Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1
Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position Controller Plant G p (s) DV E i 168 K P s(0.3s1) V See how the closed-loop poles move as proportional gain K P varies from 0 to. Find closed-loop characteristic equation: 0.09 ME375 Root Locus - 2
Servo Table Example (cont.) ME375 Root Locus - 3
Motivation Example 1 Revisit the DC motor positioning system with proportional control. corresponding block diagram is: Plant G p (s) R(s) + Controller U(s) 27.6 K P ss ( 57.5) Y(s) Its Sketch the closed-loop poles as the controller gain K P varies from 0 to. Find closed-loop characteristic equation: 2 s s K P 57.5 27.6 0 ME375 Root Locus - 4
Example 1 Formulate an expression for the roots of the characteristic equation: Find the roots for K P = 0 and K P : Find K P when the roots are repeated. ME375 Root Locus - 5
Example 1 Sketch the root locus: Img. Axis 30 20 10 Real Axis -60-50 -40-30 -20-10 0 0-10 -20-30 ME375 Root Locus - 6
Example 2 Using the same plant as in Example 1, try a different controller choice: Plant G p (s) R(s) + Controller U(s) K d (s + 80) 27.6 ss ( 57.5) Y(s) Sketch the root locus of the closed-loop poles as the controller gain K d varies from 0to. Find closed-loop characteristic equation: ME375 Root Locus - 7
Example 2 Formulate an expression for the roots of the characteristic equation: Find the roots for K d = 0 and K d : Find repeated roots. ME375 Root Locus - 8
Example 2 Repeated roots (cont.): ME375 Root Locus - 9
Example 2 Sketch the root locus: Imag Axis 40 30 20 10 Real Axis 0-10 -20-30 -40-140 -120-100 -80-60 -40-20 0 ME375 Root Locus - 10
Closed-Loop Characteristic Roots (CL Poles) Reference Input R(s) + Error E(s) K P Control Input U(s) G p (s) Plant Output Y(s) H(s) The closed-loop transfer function G CL (s) is: G () CL s Img. j The closed-loop characteristic equation is: Real j ME375 Root Locus - 11
Definitions Root Locus Root Locus plotting is the method of determining the roots of the following equation on the complex plane when the parameter K varies from 0 : Ns () 1 KGOL ( s) 0 or 1K 0 Ds () where N(s) and D(s) are known polynomials in factorized form: N() s ( s z )( s z ) ( s z ) 1 2 D() s ( s p )( s p ) ( s p ) 1 2 The N Z roots of the polynomial N(s),z 1,z 2,,z Nz, are called the finite openloop zeros. The N P roots of the polynomial D(s), p 1,p 2,,p Np, are called the finite open-loop poles. N z N P ME375 Root Locus - 12
Root Locus Methods of obtaining root locus: Given a value of K, numerically solve the 1 + KG OL (s) = 0 equation for a set of roots. Repeat this for a set of K values and plot the corresponding roots on the complex plane. (This is what we did in the last in-class exercise.) Use MATLAB. In MATLAB use the commands rlocus and rlocfind. You can use on-line help to find the usage for these commands. 16 1 K P003. s( 0. 0174s1) >> op_num=[0.48]; >> op_den=[0.0174 1 0]; >> Gol=tf(op_num,op_den); >> rlocus(gol) >> [K, poles]=rlocfind(gol); 0.48 0 1KP 0. 0174s Apply root locus sketching rules to obtain an approximate root locus plot. 2 s 0 ME375 Root Locus - 13
Example 3 A feedback control system is proposed. The corresponding block diagram is: Controller Plant G p (s) R(s) + K U(s) 1 ( s 4) ss ( 2) Y(s) Sketch the root locus of the closed-loop poles as the controller gain K varies from 0to. Find closed-loop characteristic equation: ME375 Root Locus - 14
Example 3 Step 1: Formulate the (closed-loop) characteristic equation into the standard form for sketching root locus: Step 2: Find the open-loop zeros, z i, and the open-loop poles, p i : ME375 Root Locus - 15
Example 3 Step 3: Determine locations of repeated roots, if any. ME375 Root Locus - 16
