Controller Design by Root-Locus Method PDF - Fall 2024

Controller Design by Root-Locus Method MCE 309 - System Dynamics and Control Dr. Tolga Özaslan Mechanical Engineering Department Yıldırım Beyazıt University...

Controller Design by Root-Locus Method MCE 309 - System Dynamics and Control Dr. Tolga Özaslan Mechanical Engineering Department Yıldırım Beyazıt University Fall 2024 Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 1 / 147  Introduction Introduction to Root-Locus Ref : https://www.merriam-webster.com/dictionary/locus Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 2 / 147  Introduction Introduction to Root-Locus Transient response of a closed-loop system is determined by the loci of its closed-loop poles In case there is an adjustable/variable loop gain, e.g. K , the loci of poles change with variations in the gain We would like to know how the poles move as the gain changes before we make a design decision It is sometimes possible to move the closed-loop poles to desired locations on the s-plane by simply changing the loop gain, K or Kh , as we have seen in the servo system with velocity feedback On the other hand, moving the poles to desired locations might not be possible by simply changing a gain. In such cases we need to add a compensator to the system. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 3 / 147  Introduction Introduction to Root-Locus Finding the poles of a system for different gains might not be that easy and quick, especially if the degree of polynomials is high, > 3 We often resort to software packages that does this for us such as Matlab, Python control toolboxes Regardless, finding the loci of poles for every gain might not pay off the effort, since the roots may change abruptly with small variations in the gain W. R. Evans proposed a technique called root-locus method which is used very extensively in control system design and is yet very simple This method plots the paths the poles follow on the s-plane due to variations in the gain. The plot is usually generated for values of the gain ranging from 0 to ∞ for negative feedback control system Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 4 / 147  Introduction Introduction to Root-Locus The effect of variations in the gain on the loci of closed-loop poles can be predicted using the root-locus method Although there are many computer software tools that can generate accurate root-locus plots for us, it is desirable for an engineer to have a good understanding of this method A software program won’t tell you where you should put a new pole or zero in order to stabilize an unstable system Performance criterion requires introduction of new poles/zeros or move the existing ones to more suitable locations Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 5 / 147  Complex Numbers Complex Numbers - Recap Complex numbers live in the complex space, represented as C There are two representations of complex numbers Complex-Cartesian coordinates s = σ + jω σ, ω ∈ R Polar coordinates s = r e jθ r, θ ∈ R The exponential term expands to e jω = cos(ω) + j sin(ω) √ and r = σ 2 + ω 2 and θ = atan2(ω, σ) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 6 / 147  Complex Numbers Complex Numbers - Recap Depending on the operations we carry over complex numbers, one of the representations might be preferable Summation - subtraction : Complex-Cartesian representation wins! s1 ± s2 = (σ1 + jω1 ) ± (σ2 + jω2 ) = (σ1 ± σ2 ) + j(ω1 ± ω2 ) Multiplication - division : Polar representation wins! s1 · s2 = r1 e jθ1 · r2 e jθ2 = (r1 · r2 ) e j(θ1 +θ2 )  .  s1 /s2 = r1 e jθ1 r2 e jθ2 = (r1 /r2 ) e j(θ1 −θ2 ) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 7 / 147  Complex Numbers Complex Numbers - Recap Magnitude and phase (angle) of a complex number given as s = σ + jω = re jθ are defined as Complex-cartersian repr. Polar repr. p |s| = σ 2 + ω 2 |s| = r s = atan2(ω, σ) s=θ Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 8 / 147  Complex Numbers Complex Numbers - Recap Suppose we are given a complex function G (s) which is a fraction of two polynomials (s − z1 )(s − z2 ) · · · (s − zm ) G (s) = (s − p1 )(s − p2 ) · · · (s − pn ) The magnitude of G (s) is obtained as |s − z1 | |s − z2 | · · · |s − zm | |G (s)| = |s − p1 | |s − p2 | · · · |s − pn | and its phase is obtained as   G (s) = s − z1 + s − z2 + · · · + s − zm   − s − p1 + s − p2 + · · · + s − pm These relations become obvious when written in polar coordinates Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 9 / 147  Motivation Effect of Unknown Variable Gain Example : Consider the following unity feedback (i.e. H(s) = 1) system Given the open-loop transfer function G (s)H(s), s (s − 0) G (s)H(s) = = s3 2 + 2s + 4 (s + 2.594)(s − 0.297 ± j1.206) find the effect of variable gain K ∈ [0, ∞) (note that both num. and denom. are factored to show zero and poles) We need to answer the following questions 1 What values of K satisfy my system performance requirements? 