Ch.08 Root Locus Techniques

18/03/2016 System Dynamics and Control 8.01 Root Locus Techniques 08 Root Locus Techniques HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 8.01 Root Locus Techniques §1.Introduction Root Locus Technique was discovered by Evans in 1948 The Control System Problem Closed-loop system 𝐾𝐺(𝑠) Closed-loop TF: 𝑇 𝑠 = + 𝐾𝐺 𝑠 𝐻(𝑠) Equivalent transfer function Open-loop TF: 𝐾𝐺 𝑠 𝐻(𝑠) - From open-loop TF 𝐾𝐺(𝑠)𝐻(𝑠) • can determine the poles of 𝐾𝐺(𝑠)𝐻(𝑠) • variations in 𝐾 not affect the location of any pole of this TF - From the closed-loop TF 𝑇(𝑠) • cannot determine the poles of 𝑇(𝑠) unless factor the denominator • the poles of 𝑇(𝑠) change with 𝐾 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.01 Nguyen Tan Tien Root Locus Techniques §1.Introduction Vector Representation of Complex Numbers Any complex number, 𝑠 = 𝜎 + 𝑗𝜔, described in Cartesian coordinates can be graphically represented by a vector or in polar form 𝑀∠𝜃 System Dynamics and Control 8.01 Root Locus Techniques Learning Outcome After completing this chapter, the student will be able to • Define a root locus • State the properties of a root locus • Sketch a root locus • Find the coordinates of points on the root locus and their associated gains • Use the root locus to design a parameter value to meet a transient response specification for systems of order and higher • Sketch the root locus for positive-feedback systems • Find the root sensitivity for points along the root locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.01 Nguyen Tan Tien Root Locus Techniques §1.Introduction The Control System Problem Closed-loop system 𝐾𝐺(𝑠) + 𝐾𝐺 𝑠 𝐻(𝑠) Closed-loop TF: 𝑇 𝑠 = Equivalent transfer function Open-loop TF: 𝐾𝐺 𝑠 𝐻(𝑠) - The poles of 𝑇(𝑠) are a function of 𝐾 → find the poles for specific values of 𝐾 to estimate the system’s transient response and stability - The root-locus method displays the location of the poles of the closed-loop TF as a function of the gain factor 𝐾 of the openloop TF HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.01 Nguyen Tan Tien Root Locus Techniques §1.Introduction ⟹ 𝑠 + 𝑎 is a complex number and can be represented by a vector drawn from the zero of the function to the point 𝑠 Ex (𝑠 + 7)ȁ𝑠→5+𝑗2 is a complex number drawn from the zero of the function, −7, to the point 𝑠, which is + 𝑗2 If 𝐹(𝑠) = 𝑠 + 𝑎 then substituting the complex number 𝑠 = 𝜎 + 𝑗𝜔 yields 𝐹(𝑠) = (𝜎 + 𝑎) + 𝑗𝜔, another complex number 𝐹(𝑠) has a zero at −𝑎 If translate the vector 𝑎 units to the left → alternate representation of the complex number that originates at the zero −𝑎 and terminates on the point 𝑠 = 𝜎 + 𝑗𝜔 ⟹ 𝑠 + 𝑎 is a complex number and can be represented by a vector drawn from the zero of the function to the point 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.01 Root Locus Techniques §1.Introduction Apply the concepts to a complicated function ς𝑚 (𝑠+𝑧𝑖 ) ς numerator’s complex factors 𝐹 𝑠 = ς𝑛𝑖=1 = ς denominator’s complex factors 𝑗=1(𝑠+𝑝𝑗 ) System Dynamics and Control §1.Introduction - Ex.8.1 Root Locus Techniques Evaluation of a Complex Function via Vectors 𝑠+1 Find 𝐹(𝑠) at the point 𝑠 = −3 + 𝑗4 𝑠(𝑠 + 2) The vector originating at Given 𝐹 𝑠 = (8.4) The function defines the complex arithmetic to be performed in order to evaluate 𝐹(𝑠) at any point, 𝑠 Since each complex factor can be thought of as a vector • the magnitude, 𝑀, of 𝐹(𝑠) ς𝑚 ȁ(𝑠+𝑧𝑖 )ȁ ς zero lengths 𝑀=ς = 𝑖=1 (8.5) pole lengths ς𝑛 Solution 20∠116.60 5∠126.90 • the zero at −1 • the pole at • the pole at −2 17∠104.00 Therefore 20 𝑀∠𝜃 = ∠(116.60 − 126.90 − 104.00 ) × 17 = 0.217∠ −114.30 𝑗=1 ȁ(𝑠+𝑝𝑗 )ȁ • the angular, 𝜃, of 𝐹(𝑠) 𝜃 = ∑zero angles − ∑pole angles 𝑛 = ∑𝑚 𝑖=1 ∠(𝑠 + 𝑧𝑖 ) − ∑𝑗=1 ∠(𝑠 + 𝑝𝑗 ) 8.01 (8.6) ς zero 𝑀=ς ȁ(𝑠+𝑧𝑖 )ȁ lengths ς𝑚 = 𝑖=1 pole lengths ς𝑛𝑗=1 ȁ(𝑠+𝑝𝑗)ȁ (8.5) 𝑛 𝜃 = ∑zero angles − ∑pole angles = ∑𝑚 𝑖=1 ∠(𝑠 + 𝑧𝑖 ) − ∑𝑗=1 ∠(𝑠 + 𝑝𝑗 ) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.01 Nguyen Tan Tien Root Locus Techniques HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control (8.6) Nguyen Tan Tien 8.10 Root Locus Techniques §1.Introduction Skill-Assessment Ex.8.1 (𝑠 + 2)(𝑠 + 4) Problem Given 𝐹 𝑠 = 𝑠(𝑠 + 3)(𝑠 + 6) Find 𝐹(𝑠) at the point 𝑠 = −7 + 𝑗9 the following ways a.Directly substituting the point into 𝐹(𝑠) b.Calculating the result using vectors Solution a.Directly substituting the point into 𝐹(𝑠) −7 + 𝑗9 + −7 + 𝑗9 + 𝐹 −7 + 𝑗9 = (−7 + 𝑗9)(−7 + 𝑗9 + 3)(−7 + 𝑗9 + 6) (−5 + 𝑗9)(−3 + 𝑗9) = (−7 + 𝑗9)(−7 + 𝑗9)(−1 + 𝑗9) −66 − 𝑗72 = = −0.0339 − 𝑗0.0899 944 − 𝑗378 = 0.096∠ − 110.70 §1.Introduction b.Calculating the result using vectors 𝑀2 𝑀4 𝐹 −7 + 𝑗9 = 𝑀1𝑀3 𝑀5 𝑗𝜔 (−7 + 𝑗9) (−3 + 𝑗9)(−5 + 𝑗9) = (−1 + 𝑗9)(−4 + 𝑗9)(−7 + 𝑗9) −66 − 𝑗72 = 944 − 𝑗378 = −0.0339 − 𝑗0.0899 = 0.096∠ − 110.70 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.11 Nguyen Tan Tien Root Locus Techniques §1.Introduction TryIt 8.1 Use the following MATLAB statements to solve the problem given in SkillAssessment Ex.8.1 s=-7+9j; G=(s+2)*(s+4)/ (s*(s+3)*(s+6)); Theta=(180/pi)*angle(G) M=abs(G) 𝑀1 𝑀2 𝑀3 𝑀4 𝑀5 𝜎 −7 −6 −5 −4 −3 −2 −1 System Dynamics and Control Nguyen Tan Tien 8.12 Root Locus Techniques §2.Defining the Root Locus 𝐺 𝑠 = (𝑠 + 2)(𝑠 + 4) 𝑠(𝑠 + 3)(𝑠 + 6) Block diagram Security cameras with auto tracking can be used to follow moving objects automatically Theta = -110.6881 M= 0.0961 HCM City Univ of Technology, Faculty of Mechanical Engineering Closed-loop transfer function - The variation of pole location for different values of gain, 𝐾 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.13 Root Locus Techniques System Dynamics and Control 8.14 Root Locus Techniques §2.Defining the Root Locus - The individual closed-loop pole locations are removed and their paths are represented with solid lines It is this representation of the paths of the closed-loop poles as the gain is varied that we call a root locus The discussion will be limited to positive gain, or 𝐾 ≥ §2.Defining the Root Locus - The root locus shows the changes in the transient response • 𝐾 < 25: the system is overdamped • 𝐾 = 25: the poles are real and multiple and hence critically damped • 𝐾 > 25: the system is underdamped HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.15 Nguyen Tan Tien Root Locus Techniques System Dynamics and Control 8.16 §3.Properties of the Root Locus - The closed-loop transfer function for the system 𝐾𝐺(𝑠) (8.12) 𝑇 𝑠 = + 𝐾𝐺 𝑠 𝐻(𝑠) - The pole, 𝑠, is the roots of the characteristics equation 𝐾𝐺 𝑠 𝐻 𝑠 = −1 = 1∠ 2𝑘 + 1800 𝑘 = 0,±1, … (8.13) - Alternately, value of 𝑠 is a closed-loop pole if ȁ𝐾𝐺 𝑠 𝐻 𝑠 ȁ = (8.14) and ∠𝐾𝐺 𝑠 𝐻 𝑠 = 2𝑘 + 1800 (8.15) (8.16) 𝐾= 𝐺 𝑠 ȁ𝐻 𝑠 ȁ The magnitude of 𝐾𝐺(𝑠)𝐻(𝑠) must be unity, implying that the value of 𝐾 is the reciprocal of the magnitude of 𝐺(𝑠)𝐻(𝑠) when the pole value is substituted for 𝑠 §3.Properties of the Root Locus - Ex HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.17 Nguyen Tan Tien Root Locus Techniques = −1 −9.47 −9.47 + 10 𝐾 = 5, 𝑝 = −0.53: 𝐾𝐺 𝑠 𝐻 𝑠 = = −1 −0.53 −0.53 + 10 10 𝐾 = 10, 𝑝 = −8.87: 𝐾𝐺 𝑠 𝐻 𝑠 = = −1 −8.87 −8.87 + 10 … 𝐾 = 5, 𝑝 = −9.47: 𝐾𝐺 𝑠 𝐻 𝑠 = System Dynamics and Control 8.18 §3.Properties of the Root Locus 𝑗𝜔 𝑗𝜔 Nguyen Tan Tien (−2 + 𝑗 2/2) −4 Nguyen Tan Tien Root Locus Techniques Consider the point −2 + 𝑗( 2/2) ∑𝑛1 𝜃𝑧𝑖 − ∑𝑚 𝜃𝑝𝑖 = (2𝑘 + 1)180 ⟹ 𝜃1 + 𝜃2 − 𝜃3 − 𝜃4 = 1800 (−2 + 𝑗3) HCM City Univ of Technology, Faculty of Mechanical Engineering Root Locus Techniques The pole location varies with 𝐾 𝐾 𝐾𝐺 𝑠 𝐻 𝑠 = 𝑠(𝑠 + 10) §3.Properties of the Root Locus - Ex The open-loop and closed-loop TF 𝐾(𝑠 + 3)(𝑠 + 4) (8.18) 𝐾𝐺 𝑠 𝐻 𝑠 = (𝑠 + 1)(𝑠 + 2) 𝐿1 𝐿2 𝐿3 𝐿4 𝐾(𝑠 + 3)(𝑠 + 4) 𝜃1 𝜃2 𝜃3 𝜃4 (8.19) 𝑇 𝑠 = 𝜎 + 𝐾 𝑠2 + + 7𝐾 𝑠 + (2 + 12𝐾) −4 −3 −2 −1 Consider the point −2 + 𝑗3 If this point is a closed-loop pole, then the angles of the zeros minus the angles of the poles must equal an odd multiple of 1800 𝜃1 + 𝜃2 − 𝜃3 − 𝜃4 = 56.310 + 71.570 − 900 − 108.430 = −70.550 ⟹ −2 + 𝑗3 is not a point on the root locus, or alternatively, − + 𝑗3 is not a closed-loop pole for any gain Nguyen Tan Tien (8.20) −2 + 𝑗( 2/2) is a point on the root locus for some value of gain 𝐾 1 ς pole lengths (8.21) 𝐾= = = ȁ𝐺(𝑠)ȁȁ𝐻(𝑠)ȁ 𝑀 ς zero lengths 𝜃4 𝜃1 𝜃2 𝜃3 −3 −2 −1 ⟹ 𝜎 The gain 𝐾 for the point −2 + 𝑗( 2/2) ς pole lengths 𝐿3 𝐿4 ( 2/2) × 1.22 𝐾= = = = 0.33 ς zero lengths 𝐿1 𝐿2 2.12 × 1.22 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.19 Root Locus Techniques System Dynamics and Control 8.20 Root Locus Techniques §3.Properties of the Root Locus Skill-Assessment Ex.8.2 Problem Given a unity feedback system that has the forward TF 𝐾(𝑠 + 2) 𝐺 𝑠 = (𝑠 + 4𝑠 + 13) the following a.Calculate the angle of 𝐺(𝑠) at the point (−3 + 𝑗0) by finding the algebraic sum of angles of the vectors drawn from the zeros and poles of 𝐺(𝑠) to the given point b.Determine if the point specified in a is on the root locus c.If the point specified in a is on the root locus, find the gain, 𝐾, using the lengths of the vectors §3.Properties of the Root Locus Solution a.Directly substituting the point into 𝐹(𝑠) 𝐾(𝑠 + 2) 𝐾(𝑠 + 2) 𝐺 𝑠 = = (𝑠 + 4𝑠 + 13) [𝑠 − −1 + 𝑗3 ][𝑠 − −1 − 𝑗3 ] 𝑗𝜔 𝜃2 From the diagram 𝑗3 ∑angles = 1800 − 𝜃1 − 𝜃2 𝑗2 𝜃1 = 900 + 𝑡𝑎𝑛−1 (1/3) = 108.430 𝑗1 (−3 + 𝑗0) 𝜃2 = 2700 − 𝑡𝑎𝑛−1 (1/3) = 251.570 𝜎 ⟹ ∑angles = 1800 −4 −2 −1 −𝑗1 b.Since the angle is 1800, the point (−3 + 𝑗0) −𝑗2 is on the locus pole lengths 𝜃1 ς pole lengths −𝑗3 c.