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MTRN3210 Feedback Control Systems Week 5 Lecture 1 Root Locus
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MTRN3210 Feedback Control Systems Week 5 – Lecture 1 Mohammad Deghat – 2026 T1 WARNING Uploading course materials, including lecture slides, problem sets, and assessments to document-sharing websites is against UNSW’s Code of Conduct & Values. Breaches of this Code will be managed by the University in accordance with relevant policies and procedures. Do not remove this notice. 2 MTRN3210 Plan of the lecture: Root Locus • Recommended reading: Chap. 5 of FPE This content is protected and may not be shared, uploaded, or distributed. Motivation We learned in Week 1 that LTI system models (which are nth order differential equation) can be modeled by nth order algebraic equations using Laplace transform. We then learned in W2-3 that a system is stable if all closed-loop poles are in the LHP. We learned in W4 that Routh-Hurwitz criterion can determine the number of poles in the LHP, RHP and on the imaginary axis. So we could easily analyse and/or design a feedback control system. In this lecture, we learn a graphical technique that shows how the location of the poles of a closed-loop system are modified when one of the system’s parameters is changed. π π (π π ) 4 MTRN3210 + Σ − πΎπΎ πΊπΊ(π π ) ππ(π π ) Motivation In this lecture, we learn a graphical technique that shows how the location of the poles of a closed-loop system are modified when one of the system’s parameters is changed. π π (π π ) + Σ − πΎπΎ πΊπΊ(π π ) ππ(π π ) This method is called the root locus and was developed by Walter R. Evans in 1948. With the development of numerical software, the need for root locus plotting rules for detailed plotting has diminished. However, it is essential for a control designer to understand how varying a gain or adding a pole or zero will influence the stability and performance of a closed-loop system. Walter R. Evans 5 MTRN3210 Root Locus of a Basic Feedback System π π (π π ) + Σ − πΎπΎ πΊπΊ(π π ) ππ(π π ) π»π»(π π ) The transfer function of this feedback system is ππ π π πΎπΎπΎπΎ π π = π π π π 1 + πΎπΎπΎπΎ π π π»π»(π π ) and the characteristic equation is 1 + πΎπΎπΎπΎ π π π»π»(π π ) = 0. We define the function πΏπΏ π π as πΏπΏ π π β πΊπΊ π π π»π» π π = 6 ππ π π ππ π π where ππ(π π ) and ππ(π π ) are polynomials in π π . Notation: • πΏπΏ(π π ) is a rational function. • ππ(π π ) is a monic polynomial (the coefficient of the highest power of s is 1) of degree ππ. • ππ(π π ) is a monic polynomial of degree ππ. • ππ ≥ ππ (this condition holds for all physical systems). MTRN3210 Root Locus of a Basic Feedback System π π (π π ) + Σ − πΎπΎ πΊπΊ(π π ) The characteristic equation can be written as 1 + πΎπΎ ππ π π ππ π π =0 ⇒ ππ(π π ) π»π»(π π ) ππ(π π ) + πΎπΎπΎπΎ π π = 0 Because the solutions to the above equation are the roots of the characteristic equation and are thus closed-loop poles of the system, the root-locus method can be thought of as a method for inferring dynamic properties of the closed-loop system as the parameter πΎπΎ changes. 