AN INTRODUCTION TO CONTROL AND STRUCTURAL CONTROL

by: Oreste S. Bursi, Professor of Structural Dynamics and Control Dept. of Mechanical and Structural Engineering University of Trento, Trento, I

Doctoral School in Engineering of Civil and Mechanical Structural Systems

University of Trento, Italy

Spring 2008

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Content Page PART A: SINGLE INPUT SINGLE OUTPUT (SISO) SYSTEM CONTROL A1. Introduction 3 A2. Linear Dynamics of SISO Systems 4 A3. An Introduction to Control System Design 12 A4. Frequency Domain Design 27 A5. Discrete-Time Control System Design 40 PART B: MULTIVARIABLE SYSTEM CONTROL B1. Introduction 46 B2. Linear Dynamics of Multivariable Systems 48 B3. State Feedback Control 51 B4. Adaptive Control 62 APPENDICES

1. Table of Laplace (s) Transform Pairs 71

2. Table of s – z Transform Pairs 72

3. Summary of Linear Algebra 73

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PART A: SINGLE INPUT SINGLE OUTPUT (SISO) SYSTEM CONTROL

A1 INTRODUCTION • Most of this course contains material specific to approximately linear control problems, often

encountered when we wish to control test machines for civil and structural engineering applications, using servohydraulic actuators.

• It will be assumed that you have a basic background knowledge of system dynamics (differential equation models, system response, etc).

• Part A of the course introduces some of the techniques used to control systems with a single-input and a single output – ie SISO systems.

• Part B of the course introduces some of the techniques used to control systems with more than one input and output – ie multivariable systems.

• Robust, well-designed, linear controllers will often yield acceptable responses from nonlinear and parametrically uncertain test systems. This is the subject matter of sections A1 – B3.

• In the most demanding cases, nonlinearities and parameter variations can cause even well-designed linear controllers to yield inadequate responses. In such cases, we might use an adaptive controller of the type examined in section B4.

• Practical problems for the designer are highlighted and solutions presented. Such problems include the presence of stiction, signal noise, non-linear dynamics, load disturbances, parameter variations and signal aliasing.

• The emphasis of this course is on control system design, rather than control system analysis. Therefore, analytic developments are mainly left to the student’s own reading. Some introductory textbooks on the subject are listed below.

1) Close C.M., Frederick D.H. and Newell J.C. Modelling and Analysis of Dynamic Systems, Third

Edition, Wiley & Sons (USA)

2) Dorf R C and Bishop R H Modern Control Systems, Addison-Wesley (Reading MA)

3) Chu S Y, Soong A M and Reinhorn Active, Hybrid, and Semi-active Structural Control: A Design and Implementation Handbook Wiley (UK)

4) Gawronski W K Advanced Structural Dynamics and Active Control of Structures Springer-Verlag (USA)

5) Bursi, O.S. Computational techniques for simulation of monolithic and heterogeneous structural dynamic systems, in Modern Testing Techniques for Structural Systems Dynamics and Control , Edited by O. S. Bursi and D. J. Wagg, CISM-SpringerWienNewYork 2008.

Notes: 1) is comprehensive with some elements of control!

2) is a modern introductory book with many MATLAB/SIMULINK examples.

3) gives theoretical and practical information on Structural Control.

4) describes new areas of structural dynamics and control

5) provides computation techniques for simulation of dynamic systems.

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A2 LINEAR DYNAMICS OF SISO SYSTEMS A2.1 Linear Differential Equation Models

• The starting point of a typical control design is to obtain a linear dynamic model of the system to be controlled.

• By convention, the system to be controlled is called the plant.

• The simplest plant dynamic models will usually suffice - based upon linear differential equations with constant coefficients.

• The coefficients are called the plant parameters.

• The plant dynamics usually include those of the input actuation system (eg electrical or servohydraulic drive, amplifier and associated electronics) and the transducer system (eg LVDT, encoder).

• The bandwidths (speeds of response) of actuation and transducer systems are usually far in excess of the mechanical components in the plant. Consequently, the actuation and transducer systems can be simply represented by static gains in the plant model.

• A few examples of differential equation models are now presented.