𝐾 = ς zero lengths 12 + 32 × 12 + 32 = = 10 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.21 Nguyen Tan Tien Root Locus Techniques §3.Properties of the Root Locus TryIt 8.2 Use MATLAB and the following statements to solve Skill-Assessment Ex.8.2 s=-3+0j; G=(s+2)/(s^2+4*s+13); Theta=(180/pi)*angle(G) M=abs(G); K=1/M 𝐺 𝑠 = s=-3+0j; G=(s+2)*(s^2+4*s+13); Theta=(180/pi)*angle(G) M=abs(G) System Dynamics and Control 8.23 Nguyen Tan Tien Root Locus Techniques §4.Sketching the Root Locus 𝑁𝐺 (𝑠) 𝐾𝑁𝐺 (𝑠) 𝐷𝐺 (𝑠) = 𝑁𝐺 (𝑠) 𝑁𝐻 (𝑠) 𝐷𝐺 𝑠 𝐷𝐻 𝑠 + 𝐾𝑁𝐺 (𝑠)𝑁𝐻 (𝑠) 1+𝐾 𝐷𝐺 (𝑠) 𝐷𝐻 (𝑠) 𝐾𝑁𝐺 (𝑠) 𝐾𝑁𝐺 (𝑠) 𝐾 → 0: 𝑇 𝑠 = ≈ 𝐷𝐺 𝑠 𝐷𝐻 𝑠 + 𝐾𝑁𝐺 (𝑠)𝑁𝐻 (𝑠) 𝐷𝐺 𝑠 𝐷𝐻 𝑠 + 𝜖 𝐾𝐺(𝑠) = + 𝐾𝐺 𝑠 𝐻(𝑠) 𝐾 ⟹the closed-loop system poles at small gains approach the combined poles of 𝐺(𝑠) and 𝐻(𝑠): the root locus begins at the poles of 𝐺(𝑠)𝐻(𝑠), the open-loop TF 𝐾𝑁𝐺 (𝑠) 𝐾𝑁𝐺 (𝑠) 𝐾 → ∞: 𝑇 𝑠 = ≈ 𝐷𝐺 𝑠 𝐷𝐻 𝑠 + 𝐾𝑁𝐺 (𝑠)𝑁𝐻 (𝑠) 𝜖 + 𝐾𝑁𝐺 (𝑠)𝑁𝐻 (𝑠) ⟹the closed-loop system poles at large gains approach the combined zeros of 𝐺(𝑠) and 𝐻 𝑠 : the root locus ends at the zeros of 𝐺(𝑠)𝐻(𝑠), the open-loop TF HCM City Univ of Technology, Faculty of Mechanical Engineering Root Locus Techniques HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.24 Nguyen Tan Tien Root Locus Techniques §4.Sketching the Root Locus 4.Starting and ending points The root locus begins at the finite and infinite poles of 𝐺(𝑠)𝐻(𝑠) and ends at the finite and infinite zeros of 𝐺(𝑠)𝐻(𝑠) 𝑇 𝑠 = 8.22 §4.Sketching the Root Locus The following five rules allow us to sketch the root locus 1.Number of branches The number of branches of the root locus equals the number of closed-loop poles 2.Symmetry The root locus is symmetrical about the real axis 3.Real-axis segments On the real axis, for 𝐾 > the root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros Ex (𝑠 + 2)(𝑠 + 4) 𝑠(𝑠 + 3)(𝑠 + 6) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien Nguyen Tan Tien 5.Behavior at infinity The root locus approaches straight lines as asymptotes as the locus approaches infinity Further, the equation of the asymptotes is given by the real-axis intercept, 𝜎𝑎 and angle, 𝜃𝑎 as follows finite poles ∑𝑛1 𝑝𝑖 − ∑𝑚 ∑finite poles−∑finite zeros 𝑧𝑖 (8.27) 𝜎𝑎 = = ⋕ finite poles− ⋕ finite zeros 𝑛−𝑚 (2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋 (8.28) 𝜃𝑎 = = ⋕ finite poles− ⋕ finite zeros 𝑛−𝑚 where 𝑘 = 0, ±1, ±2, … and the angle is given in radians with respect to the positive extension of the real axis HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.25 Root Locus Techniques §4.Sketching the Root Locus - Ex.8.2 Sketching a Root Locus with Asymptotes Sketch the root locus for the system Solution System Dynamics and Control 8.26 𝑘, 𝜃𝑎 = 𝜋 5𝜋 0; , 1; 𝜋 , 2; 3 𝜎𝑎 = − Find the asymptotes The real-axis intercept 𝑛 ∑𝑚 − − − − (−3) 𝑝𝑖 − ∑1 𝑧𝑖 𝜎𝑎 = = =− 𝑚−𝑛 4−1 The angles of line that intersect at −4/3 (2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋 𝜃𝑎 = = = 𝑚−𝑛 4−1 ⟹ 𝑘, 𝜃𝑎 = [ 0; 𝜋/3 , 1; 𝜋 , 2; 5𝜋/3 ] HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.27 Nguyen Tan Tien Root Locus Techniques Root Locus Techniques §4.Sketching the Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.28 §4.Sketching the Root Locus Skill-Assessment Ex.8.3 Problem Sketch the root locus and its asymptotes for a unity feedback system that has the forward TF 𝐾 𝐺 𝑠 = (𝑠 + 2)(𝑠 + 4)(𝑠 + 6) Solution Find the asymptotes The real-axis intercept 𝑛 ∑𝑚 −2 − − − (0) 𝑝𝑖 − ∑1 𝑧𝑖 𝜎𝑎 = = = −4 𝑚−𝑛 3−0 The angles of line that intersect at −4 (2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋 𝜃𝑎 = = = 𝑚−𝑛 3−0 ⟹ 𝑘, 𝜃𝑎 = [ 0; 𝜋/3 , 1; 𝜋 , 2; 5𝜋/3 ] §4.Sketching the Root Locus 𝜎𝑎 = −4 𝑘, 𝜃𝑎 = [ 0; 𝜋/3 , 1; 𝜋 , 2; 5𝜋/3 ] HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.29 Nguyen Tan Tien Root Locus Techniques System Dynamics and Control 8.30 Nguyen Tan Tien Root Locus Techniques Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Real-Axis Breakaway and Break-In Points breakaway point: −𝜎1 break-in point: 𝜎2 At the breakaway or break-in point, the branches of the root locus form an angle of 1800 /𝑛 with the real axis 𝑛: the number of closed-loop poles arriving at or departing from the single breakaway or break-in point on the real axis (Kuo, 1991) For the two poles shown, the branches at the breakaway point form 1800/𝑛 = 1800 angles with the real axis §5.Refining the Sketch - Find breakaway and break-in point by maximize/minimize the gain HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering (8.31) 𝐺 𝑠 𝐻(𝑠) Along the real axis Eq (8.31) becomes (8.32) 𝐾=− 𝐺 𝜎 𝐻(𝜎) To find