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Multivariable Control. Root Locus . By Kirsten Mølgaard Nielsen. Outline. Root Locus Design Method Root locus definition The phase condition Test points, and sketching a root locus Dynamic Compensation Lead compensation Lag compensation. Introduction.
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Multivariable Control Root Locus By Kirsten Mølgaard Nielsen
Outline • Root Locus Design Method • Root locus • definition • The phase condition • Test points, and sketching a root locus • Dynamic Compensation • Lead compensation • Lag compensation
Introduction Closed loop transfer function Characteristic equation, roots are poles in T(s)
Introduction • Dynamic features depend on the pole locations. • For example, time constant t, rise time tr, and overshoot Mp (1st order) (2nd order)
Introduction Root locus • Determination of the closed loop pole locations under varying K. • For example, K could be the control gain. The characteristic equation can be written in various ways
Introduction • Root locus of a motor position control (example)
Introduction break-away point
Introduction Some root loci examples (K from zero to infinity)
Sketching a Root Locus • Definition 1 • The root locus is the values of s for which 1+KL(s)=0 is satisfied as K varies from 0 to infinity (pos.). • Definition 2 • The root locus is the points in the s-plane where the phase of L(s) is 180°. • The angle to a point on the root locus from zero number i is yi. • The angle to a point on the root locus from pole number i is fi. • Therefore,
Sketching a Root Locus s0 f1 p y1 f2 p* A root locus can be plotted using Matlab: rlocus(sysL)
Root Locus characteristics • Rule 1 (of 6) • The n branches of the locus start at the poles L(s) and m of these branches end on the zeros of L(s).
Root Locus characteristics • Rule 2 (of 6) • The loci on the real axis (real-axis part) are to the left of an odd number of poles plus zeros. • Notice, if we take a test point s0 on the real axis : • The angle of complex poles cancel each other. • Angles from real poles or zeros are 0° if s0 are to the right. • Angles from real poles or zeros are 180° if s0 are to the left. • Total angle = 180° + 360° l
Root Locus characteristics • Rule 3 (of 6) • For large s and K, n-m branches of the loci are asymptotic to lines at angles fl radiating out from a point s = a on the real axis. fl a For example
Root Locus characteristics n-m=1 n-m=2 n-m=3 n-m=4 For ex., n-m=3
Root Locus 3 poles: q = 3 • Rule 4: The angles of departure (arrival) of a branch of the locus from a pole (zero) of multiplicity q. f1,dep = 60 deg. f2,dep = 180 deg. f3,dep = 300 deg.
Root Locus characteristics • Rule 5 (of 6) • The locus crosses the jw axis at points where the Routh criterion shows a transition from roots in the left half-plane to roots in the right half-plane. • Routh: A system is stable if and only if all the elements in the first column of the Routh array are positive.
Root Locus characteristics • Rule 6 (of 6) • The locus will have multiplicative roots of q at points on the locus where (1) applies. • The branches will approach a point of q roots at angles separated by (2) and will depart at angles with the same separation. Characteristic equation
Example Root locus for double integrator with P-control Rule 1: The locus has two branches that starts in s=0. There are no zeros. Thus, the branches do not end at zeros. Rule 3: Two branches have asymptotes for s going to infinity. We get
Rule 2: No branches that the real axis. Rule 4: Same argument and conclusion as rule 3. Rule 5: The loci remain on the imaginary axis. Thus, no crossings of the jw-axis. Rule 6: Easy to see, no further multiple poles. Verification:
Example Root locus for satellite attitude control with PD-control
Selecting the Parameter Value • The (positive) root locus • A plot of all possible locations of roots to the equation 1+KL(s)=0 for some real positive value of K. • The purpose of design is to select a particular value of K that will meet the specifications for static and dynamic characteristics. • For a given root locus we have (1). Thus, for some desired pole locations it is possible to find K.
