CONTROL SYSTEMS Chapter 5 : Root Locu Diagram GATE Objective & Numerical Type Solution Quetion 1 [Work Book] [GATE EC 199 IISc-Bangalore : Mark] The tranfer function of a cloed loop ytem i T () where i the path gain. The root locu plot of the ytem i ( ) 1 (A) j (B) = j = = 1 = (C) j (D) j An. (B) Sol. Given : T () ( ) 1 T () 1 1 1 1 Tranfer function for poitive feedback i given by, G () T () 1 GH ( ) ( ) On comparing equation (i) and (ii), we get G ( ), 1 H ( ) Characteritic equation i 1 (i) (ii) 1
94 Pole at.81,.618 (iii) Both pole lie at negative real axi. Locate the open loop pole and zero on the -plane, and any point on the real axi i part of root locu if the total number of open loop pole and zero to the right of the point i even (complimentary root locu). j CRL.618.81 CRL Fig. Pole-zero diagram The characteritic equation i given by, 1 G( ) H ( ) 1 1 1 (iv) Breakaway/break-in point can be calculated a, d d d ( ) ( 1) d d 1 d 1 1 From equation (iv), we get ( 1) ( 1) 1 1 1. (v) The root locu branch croe the imaginary axi i determined by applying Routh Hurwitz criterion to the characteritic equation. Routh Tabulation : 1 1 1 Row of zero 1 = 1, valid breakaway point j.618.81 The interection of root locu plot with imaginary axi i given by the value of obtained by olving the following equation. (vi) = 1, valid break-in point
By uing equation (iii), (v) and (vi) option (B) i atified. = j = 1 = =.618 1.81 1 = Hence, the correct option i (B). Quetion [Work Book] [GATE EE 1991 IIT-Madra : 1 Mark] ( a) A unity feedback ytem ha an open loop tranfer function of the form G() ; b a which ( b) of the loci hown in figure can be valid root-loci for the ytem? (A) (B) j j b a b a (C) j (D) j b a b a An. (A, C) Sol. Number of open loop pole P = at =, and b. Number of open loop zero Z = 1 at = a. The interection point of aymptote on the real axi i called centroid. Centroid i given by, Real partof polein GH ( ) ( ) Real partof zeroin GH ( ) ( ) P Z b a ( b a).. (i) 1 Angle of aymptote : (1) 18 A where,, 1,,... ( P Z 1) P Z If PZ, 1 A 9, 7
Now conider two cae. Cae I : 4 a and b = 1 4 ( a) G( ) ( b) ( 1) A point on the real axi lie on the root locu if the um of the number of open loop pole and the open loop zero on the real axi to the right hand ide of thi point i odd. j RL 1 4/ Fig. Pole-zero diagram From equation (i), 4 1 16 σ 5. Breakaway/Break-in point can be calculated a, d d Characteritic equation i given by, 1 G( ) H ( ) 4 1 ( 1) ( 1) ( 1 ) 4 4 d d 4 4 4 1 4 4 1 16 ( 8 16) ( 4) =, 4, 4 There are repeated pole at origin o there i a breakaway point at origin. 4
( 64 1 16) 18 For = 4, 48 4 8 4 So, = 4 i alo a valid breakaway point. j 1 5. 48 4/, Cae II : a =, b = 4 G( ) = 4 breakaway point ( ) ( 4) A point on the real axi lie on the root locu if the um of the number of open loop pole and the open loop zero on the real axi to the right hand ide of thi point i odd. j RL 4 Fig. Pole-zero diagram ( ba) (4) From equation (i), 1 Breakaway/Break-in point can be calculated a, d d Characteritic equation i given by, 1 GH ( ) ( ) ( 4) ( 4 ) ( ) ( ) d ( )( 8 ) ( 4 ) d ( ) 8 6 16 4 1 16 ( 116),.5 j1. For exitence of complex breakaway/break-in point it mut atify angle criteria. 5
At.5 j1., (.5 j1.) G() (.5 j1.) (.5 j1. 4) By uing calculator, G( ) = +15.5 (Not an odd multiple of 18 ) It doe not atify angle criterion. Hence, thi i not a valid breakaway point. j 4 1, Hence, the correct option are (A) and (C) both. Quetion 7 [Practice Book] Sol. [GATE IN 199 IIT-anpur : Mark] A unity feedback ytem ha open loop pole ( j ) and. It ha a ingle zero at ( 4 j). The angle of departure in degree of the root locu branch tarting from the pole ( j ) i. Figure to calculate angle of departure i hown in below. ( j) P j j Pole Zero p1 4 Z 1 z1 P 1 p Angle of departure at complex pole : The angle of departure from an open loop complex pole i given by (for ) d 18 ( pz ) z : Sum of all the angle ubtended by zero. p: Sum of all the angle ubtended by remaining pole. At pole ( j), 1 p 1 18 tan 15 p 9 1 z 1 tan 45 p p 1 p 15 9 5 P ( j) j 6