breakaway and break-in point → differentiate Eq (8.32) with respect to 𝜎 and set the derivative equal to zero Eq (8.13) → 𝐾 = − 𝐾𝐺 𝑠 𝐻 𝑠 = −1 = 1∠ 2𝑘 + 1800 𝑘 = 0,±1, … Nguyen Tan Tien (8.13) Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.31 Root Locus Techniques System Dynamics and Control 8.32 Root Locus Techniques §5.Refining the Sketch Ex.8.3 Breakaway and Break-in Points via Differentiation Find the breakaway and break-in points for the root locus of the given figure, using differential calculus Solution Using the open-loop poles and zeros 𝐾(𝑠 − 3)(𝑠 − 5) 𝐾(𝑠 − 8𝑠 + 15) 𝐾𝐺 𝑠 𝐻 𝑠 = = (𝑠 + 1)(𝑠 + 2) (𝑠 + 3𝑠 + 2) For all points along the real axis, 𝐾𝐺 𝜎 𝐻 𝜎 = −1 𝐾(𝜎2 − 8𝜎 + 15) 𝜎2 + 3𝜎 + 𝑑𝐾 11𝜎2 − 26𝜎 − 61 = −1 → 𝐾 = − → = (𝜎2 + 3𝜎 + 2) 𝜎 − 8𝜎 + 15 𝑑𝜎 (𝜎2 − 8𝜎 + 15)2 Setting the derivative equal to zero yields 𝑑𝐾 = ⟹ 𝜎 = −1.45, 𝜎 = +3.82 ⟹ 𝜎1 = −1.45, 𝜎2 = 3.82 𝑑𝜎 §5.Refining the Sketch - Find breakaway and break-in point using transition method Breakaway and break-in points satisfy the relationship HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.33 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Ex.8.4 Breakaway and Break-in Points Without Differentiation Find the breakaway and break-in points for the root locus of the given figure, without differentiating Solution Using Eq (8.37) 1 1 + = + 𝜎−3 𝜎−5 𝜎+1 𝜎+2 Simplifying 11𝜎 − 26𝜎 − 61 = ⟹ 𝜎 + 1.45 𝜎 − 3.82 = ⟹ 𝜎1 = −1.45, 𝜎2 = 3.82 ∑𝑚 1 𝜎+𝑧𝑖 = ∑𝑛1 𝜎+𝑝𝑖 𝑛 1 =෍ 𝜎 + 𝑧𝑖 𝜎 + 𝑝𝑖 (8.37) 𝑧𝑖 , 𝑝𝑖 : the negative of the zero and pole values, respectively, of 𝐺 𝜎 𝐻(𝜎) Solving Eq (8.37) for 𝜎, the real-axis values that minimize or maximize 𝐾, yields the breakaway and break-in points without differentiating 𝐾𝐺 𝑠 𝐻 𝑠 = −1 = 1∠ 2𝑘 + 1800 𝑘 = 0,±1, … System Dynamics and Control 8.34 (8.13) Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch The 𝑗𝜔-Axis Crossings Use the Routh-Hurwitz criterion to find the 𝑗𝜔-axis crossing as follows Forcing a row of zeros in the Routh table will yield the gain Going back one row to the even polynomial equation and solving for the roots yields the frequency at the imaginaryaxis crossing (8.37) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 𝑚 ෍ 8.35 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch - Ex.8.5 Frequency and Gain at Imaginary-Axis Crossing For the given system, find the frequency and gain, 𝐾, for which the root locus crosses the imaginary axis For what range of 𝐾 is the system stable? Solution The closed-loop transfer function for the system 𝐺(𝑠) 𝑇 𝑠 = + 𝐺(𝑠) 𝐾(𝑠 + 3) = 𝑠 + 7𝑠 + 14𝑠 + + 𝐾 𝑠 + 3𝐾 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.36 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch 𝑇 𝑠 = 𝐾(𝑠 + 3) 𝑠 + 7𝑠 + 14𝑠 + + 𝐾 𝑠 + 3𝐾 A complete row of zeros yields the possibility for imaginary axis roots 𝑠1 : −𝐾2 − 65𝐾 + 720 = → 𝐾 + 74.65 𝐾 − 9.65 = → 𝐾 = 9.65 Forming the even polynomial by using the 𝑠 row with 𝐾 = 9.65 𝑠 : 90 − 𝐾 𝑠 + 21𝐾 = 80.35𝑠 + 202.7 = → 𝑠 = ±𝑗1.59 The system is stable for ≤ 𝐾 ≤ 9.65 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.37 Root Locus Techniques §5.Refining the Sketch Angles of Departure and Arrival Assume a point on the root locus 𝜖 close to a complex pole/zero, the sum of angles drawn from all finite poles and zeros to this point is an odd multiple of 1800 a b System Dynamics and Control 8.38 Root Locus Techniques §5.Refining the Sketch - Ex.8.6 Angle of Departure from a Complex Pole Find the angle of departure from the complex poles and sketch the root locus Solution Calculate the sum of angles draw to a point 𝜖 close to the complex pole, −1 + 𝑗1, in the second quadrant −𝜃1 − 𝜃2 + 𝜃3 − 𝜃4 1 = −𝜃1 − 900 + 𝑡𝑎𝑛−1 + 𝑡𝑎𝑛−1 = 1800 ⟹ 𝜃1 = −251.60 = 108.40 −𝜃1 + 𝜃2 + 𝜃3 − 𝜃4 − 𝜃5 +𝜃6 = 2𝑘 + 1800 ⟹ 𝜃1 −𝜃1 + 𝜃2 + 𝜃3 − 𝜃4 − 𝜃5 + 𝜃6 = 2𝑘 +1 1800 ⟹ 𝜃2 Open-loop poles and zeros and calculation of a angle of departure; b angle of arrival HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.39 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Plotting and Calibrating the Root Locus Given the root locus ⟹ Find the exact point at which the locus crosses the 0.45 damping ratio line and the gain at that point HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.40 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch𝑗𝜔 𝜎 The system’s open-loop poles and zeros along with the 𝜁 = 0.45 line The point at radius 𝑟 = on the 𝜁 = 0.45 line 𝜃2 − 𝜃1 − 𝜃3 − 𝜃4 − 𝜃5 = −251.50≠ (2𝑘 + 1)1800 ⟹ this point is not on the root locus The gain, 𝐾, at this point 𝐴 𝐶 𝐷 ȁ𝐸ȁ 𝐾= = 1.71 ȁ𝐵ȁ ∑𝑛1 𝜃𝑧𝑖 − ∑𝑚 𝜃𝑝𝑖 = (2𝑘 + 1)180 ς pole lengths 1 𝐾= = = ȁ𝐺(𝑠)ȁȁ𝐻(𝑠)ȁ 𝑀 ς zero lengths HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.41 