ME2142/ME2142E Feedback Control Systems

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Root Locu Analyi

Root Locu Analyi Conider the cloed-loop ytem R + E - G C B H The tranient repone, and tability, of the cloed-loop ytem i determined by the value of the root of the characteritic equation G( ) H ( ) 0 or, in other word, the location of the cloed-loop pole on the =plane. The open-loop tranfer function can be written in the form G( ) H ( ) ( z K ( p )( where K i an adjutable gain, the -z and -p are the zero and pole of the open-loop tranfer function. A the gain K change, the value of the cloed-loop pole will change and thu the tranient repone, and tability. The root locu plot i a plot of the loci, or location, of the cloedloop pole on the -plane a the gain K varie from 0 to infinity. )( z p ) ( z ) ( m p n ) )

Why Root Locu? Conider the cloed-loop ytem for which K i a proportional controller: ( )( ) Quetion: )How will the ytem repond a K i varied? )Can the ytem ever be untable for ome value of K? 3)How hould K be adjuted to give a good repone? Depend upon how change in K change the value of the cloed-loop pole, or the root of the characteritic equation: GH K 0 ( )( ) 3

Root Locu Analyi Example: howing how cloed-loop pole change a K change Conider characteritic Eqn: K ( )( ) K 0 ( )( ) 0 3 rd -order ytem give 3 pole Root Locu Plot Cloed-loop loop Pole v K K P P P 3 K=7 0 0 - - 0. -0.054-0.90 -.05 K= 0 0. -0. -0.79 -.09 K=0. 0.4-0.4+j0.09-0.4-j0.09 -.6 0.7-0.38+j0.4-0.38-j0.4 -.5-0.34+j0.56-0.34-j0.56 -.3-0.4+j0.86-0.4-j0.86 -.5 4-0.0+j.9-0.0-j.9 -.80 K=0.4 K= 7 0.04+j.50 0.04-j.50-3.09 0 0.5+j.73 0.5-j.73-3.3 K=0 K=7 3 pole, thu 3 loci. 4

So why Root Locu? How do the root of K 0 change a K varie? ( )( ) We will be able to eaily ee thi if we have a Root Locu for K G( ) H ( ) ( )( ) K=7 K= K=0. K=0.4 K= K=0 K=7 5

Plotting the Root Loci The root loci are plotted either Manually, or Uing a computer program uch a OCTAVE (eay if you have the program and know how to ue it.) 6

Manual plotting Root Locu Concept The Characteritic Equation i firt written in the form G ( ) H ( ) KF ( ) where K>0 i the parameter to be varied from 0 to infinity. The root of the characteritic ti equation are the value of which atify the equation KF( ) n80 for 0 n, 3, 5, Or when F( ) n80 n, 3, 5, For any value of, the value of K can be found from KF( ) giving K F( ) 7

Manual plotting Root Locu Concept Determining the phae angle for F() Conider Then where F G( ) H ( ) KF( ) ( p P e j )( P e ( 3 4 n K ( z p KZ j ) 80 z ( ) p ( ) p ( ) 3 p ( ) 3 )( e 3 j P e ) j (+z ) p )( p 3 3 P e 4 j 4 4 ) n, 3, 5, P P P P e (+p ) -p 3 KZ 4 Im e j j( ) 3 4 4 p ( 4 ) Note that n360 n,,3, -z 0 Re 8

Manual plotting Procedure and Guideline ) Locate the pole and zero of the open-loop tranfer, G()H(), function on the plane. ) There are a many loci, or branche, a pole of the G()H(). 3) Each branch tart from a pole of G()H() and end in a zero. If there are no zero in the finite region, then the zero are at infinity. N( ) Reaon: G( ) H ( ) K 0 D( ) KN( ) 0 D( ) D ( ) 0 When K=0, and root are root of D(). When K, N ( ) 0 and root are root of N(). 9