z z1 45 d 18 (5 45 ) At pole ( j), d Hence, the anwer i. Quetion 14 [Practice Book] [GATE EE IISc-Bangalore : 5 Mark] ( ) The open loop tranfer function of a unity feedback ytem i given byg (). Sketch ( )( 1) the root locu a varie from to. Find the angle and real axi intercept of the aymptote, breakaway point and the imaginary axi croing point, if any. ( ) Sol. Given : G (), H () 1 ( )( 1) Characteritic equation i given by, 1 G( ) H ( ) ( ) 1 ( )( 1) ( 1) 1 (i) ( 1 ) Modified characteritic equation i given by, 1 G'( ) H '( ) (ii) On comparing equation (i) and (ii), we get G'( ) H'( ) ( 1 ) G '( ) H '( ) 1 1 ( 1 ) 1 1 Concept of root contour : Number of pole, P = at,.6, 9.74. Number of zero, Z =. j RL 1 RL Fig. Pole-zero diagram 7
Number of aymptote, P Z The interection point of aymptote on the real axi i called centroid. Centroid i given by, Realpartof polein G'( ) H'( ) Realpartof zeroin G'( ) H'( ) P Z (.6 9.74) 4 An. Angle of aymptote : (1) 18 A where,, 1,,... ( P Z 1) P Z If PZ, 1, Angle of aymptote are Breakaway/break-in point can be calculated a, d d A 6, 18, 6, 18,. An. d 1 d 4 1.6, 6.94 = 6.94 invalid Breakaway point j 1 Rejecting 6.94 a it doen t lie on root locu. Hence, valid break away point 1.6. An. The root locu branch croe the imaginary axi i determined by applying Routh Hurwitz criterion to the characteritic equation. Characteritic equation i given by, 1 G '( ) H '( ) 1 1 1 1 = 1.6 valid Breakaway point 64 1 Row of zero The interection of root locu plot with imaginary axi i given by the value of obtained by olving the following equation. 8
64 1 1 Auxiliary equation i formed uing row, 1 1 64 j4.69 The root contour diagram i hown in below figure. 1 1 j = 1.6 1 j4.69 j4.69 An. Quetion 7 [Practice Book] [GATE EC 9 IIT-Roorkee : Mark] The feedback configuration and the pole-zero location of G( ) are hown below. The root locu for negative value i.e. for, ha breakaway/break-in point and angle of departure at pole P (with repect to the poitive real axi) equal to j G () P An. (A) and (B) and 45 (C) and (D) and 45 (B) Sol. Given : G () j G () P Root locu for negative value of repreent the invere root locu. The characteritic equation i given by, 1 G( ) H( ) 9
1 Breakaway/break-in point can be calculated a, d 1 45 j j 1 9 d ( )() ( )() ( ) j P 4 8 Angle of departure for a complex pole 1 j i given by, d ( z p ) Invere root locu where p um of all the angle ubtended by remaining pole z um of all the angle ubtended by zero ( 45 ) ( 9 ) 45 Hence, the correct option i (B). d Note : In the original GATE quetion paper function wa G (). Thi function ha a zero in right half of -plane and pole in left half of -plane but in pole-zero diagram of G( ) the pole are given in the right half and zero in left half of -plane. To correct the quetion function G( ) ha been modified to G (). Quetion 8 [Work Book] [GATE IN 9 IIT-Roorkee : Mark] A unity feedback ytem ha the tranfer function ( b) ( ) The value of b for which the loci of all the three root of the cloed loop characteritic equation meet at a ingle point i (A) 1 (B) (C) (D) 4 9 9 9 9 An. (B) ( b) Sol. Given : G () ( ) Characteritic equation i given by, 1 G( ) 1
