Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus  1


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1 Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1
2 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 closedloop poles move as proportional gain K P varies from 0 to. Find closedloop characteristic equation: 0.09 ME375 Root Locus  2
3 Servo Table Example (cont.) ME375 Root Locus  3
4 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 closedloop poles as the controller gain K P varies from 0 to. Find closedloop characteristic equation: 2 s s K P ME375 Root Locus  4
5 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
6 Example 1 Sketch the root locus: Img. Axis Real Axis ME375 Root Locus  6
7 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 closedloop poles as the controller gain K d varies from 0to. Find closedloop characteristic equation: ME375 Root Locus  7
8 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
9 Example 2 Repeated roots (cont.): ME375 Root Locus  9
10 Example 2 Sketch the root locus: Imag Axis Real Axis ME375 Root Locus  10
11 ClosedLoop 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 closedloop transfer function G CL (s) is: G () CL s Img. j The closedloop characteristic equation is: Real j ME375 Root Locus  11
12 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 openloop poles. N z N P ME375 Root Locus  12
13 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 inclass exercise.) Use MATLAB. In MATLAB use the commands rlocus and rlocfind. You can use online help to find the usage for these commands K P003. s( s1) >> op_num=[0.48]; >> op_den=[ ]; >> Gol=tf(op_num,op_den); >> rlocus(gol) >> [K, poles]=rlocfind(gol); KP s Apply root locus sketching rules to obtain an approximate root locus plot. 2 s 0 ME375 Root Locus  13
14 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 closedloop poles as the controller gain K varies from 0to. Find closedloop characteristic equation: ME375 Root Locus  14
15 Example 3 Step 1: Formulate the (closedloop) characteristic equation into the standard form for sketching root locus: Step 2: Find the openloop zeros, z i, and the openloop poles, p i : ME375 Root Locus  15
16 Example 3 Step 3: Determine locations of repeated roots, if any. ME375 Root Locus  16
17 Example 3 Step 4: Determine the imaginary axis crossings, if any. ME375 Root Locus  17
18 Example 3 Step 5: Use the information from Steps 14 to sketch the root locus. Imag Axis Real Axis ME375 Root Locus  18
19 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, ProportionalPlusIntegral (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, ProportionalPlusIntegralPlus 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
20 Effect of Proportional Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Openloop poles and zeros: 2 s s K P Repeated roots: Imaginary part of s Real part of s ME375 Root Locus  20
21 Effect of Proportional Control 1.5 Root Locus for K P 1 Imaginary Axis Real Axis ME375 Root Locus  21
22 Effect of PI Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Openloop poles and zeros: 3 2 s s KP s KI Repeated roots: Imaginary axis crossings: Imaginary part of s Real part of s ME375 Root Locus  22
23 Effect of PI Control Root Root Locus for K I I Imaginary Axis Imaginary Axis Real Axis Real Axis ME375 Root Locus  23
24 Effect of PID Control Closed Loop Characteristic Equation: Rewrite the characteristic equation: Openloop poles and zeros: s 2.5 4K s 14K s4k D P I 5 Repeated roots: Imaginary axis crossings: Imaginary part of s Real part of s ME375 Root Locus  24
25 Effect of PID Control 5 Root Locus for K D Imaginary Axis Real Axis ME375 Root Locus  25
26 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
27 Design PI Control Closed Loop Characteristic Equation: let = and n = 1 s(2s1)(0.5s1) 4 K sk s s KP s KI P I ME375 Root Locus  27
28 Step Response with PI Control 1.4 Step Response Amplitude Time (sec) ME375 Root Locus  28
29 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
30 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 D P I ME375 Root Locus  30
31 Step Response with PID Control 1.4 Step Response Amplitude Time (sec) ME375 Root Locus  31
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