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Systems with Real Components and Saturating Signals ◾ 67For the phase plane, instead of plotting position against time we plot velocity against position.. 68 ◾ Essentials of Control Tec

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Systems with Real Components and Saturating Signals ◾ 67

For the phase plane, instead of plotting position against time we plot velocity

against position

An extra change has been made in www.esscont.com/6/PhaseLim.htm

Whenever the drive reaches its limit, the color of the plot is changed, though this is

hard to see in the black-and-white image of Figure 6.3

For an explanation of the new plot, we will explore how to construct it without

the aid of a computer Firstly, let us look at the system without its drive limit To

allow the computer to give us a preview, we can “remark out” the two lines that

impose the limit By putting // in front of a line of code, it becomes a comment and

is not executed You can do this in the “step model” window

Without the limit there is no overshoot and the velocity runs off the screen, as

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68 ◾ Essentials of Control Techniques and Theory

To construct the plot by hand, we can rearrange the differential equation to set

the acceleration on the left equal to the feedback terms on the right:

 

If we are to trace the track of the (position, velocity) coordinates around the

phase plane, it would be helpful to know in which direction they might move Of

particular interest is the slope of their curve at any point,

d x

d x

We wish to look at the derivative of the velocity with respect to x, not with

respect to time, as we usually do These derivatives are closely related, however,

since for the general function f,

df dx

dt dx

df

dt x

df dt

The first thing we notice is that for all points where position and velocity are

in the same proportion, the slope is the same The lines of constant ratio all pass

through the origin

On the line x = 0, we have slope –5.

On the linex = x, we have slope –11

On the linex = − x, we have slope +1, and so on

We can make up a spider’s web of lines, with small dashes showing the

direc-tions in which the “trajectories” will cross them These lines on which the slope is

the same are termed “isoclines.” (Figure 6.5)

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Systems with Real Components and Saturating Signals ◾ 69

An interesting isocline is the line6x+5x=0 On this line the acceleration is

zero, and so the isocline represents points of zero slope; and the trajectories cross

the line horizontally

In this particular example, there are two isoclines that are worth even closer

scrutiny Consider the line x+2x=0 Here the slope is –2—exactly the same as

the slope of the line itself Once the trajectory encounters this line, it will lock on

and never leave it The same is true for the line x+3x=0, where the trajectory

slope is found to be –3

Q 6.3.1

Is it a coincidence that the “special” state variables found in Section 5.5 could be

expressed as x+2x=0 and x+3x=0, respectively?

Having mapped out the isoclines, we can steer our trajectory around the plane,

following the local slope From a variety of starting points, we can map out sets of

trajectories This is shown in Figure 6.6

The phase plane “portrait” gives a good insight into the system’s behavior,

with-out having to make any attempt to solve its equations We see that for any starting

point, the trajectory homes in on one of the special isoclines and settles without any

Slope = –5

Infinite slope

figure 6.5 Isoclines.

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70 ◾ Essentials of Control Techniques and Theory

i.e., for the same position control system, but with much reduced velocity feedback

You will find an equation similar to Equation 6.3 for the slopes on the isoclines, but

will not find any “special” isoclines The trajectories will match the “spider’s web”

image better than before, spiraling into the origin to represent system responses

which now are lightly damped sine-waves Scale the plot over a range of –1 to 1

in both position and velocity

Q 6.3.3

This is more an answer than a question Go to the book’s website Run the

simula-tion www.esscont.com/6/q6-3-2.htm, and compare it with your answer to quessimula-tion

Q 6.3.2

6.4 Phase Plane for Saturating drive

Now remember that our system is the simple one where acceleration is proportional

Isocline slope = 1

Zero slope isocline

“Special”

isoclines

figure 6.6 example of phase plane with isoclines.

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Systems with Real Components and Saturating Signals ◾ 71

the drive will reach its saturation value Between these lines the phase plane plot

will be exactly as we have already found it, as in Figure 6.7

Outside the lines, the equations are completely different The trajectories are

To fill in the mystery region in this phase plane, we must find how the system

will behave under saturated drive

figure 6.7 linear region of the phase plane.

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72 ◾ Essentials of Control Techniques and Theory

Equation 6.2 tells us that the slope of the trajectory is always given by  x x/ , so in

this case we are interested in the case where x has saturated at a value of +1 or –1,

giving the slope of the trajectories as 1/x or –1/x.

Now we see that the isoclines are no longer lines through the origin as before,

but are lines of constant x, parallel to the horizontal axis If we want to find out

the actual shape of the trajectories, we must solve for the relationship between

with a horizontal axis which is the x-axis.

