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Lecture Outline Systeem- en Regeltechniek II Previous lecture: The - - PowerPoint PPT Presentation

Mar 14, 2023 160 likes •228 views

Lecture Outline Systeem- en Regeltechniek II Previous lecture: The root locus method, analysis, design. Lecture 6 Root Locus, Frequency Response Robert Babu ska Today: Remarks on the computer session. Delft Center for Systems and

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SLIDE 1

Systeem- en Regeltechniek II

Lecture 6 – Root Locus, Frequency Response

Robert Babuˇ ska Delft Center for Systems and Control Faculty of Mechanical Engineering Delft University of Technology The Netherlands e-mail: r.babuska@dcsc.tudelft.nl www.dcsc.tudelft.nl/˜babuska tel: 015-27 85117

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 1

Lecture Outline

Previous lecture: The root locus method, analysis, design. Today:

  • Remarks on the computer session.
  • RL: additional examples.
  • Realistic PID controller.
  • Frequency response.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 2

Matlab / Simulink Computer Session

  • Do your homework:

– Read the entire handout (know what to do). – Work out by hand items a) through e) of Section 5.

  • Bring the handout with you to the computer lab.
  • Be on time, please, 10 min in advance.

You may bring your own laptop to the lab, if you want.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 3

RL: Effect of Parameter Change

Consider our DC motor in a position control loop:

  • 1. Use root locus to design a P controller for the nominal system

with: Kt = 0.5, R = 1, L = 0, b = 0.1, J = 0.01

  • 2. Use root locus to analyze how the closed-loop poles change

when the moment of inertia J of the load changes.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 4

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SLIDE 2

DC Motor: Root Locus for P Control

G(s) = θ(s) E(s) = Kt s[(Ls + R)(Js + b) + K2

t ]

Consider L = 0: G(s) = θ(s) E(s) = Kt s[(JRs + bR) + K2

t ] =

k s(s + a) with k = Kt JR, a = bR + K2

t

JR Design a proportional controller such that the closed loop has a double real pole (use rltool in Matlab).

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 5

DC Motor: Root Locus for P Control

Kp = 6.125 p1,2 = −17.5

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 6

DC Motor: Closed-Loop Step Response

Step Response Time (sec) Amplitude 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.2 0.4 0.6 0.8 1

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 7

RL for Analysis: Varying Moment of Inertia

R s ( )

  • Kp

G s ( ) Y s ( )

DC motor with varying J

L(s) = G(s)Kp process and fixed controller in series K will represent influence of varying moment of inertia J

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 8

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SLIDE 3

RL for Analysis: Varying Moment of Inertia

G(s)Kp = kn s(s + an) for the given controller gain Kp: kn = KtKp JnR , an = bR + K2

t

JnR with Jn = 0.01 Dividing Jn by factor K means multiplying an and kn by K. Characteristic equation: s2 + K(san + kn) = 0 ⇒ 1 + Ksan + kn s2 = 0

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 9

DC Motor: Root Locus for Varying J

K = 0.1 ⇓ J = 10 · Jn

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 10

Large Inertia: Closed-Loop Step Response

Step Response Time (sec) Amplitude 0.5 1 1.5 2 2.5 3 3.5 0.5 1 1.5

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 11

Proper Systems

A system G(s) = B(s) A(s) for which deg A(s) ≥ deg B(s) is called proper (has not more zeros than poles). In reality only proper systems exist! Consequence: the ‘textbook’ form of the PD controller cannot be realized: C(s) = Kp(1 + Tds)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 12

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SLIDE 4

More Realistic PD Controller

Filtered derivative: C(s) = Kp

  • 1 +

Tds (Td/N)s + 1

  • where N is typically in the range 10 – 20.

This means that an additional pole is introduced far left on the real axis.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 13

Satellite Attitude Control Revisited

Transfer function: G(s) = Θ(s) T(s) = 1 s2 Compare the RL for ideal and realistic PD controller.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 14

PID Controller Used in Practice

U(s) = Kp

  • E(s) + 1

sTi E(s) − Tds (Td/N)s + 1Y (s)

  • – Derivative action applied to −Y (s) instead of E(s).

– Anti-windup scheme used for the integral action (prevent integration when the actuator becomes saturated).

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 15

Anti-Windup Tracking Scheme

u es v

Saturation Actuator

e

  • y

+

  • K T s

p d

1 Tt Kp Ti Kp 1 s

  • Robert Babuˇ

ska Delft Center for Systems and Control, TU Delft 16

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SLIDE 5

Root Locus: Homework Assignments

  • Read Chapter 5 of the book by Franklin et al.
  • Sketch root loci of first, second and third-order systems with

real zeros and both real and complex poles by hand.

  • For Examples 5.1 through 5.8 in the book verify the results by

using Matlab.

  • Problems at the end of Chapter 5: work out problems 5.1 and

5.2 by hand and a selection of the remaining problems by using Matlab.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 17

Representations of Transfer Functions

s-plane frequency response Transfer Functions

Frequency (rad/sec) Phase (deg); Magnitude (dB) Bode Diagrams
  • 40
  • 30
  • 20
  • 10
10
  • 1
1 0 0 1 0 1
  • 1 5 0
  • 1 0 0
  • 5 0

Frequency response:

  • Bode plot
  • Nyquist plot

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 18

Frequency Response: Setting

Consider a linear time invariant system: U(s) Y (s) G(s) ? Input: u(t) = M sin ωt What is the steady-state output?

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 19

Frequency Response: Laplace Transform

u(t) = M sin ωt U(s) = L {u(t)} = Mω s2 + ω2 = Mω (s + jω)(s − jω) Y (s) = G(s)U(s) = G(s) Mω (s + jω)(s − jω) y(t) = ?

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 20

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SLIDE 6

Partial Fraction Expansion

General Laplace transform of the output: Y (s) =

n

  • i=1

mi

  • j=1

Kij (s − pi)j + K s − jω + K∗ s + jω Corresponding time signal: y(t) =

n

  • i=1

mi

  • j=1

Kijtjepit

  • transient

+ Kejωt + K∗e−jωt

  • periodic signal

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 21

Compute Coefficients K and K*

In steady state: Y (s) = K s − jω + K∗ s + jω K = Y (s)(s − jω)|s=jω = G(s) Mω (s + jω)(s − jω)(s − jω)

  • s=jω

= G(s) Mω s + jω

  • s=jω

= G(jω)M 2j

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Compute Coefficients K and K*

K = M 2j G(jω) = M 2j |G(jω)| ej∠G(jω) and K∗ = −M 2j G(−jω) = −M 2j |G(jω)| e−j∠G(jω) Kejωt + K∗e−jωt = M |G(jω)| ej(ωt+∠G(jω)) − e−j(ωt+∠G(jω)) 2j = M |G(jω)| sin(ωt + ∠G(jω))

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 23

Frequency Response: Summary

u(t) = M sin ωt y(t) ≃ M |G(jω)| sin(ωt + ∠G(jω)) |G(jω)| . . . magnitude (gain) ∠G(jω) . . . phase

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