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Skill-Assessment Ex.8.4 Problem Given a unity feedback system that has the forward TF 𝐾(𝑠 + 2) 𝐺 𝑠 = 𝑠 − 4𝑠 + 13 the following a.Sketch the root locus b.Find the imaginary-axis crossing c.Find the gain, 𝐾, at the 𝑗𝜔-axis crossing d.Find the break-in point e.Find the angle of departure from the complex poles (8.20) (8.21) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.42 Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Solution a.The sketched root locus 𝐾(𝑠 + 2) 𝐺 𝑠 = 𝑠 − 4𝑠 + 13 b.The imaginary-axis crossing 𝐺(𝑠) 𝑇 𝑠 = + 𝐺(𝑠) 𝐾(𝑠 + 2) = 𝑠 − (𝐾 − 4)𝑠 + (2𝐾 + 13) The Routh table from row of zero → 𝐾 = from 𝑠 row with 𝐾 = → 𝑠 + 21 = → 𝑠 = ±𝑗 21 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.43 Root Locus Techniques §5.Refining the Sketch c.The gain, 𝐾, at the 𝑗𝜔-axis crossing From (b) → 𝐾 = d.The break-in point From Eq.(8.32) 𝜎 − 4𝜎 + 13 𝐾=− 𝜎+2 𝑑𝐾 2𝜎 − 𝜎 + − 𝜎 − 4𝜎 + 13 ⟹ =− 𝑑𝜎 𝜎+2 𝜎 + 4𝜎 − 21 (𝜎 + 7)(𝜎 − 3) =− =− 𝜎+2 𝜎+2 𝑑𝐾 = ⟹ 𝜎 = −7 𝑑𝜎 𝐾 = −𝐺 𝜎 𝐻(𝜎) 8.44 8.45 Nguyen Tan Tien Root Locus Techniques HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.46 §6.An Example Basic Rules for Sketching the Root Locus Number of branches number of branches = number of closed-loop poles Symmetry The root locus is symmetrical about the real axis Real-axis segments The root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros Starting and ending points The root locus - begins at the finite and infinite poles of 𝐺(𝑠)𝐻(𝑠), and - ends at the finite and infinite zeros of 𝐺(𝑠)𝐻(𝑠) §6.An Example HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Root Locus Techniques §5.Refining the Sketch e.The angle of departure from the complex poles 𝑗𝜔 𝜃2 Draw vectors to a point 𝜖 close to the 𝑗3 complex pole 𝜃3 − 𝜃2 − 𝜃1 = 1800 −1 𝜃3 − 𝜃2 − 900 = 1800 𝜎 ⟹ 𝑡𝑎𝑛 −2 −1 ⟹ 𝜃2 = 𝑡𝑎𝑛−1 − 2700 𝜃1 = −233.1 −𝑗3 = 126.90 (8.32) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 8.47 Nguyen Tan Tien Root Locus Techniques §6.An Example Additional Rules for Refining the Sketch Real-axis breakaway and break-in points 𝜎𝐾 ∶ breaks away point 𝑑𝐾 𝐾=− → = → 𝜎 → 𝑚𝑎𝑥 𝜎𝐾𝑚𝑖𝑛 ∶ breaks into point 𝐺 𝜎 𝐻(𝜎) 𝑑𝜎 Calculation of 𝑗𝜔-axis crossings Using the Routh-Hurwitz criterion • Forcing a row of zeros in the Routh table will yield the gain • Going back one row to the even polynomial equation and solving for the roots yields the frequency at the imaginary-axis crossing HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Root Locus Techniques Behavior at infinity The root locus approaches straight lines as asymptotes as the locus approaches infinity The asymptotes are given by the real-axis intercept and angle in radians as follows ∑𝑛1 𝑝𝑖 − ∑𝑚 𝑧𝑖 𝜎𝑎 = 𝑛−𝑚 (2𝑘 + 1)𝜋 𝜃𝑎 = 𝑛−𝑚 where 𝑘 = 0, ±1, ±2, … System Dynamics and Control 8.48 Nguyen Tan Tien Root Locus Techniques §6.An Example Angles of departure and arrival Assume a point 𝜖 close to the complex pole or zero Add all angles drawn from all open-loop poles and zeros to this point ∑𝑛1 𝜃𝑧𝑖 − ∑𝑚 𝜃𝑝𝑖 = (2𝑘 + 1)180 Solving for the unknown angle yields the angle of departure or arrival Plotting and calibrating the root locus All points on the root locus satisfy the relationship ∠𝐺 𝑠 𝐻 𝑠 = (2𝑘 + 1)1800 The gain, 𝐾, at any point on the root locus is given by 1 ς finite pole lengths 𝐾= = − ȁ𝐺 𝑠 𝐻(𝑠)ȁ 𝑀 ς finite zero lengths HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.49 Root Locus Techniques §6.An Example - Ex.8.7 Sketching a Root Locus and Finding Critical Points System Dynamics and Control System Dynamics and Control 8.51 Nguyen Tan Tien Root Locus Techniques §6.An Example HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.52 Nguyen Tan Tien Root Locus Techniques §6.An Example b.The exact point and gain where the locus crosses the 𝑗𝜔-axis The closed-loop TF for the system 𝐺(𝑠) 𝐾(𝑠 − 4𝑠 + 20) 𝑇 𝑠 = = + 𝐺(𝑠) (1 + 𝐾)𝑠 + (6 − 4𝐾)𝑠 + (8 + 20𝐾) The Routh table A complete row of zeros yields the possibility for imaginary axis roots 𝑠1 :6 − 4𝐾 = → 𝐾 = 1.5 Forming the even polynomial by using the 𝑠 row with 𝐾 = 𝑠 : + 𝐾 𝑠 + (8 + 20𝐾) = 2.5𝑠 + 38 = → 𝑠 = ±𝑗3 The root locus crosses the 𝑗𝜔-axis at ±𝑗3.9 with a gain of 𝐾 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Root Locus Techniques First sketch the root locus - the real-axis segment 𝜎 = −2, −4 - the root locus starts at the openloop poles (−2,−4) and ends at the open-loop zeros (2 ± 𝑗4) a.The exact point and gain where the locus crosses the 0.45 damping ratio line 𝜁 Searching in polar coordinates, we find that the root locus crosses the 𝜁 = 0.45 line at 3.4∠116.70 with a gain, 𝐾 = 0.417 Sketch the root locus for the system and find the following a.The exact point and gain where the locus crosses the 0.45 damping ratio line b.The exact point and gain where the locus crosses the 𝑗𝜔-axis c.The breakaway point on the real axis d.The range of 𝐾 within which the system is stable HCM City Univ of Technology, Faculty of Mechanical Engineering 8.50 §6.An Example Solution 8.53 Nguyen Tan Tien Root Locus Techniques c.The breakaway point on the real axis From Eq.