Similarly, the trajectories to the right, where u = –1, are of the form

x=− x2/2+c

The three regions of the phase plane can now be cut and pasted together to give

the full picture of Figure 6.8

We will see that the phase plane can become a powerful tool for the design

of high performance position control systems The motor drive might be

pro-portional to the position error for small errors, but to achieve accuracy the drive

must approach its limit for a small displacement The “proportional band” is

small, and for any substantial disturbance the drive will spend much of its time

saturated The ability to design feedback on a non-linear basis is then of great

importance

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Systems with Real Components and Saturating Signals ◾ 73

Q 6.4.1

The position control system is described by the same equations as before,

 

x+5x+6x=6 w

This time, however, the damping term 5 x is given not by feedback but by

pas-sive damping in the motor, i.e.,

 

x=−5x u+

where

u= −6(x−demand)

Once again the drive u saturates at values +1 or −1 Sketch the new phase plane,

noting that the saturated trajectories will no longer be parabolae The answer can

be found on the website at www.esscont.com/6/q6-4-1.htm

Q 6.4.2

Consider the following design problem A manufacturer of pick-and-place machines

requires a robot axis It must move a one kilogram load a distance of one meter,

bringing it to rest within one second It must hold the load at the target position

with sufficient “stiffness” to resist a disturbing force, so that for a deflection of one

millimeter the motor will exert its maximum restoring force

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74 ◾ Essentials of Control Techniques and Theory

Steps in the design are as follows:

(a) What acceleration is required to achieve the one meter movement in one

second?

(b) Multiply this by an “engineering margin” of 2.5 when choosing the motor

(c) Determine feedback values for the position and velocity signals

You might first try pole assignment, but you would never guess the values of the

poles that are necessary in practice

Consider the “stiffness” criterion An error of 10–3 meters must produce an

acceleration of 10 m/s2 That implies a position gain of 10,000 Use the phase plane

to deduce a suitable velocity gain

6.5 Bang–Bang Control and Sliding Mode

In the light of a saturating input, the gain can be made infinite The proportional

regional has now shrunk to a switching line and the drive always takes one limiting

value or the other This is termed bang–bang control Clearly stability is now

impos-sible, since even at the target the drive will “buzz” between the limits, but the quality

of control can be excellent Full correcting drive is applied for the slightest error

Suppose now that we have the system

x u= ,

where the magnitude of u is constrained to be no greater than 1.

Suppose we apply a control law in the form of logic,

then Figure 6.3 will be modified to show the same parabolic trajectories as before,

with a switching line now dividing the full positive and negative drive regions This

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Systems with Real Components and Saturating Signals ◾ 75

which at this point has value 2 This will take the trajectory across the switching

line into the negative drive region, where the slope is now –2 And so we cross the

trajectory yet again The drive buzzes to and fro, holding the state on the switching

line This is termed sliding.

But the state is not glued in place On the switching line,

x x+ = 0

If we regard this as a differential equation, rather than the equation of a line,

we see that x decays as e –t

Once sliding has started, the second order system behaves like a first order

system This principle is at the heart of variable structure control.

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7 Chapter

frequency domain

Methods

7.1 Introduction

The first few chapters have been concerned with the time domain, where the

perfor-mance of the system is assessed in term of its recovery from an initial disturbance

or from a sudden change of target point Before we move on to consider discrete

time control, we should look at some of the theory that is fundamental to stability

analysis An assumption of linearity might lead us astray in choosing feedback

parameters for a system with drive limitation, but if the response is to settle to a

stable value the system must satisfy all the requirements for stability

For reasons described in the Introduction, the foundations of control theory

were laid down in terms of sinusoidal signals A sinusoidal input is applied to the

system and all the information contained in the initial transient is ignored until

the output has settled down to another sinusoid Everything is deduced from the

“gain,” the ratio of output amplitude to input, and “phase shift” of the system

If the system has no input, testing it by frequency domain methods will be

something of a problem! But then, controlling it would be difficult, too

There are “rules of thumb” that can be applied without questioning them, but

to get to grips with this theory, it is necessary to understand complex variables At

the core is the way that functions can define a “mapping” from one complex plane

to another Any course in control will always be obliged to cover these aspects,

although many engineering design tasks will benefit from a broader view

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78 ◾ Essentials of Control Techniques and Theory