Example A2.1.1 Servomotor with Angular Velocity Output

ω(t) AngularVelocityPower

Amplifier

Tachometer

u(t) Input

Voltage

τ(t) Torque

y(t) Measured Velocity cω(t)

Friction

Inertia: J Assuming static gains km and kt:

τ(t) = kmu(t) ; y(t) = ktω(t) Mechanical system:

Inertia J ; Viscous friction constant c

Newton’s second law: τ ω ω( ) ( ) ( )t c t J t− =

k u t cy tk

Jy tkm

t t( )

( ) ( )− = ⇒ ( ) ( ) ( )y t

cJ

y tk k

Ju tm t+ =

( ) ( ) ( )y tT

y tT

u t+ =1 λ

Thus the dynamics of this plant are modelled as a first-order differential equation with the constant parameters T (the time constant) and λ (the low frequency gain):

TJc

k kcm t= =; λ

5

Example A2.1.2 Suspension System

m

c k

u(t) x(t)

y(t)

LVDT

Assuming static gain kt:

y(t) = kt x(t) Mechanical system:

Mass m; Viscous damper constant c; Spring constant k Newton’s second law:

u t cx t kx t mx t( ) ( ) ( ) ( )− − =

u t cy tk

ky tk

my tkt t t

( )( ) ( ) ( )

− − = ⇒ ( ) ( ) ( ) ( )y tcm

y tkm

y tkm

u tt+ + =

( ) ( ) ( ) ( )y t y t y t u tn n n+ + =2 2 2ζω ω λω Thus the dynamics of this plant are modelled as a second-order differential equation with the constant parameters ζ (the damping ratio), ωn (the natural frequency) and λ (the low frequency gain):

ζ ω λ= = =ckm

km

kkn

t

2; ;

Example A2.1.3 Linear Actuator with Position Output

m

x(t)

y(t) Displacement Transducer

Friction Force cx t( )

Actuator Force

f(t)

Assuming static gains km and kt:

f(t) = kmu(t) ; y(t) = ktx(t) Mechanical system:

Mass m ; Viscous friction constant c Newton’s second law:

f t cx t mx t( ) ( ) ( )− = ⇒ k u t cy tk

my tkm

t t( )

( ) ( )− = ⇒ ( ) ( ) ( )y t

cm

y tk km

u tm t+ =

( ) ( ) ( )y tT

y tT

u t+ =1 λ

Thus the dynamics of this plant are modelled as a second-order differential equation with the constant parameters T (the time constant) and λ (the low frequency gain):

Tmc

k kcm t= =; λ

This system is a basic, but very useful, model for a servohydraulic actuator - as used on many civil engineering test facilities (reaction walls, reaction frames and shaking tables).

6

A2.2 The Laplace Transform • Given a continuous signal, x(t), its Laplace Transform is defined as:

0{ ( )} ( ) ( )stx t x s e x t dt

∞ −= = ∫L

• One of the most useful attributes of the Laplace Transform is that it ‘converts’

differential/integral functions of time into algebraic expressions (assuming zero initial conditions):

2

3

0

20 0

{ ( )} ( ){ ( )} ( ){ ( )} ( )

( ){ ( ) }

( ){ ( ) }

t

t t

x t sx sx t s x sx t s x s

x sx ds

x sx d ds

τ τ

τ τ τ

=

=

=

=

=

∫

∫ ∫

LLL

L

L

etc • Hence, our differential equation models are converted to algebraic models

• A table of commonly used Laplace transforms is given in Appendix 1. • Example A2.2.1 Servomotor with Angular Velocity Output

( ) ( ) ( )y tT

y tT

u t+ =1 λ

sy sT

y sT

u s( ) ( ) ( )+ =1 λ

sT

y sT

u s+⎛⎝⎜

⎞⎠⎟ =

1( ) ( )

λ

• Example A2.2.2 Suspension System ( ) ( ) ( ) ( )y t y t y t u tn n n+ + =2 2 2ζω ω λω

s y s sy s y s u sn n n2 2 22( ) ( ) ( ) ( )+ + =ζω ω λω

( )s s y s u sn n n2 2 22+ + =ζω ω λω( ) ( )

• Example A2.2.3 Linear Actuator with Position Output

( ) ( ) ( )y tT

y tT

u t+ =1 λ

s y sT

sy sT

u s2 1( ) ( ) ( )+ =

λ

s sT

y sT

u s+⎛⎝⎜

⎞⎠⎟ =

1( ) ( )

λ

7

A2.3 The Transfer Function • Rearranging the Laplace Transformed differential equation allows us to represent the plant

dynamics in the algebraic form y(s) / u(s). • This form is called the transfer function G(s) : G(s) = y(s) / u(s)

• The transfer function is the keystone to control system design. • The equation for the transfer function implies a block diagram representation of the plant

dynamics:

G(s)u(s) y(s)

• The block diagram is ‘read’ in the opposite direction to the signal flow, ie y(s) = G(s)u(s), yielding the above equation again.