7 MTRN3210 Root Locus of a Second-Order System Example: Plot the poles location of the following system for all value of πΎπΎ ≥ 0. π π (π π ) Transfer function: Characteristic equation: + Σ − πΎπΎ 1 π π (π π + 1) ππ π π πΎπΎ = 2 π π π π π π + π π + πΎπΎ π π 2 + π π + πΎπΎ = 0 The above equation is a simple quadratic equation with roots: 1 1 s=− ± 1 − 4πΎπΎ 2 2 • • • ππ(π π ) Breakaway point For 0 ≤ πΎπΎ ≤ 1/4 the roots are real and distinct (between −1 and 0) For πΎπΎ = 1/4 we have two identical roots (at −0.5) For πΎπΎ > 1/4 the roots are complex (with real part −0.5) 8 MTRN3210 −1 πΎπΎ = 0 πΎπΎ = s-plane −0.5 1 4 πΎπΎ = 0 Root Locus of a Second-Order System Example: Plot the poles location of the following system for all value of πΎπΎ ≥ 0. π π (π π ) Transfer function: Characteristic equation: + Σ − πΎπΎ 1 π π (π π + 1) ππ π π πΎπΎ = 2 π π π π π π + π π + πΎπΎ π π 2 + π π + πΎπΎ = 0 The above equation is a simple quadratic equation with roots: 1 1 s=− ± 1 − 4πΎπΎ 2 2 By changing πΎπΎ, we can move the closed-loop poles to any point along the locus. 9 MTRN3210 ππ(π π ) Breakaway point −1 πΎπΎ = 0 πΎπΎ = s-plane −0.5 1 4 πΎπΎ = 0 Determining a Root Locus 10This content is protected and may not be shared, uploaded, or distributed. MTRN9400 Definition I Formal Definition of a Root Locus The root locus is the set of values of π π for which 1 + πΎπΎπΎπΎ(π π ) = 0 is satisfied as the real parameter πΎπΎ varies from 0 to +∞. Typically, 1 + πΎπΎπΎπΎ(π π ) = 0 is the characteristic equation of the system, and in this case the roots on the locus are the closed-loop poles of that system. πΏπΏ π π β πΊπΊ π π π»π» π π = π π (π π ) 11 MTRN3210 + Σ − ππ π π ππ π π where ππ(π π ) and ππ(π π ) are polynomials in π π . πΊπΊ(π π ) πΎπΎ π»π»(π π ) πΎπΎ → ∞ s-plane −1 ππ(π π ) πΎπΎ = 0 πΎπΎ = 1 4 πΎπΎ → ∞ πΎπΎ = 0 Start and End Points πΏπΏ π π β πΊπΊ π π π»π» π π = ππ π π ππ π π where ππ(π π ) and ππ(π π ) are polynomials in π π . The characteristic equation is ππ π π + πΎπΎπΎπΎ π π = 0 • ⇒ πΎπΎ = − ππ π π ππ π π π π (π π ) + If π²π² = ππ, then ππ π π = 0 (poles of π³π³(ππ)) • Σ − If π²π² → ∞, then either ππ π π = 0 (zeros of π³π³(ππ)) or ππ → ∞ ππ(π π ) πΊπΊ(π π ) πΎπΎ π»π»(π π ) Note that we assumed the order of ππ(π π ) is greater than or equal to the order of ππ(π π ); ππ ≥ ππ Example (cont.): 1 + • • 12 MTRN3210 πΎπΎ = 0 π π π π + 1 πΎπΎ → ∞ ⇒ πΎπΎ = −π π π π + 1 πΎπΎ = 0 at the poles of πΏπΏ π π : π π = 0, π π = −1 −1 πΎπΎ = 0 If πΎπΎ → ∞, then either ππ π π = 0 (zeros of πΏπΏ(π π )) or π π → ∞ s-plane −0.5 πΎπΎ → ∞ πΎπΎ = 0 Magnitude & Phase Conditions Characteristic equation: πΎπΎπΏπΏ π π = −1 ⇒ ∠πΎπΎπΎπΎ π π = ∠ − 1 ⇒ 13 MTRN3210 1 + πΎπΎπΎπΎ(π π ) = 0 Phase condition: πΎπΎπΎπΎ π π = −1 πΎπΎ πΏπΏ π π =1 ⇒ Magnitude condition: ∠πΎπΎ + ∠πΏπΏ π π = 180β ⇒ ⇒ πΎπΎπΏπΏ π π ∠πΎπΎπΎπΎ π π = −1 + πππ ∠πΏπΏ π π = 180β Formal Definition of a Root Locus