12 elec3114

Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise

Copyright © 2004 by John Wiley & Sons. All rights reserved.

1

Design via State Space

• How to design a state-feedback controller using pole placement to meet transient response specifications

• How to design an observer for systems where the states are not available to the controller

• How to design steady-state error characteristics for systems represented in state space



2

Introduction

• State space techniques can be applied to a wider class of systemsthan transform methods, for example, to systems with nonlinearities, multi-input multi-output (MIMO) systems

• Frequency domain methods of design – cannot be used to specify all closed-loop poles of the higher-order system

• State space techniques allow to place all poles of the closed-loop system

• Frequency domain methods – allow placement of zero through zero of the lead compensator

• State space techniques – do not allow to specify zero locations

• State space techniques – more sensitive to parameter variations



3

Controller Design

• Let us consider an n-th order control system with an n-th-orderclosed-loop characteristic equation

• we need n-adjustable parameters in order to be able to set the poles to any desired location

Topology for Pole Placement

Let us consider a plant:



4

Now, we introduce feedback , where in order to set the poles to the desired location



5

Phase-variable representation for plant

Plant with state-variable feedback



6

Pole Placement for Plants in Phase-Variable Form(Method of Matching the Coefficients)

1. Represent the plant in phase-variable form.

2. Feed back each phase variable to the input of the plant through a gain, ki

3. Find the characteristic equation for the closed-loop system represented in step 2.

4. Decide upon all closed-loop pole locations and determine an equivalent characteristic equation.

5. Equate like coefficients of the characteristic equations from steps 3 and 4 and solve for ki



7

Phase-variable representation of the plant is given by:

The characteristic equation of the plant is

Feedback:



8



9

The characteristic equation can be found by inspection as:

Now if the desired characteristic equation for proper pole placement is:

Then we can find the feedback gains ki as:



10

ProblemDesign the phase-variable feedback gains to yield 9.5% overshoot and a settlingtime of 0.74 second for the given plant

Solution

• Based on the desired response, we choose two closed loop poles asp1,2 = -5.4 +/- j7.2

• We choose the third closed-loop pole as p3 = -5.1

• Then the desired characteristic equation is

ssss

451002023 +++

=

0))()(( 321 =−−− pspsps

[ ] ,100

540100010

)( ,540100010

321⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= kkkBK-AA



11

• The closed-loop system's characteristic equation is

• Equating coefficients with the desired characteristic equation

we obtain

• Hence

DBA)IC( +−= −1)( ssT



12

11.5% overshoot and a settling time of 0.8 second

- does not meet the desired specifications because the zero at -5 was not cancelled

- if the third pole is chosen at -5 then the design will meet the desired specifications

Note, that there is a large steady-state error !



13

- consists of matching the coefficients of det(sI - (A - BK)) with the coefficients of the desired characteristic equation – (can result in difficult calculations)

Pole Placement for Plants NOT in Phase-Variable Form(Method of Matching the Coefficients)

Problem



14

[ ] =⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡+−−

−−⎥

⎦

⎤⎢⎣

⎡=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−

−−⎥

⎦

⎤⎢⎣

⎡=−−

)1(12

1001

det10

1012

1001

det))(det(21

21 kkskkss BKAI



15

Controllability• If any one of the state variables cannot be controlled by the control u,

then we cannot place the poles of the system where we desire.

If an input to a system can be found that takes every state variable from a desired initial state to a desired final state, the system is said to be controllable; otherwise, the system is uncontrollable.

controllable system

uncontrollable system



16

The Controllability Matrix

• enables to determine controllability for a plant under any representationor choice of state variables

The rank of CM equals the number of linearly independent rows or columns.The rank of CM equals n if the determinant of CM is non-zero.

An nth-order plant whose state equation is

is completely controllable if the controllability matrix CM

is of rank n.



17

Observer Design

• In some applications, some of the state variables may not be available at all, or it is too costly to measure them or send them to the controller

• In that case we can estimate states and the estimated states, rather than actual states, are then fed to the controller

Observer (Estimator) is used to calculate state variables that are not accessible from the plant.

Open-loop observer Closed-loop observer



18

Plant Observer

• the speed of convergence between the actual state and the estimated state is the same as the transient response of the plant since the characteristic equation is the same. Hence we cannot use the estimated states for the controller.

