The Root Locu Method MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity
Outline The root locu method wa introduced by Evan in the 1950. It remain a popular tool for imple SISO control deign. What i a root locu? Pole & Tranient Repone (Why do we care about pole?) The Root Locu Method Problem Definition The Two Key Formula Root Locu Rule Example: Flexible Spacecraft Robotic Arm General Aviation Aircraft Helicopter Pitch Control
Deigning a Feedback Control Sytem Firt we chooe a compenator There are many ueful compenator type.we have already een Proportional and Proportional plu Integral. Thi give u a tructure, i.e., a compenator tranfer function. The tranfer function will have one or more free parameter. The root locu method i one approach to elect the free deign parameter o a to achieve deired tranient behavior.
What i a Root Locu? On the right i a negative feedback loop We wih to examine the cloed loop pole a the gain K varie A K increae from zero the 4 pole move from the open loop value & trace 4 loci At any particular value of K there are 4 cloed loop pole In thi example there i a critical value of K at which the ytem become untable. y ye - K = K = 0 G () H( ) ( + 3) ( + 1)( + )( + 4) ( + 1)( + )( + 4) G H = K G 1 = = 4 3 1+ GH + 7 + 14 + 8 + K + 3K Im at K critical y Re tability i lot
Uing Matlab >> =tf(''); >> G=(+3)/(*(+1)*(+)*(+4)); >> rlocu(g) >> grid >> [K,Pole]=rlocfind(G) Select a point in the graphic window elected_point = -0.070 + 1.4161i K = 7.379 Pole = -4.940 + 0.0000i -.5617 + 0.0000i -0.07 + 1.416i -0.07-1.416i -1-1 ) Imaginary Axi (econd Root Locu 10 0.68 0.56 0.4 0.8 0.14 0.8 8 6 0.91 4 0.975 1 10 8 6 4 0-0.975-4 0.91-6 -8 0.8 0.68 0.56 0.4 0.8 0.14-10 -10-8 -6-4 - 0 4 Real Axi (econd -1 )
Example ( + 1) 1, 1 CL 1 ( + 1) 1 + K + K G G = K H = G = + G 1+ G = 1+ K = 1 0 4 + K + K = = K ± K K Im K = K = K = 4 K = 0 Re
Tranient Repone Conider a ytem decribed by the open loop tranfer function n G K c pfd 1 = 0 + + + d + λ1 There are three way to ae ytem tranient behavior: 1. time domain (output time trajectorie). pole (or eigenvalue) location 3. frequency repone (Bode or Nyquit plot) Here we conider pole location. The pole are the root of c The root locu method i concerned with adjuting the cloed loop pole poition ( ρω 1 1 ω1 ) ( ρpωp ωp)( λ1) ( λq) d = + + + + + + = 0 Complex root occur in complex conjugate pair. In thi cae there are p+q pole
Ideal Pole Location Im α degree of tability, decay rate 1/α ideal region for cloed loop pole θ damping ratio θ = in 1 ρ Re Our goal i to deign a compenator o that the cloed loop pole lie in the haded region. We get to chooe the form of the compenator and elect it parameter.
Problem Definition The cloed loop input repone tranfer function i G yy ( ) G = 1 + GH The error repone tranfer function i (recall e= y y) 1+ GH G Ge( ) = 1 Gy( ) = 1+ GH The pole of the cloed loop ytem are the root (zero) of y - G () H( ) y 1+ GH( ) = 0
Problem Definition, Cont d Suppoe ( m ) n z z + b + + b G( ) H( ) = K = K = K d p p + a + + a m m 1 1 m 1 0 n n 1 1 n n 1 0 n, d are completely known, but K i a parameter that we can adjut. Root Locu Problem: Generate a ketch in the complex plane of with varying gain K. the cloed loop pole
Solution Strategie We will do thi two way: The eay way: Have MATLAB olve for the root for each of a pecified lit of value for K and plot them. The hard (old) way: Generate a ketch by hand. Why do it the hard way at all? We need to know how to interpret the plot. We obtain inight concerning the choice of compenator. We learn how to et the compenator parameter other than the gain K.