7.2 Sine-Wave fundamentals

Sine-waves and allied signals are the tools of the trade of classical control They can

be represented in terms of their amplitude, frequency, and phase If you have ever

struggled through a page or two of algebraic trigonometry extracting phase angles

from mixtures of sines and cosines, you will realize that some computational short

cuts are more than welcome This section is concerned with the representation of

sine-waves as imaginary exponentials, with a further extension to include the

inter-pretation of complex exponentials

DeMoivre’s theorem tells us that:

e jt =cos( )t +jsin( )t

from which we can deduce that

cos( ) (t = e jt +ejt)/2sin( ) (t = e jtejt)/ 2j (7.1)This can be shown by a power series expansion, but there is an alternative proof

(or demonstration) much closer to the heart of a control engineer It depends on

the techniques for solving linear differential equations that were touched upon in

Chapter 5 It depends also on the concept of the uniqueness of a solution that

satis-fies enough initial conditions

If y = cos(t) and if we differentiate twice with respect to t, we have:

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Frequency Domain Methods ◾ 79

7.3 Complex amplitudes

Equation 7.1 allow us to exchange the ungainly sine and cosine functions for more

manageable exponentials, but we are still faced with exponentials of both +jt and

−jt Can this be simplified?

So we can express a mixture of sines and cosines with a simple complex number

This number represents both amplitude and phase On the strict understanding

that we take the real part of any expression when describing a function of time, we

can now deal in complex amplitudes of e jt

Algebra of addition and subtraction will clearly work without any

complica-tions The real part of the sum of (a + jb)e jt and (c + jd) e jt is seen to be the sum

of the individual real parts, i.e., we can add the complex numbers (a + jb) and

(c + jd) to represent the new mixture of sines and cosines (a + c)cos(t) − (b + d)

sin(t) Beware, however, of multiplying the complex numbers to denote the product

of two sine-waves For anything of that sort you must go back to the precise

repre-sentation given in Equation 7.1

Another operation that is linear, and therefore in harmony with this

representa-tion, is differentiation It is not hard to show that

d

dt(a jb e+ ) jt = j a jb e( + ) jt

Differentiation a second time or more is still a linear operation, and so each

time that the mixture is differentiated we obtain an extra factor of j.

This has an enormous simplifying effect on the task of solving differential

equa-tions for steady soluequa-tions with sinusoidal forcing funcequa-tions In the “knife and fork”

approach we would have to assume a result of the form A cos(t) + B sin(t), substitute

this into the equations and unscramble the resulting mess of sines and cosines Let

us use an example to see the improvement

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80 ◾ Essentials of Control Techniques and Theory

The solution will be a mixture of sines and cosines of 3t, which we can represent

as the real part of Xe 3jt

The derivative of x will be the real part of 3jXe 3jt

When we differentiate a second time, we multiply this by another 3j, so the

second derivative will be the real part of –9Xe 3jt

At the same time, u can be represented as the real part of (2 – j)e 3jt

Substituting all of these into the differential equation produces a bundle of

multiples of e 3jt, and so we can take the exponential term out of the equation and

just equate the coefficients We get:

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Frequency Domain Methods ◾ 81

7.4 More Complex Still-Complex frequencies

We have tried out the use of complex numbers to represent the amplitudes and phases

of sinusoids Could we usefully consider complex frequencies too? Yes, anything goes

How can we interpret e (λ + jω)t ? Well it immediately expands to give e λt e jωt, in

other words the imaginary exponential that we now know as a sinusoid is

multi-plied by a real exponential of time If the value of λ is negative, then the envelope

of the sinusoid will decay toward zero, rather like the clang of a bell If λ is positive,

however, the amplitude will swell indefinitely

Now we can represent the value of λ + jω by a point in a plane where λ is

plotted horizontally and ω is vertical We can illustrate the functions of time

rep-resented by points in this “frequency plane” by plotting them in an array of small

panels, as in Figure 7.1

Code at www.esscont.com/7/responses.htm and www.esscont.com/7/responses2

htm has been used to produce this figure

The signal shown is plotted over three seconds The center column represents

“pure” frequencies where λ is zero The middle row is for zero frequency, where the

function is just an exponential function of time You will see that the scale has been

compressed so that λ has a range of just _+1, while the frequencies run from –4 to

4 But these are angular frequencies of radians per second, where 2π radians per

second correspond to one cycle per second

The process of differentiation is still linear, and we see that:

d

dt(a jb e+ ) (λ ω +j t) = +(a jb)(λ+j eω) (λ ω +j t).

In other words, we can consider forcing functions that are products of sinusoids

and exponentials, and can take exactly the same algebraic short cuts as before

7.5 eigenfunctions and gain

Classical control theory is concerned with linear systems That is to say, the

differ-ential equations contain constants, variables, and their derivatives of various orders,

but never the products of variables, whether states or inputs

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