zero on real axi & real axi on left i on root locu & P > Z, then there will be a break in to left of zero. Effect of adding pole to a tranfer function Suppoe we add a pole at =-4 to our previou Tranfer Function G K The new tranfer function will be: G K 4 Rule : P = 3, Z =, P Z = 3 Kreatryx. All Right Reerved. 8

9 Rule 3: The ection of real axi lying on Root Locu i hown on right Rule 4: 6, 18, Rule 5: Centroid 4 3 Rule 6: Break away point Characteritic Equation: K 3 => K 6 8 dk d , 3.15 Since Breakaway point mut lie between two ucceive pole that i it mut lie between and - o =-.845 i a valid breakaway point wherea =-3.15 i invalid. The root locu for thi ytem i hown below, Here, the root locu hift toward right & hence tability decreae. Kreatryx. All Right Reerved. 9

10 Effect of adding zeroe to a tranfer function Suppoe we add a zero to the tranfer function in the previou cae at = -1, then the Tranfer Function become: Rule : P 3, Z 1, P Z G K 1 4 Rule 3: The ection of real axi lying on Root Locu are hown below: Rule 4: 9, Rule 5: centroid.5 Rule 6: Breakaway point Characteritic Equation of Cloed Loop Sytem i: K K K => dk.9 d 1 Kreatryx. All Right Reerved. 1

11 The branche originate at Breakaway Point and tend to infinity along the aymptote. Here, the root locu hift toward left and hence tability increae. Solved Example Problem: Draw Root Locu for the ytem whoe open loop tranfer function i G K 1 Solution: Applying Contruction rule of the Root Loci Rule : P, Z 1, P Z 1 Rule 3: The ection of real axi lying on Root Locu are hown below: Rule 4: Angle of Aymptote 18 1 Rule 5: centroid 1 1 Rule 6: Break away point 1 K => K dk d =>.6, 3.4 Since, = -. Lie between two conecutive pole it i a Break away point and ince = -3.4 lie between a zero and infinity it i a break-in point. Kreatryx. All Right Reerved. 11

12 So, the root locu look like a hown below: Proof of path being a circle k b a Let x jy Kx jy b K x b jy x jy x jy a x jxy ax jay y K x b jy x ax y jxy ay For x jy lie on the root locu, angle hould be odd multiple of 18 y 1 1 xy ay tan tan 18 x b x ax y Taking tan on both ide y xy ay x b x ax y x ax y x a x b => x b y bb a x bx y ab centre b, radiu b b a So, in uch a cae the centre of circular trajectory of Root Locu lie at the zero of Open Loop Sytem. Kreatryx. All Right Reerved. 1

13 Rule 7: Interection of Root Locu with imaginary axi Root of auxiliary equation A() at k k max (i.e. Maximum value of Gain K for which the cloed loop ytem i Stable) from Routh Array give the interection of Root locu with imaginary axi. The value of k k max i obtained by equating the coefficient of 1 in the Routh Array to zero. Solved Example Problem: Find the interection of root locu with the imaginary axi and alo the interection of aymptote with imaginary axi for the ytem whoe open loop Tranfer Function i given k by G. 4 Solution: Characteritic Equation of the ytem i given a k Routh Array: K 1 48 k 6 K For tability 48 k k 48 & k 6 range k 48 At k k 48 max A 6 k 6 48 j 8 j.8 Kreatryx. All Right Reerved. 13

14 Shortcut method If the Tranfer Function i of the type G K a b Interection of RL with jω axi i given by j ab Interection of aymptote with jω axi Since we are aware about the angle of aymptote from Rule-4 and alo the x-axi intercept of the aymptote which i centroid o we can find y-intercept a hown below: tan y x tan6 y y To find Gain, k at any point on root locu geometrically Pr oduct of vector length of pole k Product of vector lengthof zeroe Suppoe, we need to find Gain k of the root locu plot for.5, then we need to find the root of characteritic equation correponding to.5,which can be done by finding interection of root locu with a traight line oriented from ve x axi in clockwie direction and paing through origin. co 1 6 Gain, l l l Pr oduct of vector length of pole p1 p p3 k Product of vector lengthof zeroe 1 Kreatryx. All Right Reerved. 14

15 Thi i hown in the figure below. Rule 8: Angle of Departure and Arrival The angle made by root locu with real axi when it depart from a complex open loop pole i called a angle of departure. The angle made by root locu with real axi when it arrive at complex open loop zero i called a angle of arrival. angle of departure 18 GH' D angle of arrival 18 GH' A GH' = angle of the function excluding the concerned pole at the pole itelf. Solved Example Problem: Plot Root Locu for the ytem whoe open loop tranfer function i given by: G H k 4 4 Solution: Rule : P = 4, Z =, P Z = 4 Rule 3: The ection of real axi lying on root locu i hown In adjoining figure Kreatryx. All Right Reerved. 15