( b) 1 ( ) ( ) ( b) ( ) ( b) d d ( b)( 4 ) ( )(1) ( ) b b 4 4b (b ) 4b [ (b ) 4 b] and (b ) 4b There i repeated pole at origin o there i a breakaway point in origin i.e.. Solving quadratic equation, we get ( b) ( b) b ( b) 4 1b9b b 4 4 ( b) b b 9 9 4 ( b) ( b) b 9 4 Cae - I : When b. 9 Breakaway point, Cae - II : When b 9 6.66 4 = 6.67 valid breakaway point Breakaway point, [Not valid] 4 The loci of all thee root can meet at a ingle point only if the breakaway point exit between b. When b. the breakaway point exit at 6.66 which lie between 9 b. In uch cae the open loop tranfer function of the ytem become a under, 9 GH () () ( ) The interection point of aymptote on the real axi i called centroid. 9 j 11
Centroid i given by, Real part of pole in GH ( ) ( ) Real part of zero in GH ( ) ( ) P Z σ 9 8.88 1 The number of aymptote, NZ 1 Angle of aymptote : (1) 18 A P Z If P Z, 1 The root locu of the ytem i hown in below figure. where,, 1,,... ( P Z 1) j A 9, 7 /9, Hence, the correct option i (B). Quetion 1 [Practice Book] An. Sol. [GATE IN 1 IIT-Delhi : Mark] The open loop tranfer function of a unity negative feedback control ytem i given by 48 G (). The angle θ, at which the root locu approache the zero of the ytem, atifie ( )( 8) (A) (C) (D) tan 4 1 1 1 1 tan 4 Open loop tranfer function of a unity feedback ytem i 48 GH () () ( )( 8) 4 8 4 4 4 8 4 16 (B) (D) 4 4 j j Number of open loop pole, P = at =, and 8. Number of open loop zero, Z = at = + j and j. Pole-zero pattern i hown below. = 6.67 breakaway point 1 4 1 tan 1 1 tan 4 1
A j j B p1 p p 8 C D z1 j Angle of arrival at complex zero : The angle of arrival for an open loop complex zero i given by (for ) a 18 ( pz ) z : Sum of all the angle ubtended by remaining zero. p: Sum of all the angle ubtended by pole. At j, 1 p 1 tan 4 4 p tan p z1 1 AC BC tan 1 tan 6 1 1 1 4 1 p p 1p p tan rad z z1 11 a 18 tan 4 rad 1 1 7 11 11 a tan tan tan 4 4 4 rad rad rad 11 11 a tan tan 4 rad 4 rad j 1 1 a tan 4 rad 7 (CCW) 4 4 (CW) Hence, the correct option i (D). 1
Quetion 1 [Work Book] [GATE EC 14 (Set ) IIT-haragpur : Mark] In the root locu plot hown in the figure, the pole/zero mark and the arrow have been removed. Which one of the following tranfer function ha thi root locu? j 1 (A) 1 ( )( 4)( 7) (B) 4 ( 1)( )( 7) (C) 7 ( 1)( )( 4) (D) ( 1)( ) ( 7)( 4) An. Sol. (B) The given root locu plot i hown below. j 1 For option (A), (B) and (C) : Since P > Z branche will tart from each of the location of open loop pole and terminate at the location of zero. The remaining P Z branche will approach to infinity. Angle of aymptote : (1) 18 A where,, 1,,... ( P Z 1) P Z If PZ, 1 A 9, 7 It mean that root locu branche will approach to infinity with angle The interection point of aymptote on the real axi i called centroid. Centroid i given by, For option (A), 9 and 7. Real partof polein GH ( ) ( ) Real partof zeroin GH ( ) ( ) 471 6 1 P Z The root locu for option (A) can be drawn a hown below. 14
j 7 6 4 1 Fig. (A) 174 For option (B), 1 The root locu for option (B) can be drawn a hown below. j 7 4 1 Fig. (B) 147 For option (C), 1 The root locu for option (C) can be drawn a hown below. j 7 4 1 Fig. (C) 741 For option (D), The root locu for option (D) can be drawn a hown below. j 7 4 1 Hence, the correct option i (B). Fig. (D) 15
Quetion 1 [Work Book] [GATE EC 14 (Set 4) IIT-haragpur : Mark] The characteritic equation of a unity negative feedback ytem i 1 G( ). The open loop tranfer function G( ) ha one pole at and two pole at 1. The root locu of the ytem for varying i hown in the figure. j.5 A 1 1 (, ) O The contant damping ratio line, for.5, interect the root locu at point A. The ditance from the origin to point A i given a.5. The value of at point A i. An.. to.41 Sol. The given root locu for varying i hown below. j.5 A, 1 1 X O Given : GH () (),.5, OA =.5. ( 1) We know that, In AXO, co θ co 6 OX OA 1 1 1 OX OAco 6.5 4 16