Definition II Magnitude condition: 14 MTRN3210 πΎπΎ πΏπΏ π π =1 Phase condition: ∠πΏπΏ π π = 180β The root locus of πΏπΏ(π π ) is the set of points in the s-plane where the phase of πΏπΏ(π π ) is 180β¦ . To test whether a point in the s-plane is on the locus, we define the angle to the test point from a zero as ππππ and the angle to the test point from a pole as ππππ and check if the phase condition holds. Formal Definition of a Root Locus Definition II Magnitude condition: πΎπΎ πΏπΏ π π =1 To test whether a point in the s-plane is on the locus, we define the angle to the test point from a zero as ππππ and the angle to the test point from a pole as ππππ and check if the phase condition holds. 1 + π π + ππ1 πΎπΎ π π + π§π§1 = 0 π π + ππ2 π π + ππ3 π π + ππ4 ∠πΏπΏ π π = ππ1 − ππ1 − ππ2 − ππ3 − ππ4 = 180β MTRN3210 ∠πΏπΏ π π = 180β The root locus of πΏπΏ(π π ) is the set of points in the s-plane where the phase of πΏπΏ(π π ) is 180β¦ . Consider the characteristic equation: 15 Phase condition: Magnitude & Phase Conditions: Example Consider the characteristic equation: Magnitude condition: πΎπΎ πΏπΏ π π 16 MTRN3210 = πΎπΎ πΏπΏ π π πΎπΎπ΅π΅1 =1 π΄π΄1 π΄π΄2 π΄π΄3 π΄π΄4 =1 1 + π π + ππ1 πΎπΎ π π + π§π§1 = 0 π π + ππ2 π π + ππ3 π π + ππ4 Phase condition: ∠πΏπΏ π π = 180β ∠πΏπΏ π π = ππ1 − ππ1 − ππ2 − ππ3 − ππ4 = 180β Formal Definition of a Root Locus Example (cont.): If a point is on the root locus it should satisfy the phase and magnitude conditions. π π (π π ) + Phase condition: Σ − πΎπΎ 1 π π (π π + 1) ππ(π π ) Point ππππ : ∠πΏπΏ π π = −∠π π 0 − ∠ π π 0 + 1 = −ππ1 − ππ2 = 180β 180β − tan−1 πΌπΌ 0.5 tan−1 πΌπΌ 0.5 Point ππππ : ∠πΏπΏ π π = −∠π π 1 − ∠ π π 1 + 1 = −ππ3 − ππ4 ≠ 180β π π 1 is not on the root locus. 17 MTRN3210 180β − tan−1 πΌπΌ 90β −1 ππ2 π π 0 πΌπΌπΌπΌ ππ1 −0.5 π π 1 = −1 + πΌπΌπΌπΌ −1 ππ4 −0.5 πΌπΌπΌπΌ ππ3 Rule 1 Rules for Determining a Positive (180β¦) Root Locus The ππ branches of the locus start at the poles of πΏπΏ(π π ) and ππ of these branches end on the zeros of πΏπΏ(π π ). The total number of branches: ππ (number of poles of πΏπΏ(π π )) Example (cont.): 1 + πΎπΎ = 0 π π π π + 1 The total number of branches: 2 • • 18 MTRN3210 πΎπΎ = 0 at the poles of πΏπΏ π π : π π = 0, π π = −1 If πΎπΎ → ∞, then π π → ∞ s-plane −1 πΎπΎ = 0 πΎπΎ = 0 Rule 2 Rules for Determining a Positive (180β¦) Root Locus The loci are on the real axis to the left of an odd number of poles and zeros If we take a test point on the real axis, we find that • the angles of the complex poles cancel each other, as would the angles from complex conjugate zeros. • angles from real poles or zeros are 0β¦ if the test point is to the right of a given pole or zero. • angles from real poles or zeros are 180β¦ if the test point is to the left of a given pole or zero. Example (cont.): s-plane 1 + πΎπΎ = 0 π π π π + 1 −1 πΎπΎ = 0 19 MTRN3210 πΎπΎ = 0 Rule 3 Rules for Determining a Positive (180β¦) Root Locus For large πΎπΎ, ππ − ππ branches of the loci approach asymptotes at angles ππβ radiating