Control Systems. Root Locus & Pole Assignment. L. Lanari


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1 Control Systems Root Locus & Pole Assignment L. Lanari
2 Outline rootlocus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS  Root Locus 2
3 Root locus r +  e k F (s) y with gain/zero/pole representation of the loop function L(s) =kf(s) =k N(s) (s D(s) = k zi ) (s pi ) since T (s) = y(s) r(s) = L(s) 1+L(s) = kf(s) 1+kF(s) openloop monic polynomials the poles of the closedloop system are the roots of the closedloop characteristic polynomial 1+k N(s) D(s) = 0 root locus equation 1 + k F (s) = 0 which can be rewritten as D(s)+kN(s) = 0 p (s,k) = 0 F (s) = 1 k if the real gain k varies from  to + the closedloop poles describe a graph which is defined as the rootlocus Lanari: CS  Root Locus 3
4 we have to learn to visualize how all the poles move simultaneously in the complex plane as k increases see the animation (not available in the PDF file) for the positive root locus of 1+kF(s) =1+k 1 s +1 a point s? belongs to the root locus p (s,k) = 0 (or equivalently s? is a pole of the closedloop system for some value of k = k? ) if and only if there exists a real value k? such that p (s?,k? ) = 0 Lanari: CS  Root Locus 4
5 Let us define L(s) =k m (s z i ) i=1 n (s p j ) m zeroes n poles with m < n n (s p j )+k j=1 j=1 m (s z i )=0 i=1 n (s p j )= k j=1 m (s z i ) complex numbers i=1 k = n s p j j=1 m s z i i=1 magnitude condition n (s p j ) j=1 m (s z i )=π + k +2hπ i=1 h Z n (s p j ) j=1 m (s z i )= i=1 (2h + 1)π for k 0 2hπ for k 0 phase condition Lanari: CS  Root Locus 5
6 phase condition from 1+k N(s) D(s) =0 positive locus (k positive) negative locus (k negative) N(s) k = ( 1) = (2h + 1)π D(s) N(s) k = (1) = 2hπ D(s) the phase condition is used to draw the locus of the roots: we do not need to solve for the highorder polynomial roots, we just need to verify if a given point of the complex plane satisfies the phase condition and therefore corresponds to a root of 1 + k F (s) = 0 for some real value of k (this value is found by using the magnitude condition) there are however some guidelines for drawing rapidly, by hand, a sketch of the root locus Since the closedloop system has the same number n of poles as the openloop, each of n the closedloop poles will move along a branch of the root locus as k varies from 0 to + (similarly for the negative locus as k varies from  to 0) the positive root locus has n branches (same for negative locus) The coefficients of the root locus equation are real the root locus is symmetric w.r.t. the real axis Lanari: CS  Root Locus 6
7 Rule 1 The n branches of the positive locus start at the openloop poles and, if k tends to +, m of the branches tend to the openloop zeros while (n  m) tend to infinity along (n  m) asymptotes positive locus k = + Im zero pole pole k = + k = 0 k = 0 Re F (s) = s +3 s(s + 2) Rule 1bis Of the n branches of the negative locus, as k varies from  to 0, m start from the m openloop zeros while the other (n  m) come from infinity along (n  m) asymptotes negative locus Im zero pole pole k =  k = 0 k = 0 Re k =  Lanari: CS  Root Locus 7
8 Rule 2 Every point on the real axis belongs either to the positive or the negative locus. A point on the real axis that leaves on its right an odd number of poles and zeros counted with their multiplicity belongs to the positive locus. All the other points belong to the negative locus. negative locus Im positive locus Re # poles/zeros on the right Lanari: CS  Root Locus 8
9 Rule 3 The (n  m) asymptotes are centered in s 0 = n m p j j=1 i=1 n m z i and form angles of (2h + 1)π n m 2hπ n m for positive locus for negative locus h = 1, 2,..., n  m example n  m = 3 h = 1 h = 3 Pos Neg h = 1 s0 h = 3 h = 1 h = 2 3¼/3 5¼/3 2¼/3 4¼/3 h = 3 7¼/3 6¼/3 h = 2 h = 2 Lanari: CS  Root Locus 9
10 we have the following cases positive locus s0 s0 s0 s0 n  m = 1 n  m = 2 n  m = 3 n  m = 4 negative locus s0 s0 s0 s0 N.B. the asymptotes are not necessarily part of the root locus Lanari: CS  Root Locus 10
11 the points of the locus are either regular points (solutions of p (s,k) = 0) or singular points (or breakaway/breakin points) solutions of being k = p(s, k) =0 p(s, k) s =0 n (s p j ) j=1 m (s z i ) i=1 s n (s p j )+k j=1 m (s z i )=0 i=1 n (s p j )+k s j=1 m (s z i )=0 i=1 from locus equation, substituted in the second gives m (s z i ) s n (s p j ) n (s p j ) s m (s z i )=0 i=1 j=1 j=1 i=1 equation of order n + m  1 for the candidates singular points (we may have solutions corresponding to complex k which are therefore not points of the root locus) Lanari: CS  Root Locus 11