Manual plotting Procedure and Guideline 4) The loci exit on the real axi only to the left of an odd number of pole and/or zero. Complex pole and/or zero have no effect becaue, for a point on the real axi, the angle involved are equal and oppoite. Conider a tet point,, on the real axi a hown. Every pair of complex conjugate pole (or zero) will contribute a pair of angle, and uch that 360 They can thu be ignored. -z -3 -p - -p - 3 -p 3 Im j j 0 -j -j Re 0

Manual plotting Procedure and Guideline 4) The loci exit on the real axi only to the left of an odd number of pole and/or zero. Complex pole and/or zero have no effect becaue, for a point on the real axi, the angle involved are equal and oppoite. Conider a tet point,, on the real axi a hown. Each real pole or zero to the left of point doe not contribute to the angle um and thu can be ignored. Each pole/zero to the right of point contribute an angle of 80. An odd number of them will thu contribute a total of n80 where n, 3, 5, -z -3 -p - -p - 3 -p 3 Im j j 0 -j -j Re

Locu exit to left of odd No. of zero/pole Example: Conider the characteritic equation GH K ( ( 3) j)( j) K ( ( z) p )( p ) KF( ) One zero at =-z=-3; One pole at =0; -p Im j Complex conjugate pole at =-p=--j, and =-p=-+j -z j p 0-3 - - 0 Re -j -p -j

Locu exit to left of odd No. of zero/pole Example: Conider the characteritic equation GH K ( ( 3) j)( j) K ( ( z) p )( p ) KF( ) Conider a tet point, S, on the real axi to the left of ONE pole at =0. Thi tet point will be part of the root locu if F ( ) n 80 Or 80 3 n Note that, and 3 0 80 0 Point S form part of the root locu. -p p +p -z Im j j j p 0-3 - - 0 Re +z +p -j 3 -j -p -p 3

Locu exit to left of odd No. of zero/pole Example: Conider the characteritic equation GH K ( ( 3) j)( j) K ( ( z) p )( p ) KF( ) Conider a tet point, S, on the real axi to the left of the zero at =-3. 3 0 80 Note that, and Thu F () 0 -p -p +p -z Im Im j j j j p 0-3 -3 - - - - 00 Re Re Point S i not part of of the root locu. Note S i to the left of an even No. of zero/pole +z +p -p -p 3 -j -j -j -j 4

Root Locu Analyi Example Conider the ytem with G( ) H ( ) K ( )( ) GH 0 ( )( ) K 0 Or K ( )( ) Conider =-0.5: K ( 0.5)( 0.5 )( 0.5 0.5(0.5)(.5) 5)( 5) 0. 375 ) Conider =-.5: K (.5)(.5 )(.5.5( 0.5)(0.5) 0.375 ) But K need to be poitive. 5

Root Locu Analyi Example Conider the ytem with GH 0 ( )( ) K 0 G( ) H ( ) K ( )( ) Or K ( )( ) Similarly, any point which doe not atify the angle condition F ( ) n80 reult in K being will complex and not a pure real value. 6

Manual plotting Procedure and Guideline 5) Becaue complex root mut occur in conjugate pair, i.e. ymmetrical about the real axi, the root-locu plot i ymmetrical about the real axi. 6) Loci which terminate at infinity approach aymptote in doing o. For a tet point S at infinity, each pole/zero contribute an equal phae angle. Thu ( Z P) n80 for n, 3, 5, -p -z -p Im 0 Re Z/P=No. of zero/pole Aymptote angle n80 Z P -p 3 7

Manual plotting Procedure and Guideline 7) All the aymptote tart from a point on the real axi with coordinate a n pi P Z Example: For m z i Im j j G ( ) H ( ) - - 0 ( )( ) Then n80 n80 Z P 0 3 60, 80, 300-3 Re Z -j -j a 3 0 3 8