Selecting the Parameter Value Example Selection of K using Matlab: [K,p]=rlocfind(sysL)
Dynamic Compensation • Some facts • We are able to determine the roots of the characteristic equation (closed loop poles) for a varying parameter K. • The location of the roots determine the dynamic characteristics (performance) of the closed loop system. • It might not be possible to achieve the desired performance with D(s) = K. • Controller design using root locus • Lead compensation (similar to PD control) • overshoot, rise time requirements • Lag compensation (similar to PI control) • steady state requirements
Dynamic Compensation PD Lead PI Lag
Lead Compensation PD P Lower damping ratio with PD !
Lead Compensation • PD control • Pure differentiation • Output noise has a great effect on u(t) • Solution: Insert a pole at a higher frequency r(t) e(t) controller D(s) u(t) plant G(s) y(t) + - + + noise
Lead Compensation • Selecting p and z • Usually, trial and error • The placement of the zero is determined by dynamic requirements (rise time, overshoot). • The exact placement of the pole is determined by conflicting interests. • suppression of output noise • effectiveness of the zero • In general, • the zero z is placed around desired closed loop wn • the pole p is placed between 5 and 20 times of z
Lead Compensation • Design approach (A) • Initial design z=wn, p=5z • Noise: additional requirement, max(p)=20 • Iterations • First design: z=wn, p=5z (neglecting the add. req.) • Second design: z=wn, p=20 • Third design: new z, p=20
Lead Compensation Possible pole location Matlab – design 1 sysG = tf([1],[1 1 0]) sysD = tf([1 7],[1 5*7]) rlocus(sysD*sysG)
Lead Compensation No possible pole location ! Matlab – design 2 sysG = tf([1],[1 1 0]) sysD = tf([1 7],[1 20]) rlocus(sysD*sysG)
Lead Compensation Possible location New zero location Matlab – design 3 sysG = tf([1],[1 1 0]) sysD = tf([1 4],[1 20]) rlocus(sysD*sysG)
Lead Compensation • Design approach (B) • Pole placement from the requirements • Noise: Additional requirement, max(p)=20. So, p=20 to minimize the effect of the pole.
Lead Compensation Pole location Matlab – design B sysG = tf([1],[1 1 0]) sysD = tf([1 5.4],[1 20]) rlocus(sysD*sysG)
Lead Compensation • Exercise • Design a lead compensation D(s) to the plant G(s) so that the dominant poles are located at s = -2 ± 2j You can use, triangle a/sin(A) = b/sin(B) r(t) e(t) controller D(s) u(t) plant G(s) y(t) + -
Lead Compensation • Answer • we have three poles and one zero f1 y f2 Notice, triangle a/sin(A) = b/sin(B)
Dynamic Compensation PD Lead PI Lag
Lag Compensation • Lag compensation • Steady state characteristics (requirements) • Position error constant Kp • Velocity error constant Kv • Let us continue using the example system • and the designed controller (lead compensation)
Lag Compensation • Calculation of error constants • Suppose we want Kv≥ 100 • We need additional gain at low frequencies !
Lag Compensation • Selecting p and z • To minimize the effect on the dominant dynamics z and p must chosen as low as possible (i.e. at low frequencies) • To minimize the settling time z and p must chosen as high as possible (i.e. at high frequencies) • Thus, the lag pole-zero location must be chosen at as high a frequency as possible without causing any major shifts in the dominant pole locations
Lag Compensation • Design approach • Increase the error constant by increasing the low frequency gain • Choose z and p somewhat below wn • Example system: • K = 1 (the proportional part has already been chosen) • z/p = 3, with z = 0.03, and p = 0.01.
Lag Compensation Notice, a slightly different pole location Lead-lag Controller
Lag Compensation z p
Lag Compensation Step
Lag Compensation Ramp
Notch compensation • Example system • We have successfully designed a lead-lag controller • Suppose the real system has a rather undampen oscillation about 50 rad/sec. • Include this oscillation in the model • Can we use the original controller ?
Notch compensation Step response using the lead-lag controller