• Example A2.3.1 Servomotor with Angular Velocity Output

sT

y sT

u s+⎛⎝⎜

⎞⎠⎟ =

1( ) ( )

λ ⇒ G s

y su s

Ts T

( )( )( )

//

= =+λ

1

u(s) y(s)λ //T

s T+ 1

• Example A2.3.2 Suspension System

( )s s y s u sn n n2 2 22+ + =ζω ω λω( ) ( ) ⇒ G s

y su s s s

n

n n( )

( )( )

= =+ +

λωζω ω

2

2 22

u(s) y(s)λωζω ω

n

n ns s

2

2 22+ +

• Example A2.3.3 Linear Actuator with Position Output

s sT

y sT

u s+⎛⎝⎜

⎞⎠⎟ =

1( ) ( )

λ ⇒ G s

y su s

Ts s T

( )( )( )

/( / )

= =+λ

1

u(s) y(s)λ /( / )

Ts s T+ 1

8

The parameters λ, T, ζ and ωn characterise the response of the plant to a given input. For example, following a step input, the responses of the 3 previous plants are:

• Servomotor with Angular Velocity Output: G sT

s T( )

//

=+λ

1

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3λ = 2 , T = 1

y(t)

u(t)

time (s)

y u= =λ 2

u =1

ts ≈ 4T = 4

Notes: The steady-state output is λ × the steady-state input. The settling-time of the transient is ~ 4T . The transient is an exponential curve.

• Suspension System: G ss s

n

n n( ) =

+ +λωζω ω

2

2 22

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

time (s)

y u= =λ 2

u = 1u(t)y(t)

t ssn

≈ ≈4 2 7

ζω.

λ = 2, ζ = 0.5, ωn = 3

Notes: The steady-state output is λ × the steady-state input. The settling-time of the transient is ~ 4/(ζωn) . The transient is a damped sinusoidal curve. The number of overshoots is (approximately) given by: ζ = 1.0 ⇒ 0 overshoots (critical damping) ζ = 0.7 ⇒ 1 overshoot (underdamping) ζ = 0.5 ⇒ 2 overshoots (underdamping)

9

• Linear Actuator with Position Output: G sT

s s T( )

/( / )

=+λ

1

0 1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

14

16

18

20λ = 2 , T = 1

y u= =λ 2

time (s)

y(t)

u(t) u =1

ts ≈ 4T = 4

Notes: The steady-state output velocity is λ × the steady-state input. The settling-time of the transient is ~ 4T . The transient is an exponential curve. A2.4 Effects of Unmodelled Terms • The linear transfer function is an idealised model of the plant dynamics.

• Unmodelled terms can sometimes have a significant effect. Examples include stiction, signal noise, non-linear dynamics, load disturbances and parameter variations.

• We examine (semi-quantitatively) the effect of such phenomena for the specific case of a simulated actuator response to a square-wave input of amplitude 0.1V and frequency 20rad/s.

• As we will see in section A3, the transfer function of a typical linear actuator, with displacement output, is estimated as:

G ss s

( )( . )

=+167

42 5

and this is used to generate the following simulation results.

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• Stiction (static friction) is often a real problem in mechanical positioning systems:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

linear

withstiction

time (s)

• Signal noise usually occurs on the transducer:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

linear

withnoise

time (s)

• Non-linear dynamics can occur anywhere in the system. Stiction is one form of non-linearity. Another form of friction is square-law (air) drag:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.09

-0.08

-0.07

-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

linear

withsquare-law

drag

time (s)

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• Load disturbances can include dynamic interactions with other parts of the machine or the environment. One common load disturbance is due to gravity:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

linear

with loaddisturbance

(gravity)

time (s)

• Parameter variations include continuous or piecewise changes in the physical quantities in the plant, eg mass, motor gain or friction constant:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

with parametervariation

(m → 2m)

linear

time (s)

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A3. AN INTRODUCTION TO CONTROL SYSTEM DESIGN

There are numerous control strategies and design techniques. We concentrate on a few basic (but effective) control strategies all of which are relevant to the control of civil engineering test systems.