in θ AX AO AX AO in 6.5 4 So, the coordinate of A i 1 j. 4 4 We know that the co-ordinate of point A of the given root locu i.e. magnitude condition. GH () () 1 Applying magnitude condition at GH () () ( 1) 1 j 4 4, we get GH () () 1 1 j j 1 4 4 4 4 GH () () 1 1 j 1 1 4 4 1 9 16 16 16 16.75 Hence, the anwer i.75. Alternatively : Graphical calculation of : j 1 j j 4 4 4 4 A P 1 X.5 O 4.4.75 From figure, AP (.75) (.4).866 17
and OA.5 Hence, the anwer i.75. Quetion [Work Book] Product of vector lengthdrawn from thepoleof GH ( ) ( ) to A Product of vector lengthdrawn from thezeroof GH ( ) ( ) to A.866.866.5.75 1 The open-loop tranfer function of a unity feedback configuration i given a value of a gain ( ) for which 1 j lie on the root locu i. An. (5.54) Sol. Pole-zero diagram i hown below. P [GATE EC 15 (Set-1) IIT-anpur : Mark] j j G( ) ( 4) ( 8)( 9). The A B C D 8 4 1 Vector Length : AP (7) () 5 BP () () 1 CP () () 8 DP (4) () Product of vector lengthdrawn from thepoleof GH ( ) ( ) to P Product of vector lengthdrawn from thezeroof GH ( ) ( ) to P 58 5.5 1 Hence, the correct option i 5.5. Alternatively, By magnitude criteria, GH () () 1 G( ) 1 j 1 j j 7j j 4j 8 5 5.5 1 Hence, the correct option i 5.5. ( 1 j) 4 ( 1 j) 8 ( 1 j) 9 1 18
Quetion [Practice Book] [GATE EE 15 (Set-1) IIT-anpur : Mark] The open loop pole of a third order unity feedback ytem are at, 1,. Let the frequency correponding to the point where the root locu of the ytem tranit to untable region be. Now uppoe we introduce a zero in the open loop tranfer function at, while keeping all the earlier open loop pole intact. Which one of the following i TRUE about the point where the root locu of the modified ytem tranit to untable region? (A) It correpond to a frequency greater than. (B) It correpond to a frequency le than. (C) It correpond to a frequency. (D) Root locu of modified ytem never tranit to untable region. An. (D) Sol. Given : Number of open loop pole, P = at =, 1 and. Open loop tranfer function will be, GH () () ( 1)( ) Root locu plot i hown in figure (a). j 1 Fig. (a) After introducing a zero in open loop tranfer function, ( ) G'( ) H'( ) ( 1)( ) Root locu plot i hown in figure (b). j 1 Fig. (b) From the above figure (b), it can be een that root locu plot never croe j-axi for all value of. Alternatively : The root locu branch croe the imaginary axi i determined by applying Routh Hurwitz criterion to the characteritic equation. Characteritic equation i given by, ( ) 19
Routh Tabulation : 1 + 1 6 For ytem to be table > i.e. for all value of ytem i table. Effect of addition of open loop zero in Root Locu (i) Root locu hift to left away from the imaginary axi. (ii) Relative tability of the ytem increae. Hence, the correct option i (D). Quetion 5 [Practice Book] An. [GATE EE 16 (Set-) IISc-Bangalore : Mark] The gain at the breakaway point of the root locu of a unity feedback ytem with open loop tranfer G( ) function ( 1)( 4) i (A) 1 (B) (C) 5 (D) 9 (A) Sol. Given : G () ( 1)( 4) img R- L R- L 1 4 Real Characteritic equation 1 ( 1)( ) ( 1)( 4) For break away point dk d ( 5 4) ( )(5) ( 54)(1) 5 54
4 will be valid breakaway point 1 A it lie between two pole. Hence the correct option i (A) IES Objective Solution Quetion 14 [Practice Book] [IES EE ] Identify the correct root locu from the figure given below referring to pole and zero at j8 and j1 repectively of G( ) H( of ) a ingle-loop control ytem. j j (A) (B) j j (C) (D) An. (B) Sol. Given : The ytem ha pole at j8 and zero at j1. Hence, the tranfer function can be written a ( 1) GH () () ( 64) Pole zero location are hown in figure below, j j1 j8 Angle of departure at pole at j8 d1 18 Where, z p j8 j1 1