12 a candidate singular s* point is a true singular point if the corresponding value k* is real n (s p j ) k = j=1 m (s z i ) the singular point s* will be a solution of the locus equation p (s*,k*) = 0 with multiplicity µ greater equal to 2 i=1 p(s, k )=(s s ) µ p (s, k ) µ 2 We also have some easy to find singular points: every openloop pole/zero with multiplicity greater than 1 is a singular point of the root locus Proof: from the candidate singular points equation Lanari: CS  Root Locus 12
13 Rule 4 Let s* be a singular point with multiplicity µ, then 2 µ branches merge in the singular point and are alternatively convergent and divergent. These branches divide the plane in equal parts. If the singular point is a multiple pole or zero of the openloop system, then the branches also alternate as positive and negative branches. example F (s) = s +3 s(s + 2) +  p(s, k) =s(s + 2) + k(s + 3) = s 2 +(2+k)s +3k p(s, k) =2s +(2+k) s r e k F (s) y candidates (s + 3) s s(s + 2) s(s + 2) s (s + 3) = s2 +6s +6 s1* = s2* = real values real values of k* singular points Lanari: CS  Root Locus 13
14 was it necessary to compute the singular points? after the real axis rule we have one pole needs to go to infinity along the asymptote one pole goes to the zero two poles get out of the openloop poles Im zero pole pole there must be a singular point Re Im µ = 2 zero pole µ = 2 pole Re at these singular points of multiplicity µ = 2 the entire round angle is divided into 2 x 2 = 4 equal angles of 90 each Lanari: CS  Root Locus 14
15 an alternative formula to determine the candidates breakaway/breakin (singular) points n j=1 1 s p j m i=1 1 s z i =0 this formula will not give us the singular points corresponding to repeated poles or zeros of the openloop (obvious singular points) repeated poles and zeros in this formula need to be taken into account in the sum with their multiplicity example F (s) = (s + 1)2 (s + 2) 4 3 Root Locus 4 s +2 2 s +1 = 2s (s + 2)(s + 1) Imaginary Axis s* = Real Axis Lanari: CS  Root Locus 15
16 In order to determine stability (all the poles should be, for the same interval of values of k, in the open left half plane) it may be important to determine for which values of k some branches cross the imaginary axis. This can be achieved by determining for which values of k the elements of the first column of the Routh table become 0 since this is when a first column term changes sign and therefore a pole crosses the imaginary axis (remember that, when the table can be built from the basic definition, the number of sign changes in the first column is equal to the number of roots with positive real part). F (s) = s +1 s(s 2)(s + 4) p(s, k) =s(s 2)(s + 4) + k(s + 1) = s 3 +2s 2 +(k 8)s + k characteristic equation of the closedloop system for k = 16 p(s, 16) = (s + p)(s 2 + ω 2 ) p = 2! 2 = 8 Imaginary Axis Root Locus Routh table 1 k 8 2 k k 16 k Real Axis Lanari: CS  Root Locus 16
17 RL as a design tool the basic idea is based on the positive root locus behavior for high values of the gain k (n  m) branches tend at infinity along (n  m) asymptotes the remaining m branches tend to the m openloop zeros therefore if the zeros are in the open left halfplane (i.e. have negative real part) and the asymptotes (for the positive root locus) always stay to the left of the imaginary axis then for sufficiently high values of the gain all the poles will be to the left of the imaginary axis that is a high gain will stabilize the system the asymptotes and the openloop zeros attract the positive branches A system with all its zeros, if any, in the open left halfplane is said to be minimum phase equivalently A system having no zeros with positive or null real part is said to be minimum phase Lanari: CS  Root Locus 17
18 we can have either the n  m = 1 asymptote (positive locus) s0 center of asymptotes is not important since we look at the highgain behavior if the openloop system is minimum phase and has (relative degree) n  m = 1 then for sufficiently high values of k the closedloop system is asymptotically stable or the n  m = 2 asymptote (positive locus) s0 center of asymptotes has to be negative if the openloop system is minimum phase and has (relative degree) n  m = 2 with a negative center of asymptotes (positive locus) then for sufficiently high values of k the closedloop system is asymptotically stable Lanari: CS  Root Locus 18