Reference is made to continuous- and discrete-time control strategies. Continuous-time strategies can be implemented directly in the form of analogue controllers, although they can also be transformed into a discrete-time format (see A7). Such strategies can be implemented directly in the form of discrete-time controllers, based on PC, DSP or digital hardwired technologies. A list of the design approaches covered in this course is as follows: • Polynomial Methods: included in this section A4 (with four sub-sections):

A4.1 The Standard Closed-Loop Block Diagram A4.2 Closed-Loop Terminology A4.3 Closed-Loop Performance Criteria

A4.4 Basic Controller Design using Polynomial Methods

• Frequency Domain Design (A6): based on gain/phase plots of the transfer functions. Variants include Bode Plots, Nyquist Plots and Nichols Plots. Directly suited to continuous-time systems, but can be extended to discrete-time systems.

• Design via the State Equation (B1, B2, B3): The previously listed design methods are characterised by the system under control having just one input and one output – ie single-input/single-output (SISO) systems. In these sections we introduce controller design for systems with one or more input/output pair (ie multivariable systems). Suited to both continuous- and discrete-time systems.

• Adaptive Control (B4): suitable for SISO, multivariable, continuous- and discrete-time systems, when the dynamic characteristics of the system are poorly known and/or significant changes can occur in these characteristics.

13

A3.1 The Standard Closed-Loop Block Diagram

• Virtually all closed-loop schemes can be represented by the following standard diagram:

Gc(s) u(s)

Gp(s) y(s) r(s)

H(s)

+

-

CONTROLLER PLANT

e’(s)

f(s)

• We see that the plant is now described by the transfer function:

G sy su sp ( )

( )( )

=

• The plant input and output are still labelled u(s) and y(s), respectively.

• The design problem is to determine the controller transfer functions Gc(s) and H(s) so that y(t) → r(t) in a desirable manner. The signal, r(t), is called the reference (or demand) signal.

• The desirable characteristics for y(t) → r(t) are described in Section A4.3.

• The feedback signal is labelled f(s) .

A3.2 Closed-Loop Terminology

• The error signal is e(s) = r(s) - y(s) . Note: in the block diagram a signal e’(s) = r(s) - f(s) is shown; this ‘error-like’ signal must not be confused with e(s).

• The closed-loop system is said to have unity feedback if H(s) = 1. Only then can we write: e’(s) = e(s) .

• The transfer function Gc(s) is called the controller forward dynamics and H(s) is called the controller feedback dynamics.

• Often, the blocks Gc(s), Gp(s) are combined into the forward loop dynamics: G(s) = Gc(s)Gp(s) .

• Combining the forward loop dynamics with the controller feedback dynamics yields the rather confusingly named open-loop transfer function (OLTF): f(s)/e’(s) = G(s)H(s) .

• Most importantly, the closed-loop transfer function (CLTF) is:

y sr s

G sG s H s

( )( )

( )( ) ( )

=+1

• The CLTF denominator is called the closed-loop characteristic polynomial, and the equation:

1 0+ =G s H s( ) ( )

is the closed-loop characteristic equation (CLCE).

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• The CLCE has n (complex) roots in the s-plane.

• Zeros of the CLTF. These are the roots of the CLTF numerator polynomial. Traditionally, zeros do not play much of a part in the controller design - that is considered to be the role of the roots of the CLCE. However, zeros can have a profound effect on CL system transient response - especially on the overshoots. In particular, the number of CLTF zeros should be kept to a minimum, in order to prevent (otherwise unpredicted) overshoots. As we will see in section A4.4, some common control strategies create unwanted zeros. As a motivating example, below there are two CLTF step responses. Each has the same critically-damped CL characteristic polynomial (ζ = 1; no overshoots expected), but one has a zero at s = -1 which causes the unexpected overshoot:

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

time (s)

y t CLTF is ss s

( ): ( )16 18 162+

+ +

y t CLTF iss s

( ): 168 162 + +

CL: ωn = 4rad/s, ζ = 1, ts ≈ 1.0s

A3.3 Closed-Loop Performance Criteria

Statement of the standard control problem: The following diagram gives a qualitative idea of a successful control response:

Determine u so that y follows a reference vector, r, in a well-defined and stable manner. Also ensure that the effects of disturbances are rejected in the steady-state.

t

r

y

Effect of disturbance removed

Desirable transient behaviour

Zero steady-state error

Application of disturbance

15

Therefore, in order of priority, the main performance criteria for a closed-loop system are: • Stability • Relative Stability: Closed-Loop (CL) Transient Behaviour and Dominant Roots • Steady-State Behaviour • Disturbance Rejection The criteria are investigated in the context of the controller designs of section A4.4. Here are some brief introductory comments on the criteria: Stability

The closed-loop system must be stable under all circumstances. This is governed by the roots of the CLCE, which must all be in the left-half s-plane (LHP):

s-plane s-plane

: CLCE roots (3rd-order in this example)

Stable CL System Unstable CL System

Im

Re

Im

Re

Relative Stability: CL Transient Behaviour and Dominant Roots

We need to locate the CLCE roots within specific regions inside the LHP in order to achieve desirable transient performance. Typical measures of transient performance are settling time and number of overshoots: these are largely governed by the distribution of the dominant (CLCE) roots. In nearly all cases there is only one of two possibilities:

-ζωn

s-plane s-plane

: CLCE roots (4th-order in this example)

1st-Order Dominance 2nd-Order Underdamped Dominance

Im

Re

Im

Re

+jωd

-jωd -1/T

Damped natural frequency:ω ω ζd n= −1 2

First-order dominant CL systems behave as if they were solely first-order, with an exponential transient settling in a time ts ≈ 4T . There are no overshoots.

16

Second-order underdamped dominant CL systems behave as if they were solely second-order, with an exponentially decaying sinusoidal transient. The settling time is ts ≈ 4/(ζωn) and, as a rule of thumb: ζ = 1 (critical damping) 0 overshoot ζ = 0.7 1 overshoot ζ = 0.5 2 overshoots Steady-State (ss) Behaviour

This is governed by the characteristics of the reference signal r(s) and the number of integrators in the OLTF (ie the number of factors of s in the denominator of the OLTF). This number is called the CL system type, and is usually 0, 1 or 2. For example the basic servohydraulic system OLTF under various forms of control (see section A4.4) can have the following structures and types:

2

2

2

1 integrator Type 1( )

(1 )1 integrator Type 1

( )( )(1 )

2 integrators Type 2( )

( 1/ )2 integrators Type 2

( )

p

p d

p i d

p i d

bks s abk k s

s s ab k s k k s

s s abk s T T s

s s a

⇒++

⇒++ +

⇒+

+ +⇒

+

If the CL system has unity feedback, then:

1. Type 0 OLTFs give zero ss error when r is an impulse.

2. Type 1 OLTFs give zero ss error when r is an impulse or a step.

3. Type 2 OLTFs give zero ss error when r is an impulse, step, or a ramp.

If the CL system has non-unity feedback, then statements 1 and 2 still hold, as long as H(s) = 1 when s = 0 (this is the case with many controllers). In general, the CL ss error can be calculated from the Final Value Theorem:

e se s sr sG s H s G s

G s H sss

= =+ −

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥=

=

[ ( )] ( )( ) ( ) ( )

( ) ( )00

11

17

Example A3.3.1

An actuator system under Proportional control (see section A4.4) has unity-feedback with an OLTF:

bks s a

p

( )+

We determine the ss error following a ramp reference signal r t rt( ) = . (We expect the error to be non-zero). Now r s r s( ) /= 2 , so:

e se ssrs bk s s a

er s

s s a bk s s a

er s a

s s a bka

bkr

sp

s

ps

ps

p

= =+ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=+ + +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=+

+ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

=

=

=

=

[ ( )]/ [ ( )]

/[ ( ) ] / [ ( )]

( )( )

0 2

0

0

0

11

The ss error is a/(bkp) × the slope of the reference ramp. Disturbance Rejection

Effects of disturbances are analysed in the same manner as ss errors. The controller should negate the effect of disturbances, at least in the steady-state. Usually, the precise nature of the disturbances is unknown, but their structure can be estimated, eg the disturbances are constant, cyclic, random, etc. The linear actuator system illustrates the start of a typical analysis. Example A3.3.2