Root Locu Method ~ 1 Magnitude equation d() + Kn() 1+ G ( ) H ( ) = 0 = 0 d() + Kn() = 0 d () n () j(k+ 1) π 1+ G( ) H( ) = 0 K = 1 = e, k = 0, ± 1, ±, d () Thi mean n () n () K = 1 and K = (k+ 1) π d () d () for K 0 : K n () n () = 1 and = (k+ 1) π d () d () Angle equation
Root Locu Method ~ Our goal i to find value of that atify both of thee equation. Note that for any given, the magnitude equation i atified for ome value of K, i.e., K = d n Note that the angle equation doe not depend on K at all. Strategy: Firt, find value of that atify the angle equation. Second, calibrate the plot uing the magnitude equatio n.
Root Locu ~ 3 Uing the Angle Formula G H = ( + 3)( + 4) ( + 1)( + ) tet point = + j3 Im n () = (k + 1) π d () n d = (k+ 1) π θ 4 θ 3 θ θ 1 4 3 1 ( k ) 3 4 1 3 4 1 Re θ + θ θ + θ = 70.55 + 1 π for any integer = + j3 i not a point on the root locu The point = + i / i θ + θ θ + θ = 180 k
Baic Rule ~ 1 1. Number of branche: The number of branche of the root locu equal the number of open loop pole. + i the order of theorder of the polynomial d Kn d. Symmetry: The root locu i ymmetric about the real axi. Pole occur in complex conjugate pair. 3. Starting & ending point: The root locu begin at the open loop pole and end at the finite and infinite open loop d + Kn = 0 d =0 a K 0 1 K 0 zero. d + n = n =0 a K if i bounded
Example: ( + 3) ( + 1)( + )( + 4) G H = K Im Re
Baic Rule ~ 4. Real-axi egment: For K > 0, real axi egment to the left of an odd number of finite real axi pole and/or zero are part of the root locu. Im θ θ Re Tet point 1
Baic Rule ~ 3 5. Behavior at infinity: The root locu approache infinity along aymptote with angle: (k + 1) π θ =, k = 0, ± 1, ±, ± 3, # finite pole # finite zero Furthermore, thee aymptote interect the real axi at a common point given by σ = finite pole finite zero # finite pole # finite zero
Baic Rule ~ 4 Angle part i eay: n m = i= 1 i i= 1 d n ( z ) ( p ) jθ Take = ρe. For ρ, λ = θ Then m n z p mθ nθ i= 1 i i= 1 n So = ( k+ 1) π ( m n) θ = ( k+ 1) π d i i i
Baic Rule ~ 4 6. Real axi breakaway and break-in point: The root locu break away from the real axi where the gain i a (local) maximum on the real axi, and break into the real axi where it i a local minimum. To locate candidate break point d imply plot K( ) = on the axi egment n d d or olve = 0 d n 7. jω-axi croing: Ue Routh tet to determine value of K for which loci cro imaginary axi.
Routh Stability Tet It i deired to determine the number of right hand plane root of a polynomial, ay: 4 3 3 1 0 4 1 a a0 3 a3 a1 0 1 0 + a + a + a+ a = 0 4 1 a a0 3 a3 a1 0 b1 b b3 1 c1 c 0 d1 b 3 1 3 1 a3 a3 c 1 1 a 1 a0 a a a 0 =, b =,... = a b a 3 1 b 1 3,... The number of right half plane pole i equal to the number of ign change in the firt column. a
Uing MATLAB The baic MATLAB function are: rlocu rlocu(y) calculate and plot the root locu of the open-loop SISO model y. rlocfind [K,POLES] = RLOCFIND(SYS) i ued for interactive gain election from the root locu plot of the SISO ytem SYS generated by RLOCUS. RLOCFIND put up a crohair curor in the graphic window which i ued to elect a pole location on an exiting root locu. The root locu gain aociated with thi point i returned in K and all the ytem pole for thi gain are returned in POLES. iotool When invoked without input argument, iotool open a SISO Deign GUI for interactive compenator deign. Thi GUI allow you to deign a ingle-input/ingle-output (SISO) compenator uing root locu and Bode diagram technique.
