Pedulum Experiment

Pendulum Controller Experiment

Laboratory Report

Submitted

2007

Pendulum Controller Experiment

Laboratory Report

Summary

This is an investigation into the use of state-space dynamic models using state-feedback and pole-placement techniques. State-space design is a major method of designing feedback control systems. Hence, it is very important to understand the principle behind it and have knowledge of their limitations.

The following experiments look at the control of a crane and an inverted pendulum. Their step responses using various feedback gains of the state variables are analysed and compared to a linear model created in MATLAB.

It was found that the system poles were very sensitive to the feedback gain values. The linear models devised gave good approximations to stable crane systems but could not be used to model systems on the brink of becoming unstable or when large displacements were involved. The non-linear friction acting on the inverted pendulum caused limit cycling.

Table of Contents

Summary 2

Table of Contents 3

1. Introduction 4

2. Apparatus and Method 4

3. Theory 4

From Appendix I: section A.2 Inverted pendulum model 4

From Appendix I: A.5 Closed-loop characteristic equation 5

4. Results, Observations and Calculations 7

4.1 Crane Control: 7

4.1.1 Friction and Stiction 7

4.1.2 Carriage Controller: with only CP and CV feedback 7

4.1.3 Carriage Controller: with CP, CV and PP feedback 8

4.1.4 Pole-placement 8

4.1.5 Variation of p2 9

4.2. Inverted Pendulum: 9

4.2.1 No carriage feedback 9

4.2.2 Pole placement 10

4.2.3 Limit Cycles 10

4.2.4 No pendulum feedback 10

5. Analysis 10

4.1 Crane Control: 11

Figure 1: Crane system step response with p = [0.35 0.14 0.32 0.01] 11

Figure 2: Crane system step response with p = [0.84 0.77 0.29 0.63] 11

Figure 3: Crane system step response with p = [0.84 0.46 0.29 0.63] 11

4.2 Inverted Pendulum: 12

Figure 4: Inverted pendulum system step response with p = [0.25 0.32 0.35 0.28] 12

Figure 5: Inverted pendulum system step response with p = [0.23 0.20 0.63 0.40] 12

Figure 6: Inverted pendulum system step response with p = [0.23 0.78 0.63 0.40] 12

6. Conclusion 13

APPENDICES............................................................................................................12

Figure 1: Crane system step response with p = [0.35 0.14 0.32 0.01]..............................13

Figure 2: Crane system step response with p = [0.84 0.77 0.29 0.63] .............................14

Figure 3: Crane system step response with p = [0.84 0.46 0.29 0.63] .............................15

Figure 4: Inverted pendulum system step response with p = [0.25 0.32 0.35 0.28].........16

Figure 5: Inverted pendulum system step response with p = [0.23 0.20 0.63 0.40] .........17

Figure 6: Inverted pendulum system step response with p = [0.23 0.78 0.63 0.40] .........18

Appendix I: Laboratory Handout..............................................................................19

1. Introduction

State-space design is a major method of designing feedback control systems. It organizes the differential equations describing a dynamic system as a set of first-order vector differential equations and the solution can then be visualised as a trajectory of the state vector in space. Dynamic models are used in many areas of engineering and it is therefore important to understand the principles of analysing and ultimately, how to designing them.

The following experiments will aim to use the control of a crane/inverted pendulum to illustrate:

State-space dynamic models

State-feedback and pole-placement

Limit cycles and describing functions

2. Apparatus and Method

See laboratory handout (Appendix I).

3. Theory

(See Appendix I for detailed theory)

From Appendix I: section A.2 Inverted pendulum model

If we assume equations (14) and (15) in Appendix I are correct, they will linearize to equations (16) and (17). These are two simultaneous 2nd order o.d.e.'s equivalent to a 4th order o.d.e. in either x or, but they can also be written as a first order o.d.e. as follows.

Solving for and, we get

(15)

(17)

(a)

(14)

(16)