19 example F (s) = (s + 1)2 (s 1) 3 nf + = 3 3 poles with positive real part Root Locus Nyquist Diagram Imaginary Axis Imaginary Axis Real Axis Real Axis positive root locus Nyquist plot Ncc = nf + = 3 verified Lanari: CS  Root Locus 19
20 example F (s) = (s + 1) 2 (s 1) 2 (s + 10) 2 Nyquist plot of 100 F (s) (to make the plot more visible) Root Locus Nyquist Diagram Imaginary Axis Imaginary Axis Real Axis Real Axis Ncc = nf + = 2 Lanari: CS  Root Locus 20
21 what if n  m = 2 but the center of asymptotes is nonnegative? Let us assume we have F (s) with n  m = 2 and center of asymptotes s0 (nonnegative) and see the effect of adding a zero/pole pair (za, pa) both negative s z F (s) a the new center of asymptotes is given by (with n  m = 2) s p a za < 0 pa < 0 s 0 = new center of asymptotes n p j + p a j=1 n m m z i z a i=1 = s 0 + p a z a n m old center of asymptotes center of asymptotes variation since n  m remains 2, we can choose the additional negative pole pa and zero za such that the new center of asymptotes becomes negative once we have made the center of asymptotes negative with the addition of a pole and a zero, everything we said before applies NB for the variation to be negative the pole needs to be to the left of the zero Lanari: CS  Root Locus 21
22 note that s z a s p a with pa < za < 0 can be rewritten as s z a = z a s s p a p a s = z a (1 s/z a ) p a (1 s/p a ) 0 < z a p a < 1 and therefore, being the cutoff frequency of the zero smaller than the one of the pole, the particular pole/zero pair is equivalent to a positive gain smaller than 1 x a lead compensator but the final controller will require also the choice of a sufficiently high gain k* > kcrit so the controller will be C (s) = Gain x Lead compensator Lanari: CS  Root Locus 22
23 what if n  m > 2 and the system is minimum phase? The idea is: turn the n  m > 2 case into a n  m = 2 one by adding zeros (s  zk) to the controller solve the stabilization problem if necessary introduce high frequency dynamics (1 + hs) to make the controller at least proper. The following result guarantees that if properly chosen these dynamics do not alter the closedloop stability Th. Consider an openloop system F (s) which results in an asymptotically stable unit feedback closedloop system. Then there exists a sufficiently small h > 0 such that the closedloop system having as openloop remains asymptotically stable. 1 F (s) 1+τ h s Lanari: CS  Root Locus 23
24 if the closedloop system +  F (s) then the closedloop system τ h s F (s) is asymptotically stable note that 1/(1 + hs) does not change the gain of the openloop represents some highfrequency dynamics is also asymptotically stable provided h > 0 is sufficiently small high frequency modification! = 0  Im informal proof 1 1+τ h s alters the Bode plots only at high frequency alters the Nyquist plot only at high frequency 1! = +! =  Re no effect on closedloop stability! = 0 + Lanari: CS  Root Locus 24
25 Case n  m = 3 & minimum phase system: possible algorithm 1) add a negative zero (s  zk) so to obtain n  m = 2. a) if s0 < 0 then choose k* > kcrit to guarantee asymptotic stability of the closedloop system. Go to point 2) b) if s0 > 0 then choose a pole/zero pair pa < za < 0 to make the new center of the asymptotes s0 negative. Go to case a) 2) if the resulting controller is improper, add a high frequency dynamics term (1 + hs) where the time constant h can be chosen applying the Routh criterion in order to guarantee that the overall closedloop system is asymptotically stable Note that step 2) may not be necessary since we may apply this result once the static specifications have been met i.e. we may just need to stabilize the extended plant ˆP (s) and therefore the controller may already have an excess of poles w.r.t. the number of zeros Lanari: CS  Root Locus 25
26 other use of the RL MassSpringDamper: we want to explore the influence of some parameters on the system s dynamics represented by the following characteristic polynomial ms 2 + µs + k spring stiffness k [0, + ) p(s, k) =(ms + µ)s + k = D(s)+kN(s) Im k = 0 no spring, poles in (0, µ/m) singular point  µ/m 0 Re p(s, k) s =2ms + µ =0 s* = µ/2m at k = (ms + µ)s = µ2 4m i.e. µ =2 k m as k increases (spring more and more stiff) poles are complex with constant real part while imaginary part becomes larger k2 > k1 k normalized plots (the steadystate value changes with k) Lanari: CS  Root Locus 26