The linear actuator is subject to a constant disturbance force d(t), (eg due to gravity when the axis is inclined to the horizontal), so that the equation of motion and Laplace Transform development is as follows:

k u t cx t d t mx t y t k x tk km

u tcm

y tkm

d t y t

s s c m y s k m k u s d s

y sk m

s s c mk u s d s

m t

m t t

t m

tm

( ) ( ) ( ) ( ) ; ( ) ( )

( ) ( ) ( ) ( )

( / ) ( ) ( / )[ ( ) ( )]

( )/

( / )[ ( ) ( )]

− − = =

− − =

+ = −

=+

−

The corresponding block diagram is:

k ms s c m

t /( / )+

kmu(s)

d(s)

y(s)+ -

and the ss error due to d(s) can be determined within the CL structure, in much the same manner as the ss error due to r(s).

18

A3.4 Basic Controller Design

The strategy for controller design is now summarised using the content of the preceding sections:

1. Determine the plant transfer function, and its parameters.

2. Determine the required CL performance criteria.

3. Make an engineering judgement on the simplest controller to perform the job (see below).

4. Determine the chosen controller parameters (gains) that satisfy the CL performance criteria.

5. Simulate the CL performance as a check on your design. (Optional, but sometimes essential).

6. If appropriate, implement the discrete form of the controller (see section A7)

The main content of this sub-section is associated with steps 3 and 4. This material is specific to the servomotor with angular position output. The results of discrete controller design implementations are included here as well, thereby pre-empting section A7.

A3.4.1 Proportional (P) Control

• This is the simplest linear controller:

G s k H sc p( ) ; ( )= = 1

• The parameter kp is called the proportional gain: the design problem is to find a suitable value for this gain. Since P control has unity feedback, we can write:

u s G s e s k e s u t k e tc p p( ) ( ) ( ) ( ) ( ) ( )= = ⇒ =

• A P control design is investigated in the context of the servomotor angular positioning system whose transfer function was identified in section A3.2:

G ss sp ( )( . )

≈+

1702 0

• The plant time constant is T = 1/2.0 = 0.5, so that the estimated plant settling time is 4T = 2.0. Therefore, reasonable CL performance criteria are ts = 1.0 together with no overshoot and zero ss error.

19

• The CL performance criteria translate to ζ = 1 and ωn = 4/ts = 4rad/s, ie a desired CLCE:

s s2 8 16 0+ + =

• The actual CLCE is:

1170

20+

+=k

s sp ( )

s s k p2 2 170 0+ + =

• Comparing the actual and desired CLCE’s immediately shows that the CL performance criterion cannot be satisfied by P control: the coefficients of s1 are incompatible.

• Hence, with P control a compromise must be reached. Relaxing the criterion on ts , but not on ζ, gives the desired CLCE: s sn n

2 22 0+ + =ω ω

Comparing CLCEs: ωn = =2 2 10/ . rad / s ⇒ k pn= =

ω 2

1700 0059.

Corresponding CL implementation results are:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

r

y

u

kp = 0.0059

There is no response from y! The low value of kp generates insufficient u to overcome stiction.

• Reducing the desired damping ratio to 0.5 (2 overshoots) in an attempt to boost the control effort gives a desired CLCE: s sn n

2 2 0+ + =ω ω

Comparing CLCE’s: ωn = 2rad / s ⇒ k pn= =

ω 2

1700 0235.

20

Corresponding CL implementation results are:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

y

r

u

kp = 0.0235

stictionerror

Two overshoots are expected. The second one is ‘hidden’ by the stiction problem. Stiction still

induces a large ss error, even though it is predicted to be zero (since the OLTF is type 1 and we have a step reference signal).

• Further increases in gain reduce the stiction error, but at the expense of an excessive number of overshoots For example, setting kp = 0.15 yields:

ωn = × =170 015 505. . rad / s

ζ = 2/(2ωn) = 0.198

ts ≈ 4/(ζωn) = 4.0s

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

y

r

u

kp = 0.15

• Clearly, some extra damping is required, together with the ability to take account of the stiction

error.