16 Rule 4: 45, 135, 5, Rule 5: centroid 4 Rule 6: Breakaway point K K dk d = => =, j.45 Note: To check validity of complex break point ue angle criterion. Rule 7: Routh Array K K 1 8 8k 6 K For tability 8 8k k 6 & K > 6 range k 6 For k=6 A.E.= A 6 k 6 6 => j3.16 Interection of aymptote with jω axi y tan 45 j Kreatryx. All Right Reerved. 16

17 Rule 8: Angle of departure tan p1 18 tan p tan 63.4 p3 4 Sum of Angle from zeroe - Sum of Angle from Pole z p D Ele K GH 4 j 4 4 j 4 j K 4 j 4 j8 j GH' tan tan D So, the root locu of the ytem look like a hown below: tan 1 tan 1 9 = 7 Kreatryx. All Right Reerved. 17

18 Problem: Plot Root Locu for the ytem whoe Open Loop Tranfer Function i given by G k 5.5 Solution: Rule : P =, Z =, P Z = Rule 3: The ection of real axi lying on Root Locu i hown below: Since P-Z =, Rule 4 and Rule 5 are of no ue a there are no aymptote. Rule 6: Breakaway point Characteritic Equation of the ytem i: => k 5.5 k 5 dk d =>.4, 3.6 Here, ince Breakaway point mut lie between two conecutive pole o = -.4 i a valid Breakaway Pont wherea = 3.6 i an invalid point. Rule 7: Routh Array: 1 k 1.5 k 5k 1 1 k 5k k 5k 1 For tability 1 k k 1 & 1.5 k k.75 & 5k 1 k. range. k.75 Kreatryx. All Right Reerved. 18

19 k k.75 max A.E. = A 1 k 5k 1 j1.5 Rule 8: Angle of arrival G.5 k 1 j 1 j at 1 j A 3 j.5 j 1 j 1 j GH' tan 1 4 j tan 1 tan GH' 1 9 tan tan The Root Locu for the ytem look like a hown below: Kreatryx. All Right Reerved. 19

20 Analyi of ytem having dead time or tranportation lag Dead Time: It i one of the form of non linearity and i defined a the time in which a ytem doe not repond to change in input. It i approximated a a zero in RHS plane. Tranfer function having pole or zeroe in RHS plane are called a non minimum phae function. For curve 1 Output y(t) = input x(t) For curve y t x t T Applying Laplace Tranform both ide Y e T Y X e T Time domain approximation T y t x t T x t Tx t x t...! y t x t Tx t Y X TX T e T 1 T Ex: G ke k X 1 T X e Kreatryx. All Right Reerved.

21 Complimentary Root Locu (CRL) Angle Condition: G H q 18 Magnitude Condition: G H 1 Contruction Rule Rule 1: The CRL i ymmetrical about real axi GH 1 Rule : Number of branche terminating at infinity i (P Z). Rule 3: A point on real axi i aid to be on CRL if to the right ide of the point the um of open loop pole & zeroe i even. Rule 4: Angle of aymptote q18 P Z, q 1,,3... Rule 5: Centroid: ame a Root locu Rule 6: Break point: Same a RL Rule 7: Interection of CRL with jω axi ame a root locu. Rule 8: Angle of departure & arrival. D A Where GH' Z P Solved Example Problem: Plot Root Locu for the ytem whoe Open Loop Tranfer Function i Ke G 3 Kreatryx. All Right Reerved. 1

22 Solution: G k 1 k So, here the gain become negative and we thu plot the Complimentary Root Locu. Rule : P =, Z = 1, P Z = 1 Rule 3: The ection of real axi lying on Root Locu i hown in the figure below: Rule 6: Breakaway point Characteritic Equation of the ytem i: K k 1 => k dk 31 d = 1 3 => =3,-1 Since Breakaway point mut lie between two conecutive pole, o it mut lie between and -3 and hence =-1 i a valid breakaway point. Rule 7: Interection of Root Locu with Imaginary Axi 3 k k Characteritic Equation: Kreatryx. All Right Reerved.

23 Routh Array: 1 k 1 3 k k For tability 3-k> => k<3 & k> Range <k<3 k k 3 max A.E.= A k j1.73 Root Contour Root contour are multiple root locu diagram obtained by varying multiple parameter in a tranfer function diagram in ame plane. G H k 1 8 Cae 1: let, and gain K i varied G k 1 8 Kreatryx. All Right Reerved. 3

24 Cae : In thi cae, gain K=1 and i varied For plotting Root Locu with repect to, we mut firt manipulate Characteritic Equation in uch a way that act a gain of the ytem. k k k 1G H G H Put k = 1 G H k 1 8 k 9 9 => 1 8 k Kreatryx. All Right Reerved. 4