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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

Dr Branislav Hredzak
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.

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Introduction

• State space techniques can be applied to a wider class of systems
than 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


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Controller Design
• Let us consider an n-th order control system with an n-th-order
closed-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


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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


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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:


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Problem
Design the phase-variable feedback gains to yield 9.5% overshoot and a settling
time of 0.74 second for the given plant

20 s + 100
=
s 3 + 5s 2 + 4 s
Solution
⎡0 1 0 ⎤ ⎛ ⎡0 1 0 ⎤ ⎡0 ⎤ ⎞
⎜ ⎟
A = ⎢0 0 1⎥, ( A - BK ) = ⎜ ⎢0 0 1⎥ − ⎢0⎥[k1
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ k2 k 3 ]⎟,
⎜⎢ ⎟
⎣0 4 5 ⎥
⎢ ⎦ ⎝ ⎣0 4 5⎥ ⎢1⎥
⎦ ⎣ ⎦ ⎠

• Based on the desired response, we choose two closed loop poles as
p1,2 = -5.4 +/- j7.2

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

• Then the desired characteristic equation is
( s − p1 )( s − p 2 )( s − p3 ) = 0


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• The closed-loop system's characteristic equation is

• Equating coefficients with the desired characteristic equation

we obtain

• Hence

T ( s ) = C( sI − A) −1 B + D


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11.5% overshoot and a settling time
of 0.8 second

- does not meet the desired
Note, that there is a large specifications because the zero at -5
steady-state error ! was not cancelled

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


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Pole Placement for Plants NOT in Phase-Variable Form

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

Problem


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⎛ ⎡1 0⎤ ⎛ ⎡− 2 1 ⎤ ⎡0⎤ ⎞⎞ ⎛ ⎡1 0⎤ ⎡ − 2 1 ⎤⎞
det( sI − ( A − BK )) = det⎜ s ⎢ ⎥ −⎜ − [k k 2 ]⎟ ⎟ = det⎜ s ⎢
⎜ 0 1⎥ − ⎢− k
⎟=
⎜ ⎣0 1⎦ ⎜ ⎢ 0 − 1⎥ ⎢1⎥ 1
⎝ ⎝⎣ ⎦ ⎣ ⎦
⎟⎟
⎠⎠ ⎝ ⎣ ⎦ ⎣ 1 − (k 2 + 1)⎥ ⎟
⎦⎠


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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.

controllable uncontrollable
system system

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.


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The Controllability Matrix

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

An nth-order plant whose state equation is
is completely controllable if the controllability matrix CM

is of rank n.

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.


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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


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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


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Closed loop observer with feedback

• 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


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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.


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

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


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If the desired characteristic equation of the observer is

Then, we can find li’s as:

Observer Design for Plants in Observer-Canonical Form

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.


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Solution

The state equations for the estimated plant are

The observer error is

Characteristic polynomial is


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Characteristic polynomial is

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

l1 = 112, l2 = 2483, l3 = 49990

… simulation response to r(t) = 100t


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Observer Design for Plants NOT in Observer-Canonical Form

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

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

Solution Plant in phase variable form will be


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

l1 = −38.397 l2 = 35.506 Dr Branislav Hredzak

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

Observable Unobservable


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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)


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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


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• we now have an additional
pole to place.


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

Solution


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Using the requirements for settling time and percent overshoot, we find that the
desired 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

The characteristic polynomial for the system with integral action is

det(sI - )=


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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,

T ( s ) = C( sI − A) −1 B + D =


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The steady-state error for a unit step input:

e(∞) = 1 + CA −1B


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Name: _______________________

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.


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