Flexible Spacecraft
Flexible Spacecraft Θ K + + 1 1 G H Θ = = K ( + 4)( + + ) ( + + 1) ( + 4)( + 1± j) ( + 0.5 ± 0.866) K j 1 4 With One Flex Mode Rigid = Θ + 4 G H K K + 4 Θ
Spacecraft = + + + + + + 4 3 dcl 1 K 1 6K 10K 8K 1 1+ 6 8 4 K K 1+ 3 K K 10 0 Rigid Im Re 1 0 1 3K + 6K 1+ K ( + ) K 1 3K 6K 8K 1 3K + 6K + K K + K 1 3K + 6K 1 0, 3 6 0, 0 8K 0 0 Alway poitive (prove it) One Flex Mode Im Re untable egment
Example Robotic Arm Compenator Amplifier Motor/ Speed Controller Θ K + 0.1 4 5 1 Θ + 4 + 11+ 30.5 + 0.1 + 0.1 G = 0K = 0K ( + 4)( + 11 + 30.5) ( + 4)( + 5.5)
Robotic Arm Im Im Re Re? Im Aymptote @ 45, 135, 5, 315 Centroid @ -3.75 Re
Robotic Arm 3 Plot K d =, 4, 0.1 n Max, breakaway point at -1.117 K 5 4 3 Min, breakin point at -.1613-4 -3 - -1 1
Robotic Arm 4 >> =tf(''); >> G=0*(+0.1)/(^*(+4)*(+5.5)^); >> rlocu(g) >> rlocfind(g) Select a point in the graphic window elected_point = -0.161 + 0.0104i an =.111 elected_point = -0.0415 +.7640i an = 4.9776
Robotic Arm 5 G cl = ( + ) 0K 0.1 5 4 3 15 74.5 11 0K K + + + + + Routh Table 5 4 3 K K 1 0 1 74 0K 15 11 K 65.9 19.86K 11 4.5 71K 89.76K 11 4.5K K Governing Inequality 11 4.5 0, 71 89.76 0, 0 K > K K > K > 11 K < = 6.769 4.5 71 K < = 5.308 89.76
Example: Piper Dakota Pitch Control e, error θ - G c δ e G ( p ) θ elevator angle, δ ( ) Plant: G = 160 p Lead compenator: G e pitch angle, θ ( +.5)( + 0.7) ( + 5+ 40)( + 0.03+ 0.06) c ( ) + 3 = K + 0 Franklin & Powell 3 rd Ed
Piper Dakota Im What do we expect? Im Re Re
Piper Dakota >> =tf(''); >> Gp=160*(+.5)*(+0.7)/((^+5*+40)*(^+0.03*+0.06)); >> Gc=(+3)/(+0); >> G=Gc*Gp; >> rlocu(g) >> grid >> [K,Pole]=rlocfind(G) Select a point in the graphic window elected_point = -0.703 + 0.4803i K = 1.5395 Pole = -7.836 +1.968i -7.836-1.968i -7.9513-0.7067 + 0.4888i -0.7067-0.4888i >> Gcl=1.5*G/(1+1.5*G); >> tep(gcl)
Piper Dakota
Piper Dakota Open loop repone to elevator Cloed loop repone with: K = 1.5
Example: Helicopter Pitch Control Notice untable dynamic pilot diturbance helicopter - K1 + 1+ 1 - ( + ) 5 0.03 ( + 0.4)( 0.36+ 0.16) tabilizer + 1 K Pick the inner loop pitch control + 9 gain K o that the dominant inner loop pole have a damping ratio of 0.707. Select an outer loop gain (tick enitivity) to place the pole. Determine ultimate error in repone to unit tep diturbance.
Inner (Stabilization) Loop 1+ GH = 1+ K ( ) ( + 0.4)( 0.36+ 0.16) ( ) ( + 9) 5 + 0.03 + 1 The inner loop root locu i hown on the right. Chooe a gain K = 1.5. The inner loop reolve to: G ( ) Gp = 1 + 1.5GG = c p 5( + 0.03)( + 9) ( + 3.81± j3.53)( + 1.37)( + 0.0458)
Outer Loop Deign pilot Inner loop - K1 + 1+ 1 5( + 0.03)( + 9) ( + 0.3.81± j3.53)( + 1.37)( + 0.0458)
Outer Loop On the bai of the root locu on the right, chooe a gain of 1. The cloed loop pole are: 11.91 3.9 ± j3.50 0.633 ± j0.646 0.0314
Diturbance Repone Error G de G = 1 + G = pilot G 5( + 0.03)( + 9) ( + 11.91)( + 3.9 ± j3.50)( + 0.633 ± j0.646)( + 0.0314) 1 lim Gde = 0.7987 0
Summary How pole characterize tranient repone Oberving the influence of gain on cloed loop pole uing root locu plot Sketching the root locu: The magnitude and gain formula Baic rule of root locu ketching Uing MATLAB Example