(8.32) 𝜎 + 6𝜎 + 𝐾=− =− 𝐺 𝜎 𝐻 𝜎 𝜎 − 4𝜎 + 20 𝑑𝐾 2𝜎 + 𝜎 − 4𝜎 + 20 − 2𝜎 − (𝜎 + 6𝜎 + 8) ⟹ =− 𝑑𝜎 𝜎 − 4𝜎 + 20 10𝜎 − 24𝜎 − 152 (𝜎 − 5.279)(𝜎 + 2.879) = = 𝜎 − 4𝜎 + 20 𝜎 − 4𝜎 + 20 𝑑𝐾 = ⟹ 𝜎 = −2.879 𝑑𝜎 d.From the answer to b, the system is stable for ≤ 𝐾 ≤ 1.5 𝐾=− 𝐺 𝜎 𝐻(𝜎) (8.32) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.54 Nguyen Tan Tien Root Locus Techniques §6.An Example Run ch8p1 in Appendix B Learn how to use MATLAB to • plot and title a root locus • overlay constant 𝜁 and 𝜔𝑛 curves • zoom into and zoom out from a root locus • interact with the root locus to find critical points as well as gains at those points ã solve Ex.8.7 Đ6.An Example Skill-Assessment Ex.8.5 Problem Given a unity feedback system that has the forward TF 𝐾(𝑠 − 2)(𝑠 − 4) 𝐺 𝑠 = 𝑠 + 6𝑠 + 25 the following a.Sketch the root locus b.Find the imaginary-axis crossing c.Find the gain, 𝐾, at the 𝑗𝜔-axis crossing d.Find the break-in point e.Find the point where the locus crosses the 0.5 damping ratio line f Find the gain at the point where the locus crosses the 0.5 damping ratio line g.Find the range of gain, 𝐾, for which the system is stable HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 18/03/2016 System Dynamics and Control 8.55 Root Locus Techniques §6.An Example 𝐾(𝑠 − 2)(𝑠 − 4) Solution a.Sketch the root locus 𝐺 𝑠 = 𝑠 + 6𝑠 + 25 - the real-axis segment 𝜎 = 2, - the root locus starts at the open-loop poles (−3 ± 𝑗4) and ends at the open-loop zeros (2, 4) b.Find the imaginary-axis crossing 𝐾(𝑠 − 2)(𝑠 − 4) 𝑇 𝑠 = (1 + 𝐾)𝑠2 + 6(1 − 𝐾)𝑠 + (25 + 8𝐾) The Routh table from row of zero → 𝐾 = from 𝑠 row with 𝐾 = → 2𝑠2 + 33 = → 𝑠 = ±𝑗 11.5 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.57 Nguyen Tan Tien Root Locus Techniques System Dynamics and Control 8.56 Root Locus Techniques §6.An Example c.Find the gain, 𝐾, at the 𝑗𝜔-axis crossing From the answer to b, the gain 𝐾 = d.Find the break-away point From Eq.(8.32) 𝐾=− 𝐺 𝜎 𝐻 𝜎 𝜎 + 6𝜎 + 25 =− 𝜎 − 6𝜎 + 𝑑𝐾 12𝜎 + 34𝜎 − 198 ⟹ = 𝑑𝜎 𝜎 − 6𝜎 + (𝜎 − 2.885)(𝜎 + 5.719) = 𝜎 − 6𝜎 + 𝑑𝐾 = ⟹ 𝜎 = 2.885 𝑑𝜎 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.58 Nguyen Tan Tien Root Locus Techniques §6.An Example e.Find the point where the locus crosses the 0.5 damping ratio line Searching along 𝜁 = 0.5 for the 1800 point we find 𝑠 = −2.42 + 𝑗4.18 f Find the gain at the point where the locus crosses the 0.5 damping ratio line For the result in part e, 𝐾 = 0.108 g Find the range of gain, 𝐾, for which the system is stable Using the result from part c and the root locus, 𝐾 < §7.Transient Response Design via Gain Adjustment - The formulas describing percent overshoot, settling time, and peak time were derived only for a system with two closed-loop complex poles and no closed-loop zeros - Conditions for 2nd-order Approximations • Higher-order poles are much farther into the left half of the 𝑠plane than the dominant 2nd-order pair of poles HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.59 Nguyen Tan Tien Root Locus Techniques • Closed-loop zeros near closed-loop 2nd-order poles are canceled by proximity of higher-order closed-loop poles • Closed-loop zeros not canceled by proximity of higher-order closed-loop poles are far from closed-loop 2nd-order poles System Dynamics and Control 8.60 Nguyen Tan Tien Root Locus Techniques §7.Transient Response Design via Gain Adjustment Design Procedure for Higher Order Systems 1.Sketch the root locus for the given system 2.Assume 2nd-order system without any zeros, find the gain, 𝐾, to meet transient response specification 3.Justify the 2nd-order assumption by - evaluating that all higher-order poles are much farther from the 𝑗𝜔-axis than the dominant 2nd-order pair (5 times farther) - verifying that closed-loop zeros are approximately canceled by higher-order poles, or be sure that the zero is far removed from the dominant 2nd-order pole pair to yield approximately the same response obtained without the finite zero 4.If the assumptions cannot be justified, your solution will have to be simulated in order to be sure it meets the transient response specification §7.Transient Response Design via Gain Adjustment - Ex.8.8 Third-Order System Gain Design HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Given the system, design the value of gain, 𝐾, to yield 1.52% overshoot Also estimate the settling time, peak time, and steady-state error Nguyen Tan Tien 10 18/03/2016 System Dynamics and Control 8.61 Root Locus Techniques §7.Transient Response Design via Gain Adjustment - Solution HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.63 Nguyen Tan Tien Root Locus Techniques To test our assumption of a 2nd-order system, calculate the location of the 3rd pole When the gain is set to meet the transient response, we have also designed the steady-state error For the example, the steady-state error specification is given by 𝐾𝑣 and is calculated 𝐾(𝑠 + 1.5) 𝐾 × 1.5 𝐾𝑣 = lim 𝑠𝐺 𝑠 = lim 𝑠 = 𝑠→0 𝑠→0 𝑠(𝑠 + 1)(𝑠 + 10) × 10 System Dynamics and Control 8.65 8.62 Root Locus Techniques §7.Transient Response Design via Gain Adjustment Use the root locus program to search along the 0.8 damping ratio line for the point where the root locus crosses 1.52 percent overshoot line Three points satisfy this criterion: −0.87 ± 𝑗0.66 (𝐾 = 7.36), −1.19 ± 𝑗0.90 (𝐾 = 12.79), −4.6 ± 𝑗3.45 (𝐾 = 39.64) §7.Transient Response Design via Gain Adjustment For each point the settling time and peak time are evaluated using 𝜋 𝑇𝑠 = 𝑇𝑝 = 𝜁𝜔𝑛 𝜔𝑛 − 𝜁 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien Root Locus Techniques HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.64 Nguyen Tan Tien Root Locus Techniques §7.Transient Response Design via Gain Adjustment Run ch8p2 in Appendix B Learn how to use MATLAB to • enter a value of percent overshoot from the keyboard • draw the root locus and overlay the percent overshoot line requested • interact with MATLAB and select the point of intersection of the root locus with the requested percent overshoot line • interact with the root locus to find critical points as well as gains at those points • solve Ex.8.8 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.66 Nguyen Tan Tien Root Locus Techniques §7.Transient Response Design via Gain Adjustment Explore the SISO Design Tool in Appendix E The SISO Design Tool is a convenient and intuitive way to obtain, view, and interact with a system’s root locus Section D.7 describes the advantages of using the tool, while Section D.8 describes how to use it For practice, you may want to apply the SISO Design Tool to some of the problems at the end of this chapter §7.Transient Response Design via Gain Adjustment Skill-Assessment Ex.8.6 Problem Given a unity feedback system that has the forwardpath TF 𝐾 𝐺 𝑠 = (𝑠 + 2)(𝑠 + 4)(𝑠 + 6) the following a.Sketch the root locus b.Using a second-order approximation, design the value of 𝐾 to yield 10% overshoot for a unit-step input c.Estimate the settling time, peak time, rise time, and steady-state error for the value of 𝐾 designed in (b) d.Determine the validity of your 2nd-order approximation HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 11 18/03/2016 System Dynamics and Control 8.67 Root Locus Techniques §7.Transient Response Design via Gain Adjustment Solution a.Sketch the root locus b.Design the value of 𝐾 to yield 10%𝑂𝑆 Searching along the 𝜁 = 0.591 (10%𝑂𝑆) line for the 1800 point yields −2.028 + 𝑗2.768 with 𝐾 = 45.55 c.Estimate 𝑇𝑠 ,𝑇𝑝 ,𝑇𝑟 ,𝑒𝑠𝑡𝑒𝑝 (∞) 4 𝑇𝑠 = = = 1.97𝑠 ȁ𝑅𝑒ȁ 2.028 𝜋 𝜋 𝑇𝑝 = = = 1.13𝑠 ȁ𝐼𝑚ȁ 2.768 1.8346 𝜔𝑛𝑇𝑟ቚ ≈ 1.8346 ⟹ 𝑇𝑟 = = 0.53𝑠 𝜁=0.591 2.2282 + 2.7682 System Dynamics and Control 8.68 Root Locus Techniques §7.Transient Response Design via Gain Adjustment 𝑇𝑠 = 1.97𝑠 𝑇𝑝 = 1.13𝑠 𝑇𝑟 = 0.53𝑠 System is Type 𝐾 45.55 𝐾𝑝 = = = 0.949 2×4×6 48 1 ⟹ 𝑒𝑠𝑡𝑒𝑝 (∞) = = = 0.51 + 𝐾𝑝 + 0.949 d.Determine the validity of your 2nd-order approximation Searching the real axis to the left of −6 for the point whose gain is 45.55, we find −7.94 Comparing this value to the real part of the dominant pole, −2.028, we find that it is not five times further The 2nd-order approximation is not valid 𝜔𝑛 𝑇𝑟 ≈ 1.76𝜁 − 0.417𝜁 + 1.039𝜁 + HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.69 Nguyen Tan Tien Root Locus Techniques §8.Generalized Root Locus Consider the system requiring a root locus calibrated with 𝑝1 as a parameter The open-loop TF 𝐾𝐺 𝑠 𝐻 𝑠 = 10 (𝑠 + 2)(𝑠 + 𝑝1 ) The closed-loop TF 10 𝐾𝐺 𝑠 10 = = 𝑠 +2𝑠+ 10 𝑝 (𝑠 + 2) 1+𝐾𝐺 𝑠 𝐻(𝑠) 𝑠 +2𝑠+10+𝑝1(𝑠+2) 1+ 21 𝑠 +2𝑠+ 10 𝑝1 (𝑠 + 2) Implies that we have the system 𝐾𝐺 𝑠 𝐻(𝑠) = (8.62) 𝑠 + 2𝑠 + 10 The root locus can now be sketched as a function of 𝑝1 , assuming the open-loop system (8.62) 𝑇 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.71 Nguyen Tan Tien Root Locus Techniques §9.Root Locus for Positive-Feedback Systems - The properties of the root locus