Example 3 Step 4: Determine the imaginary axis crossings, if any. ME375 Root Locus - 17
Example 3 Step 5: Use the information from Steps 1-4 to sketch the root locus. Imag Axis 4 3 2 1 Real Axis 0-1 -2-3 -4-6 -5-4 -3-2 -1 0 ME375 Root Locus - 18
Revisit PID Control Examples Reference Input R(s) + Error E(s) Controller G C (s) Control Input U(s) Sensor H (s) Disturbance D(s) + + Plant G P (s) 4 (2s 1)(0.5s 1) Output Y(s) 1 K P () s (2s 1)(0.5s1) 4K (A) Proportional (P) control: G C (s) = K P 4 (B) Add Integral action, Proportional-Plus-Integral (PI) control: 1 K s K G s K K s s P I C() P I G YR G YR 4 Ks P KI () s s(2s1)(0.5s1) 4 K sk (C) Add Derivative action, Proportional-Plus-Integral-Plus Derivative (PID) control: 2 1 KDs KPsKI C() P I D G s K K K s s s G YR 2 4 KDs KPsKI () s s(2s1)(0.5s1) 4 K s K sk P I P 2 D P I ME375 Root Locus - 19
Effect of Proportional Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Open-loop poles and zeros: 2 s s K P 2.5 14 0 5 Repeated roots: Imaginary part of s 4 3 2 1 0-1 -2-3 -4-5 -8-7 -6-5 -4-3 -2-1 0 1 2 Real part of s ME375 Root Locus - 20
Effect of Proportional Control 1.5 Root Locus for K P 1 Imaginary Axis 0.5 0-0.5-1 -1.5-2.5-2 -1.5-1 -0.5 0 0.5 Real Axis ME375 Root Locus - 21
Effect of PI Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Open-loop poles and zeros: 3 2 s s KP s KI 2.5 14 4 0 5 Repeated roots: Imaginary axis crossings: Imaginary part of s 4 3 2 1 0-1 -2-3 -4-5 -8-7 -6-5 -4-3 -2-1 0 1 2 Real part of s ME375 Root Locus - 22
Effect of PI Control 30 30 Root Root Locus for K I I 20 20 Imaginary Axis Imaginary Axis 10 0-10 10 0-10 -20-20 -30-30 -20-10 0 10 20 Real Axis -30-30 -20-10 0 10 20 30 Real Axis ME375 Root Locus - 23
Effect of PID Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Open-loop poles and zeros: s 2.5 4K s 14K s4k 0 3 2 D P I 5 Repeated roots: Imaginary axis crossings: Imaginary part of s 4 3 2 1 0-1 -2-3 -4-5 -8-7 -6-5 -4-3 -2-1 0 1 2 Real part of s ME375 Root Locus - 24
Effect of PID Control 5 Root Locus for K D Imaginary Axis 0-5 -8-6 -4-2 0 Real Axis ME375 Root Locus - 25
PI Control Design Objective: Design a system that has zero steady state error for step inputs with %OS < 10% and T S (2%) < 6 [sec]. Reference Input R(s) + Error E(s) Controller G C (s) 1 Control Input U(s) Sensor H (s) Disturbance D(s) + + Plant G P (s) 4 (2s 1)(0.5s 1) Output Y(s) 1 K s K G s K K s s Closed loop transfer functions: G P I C() P I YR Ks P KI 4 G () () (2 1)(0.5 1) 4 C s GP s s s s KPs KI () s 1 G ( s) G ( s) H( s) KPs KI 4 s(2s1)(0.5s1) 4 1 K sk s (2s1)(0.5s1) C P P I ME375 Root Locus - 26
Design PI Control Closed Loop Characteristic Equation: let = 0.707 and n = 1 s(2s1)(0.5s1) 4 K sk 0 3 2 s s KP s KI 2.5 14 4 0 P I ME375 Root Locus - 27
Step Response with PI Control 1.4 Step Response 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 Time (sec) ME375 Root Locus - 28
PID Control Design Objective: Design a system that has zero steady state error for step inputs with %OS < 10% and T S (2%) < 6 [sec]. Reference Input R(s) + Error E(s) Controller G C (s) 1 Control Input U(s) Sensor H (s) Disturbance D(s) + + Plant G P (s) 4 (2s 1)(0.5s 1) Output Y(s) Closed loop transfer functions: G YR 2 1 KDs KPsKI C() P I D G s K K K s s s K s K sk 4 2 D P I 2 G () () (2 1)(0.5 1) 4 C s GP s s s s KDs KPsKI 2 2 C P KDs KPsKI 4 D P I () s 1 G ( s) G ( s) H( s) s(2s1)(0.5s1) 4 1 K s K sk s (2s1)(0.5s1) ME375 Root Locus - 29
Design PID Control Closed Loop Characteristic Equation: let = 0.707, n = 1, and a = 5 s s s K s K s K 2 (2 1)(0.5 1) 4 D P I 0 s 2.5 4K s 14K s4k 0 3 2 D P I ME375 Root Locus - 30
Step Response with PID Control 1.4 Step Response 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 Time (sec) ME375 Root Locus - 31