2 How is the performance of my system affected as K varies? Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 10 / 147  Motivation Effect of Unknown Variable Gain The closed-loop transfer function is given as KG (s) Ks = 3 2 1 + KG (s)H(s) s + 2s + Ks + 4 The closed-loop poles of the system are the solutions of the characteristic equation found as s 3 + 2s 2 + Ks + 4 = 0   k = 0 : −2.594 ( 0.297, −1.206) ( 0.297, 1.206) k = 1 : −2.315 ( 0.157, −1.305) ( 0.157, 1.305) k = 2 : −2.000 (−0.000, −1.414) ( 0.000, 1.414) k = 4 : −1.296 (−0.352, −1.721) (−0.352, 1.721) k = 8 :  −0.556 (−0.722, −2.584) (−0.722, 2.584) k = 16 : −0.257 (−0.871, −3.846) (−0.871, 3.846) k = 32 : −0.126 (−0.937, −5.558) (−0.937, 5.558) k = 64 : −0.063 (−0.969, −7.933) (−0.969, 7.933) k = ∞ : −0.000 (−1.000, −∞ ) (−1.000, ∞ ) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 11 / 147  Motivation Effect of Unknown Variable Gain Observe the following At K = 0, closed-loop poles equal open-loop poles since characteristic equation becomes s 3 + 2s 2 + Ks + 4 → s 3 + 2s 2 + 4 As K → ∞, closed-loop poles move towards open-loop zeros since characteristic equation approaches s 3 + 2s 2 + Ks + 4 → Ks which is the numerators of G (s)H(s). Hence closed-loop poles equal open-loop zeros, i.e. z1 = 0 z2,3 = ∞ Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 12 / 147  Motivation Effect of Unknown Variable Gain We are given that G (s) and H(a) are fractions of polynomials NG (s) NH (s) G (s) = , H(s) = DG (s) DH (s) The closed-loop transfer function with variable gain K can be written as NG KG KDG K (NG DH ) = NG NH = 1 + K GH 1+KD DH DG DH + K (NG NH ) G The characteristic equation then becomes DG DH + K (NG NH ) = 0 Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 13 / 147  Motivation Effect of Unknown Variable Gain KG K (NG DH ) = 1 + K GH DG DH + K (NG NH ) Note that the closed-loop poles depend both on open-loop poles and open-loop zeros NG NH = 0 → OL zeros , DG DH = 0 → OL poles The closed-loop poles change with varying K. lim DG DH + K (NG NH ) ≈ DG DH K →0 hence closed-loop poles approach open-loop poles. Similarly lim DG DH + K (NG NH ) ≈ K (NG NH ) K →∞ hence closed-loop poles approach open-loop zeros Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 14 / 147  Motivation Effect of Unknown Variable Gain Open-loop transfer function was s G (s)H(s) = s3 + 2s 2 + 4 and had 3 poles and zeros at p1 = −2.594 , p2,3 = 0.297 ± j1.206 z1 = 0 , z2,3 → ∞ ? As K : 0 → ∞, closed-loop poles move from open-loop poles to open-loop zeros as seen in the plot Hence for plotting root-locus we first need to find open-loop poles and zeros Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 15 / 147  Motivation Guidelines Rule #1: The number of branches equals the maximum among the numbers of finite open-loop poles and zeros (Note that the number of finite and infinite poles and zeros of a complex function are always the same) Rule #2: As K : 0 → ∞ each closed-loop pole moves from an open-loop pole towards an open-loop zero. In case the number of finite poles and zeros are not equal, the corresponding branch(es) extend(s) from/to ∞ resp. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 16 / 147  Motivation Locating Closed-Loop Poles From the example we learned that the closed-loop poles move from open-loop poles to open-loop zeros as K : 0 → ∞. By simply adjusting K we can change the response of our system Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 17 / 147  Motivation Locating Closed-Loop Poles The requirements might be on damping ratio, natural frequency or settling time Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 18 / 147  Motivation Locating Closed-Loop Poles The requirements might be on damping ratio, natural frequency or settling time Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 19 / 147  Motivation Locating Closed-Loop Poles The requirements might be on damping ratio, natural frequency or settling time Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 20 / 147  Motivation Limiting Facts Example : Consider the system with open-loop transfer function given as K G (s)H(s) = (s + 1)(s − 2) The closed-loop poles move from s = −1 and s = 2 to ∞ as K : 0 → ∞ There is no K such that both poles are on the left-hand s-plane, hence we cannot stabilize this system by simply changing K , since for every K we have either one or two unstable closed-loop poles. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 21 / 147  Motivation Limiting Facts Example : One solution is to add a zero on the left-hand s-plane such as at s = −2 which makes the open-loop transfer function (s + 2) G (s)H(s) = K (s + 1)(s − 2) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 22 / 147  Motivation Limiting Facts After adding an open-loop zero at s = −2, now there exists values of K such that both closed-loop poles are stable (i.e. on the left-hand plane) For small values of K , we still have an unstable closed-loop pole (red branch on positive real axis) One of the closed-loop poles is always stable (green branch) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 23 / 147  Motivation Guidelines By introducing poles and/or zero the root-locus can be changed. This way an unstable system can be made stable We learn how to plot root-locus manually in order to understand the effects of additional poles and/or zeros The integral and differential components of a PID controller does nothing but add poles and zeros Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 24 / 147  Finding Root-Locus Finding Root-Locus Given a negative feedback system with transfer function C (s) G (s) = R(s) 1 + G (s)H(s) The characteristic equation of this is obtained as 1 + G (s)H(s) = 0 → G (s)H(s) = −1 Note that the right-hand-side is actually · · · = −1 + j0. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 25 / 147  Finding Root-Locus Finding Root-Locus G (s)H(s) = −1 G (s) and H(s) are assumed to be ratios of polynomials This equality implies two equalities |G (s)H(s)| = 1 G (s)H(s) = 180◦ (2k + 1) k ∈Z All s values that satisfy the magnitude and angle conditions are the roots of the characteristic equation G (s)H(s) = −1 (i.e. · · · = −1 + j · 0) These points in the s-plane are the closed-loop poles of the given system In case we have a variable gain parameter, K , then we can find the poles of the closed-loop system for all K , which we call as root-locus plot. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 26 / 147  Finding Root-Locus Finding Root-Locus Since both G (s) and H(s) are fractions of polynomials 1 + G (s)H(s) = 0 can be written as K (s − z1 )(s − z2 ) · · · (s − zm ) 1+ =0 (s − p1 )(s − p2 ) · · · (s − pn ) The closed loop gain K ≥ 0 ∈ R (i.e. K ∈ [0, ∞)) The magnitude and angle constraints take the form K |s − z1 | |s − z2 | · · · |s − zm | =1 |s − p1 | |s − p2 | · · · |s − pn |   s − z1 + s − z2 + · · · + s − zm −   s − p1 + s − p2 + · · · + s − pn = 180o · (2k + 1) , k ∈ Z Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 27 / 147  Finding Root-Locus Finding Root-Locus K (s − z1 )(s − z2 ) · · · (s − zm ) 1+ =0 (s − p1 )(s − p2 ) · · · (s − pn ) As seen in this equations, the characteristic equation is a function of the open-loop poles and zeros of the system Hence we need to know open-loop poles and zeros in order to be able to draw loci of closed-loop poles As an example, consider the following open-loop transfer function K (s − z1 ) G (s)H(s) = (s − p1 )(s − p2 )(s − p3 )(s − p4 ) where z1 , p1 , p4 ∈ R and p2 , p3 ∈ C are conjugates Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 28 / 147  Finding Root-Locus Finding Root-Locus K (s − z1 ) G (s)H(s) = (s − p1 )(s − p2 )(s − p3 )(s − p4 ) Vectors are formed by connecting each open-loop pole and zero with the test point s Lengths of vectors are magnitudes of each factor in G (s)H(s) evaluated at test point s Slope/orientation of each vector is measured from the positive real axis Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 29 / 147  Finding Root-Locus Finding Root-Locus K (s − z1 ) G (s)H(s) = (s − p1 )(s − p2 )(s − p3 )(s − p4 ) Magnitude and angle equations due to the complex characteristic equation take the forms KB1 |G (s)H(s)| = A1 A2 A3 A4 G (s)H(s) = φ1 − θ1 − θ2 − θ3 − θ4 Note that angles of zeros are summed and those of poles are subtracted Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 30 / 147  Finding Root-Locus Finding Root-Locus Magnitudes and angles for a different test point Poles p2 and p3 are located symmetrically about the real axis since these are conjugates Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 31 / 147  Plotting Root-Locus Plotting Root-Locus Example : For the given negative feedback system, draw the root-locus diagram and find K such that ζ = 0.5 K G (s) = s(s + 1)(s + 2) H(s) = 1 The angle condition becomes K G (s)H(s) = s(s+1)(s+2) =− s − s +1− s +2 = 180o (2k + 1) k ∈ Z Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 32 / 147  Plotting Root-Locus Plotting Root-Locus Example : For the negative feedback system, draw the root-locus and find K such that ζ = 0.5 K G (s) = s(s + 1)(s + 2) H(s) = 1 The magnitude condition becomes K |G (s)H(s)| = s(s + 1)(s + 2) K = |s| |s + 1| |s + 2| We are looking for all the points on s-plane that satisfy the angle condition. For each such point we then find K that satisfies the magnitude condition. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 33 / 147  Plotting Root-Locus Plotting Root-Locus K Open-loop poles of G (s)H(s) = s(s+1)(s+2) are located at p1 = 0 , p2 = −1 , p3 = −2 and there are no finite zeros General practice is to denote open-loop poles with an × and poles with an o on the root-locus plot We follow a typical procedure for drawing the root locus Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 34 / 147  Plotting Root-Locus Plotting Root-Locus 1 - Determine the root loci on the real axis : We first check if the angle condition can be satisfied to the right of the greatest pole/zero located on the real line, i.e. p1 < s. Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 35 / 147  Plotting Root-Locus Plotting Root-Locus 1 - Determine the root loci on the real axis : Check if the angle condition can be satisfied for p2 < s < p1 and s ∈ R where p1 = 0 and p2 = −1 We find that s = 180o , s + 1 = s + 2 = 0o Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 37 / 147  Plotting Root-Locus Plotting Root-Locus Plugging in s = s + 1 = s + 2 = 0o into the angle constraint we get − s − s +1− s +2= −0 − 0 − 0 = 180o (2k + 1) k ∈Z But there is no k that satisfies this equality. Hence there is no closed-loop pole to the right of pole at s = 0 Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 36 / 147  Plotting Root-Locus Plotting Root-Locus Plugging in s = 180o , s + 1 = s + 2 = 0o into the angle constraint we get − s − s +1− s +2= −180 − 0 − 0 = 180o (2k + 1) k ∈Z Since k = −1 satisfies this equality, the segment [p2 , p1 ] on the real axis is a part of root-locus Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 38 / 147  Plotting Root-Locus Plotting Root-Locus 1 - Determine the root loci on the real axis : Check if the angle condition can be satisfied for p3 < s < p2 and s ∈ R where p2 = −1 and p3 = −2 We find that s = s + 1 = 180o , s + 2 = 0o Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 39 / 147  Plotting Root-Locus Plotting Root-Locus Plugging in s = s + 1 = 180o , s + 2 = 0o into the angle constraint we get − s − s +1− s +2= −180 − 180 − 0 = 180o (2k + 1) k ∈Z But there is no k that satisfies this equality. Hence there is no closed-loop pole to the right of pole at in range (p3 , p2 ) on real axis Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 40 / 147  Plotting Root-Locus Plotting Root-Locus 1 - Determine the root loci on the real axis : Check if the angle condition can be satisfied for s < p3 and s ∈ R where p3 = −2 We find that s = s + 1 = s + 2 = 180o Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 41 / 147  Plotting Root-Locus Plotting Root-Locus Plugging in s = s + 1 = s + 2 = 180o into the angle constraint we get − s − s +1− s +2= −180 − 180 − 180 = 180o (2k + 1) k ∈Z Since k = −2 satisfies this equality, the segment [−∞, p3 ] on the real axis is a part of root-locus Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 42 / 147  Plotting Root-Locus Guidelines Rule #3: For a negative feedback system, left of odd number of poles/zeros along the real axis is a part of root-locus In case there is no more finite pole/zero to the left of an odd number of pole/zero, then that branch extends to −∞ (indeed it extends to a pole/zero located at −∞) Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 43 / 147  Plotting Root-Locus Plotting Root-Locus 2 - Determine the asymptotes of the root loci : The asymptote of a function is the value it attains as the free parameter approaches to ∞. In our case s is the free variable and G (s) is the function value of which we are interested in as s → ∞ K K lim G (s)H(s) = lim ≈ lim s→∞ s→∞ s(s + 1)(s + 2) s→∞ s 3 The approximate form of G (s)H(s) can be used to find the loci of closed-loop poles as s → ∞ since small offsets have insignificant effect on G (s)H(s) Next we write the angle condition as follows lims→∞ G (s)H(s) ≈ lims→∞ K s3 = −3 s = 180o (2k + 1) k ∈Z Dr. Tolga Özaslan (AYBU) Root-Locus Fall 2024 44 / 147