• to increase the speed of convergence between the actual and estimated states, we use feedback



19

• when designing an observer, the observer canonical form yields the easy solution for the observer gains

• the observer has to be faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector

• the design of the observer is separate from the design of the controller

Closed loop observer with feedback



20

Let

The design then consists of solving for the values of Lc to yield a desired characteristic equation.

The characteristic equation is det[sI - (A - LC)].

Then, we select the eigenvalues of the observer to yield stability and a desired transient response that is faster ( about 10 times) than the controlled closed-loop response.



21

Let us consider nth-order plant represented in observer canonical form:

The characteristic equation for (A-LC) is det[sI - (A - LC)] :



22

If the desired characteristic equation of the observer is

Then, we can find li’s as:

Problem Design an observer for the plant represented in observer canonical form. The observer will respond 10 times faster than the closed loop control system with poles at -1+/-j2 and -10.

Observer Design for Plants in Observer-Canonical Form(Method of Matching the Coefficients)



23

Solution

The state equations for the estimated plant are

The observer error is

Characteristic polynomial is



24

The closed loop controlled system has poles at -1+/-j2 and -10.

We choose observer poles 10 times faster, -10+/-j20 and 100, then the desired polynomial is

After equating coefficients, we can find

49990,2483,112 321 === lll

Characteristic polynomial is

… simulation response to r(t) = 100t



25

Observer Design for Plants NOT in Observer-Canonical Form(Method of Matching the Coefficients)

• match the coefficients of det[sI - (A - LC)] with the coefficients of the desired characteristic polynomial (can yield difficult calculations for higher-order systems)

Solution Plant in phase variable form will be

Problem Design an observer for the phase variables with a transient response described by ζ= 0.7 and ωn = 100.



26

Comparing the coefficients we can find the values of l1 and l2

506.35 397.38 21 =−= ll



27



28

Observability• If any state variable has no effect upon the output, then we cannot evaluate

this state variable by observing the output

Observable Unobservable



29

The Observability Matrix

- enables to determine observability for systems under any representation or choice of state variables

Plant

is completely observable if the observability matrix OM,

is of rank n (i.e., the determinant of OM is non-zero)



30

Steady-State Error Design via Integral Control• enables to design a system for zero steady-state error for a step input as

well as design the desired transient response



31

• we now have an additional pole to place.



32

Problema. Design a controller with integral control to yield a 10% overshoot and a settling time of 0.5 second.

Solution



33

Using the requirements for settling time and percent overshoot, we find that thedesired second order poles are s1,2= -8+/-j10.9 and desired characteristic polynomial is

We choose the third pole at -100 (real part more than 5 times greater than the desired second order poles s1,2= -8+/-j10.9).

Hence, the desired 3rd order characteristic equation is

det(sI - )=

The characteristic polynomial for the system with integral action is



34

Now we match the coefficients of the desired 3rd order characteristic equation

with the characteristic polynomial for the system with integral action

and we can find

Then,

=+−= − DBA)IC( 1)( ssT



35

The steady-state error for a unit step input:

BCA 1−+=∞ 1)(e



36Name: _______________________

Student ID: ___________________

Signature: ____________________

THE UNIVERSITY OF NEW SOUTH WALES

School of Electrical Engineering & Telecommunications

FINAL EXAMINATION

Session 2 2010

ELEC3114 Control Systems

TIME ALLOWED: 3 hours TOTAL MARKS: 100 TOTAL NUMBER OF QUESTIONS: 4

THIS EXAM CONTRIBUTES 65% TO THE TOTAL COURSE ASSESSMENT. Reading Time: 10 minutes.

This paper contains 9 pages .

Candidates must ATTEMPT ALL 4 questions.

Answer each question in a separate answer book.

Marks for each question are indicated beside the question.

This paper MAY be retained by the candidate.

Print your name, student ID and question number on the front page of each answer book.

Authorised examination materials:

Drawing instruments may be brought into the examination room.

Candidates should use their own UNSW-approved electronic calculators.

This is a closed book examination.

Assumptions made in answering the questions should be stated explicitly.

All answers must be written in ink. Except where they are expressly required, pencils may

only be used for drawing, sketching or graphical work.

12 elec3114

Education