27 damping µ [0, + ) p(s, µ) =ms 2 + k + µs = D(s)+µN(s) the MSD poles move, as µ increases, as the positive root locus of µ = 0 no damper, pure imaginary poles µ N(s) D(s) = µ s ms 2 + k same singular point as before Root locus Bode note how, as µ increases and the poles become real, how a dominant (slow) dynamics arises Imaginary axis magnitude (db) Real axis frequency (rad/sec) Step fucntion amplitude time Lanari: CS  Root Locus 27
28 Pole assignment r (t) y (t) +  C (s) P (s) plant P (s) = N P (s) D P (s) = b ms m + b m 1 s m b 1 s + b 0 s n + a n 1 s n a 1 s + a 0 strictly proper monic (coefficient = 1) controller C(s) = N C(s) D C (s) = d rs r + d r 1 s r d 1 s + d 0 s r + c r 1 s r c 1 s + c 0 proper monic {d r, d r 1,...,d 1, d 0, c r 1,...,c 1, c 0 } unknowns Lanari: CS  Root Locus 28
29 closedloop characteristic polynomial d CL (s) =D P (s)d C (s)+n P (s)n C (s) degree n r m r also monic degree n + r we want to assign all the closedloop poles to be p 1, p 2,...,p n+r desired closedloop poles which can be seen as the solutions of an n + r order polynomial d CL(s) =(s p 1)(s p 2)...(s p n+r) The problem can be stated as: we need to determine the 2r  1 unknowns ci and dj (i = 1,..., r  1; j = 1,..., r) such that D P (s)d C (s) + N P (s)n C (s) = d CL(s) Diophantine equation Lanari: CS  Root Locus 29
30 Result if NP (s) and DP (s) are coprime and r = n  1 we can always solve and have a unique solution Remarks we can arbitrarily assign the 2n  1 closedloop poles closedloop system zeros: the zeros depend upon where we consider the input entering in the control system and which output we decide to monitor. For example in a unit feedback system, the output disturbance to controlled output transfer function S (s) and the reference to controlled output one T (s) will have the same poles (in there are no hidden dynamics in the open loop system) but different zeros the closedloop zeros coincide with the openloop ones when the openloop system has no hidden dynamics T (s) = N T (s) D T (s) = F (s) 1+F (s) = N F (s) D F (s)+n F (s) N F (s) =N C (s)n P (s) Lanari: CS  Root Locus 30
31 since the choice of the controller C (s) is uniquely determined by the previous algorithm, the controller zeros are a consequence of the pole assignment technique and so are the closedloop zeros note that if we choose as desired closedloop poles some openloop zeros then at closedloop we have a cancellation. Moreover a closedloop cancellation (for finite values of the open loop gain) can be originated only by an openloop cancellation, which being NP (s) and DP (s) coprime, is generated by the series interconnection plant/controller. Therefore if we choose as desired closedloop poles some openloop zeros these will necessarily be also poles of the controller from the above comment it is clear that we are not going to choose a desired closedloop pole coincident with a nonminimum phase zero of the openloop. Lanari: CS  Root Locus 31
32 example plant P (s) = (s 1) s(s 2) n = 2 C(s) = as + b s + c d CL(s) =(s + 1) 3 = s 3 +3s 2 +3s +1 desired closedloop poles: 3 in 1 d CL (s) =s(s 2)(s + c)+(s 1)(as + b) =s 3 +(c 2+a)s 2 +(b a 2c)s b equating a + c 2 = 3 a + b 2c = 3 b = 1 a = 14 b = 1 c = 9 14 ) C(s) = 14s 1 s 9 = 14(s 1 (s 9) closedloop complementary sensitivity T (s) = C(s)P (s) 1+C(s)P (s) = 14(s 1)(s 1 14 ) (s + 1) 3 Lanari: CS  Root Locus 32
33 the plant has a real positive pole so the openloop loop shaping technique cannot be used the plant has a real positive zero (so it is nonminimum phase) and therefore the highgain principle seen in the root locus cannot be applied the resulting controller, in this example, is unstable and nonminimum phase: we have no control over the final structure of the fixed dimension (r = n  1) controller root locus of F (s) = (s 1 14 ) (s 9) (s 1) s(s 2) 3 Root Locus for k = 14 we obtain the three closedloop poles at the desired location Imaginary Axis Real Axis Lanari: CS  Root Locus 33
34 note that using the basic tracing rules we can also have the following compatible positive root locus Im not compatible with the extra knowledge that the closedloop system is asymptotically stable Re closedloop system stable for some values of the gain but not compatible with the extra knowledge that the closedloop system has all three poles in 1 for some value of k Im Re Lanari: CS  Root Locus 34
35 Vocabulary English root locus singular point (breakaway/breakin) locus branch minimum phase center of asymptotes Italiano luogo delle radici punto singolare ramo del luogo a fase minima centro degli asintoti Lanari: CS  Root Locus 35
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