out from the point π π = πΌπΌ on the real axis, where Number of asymptotes: ππ − ππ Angle of asymptotes: Centre of asymptotes: Example (cont.): 180β + 360β (β − 1) ππβ = , ππ − ππ πΌπΌ = ∑ ππππ − ∑ π§π§ππ ππ − ππ πΎπΎ 1 + = 0 π π π π + 1 Number of asymptotes: ππ − ππ = 2 − 0 = 2 Angle of asymptotes: ππβ = Centre of asymptotes: 20 MTRN3210 β = 1,2, β― , ππ − ππ 180β + 360β (β − 1) = 90β , 270β ππ − ππ ∑ ππππ − ∑ π§π§ππ 0 + (−1) = −0.5 = πΌπΌ = 2 ππ − ππ s-plane −1 πΎπΎ = 0 −0.5 πΎπΎ = 0 Rules for Determining a Positive (180β¦) Root Locus Rule 4 The angle of departure of a branch of the locus from a pole and the angle of arrival to a zero is given by the phase condition: ∠πΏπΏ π π = 180β οΏ½ ππππ − οΏ½ ππππ = 180β ⇒ Angle to zeros Example (cont.): 1 + Angle to poles πΎπΎ = 0 π π π π + 1 s-plane We do not really need to calculate these angles for this example and they are already known. We do it just to practice. −1 ∠πΏπΏ π π = −∠π π − ∠(π π + 1) = 180β ππππππππ, π π = −∠(π π + 1) − 180β = 0 − 180β = ±180β β β β ππππππππ, π π +1 = −∠π π − 180 = 180 − 180 = 0 21 MTRN3210 β πΎπΎ = 0 −0.5 πΎπΎ = 0 Rule 5 Rules for Determining a Positive (180β¦) Root Locus The break in and break away point(s) can be calculated by solving the following equation: ππππ =0 ππππ Note that 1 + πΎπΎπΏπΏ π π = 0 and therefore πΎπΎ = −1/πΏπΏ π π . Example (cont.): 1 + πΎπΎ = 0 π π π π + 1 s-plane ⇒ ππππ = −2π π − 1 = 0 ππππ πΎπΎ = −π π π π + 1 = −π π 2 − π π ⇒ π π = − 1 1 2 We should substitute π π = − in the characteristic 2 equation and find πΎπΎ: πΎπΎ = −0.5 −0.5 + 1 = 1/4. 22 MTRN3210 −1 πΎπΎ = 0 πΎπΎ = −0.5 1 4 πΎπΎ = 0 Rule 6 Rules for Determining a Positive (180β¦) Root Locus To find the crossing point(s) with the imaginary axis, we set π π = ππππ and solve the characteristic equation. We can find both πΎπΎ and the crossing point(s). Example (cont.): 1 + πΎπΎ = 0 π π π π + 1 Let π π = ππππ s-plane ⇒ πΎπΎ = −π π π π + 1 = −π π 2 − π π ⇒ πΎπΎ = − ππππ 2 − ππππ = ππ2 − ππππ Imaginary part: ππππ = 0 ⇒ ππ = 0 Real part: 23 MTRN3210 2 πΎπΎ = ππ = 0 −1 πΎπΎ = 0 πΎπΎ = −0.5 1 4 πΎπΎ = 0 Example 2 Drawing the root locus for ππ < π²π² < ∞: π π (π π ) + − Σ πΎπΎ π π + 1 π π (π π − 1)(π π + 6) Example: ππ(π π ) πΏπΏ π π = π π + 1 π π (π π − 1)(π π + 6) • Number of branches: ππ = 3 The number of branches equals the number of open-loop poles: ππ = 3 Root locus branches start at the open-loop poles at πΎπΎ = 0 and end at open-loop zeros or asymptotes when πΎπΎ → ∞. A portion of the real axis will be a part of the root locus if the total number of poles and zeros to the right of that portion is odd. 24 MTRN3210 s-plane −6 Example 2 Drawing the root locus for ππ < π²π² < ∞: π π (π π ) • • • • + − Σ πΎπΎ π π + 1 π π (π π − 1)(π π + 6) Number of asymptotes: ππ − ππ Centroid of the loci: Angle of asymptotes: Example: πΌπΌ = ππ π§π§ ∑1 ππ ππππ −∑ππ 1 π§π§ππ ππππ −πππ§π§ 180β + 360β (β − 1) ππβ = , ππ − ππ β = 1,2, β― , ππ − ππ Departure and arrival angles: Obvious for this example. 