Manual plotting Procedure and Guideline 8) Break-in and breakaway point (on real axi) At a break-in point, value of K increae a the loci move onto the real axi and away from the break-in point. At a breakaway point, value of K increae along the real axi from both ide and reach maximum at the breakaway point. Along the real axi,. Thu, at the break-in or breakaway point, d K d provided K>0 and exit on the root loci. 0 Break-in pt. Breakaway pt. 9

Manual plotting Procedure and Guideline 9) Imaginary Axi Croing Two approache: a) Ue Routh Criteria to determine the value of K at which the ytem i critically table. Thi i indicated by a value of zero in the firt column but with no ign change in the firt column of the Routh Array. b) Since the root are on the imaginary axi, by letting j in the characteritic equation and olve for and K. Thi i done by equating both the real and imaginary part of the characteritic equation to zero. Im Axi Croing 0

Manual plotting Procedure and Guideline 0) Angle of Departure from complex pole and Angle of Arrival from complex zero. Angle of Departure

Manual plotting Procedure and Guideline Angle of Arrival

Manual plotting Procedure and Guideline Thee angle are determined by taking a tet point very cloe to the complex pole, or zero, and applying the angular criteria. -p Im For diagram on right, ( ) n80 n, 3, -z 0 Re Or 80 n -p 3

Manual plotting Procedure and Guideline Suppoe 90, 60 Angle of Departure Then uing 80 n 60 90 n80 50 or 0 4

End 5

Example Root Locu Plot Example: Conider the characteritic equation 3 K 0 Rewriting a KF ( ) 0 K K ( ) ( j )( j ) 0 3 pole ; No zero at p 0; p ( j ); p3 ( j ) 6

Example Root Locu Plot 3 pole at p p p 3 0; ( ( j j ); ) 7

Example Root Locu Plot 4. The loci exit on the real axi only to the left of an odd number of pole and/or zero. 8

Example Root Locu Plot 4. The loci exit on the real axi only to the left of an odd number of pole and/or zero. 3. Each branch tart from an open-loop ppole and end at an open-loop zero. 9

Example Root Locu Plot 6. 3 pole. No zero. Therefore 3 aymptote. n80 Z P n80 0 3 60, 80, 300 7. Aymptote t intercept t on real axi. a n m pi P Z ( ) 0 3 0 3 z i 30

Example Root Locu Plot 8. No break-in or breakaway point. 9. Imaginary axi croing. K 3 3 0 3 K 0 ( j) 3 ( j) ( j) K 0 3 ( j j ) ( K ) 0 3 0 K 0 K 4 3

Example Root Locu Plot 0. Angle of departure 80 3 n 35 90 n 80 90 35 n80 45 for n 3

Example Root Locu Plot 0. Angle of departure 80 3 n 35 90 n 80 90 35 n80 45 for n 33

Example Root Locu Plot OCTAVE Program >> gh=tf([],[ 0]) Tranfer function: ----------------- ^3 + ^ + >> rlocu(gh) Pole oeat 0, j 34

Root locu ketching Pole at 0, 35

Example root locu plot Pole at 0, Zero at 3, 4 36

Example Root Locu Plot OCTAVE Program >> h=zp([-3-4],[0 -],[]) Zero/pole/gain: (+3) (+4) ----------- (+) >> rlocu(h) >> Pole oeat =0, - Zero at =-3, -4 37

Example root locu plot 38

Example root locu plot Pole at 0,, j Zero at 39

Example Root Locu Plot OCTAVE Program >> gh=zp(-,[0 - -+i --i],) ) >> rlocu(gh) Note: pecifying i a pole or zero a a+bi mean locating that pole or zero at a+jb. Pole at 0,, j Zero at 40

Example root locu plot Pole at 0, 4, j 4

Example Root Locu Plot OCTAVE Program >> gh=zp([],[0-4 -+i --i],) ) >> rlocu(gh) Pole at 0, 4, j 4

End 43