• Damping is improved by incorporation of derivative action in the controller.

• The stiction error can be reduced to zero by the incorporation of integral action in the controller.

21

A3.4.2 Proportional-plus-Derivative Feedback (P+DFB) Control

• P+DFB introduces a derivative term in the feedback loop:

G s k H s k sc p d( ) ; ( )= = +1

• The parameter kd is the derivative gain.

• An alternative to P+DFB is PD (proportional - plus - derivative control), where: G s k k s H sc p d( ) ; ( )= + = 1

• The advantage of P+DFB over PD control is that the former introduces no CLTF zeros. In the case of P+DFB of the servomotor:

y sr s

ks s

k k ss s

ks k k s k

p

p d

p

p d p

( )( )

( )( )

( )( )

=+

++

+

=+ + +

1702

1170 1

2

1702 170 1702

However, the PD controller yields an undesirable zero at s = - kp/kd :

y sr s

k k ss s

k k ss s

k k ss k s k

p d

p d

p d

d p

( )( )

( )( )( )( )

( )( )

=

++

+++

=+

+ + +

1702

1170

2

1702 170 1702

• The disadvantage of P+DFB compared with PD is that it has non-unity feedback. However, the lack of a CLTF zero far outweighs this disadvantage.

• The P+DFB control signal is:

u s G s e s k e s k r s f sf s H s y s k s y su s k r s k s y s k e s k k sy su t k e t k k y t

c p p

d

p d p p d

p p d

( ) ( ) ' ( ) ' ( ) [ ( ) ( )]( ) ( ) ( ) ( ) ( )( ) [ ( ) ( ) ( )] ( ) ( )( ) ( ) ( )

= = = −

= = += − + = −

= −

11

• We return to the original CL performance criteria of ζ = 1 and ωn = 4/ts = 4rad/s, ie a desired CLCE:

s s2 8 16 0+ + =

• From above, the actual CLCE is:

s k k s kp d p2 2 170 170 0+ + + =( )

• Comparing CLCE coefficients:

k p = =16170

0 094. kkd

p=

−=

( ).

8 2170

0 375

22

Corresponding CL implementation results are:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

r

y

u

kp = 0.094, kd = 0.375

The damping and settling-time characteristics are now as required. However, the stiction-

induced ss error is still present, and this final problem is addressed by the incorporation of forward-loop integral action.

A3.4.3 Proportional - plus - Integral - plus - Derivative Feedback (P+I+DFB) Control

• P+I+DFB introduces an integral term in the forward loop whilst retaining the derivative action in the feedback loop:

G s kks

k s k ks

H s k s k s k

c pi p i p

d d d

( )( / )

( ) ( / )

= + =+

= + = +1 1

so that the combined controller transfer functions have one pole at s = 0 (an integrator) and 2 zeros at s = -ki/kp and s = -1/kd .

• The parameter ki is the integral gain.

• The P+I+DFB control signal is:

[ ]

[ ]

u s kks

e s kks

r s f s

f s k s y s

u s kks

e s k sy s

u s k e sks

e s k k sy s k k y s

u t k e t k e t dt k k

pi

pi

d

pi

d

pi

p d i d

p i

t

p d

( ) ' ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

= +⎡⎣⎢

⎤⎦⎥

= +⎡⎣⎢

⎤⎦⎥

−

= +

= +⎡⎣⎢

⎤⎦⎥

−

= + − −

= + −∫

1

0y t k k y ti d( ) ( )−

An alternative to P+I+DFB is the very commonly used PID (proportional - plus - integral - plus - derivative control). As we will see in the next section, PID control has the distinct disadvantage of introducing two CL zeros.

23

• However, P+I+DFB only introduces only one CLTF zero; in the case of P+I+DFB of the servomotor:

y sr s

k s ks sk s k k s

s s

y sr s

k s k ks k k s k k k s k

p i

p i d

p i p

p d p i d i

( )( )

( )( )

( )( )( )

( )( )

( / )( ) ( )

=

++

++ ++

=+

+ + + + +

1702

1170 1

2

1702 170 170 170 170

2

2

3 2

• Hence, in the case of P+I+DFB of the servo, the CLCE is:

s k k s k k k s kp d p i d i3 22 170 170 170 170 0+ + + + + =( ) ( )

• A strategy to deal with this third-order CL system is:

(1) Given the desired CLCE should have a 1st-order dominant root at s = -4, locate the CLTF zeros s = -ki/kp , s = -1/kd approximately 50% further into the LHP.