p p1 9 z z 1z 9 9 9 9 d1 18 ( 9 ) 9 d at pole at 8 d 9 j Hence, the root locu i hown below, in which branche of root locu are tarting from pole and are terminating on zero. j d1 9 ACW CW d 9 Hence, the correct option i (B). Quetion 19 [Practice Book] [IES EE 4] A control ytem ha G( ) H ( ) ( ). What i the number of breakaway ( 4)( 4 ) point in the root locu diagram? (A) One (B) Two (C) Three (D) Zero An. (C) Sol. Given : G( ) H ( ) ( ) ( 4)( 4 ) Characteritic equation i given by, 1 G( ) H ( ) 1 ( 4)( 4) 4 8 6 8 4 ( 8 6 8 ) Breakaway point can be calculated a, d d d (4 4 7 8) d Breakaway point i and two complex breakaway point are j.45. Exitence of complex breakaway point can be calculated uing angle condition. Angle condition :
GH n n () () ( 1)18,1,... GH () () ( j.45) ( j.45) 4 ( j.45) 4( j.45) GH () () 18 [Multiple of So j.45 are the valid breakaway point. At, 64 [poitive] So i alo a valid breakaway point. j j.45 18 ] 4 = valid breakaway point j.45 The number of breakaway point in the root locu diagram i three. The root locu plot i hown in below figure. = + j.45 valid breakaway point j j.45 4 = valid breakaway point j.45 Hence, the correct option i (C). Quetion 4 [Practice Book] [IES EC 1] If the gain of the ytem i reduced to zero value, the root of the ytem in the -plane (A) Coincide with zero (B) Move away from the zero (C) Move away from the pole (D) Coincide with the pole An. (D) Sol. Let u aume G () ( a) and a are open loop pole Characteritic equation i a = j.45 valid breakaway point
Root of characteritic equation are, 1 a a When =, the root are 1 and a i.e. they coincide with the open loop pole of ytem. Hence, the correct option i (D). Alternatively : Root locu tart with open loop tranfer function pole where = and terminate to zero where. Quetion 41 [Practice Book] [IES EC 11] Given the root locu of ytem G ().77? 4 1 what will be the gain for obtaining the damping ratio j 1 An. (A) 1 4 (B) Sol. Given : G () (B) 5 4 4 1 and 1 (C) 4 (D) 11 4 Number of open loop pole, P = at = 1 and. Number of open loop zero, Z =. The interection point of aymptote on the real axi i called centroid. Centroid i given by, real part of pole of GH ( ) ( ) real part of zero of GH ( ) ( ) P Z 1 Characteritic equation i given by, 1 G( ) 4 1 ( 1)( ) 4 4.(i) For econd-order ytem, characteritic equation i given by,.(ii) n n 4
On comparing equation (i) and (ii), we get n 4 n n 8 n 4 n 4 8 5 4 Hence, the correct option i (B). Alternatively : Graphical analyi of : B j 5 C A 45 D 1 1.77 co 45 1 AB tan 45 1 OA AB OA CB 1 5 BD 1 5 Product of vector lengthdrawn from the poleof GH ( ) ( ) to B 4 Product of vector lengthdrawn from thezeroof GH ( ) ( ) to B 5 5 4 5 1 5 4 Note : If zero doe not exit in tranfer function then we conider unity length. Hence, the correct option i (B). 5
Quetion 47 [Practice Book] [IES EC 11] Loop tranfer function of unity feedback ytem i the ytem i j G () ( 64) ( 16). The correct root locu diagram for j (A) (B) j j (C) (D) An. (D) Sol. ( 64) Given : G () ( 16) Number of pole, P =, Number of zero, Z = Total branche of root locu = Number of aymptote = P Z = 1 Angle of aymptote 18 The ytem ha pole at j4 and at and zero at j8. Pole zero location are hown in figure below, Angle of departure at pole at j4 j j8 d1 18 j4 Where, z p p p p 9 9 18 z z 1z 9 9 18 18 j4 j8 d1 18 ( 18 ) d at pole at 4 d j 6
Hence, the root locu i hown below, in which branche of root locu are tarting from pole and are terminating on zero, one pole will be terminating at zero preent in infinite -plane following the aymptote of 18. j 18 7