25 MTRN3210 ππ(π π ) πΏπΏ π π = π π + 1 π π (π π − 1)(π π + 6) • Number of branches: ππ = 3 • Number of asymptotes : 3 − 2 = 2 • Centroid of the loci: 0 + 1 − 6 − (−1) = −2 πΌπΌ = 2 • Angle of asymptotes 180Λ = 90Λ ππ1 = 2 180Λ + 360Λ = 270Λ ππ2 = 2 s-plane −6 −2 Example 2 Drawing the root locus for ππ < π²π² < ∞: Example: Breakaway (Break in) points or the points of multiple roots are the solution of the following equation πππΎπΎ =0 πππ π where πΎπΎ is expressed as a function of π π from the characteristic equation. This is a necessary but not sufficient condition. One has to check if the solutions lie on the root locus (which means πΎπΎ ≥ 0). π π π π − 1 π π + 6 π π 3 + 5π π 2 − 6π π =− πΎπΎ = − π π + 1 π π + 1 ππππ 3π π 2 + 10π π − 6 π π + 1 − π π 3 + 5π π 2 − 6π π =− =0 ππππ π π + 1 2 2π π 3 + 8π π 2 + 10π π − 6 ⇒− =0 π π + 1 2 26 MTRN3210 π π = 0.43, πΎπΎ = 1.1 Other roots are complex πΏπΏ π π = π π + 1 π π (π π − 1)(π π + 6) • Number of branches: ππ = 3 • Number of asymptotes : 3 − 2 = 2 • Centroid of the loci: 0 + 1 − 6 − (−1) = −2 πΌπΌ = 2 • Angle of asymptotes 180Λ = 90Λ ππ1 = 2 180Λ + 360Λ = 270Λ ππ2 = 2 s-plane −6 −2 Example 2 Example: Numerical plot: πΏπΏ π π = π π π π − 1 π π + 6 π π 3 + 5π π 2 − 6π π =− πΎπΎ = − π π + 1 π π + 1 ππππ 3π π 2 + 10π π − 6 π π + 1 − π π 3 + 5π π 2 − 6π π =− =0 ππππ π π + 1 2 2π π 3 + 8π π 2 + 10π π − 6 ⇒− =0 π π + 1 2 27 MTRN3210 π π = 0.43, πΎπΎ = 1.1 Other roots are complex π π + 1 π π (π π − 1)(π π + 6) • Number of branches: ππ = 3 • Number of asymptotes : 3 − 2 = 2 • Centroid of the loci: 0 + 1 − 6 − (−1) = −2 πΌπΌ = 2 • Angle of asymptotes 180Λ = 90Λ ππ1 = 2 180Λ + 360Λ = 270Λ ππ2 = 2 s-plane −6 −2 Summary: Rule 2 Rule 1 We learned how to sketch root locus by hand and MATLAB. The ππ branches of the locus start at the poles of πΏπΏ(π π ) and ππ of these branches end on the zeros of πΏπΏ(π π ). The loci are on the real axis to the left of an odd number of poles and zeros Rule 3 For large π π and πΎπΎ, ππ − ππ branches of the loci are asymptotic to lines at angles ππβ radiating out from the point π π = πΌπΌ on the real axis, where Number of asymptotes: Angle of asymptotes: Centre of asymptotes: 28 MTRN3210 ππ − ππ 180β + 360β (β − 1) , ππβ = ππ − ππ πΌπΌ = ∑ ππππ − ∑ π§π§ππ ππ − ππ β = 1,2, β― , ππ − ππ Summary: Rule 4 We learned how to sketch root locus by hand and MATLAB. The angle of departure of a branch of the locus from a pole and the angle of arrival to a zero is given by the phase condition: ⇒ ∠πΏπΏ π π = 180β οΏ½ ππ − οΏ½ ππ = 180β ππ ππ Rule 5 ππππ =0 ππππ Note that 1 + πΎπΎπΏπΏ π π = 0 and therefore πΎπΎ = −1/πΏπΏ π π . Rule 6 The break in and break away point(s) can be calculated by solving the following equation: To find the crossing point(s) with the imaginary axis, we set π π = ππππ and solve the characteristic equation. We can find both πΎπΎ and the crossing point(s). 29 MTRN3210 In-class exercise 1. Go to www.kahoot.it 2. Enter the PIN 3. Choose a nickname 30 MTRN3210
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