change dramatically if the feedback signal is added to the input rather than subtracted 𝐾𝐺(𝑠) (8.63) − 𝐾𝐺 𝑠 𝐻(𝑠) - We now retrace the development of the root locus for the denominator of Eq (8.63) Obviously, a pole, 𝑠, exists when 𝐾𝐺 𝑠 𝐻 𝑠 = = 1∠𝑘3600 𝑘 = 0, ±1, ±2, … (8.64) the root locus for positive-feedback systems consists of all points on the 𝑠-plane where the angle of 𝐾𝐺 𝑠 𝐻 𝑠 = 𝑘3600 →Refer the textbook - The closed-loop TF 𝑇 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.70 Nguyen Tan Tien Root Locus Techniques §8.Generalized Root Locus Skill-Assessment Ex.8.7 Problem Sketch the root locus for variations in the value of 𝑝1, for a unity feedback system that has the following forward TF 100 𝐺 𝑠 = 𝑠(𝑠 + 𝑝1 ) Solution Find the closed-loop TF and put it the form that yields 𝑝1 as the root locus variable 𝐺 𝑠 100 𝑇 𝑠 = = 1+𝐺 𝑠 𝑠 + 𝑝1 𝑠 + 100 100 100 = = 𝑠 +𝑝100 1𝑠 (𝑠 + 100) + 𝑝1𝑠 + 𝑠2 + 100 𝑝1 𝑠 Hence, 𝐾𝐺 𝑠 𝐻 𝑠 = 𝑠 + 100 ⟹ the root locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.72 Nguyen Tan Tien Root Locus Techniques §10.Pole Sensitivity The root locus exhibits a nonlinear relationship between gain and pole location Along some sections of the root locus • very small changes in gain yield very large changes in pole location and hence performance → high sensitivity to changes in gain • very large changes in gain yield very small changes in pole location → low sensitivity to changes in gain We prefer systems with low sensitivity to changes in gain HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 12 18/03/2016 System Dynamics and Control 8.73 Root Locus Techniques System Dynamics and Control 8.74 Root Locus Techniques §10.Pole Sensitivity - The sensitivity of a closed-loop pole, 𝑠, to gain, 𝐾 𝐾 𝛿𝑠 (8.69) 𝑆𝑠:𝐾 = × 𝑠 𝛿𝐾 𝑠 : the current pole location 𝐾: the current gain - Converting the partials to finite increments, the actual change in the closed-loop poles can be approximated Δ𝐾 (8.70) Δ𝑠 = 𝑠 × 𝑆𝑠:𝐾 × 𝐾 Δ𝑠 : the change in pole location Δ𝐾/𝐾 : the fractional change in the gain, 𝐾 §10.Pole Sensitivity - Ex.8.10 Root Sensitivity of a Closed-Loop System to Gain Variations HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.75 Nguyen Tan Tien Root Locus Techniques §10.Pole Sensitivity 𝑆𝑠:𝐾ቚ 𝑠=−9.47 = 𝐾 −1 −1 ቤ = −0.059 𝑠 2𝑠 + 10 𝑠=−9.47 −9.47 × (−9.47) + 10 The change in the pole location for a 10% change in 𝐾 Δ𝐾 Δ𝑠 = 𝑠 𝑆𝑠:𝐾 = −9.47 × −0.059 × 10% = 0.056 𝐾 𝐾 −1 50 −1 𝑆𝑠:𝐾 ቚ = ቤ 𝑠 2𝑠 + 10 𝑠=−9.47 −5 + 𝑗5 × (−5 + 𝑗5) + 10 𝑠=−5+𝑗5 = 1/(1 + 𝑗1) = (1/ 2)∠ − 450 The change in the pole location for a 10% change in 𝐾 Δ𝐾 Δ𝑠 = 𝑠 𝑆𝑠:𝐾 = −5 + 𝑗5 × [1/(1 + 𝑗1)] × 10% = −𝑗5 𝐾 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.77 Nguyen Tan Tien Root Locus Techniques §10.Pole Sensitivity 𝜕𝑠 𝑠+1 =− 𝜕𝐾 2𝑠 + 𝐾 + 𝐾 𝜕𝑠 −𝐾(𝑠 + 1) ⟹ 𝑆𝑠:𝐾 = = 𝑠 𝜕𝐾 𝑠(2𝑠 + 𝐾 + 2) −10(𝑠 + 1) ⟹ 𝑆𝑠:𝐾 ቚ = 𝑠(𝑠 + 11) 𝐾=20 The closed-loop poles when 𝐾 = 20 𝐷(𝑠)ቚ = 𝑠2 + 22𝑠 + 20 = (𝑠 + 21.05)(𝑠 + 0.95) = Find the root sensitivity of the system at 𝑠 = −9.47 and −5 + 𝑗5 Also calculate the change in the pole location for a 10% change in 𝐾 Solution Differentiating the closed-loop TF denominator, w.r.t 𝐾 𝜕𝑠 𝜕𝑠 𝜕𝑠 −1 2𝑠 + 10 +1=0⟹ = 𝜕𝐾 𝜕𝐾 𝜕𝐾 2𝑠 + 10 The sensitivity 𝐾 𝛿𝑠 𝐾 −1 𝑆𝑠:𝐾 = = 𝑠 𝛿𝐾 𝑠 2𝑠 + 10 System Dynamics and Control 8.76 Nguyen Tan Tien Root Locus Techniques §10.Pole Sensitivity Skill-Assessment Ex.8.9 Problem A negative unity feedback system has the forward TF 𝐾(𝑠 + 1) 𝐺 𝑠 = 𝑠(𝑠 + 2) If 𝐾 is set to 20, find the changes in closed-loop pole location for a 5% change in 𝐾 Solution The closed-loop TF 𝐾(𝑠 + 1) 𝑇 𝑠 = 𝑠 + 𝐾+2 𝑠+𝐾 Differentiating the denominator with respect to 𝐾 𝜕𝑠 𝜕𝑠 𝜕𝑠 2𝑠 + 𝐾 + + 𝑠 + = 2𝑠 + 𝐾 + +𝑠 +1 = 𝜕𝐾 𝜕𝐾 𝜕𝐾 𝜕𝑠 𝑠+1 ⟹ =− 𝜕𝐾 2𝑠 + 𝐾 + HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.78 Nguyen Tan Tien Root Locus Techniques §11 Case Study 𝐾=20 For the pole at −21.05 Δ𝐾 −10 × −21.05 + Δ𝑠 = 𝑠 𝑆𝑠:𝐾 = −21.05 0.05 = −0.9975 𝐾 −21.05× −21.05+11 For the pole at −0.95 Δ𝐾 −10 × −0.95 + Δ𝑠 = 𝑠 𝑆𝑠:𝐾 = −0.95 0.05 = −0.0025 𝐾 −0.95 × −0.95 + 11 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 13 ... Root Locus Techniques Root Locus Techniques §4.Sketching the Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 8.28 §4.Sketching the Root Locus. .. 18/03/2016 System Dynamics and Control 8.25 Root Locus Techniques §4.Sketching the Root Locus - Ex.8.2 Sketching a Root Locus with Asymptotes Sketch the root locus for the system Solution System Dynamics... Dynamics and Control 8.29 Nguyen Tan Tien Root Locus Techniques System Dynamics and Control 8.30 Nguyen Tan Tien Root Locus Techniques Nguyen Tan Tien Root Locus Techniques §5.Refining the Sketch Real-Axis