(2) The gain k = 170kpkd is given by:

[product of distances from dominant CLCE root to all plant+controller poles][product of distances from dominant CLCE root to all plant+controller zeros]

k =

Thus, (1) and (2) are sufficient to derive all the required gains.

• Part (2) of the above strategy is called the Magnitude Criterion, and is one of the basic criteria for a much more general design method called the Roots’ Loci Method. See section A5.

Applying the strategy to our servo problem:

(1) The 1st-order dominant CLCE root at s = -4 ensures the CL settling-time is ts ≈ 4/4 = 1.0s. Let the CLTF zeros be s = -ki/kp = -1/kd = -6, so that ki = 6kp and:

kd = =1 6 0167/ .

(2) The plant has 2 poles at s = 0, -2 and no zeros. The controller has 1 pole at s = 0 and 2 zeros at s = -6 . Hence:

k = 170kpkd = [(4-0)×(4-0)×(4-2)]/[(6-4) ×(6-4)] = 8

k p = ×=

8170 0167

0 282.

. ki = × =6 0 282 169. .

24

Corresponding CL implementation results are:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

yr

u

kp = 0.282, ki = 1.69, kd = 0.167

• The desired CL response has been nearly achieved. A process of ‘manual retuning’ - increasing the derivative gain to reduce the overshoot - gives us the desired response:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

ry

u

kp = 0.282, ki = 1.69, kd = 0.23

25

A3.4.4 Proportional - plus- Integral - plus - Derivative (PID) Control

• PID introduces integral and derivative terms in the forward loop only:

G s kT s

T s H sc pi

d( ) ; ( )= + +⎡

⎣⎢

⎤

⎦⎥ =1

11

where Ti is the integral ‘time constant’ and Td is the derivative ‘time constant’.

• The PID control signal is:

u s kTs

T s e s

u t k e tT

e t dt T e t

pi

d

pi

d

t

( ) ( )

( ) ( ) ( ) ( )

= + +⎡

⎣⎢

⎤

⎦⎥

= + +∫

11

10

• PID introduces two zeros into the CLTF:

[ ]

[ ]

[ ][ ]

y su s

k TT s Ts Tss s

k TT s Ts Tss s

y su s

k TT s Ts Tss s k TT s Ts Ts

y su s

k TT s Ts

p i d i i

p i d i i

p i d i i

p i d i i

p i d i

( )( )

( ) / ( )( )

( ) / ( )( )

( )( )

( ) / ( )( ) / ( )

( )( )

(

=

+ ++

++ ++

=+ +

+ + + +

=+

170 12

1170 1

2

170 12 170 1

170

2

2

2

2 2

2 ++ + + +

12 170 170 1703 2

)( )Ts T k TT s k Ts ki i p i d p i p

• A common form of PID gain selection is via one of the empirical Ziegler-Nichols methods. P control (alone) is implemented on the plant, and kp is increased until unstable CL oscillations appear. The corresponding gain, kp

*, and the period of oscillations, P, are noted.

• For PID control select:

k k T P T Pp p i d= = =0 6 2 8. ; / ; /*

26

• The unstable P control results for the servo system are:

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

y

r

u

kp = kp* = 0.25

P = 0.44s

• Hence the Ziegler-Nichols PID gains are:

k k T P T Pp p i d= = = = = =0 6 015 2 0 218 8 0 055. . ; / . ; / .*

and the corresponding CL response is:

0 2 4 6 8 10 12 14 16 18 20 -10

-8

-6

-4

-2

0

2

4

6

8

10

time (s)

y

r

u Potentiometer wrap-around

kp = 0.15; Ti = 0.218; Td = 0.055

As is often the case, the Ziegler-Nichols method does not result in acceptable control for electro-mechanical plant. (The result is worsened by the potentiometer wrap-around). Manual re-tuning can often be problematic, given the 3 variables in the controller, and the presence of the 2 CLTF zeros.

A final note on Section A4: Derivative action and noise All controllers containing derivative action have the potential for amplifying signal noise. The result is ‘chatter’ on the plant actuator, imposing an upper bound on the derivative gain.