Principles of Engineering Mechanics (2nd Edition) Episode 9 ppsx

The system error, x, , is formally defined by and it is the object of the control system to take corrective action and reduce this error to zero.. If a new position is required, the appr

156 Vi bration

Determine the natural frequencies for small oscilla- point A and one at point B The static deflections when tions about the equilibrium position the machine at A is installed are 5 mm at A and 2 mm at

(Hint: The small angle approximation implies that all B When the machine at B is added the deflection at B non-linear terms can be excluded Such terms are the is increased to 6 mm and that at A becomes 8 mm If a centripetal accelerations and any product of co- 7 Hz sinusoidal force of amplitude equal to 1% of the

what is the amplitude of motion of the structure at A

9.27 A structure carries two heavy machines, one at and B? Find also the natural frequencies of the system

10

Introduction to automatic control

10.1 Introduction

This chapter is devoted to an examination of

elementary mechanical control systems The

discussion will be limited to the class of systems

whose motion can be described by linear

differential equations with constant coefficients

In practice many control systems have non-linear

elements, but the overall motion can very often

be closely approximated to that of a purely linear

system The main features of all control systems

can be introduced by discussing specific examples

Let us consider the position control of a

machine tool which has only straight-line motion

Let the actual position of the tool be defined by x,

and the desired position by x i The variables x,

and xi are referred to as the system output and the

system input respectively

The system error, x, , is formally defined by

and it is the object of the control system to take

corrective action and reduce this error to zero

Assume that the tool is initially at rest and that

the system has zero error If a new position is

required, the appropriate input is applied, giving

rise to an error in position The controller then

acts, attempting to reduce the error to zero, and,

for a linear system, the motion of the tool will be

described by a linear differential equation

A human operator often forms part of a control

system As an example of this consider the case of

a man driving a car at a speed which he wishes to

remain constant at 100 k d h This constitutes a

speed-control system where the desired speed or

input, v i , is 100 k d h The output, v, , is the actual

speed of the car, and the error, v,, is the

difference between input and output

If, for example, the car meets a headwind, the

drop in speed (the error) will be noticed by the

driver who, among other things, is acting as an

error-sensing device, and he will take corrective

action by adjusting the position of the accelerator

pedal in an attempt to reduce the error to zero If the head wind is such as to cause a rapid increase

of error, the corrective action will not be the same

as that for a slow change Thus we observe that the driver’s control action takes account not only

of the magnitude of the error, v,, but also the rate

of change of error, dv,ldt

Later we shall see that in some control systems

a measure of the integral Jv,dt is useful When a human operator is part of the control process, his reaction time introduces a finite delay into the system, making it non-linear Such systems are not discussed further here

10.2 Position-control system

We can now examine in some detail a particular elementary position-control system and use it to introduce the block-diagram notation by which control systems are often represented

A rotatable radar aerial has an effective

moment of inertia 1 The aerial is driven directly

by a d.c motor which produces a torque T, equal

to kl times the motor voltage V ; thus

The motor voltage V is effectively the

difference between two voltages V , and v bwhich are applied to the two motor terminals so that

and, of course, if V , and V , were identical the

motor would have zero output torque A potentiometer-and-amplifier system produces the voltage V , proportional to the desired angular

position 6, of the aerial, the constant of

proportionality being k 2 Thus

A position transducer, attached to the aerial whose angular position is 6, (the system output) produces the voltage v bsuch that

158 Introduction to automatic control

If 6, and 6, are equal, then the position error

defined by

(10.6)

is zero and for this condition it is required that the

voltage V and hence the torque T, be zero The

voltage V , represents the desired position or

input, and the voltage V,, represents the actual

position or output The voltage V thus represents

the error, and we conclude that k2 must equal k3

and equations 10.3 to 10.6 can be combined to

give

6, = 6i - 6,

Equation 10.2 can be written as

T, = kl k2 6, (10.8)

and we see that the motor torque is proportional

to the error

Equation 10.8 represents the control action of

the system In order to determine the motion of

the system for a particular input 6 , , we need to

incorporate the dynamics of the aerial itself (In

mechanical control systems, the object whose

position or speed is being controlled is usually

referred to as the load.)

If the aerial has negligible damping, the only

torque applied to it is that from the motor; thus

T , = Id2 6,1dt2 (10.9a)

or T,,, =ID28, (10.9b)

where D is the operator ddt

Eliminating T,,, from equations 10.8 and

10.9( b) ,

k l k2 6, = I D 2 6, (10.10)

For any control system, the relationship

between input and output is of major importance

From equations 10.6 and 10.10,

kl kZ(0i- 0,) = I D 2 6,

(ID2 + kl k2) 6 = k l k , ei

IO, + kl k2 0, = kl k2 0,

( 10.1 la)

By solving equations 10.11, we can find the

output 0, as a function of time for a given function

01

Note that a purely mechanical analogue of this

system could consist of a flywheel of moment of

inertia I connected to a shaft of torsional stiffness

K = k l k 2 , as shown in Fig 10.1

10.3 Block-diagram notation

It is common practice to represent control systems in block-diagram form There are three basic elements: an adderlsubtracter, a multiplier, and a pick-off point as shown in Figs 10.2, 10.3, and 10.4

In Fig 10.3, the simplest form of the multiplier

E will be a constant, and the most complicated form can always be reduced to a ratio of two polynomials in operator D We can write

- = E 0 ,

e,

and E is called the transfer operator between 6,

and 6, Note that if 6, = E61 and 6, = Fez, as shown in Figs 10.5(a) and (b), then 6, = E F & , as shown in Fig 10.5(c)

Equations 10.2, 10.6, 10.7, and 10.9(b) are

represented by the block-diagram elements

shown in Fig 10.6 Note that there is an

implication of cause and effect: the output of a

block-diagram element is the result of applying

the input(s) In equation 10.9(b), the angular

rotation e, is the result of applying the torque T,

The equation is thus rewritten as T,(1D2) = e,,

so that Fig 10.6(d) can be drawn with T, as

input

10.4 System response 159

where w, = ( k , k2/1)”*, as shown in Fig 10.10

-

Rather than taking up the required position

e, = ao7 the load oscillates about this position with circular frequency w, This performance is Noting that the output of Fig 10.6(d) is one of clearly unsatisfactory and it is evident that some the inputs to Fig 10.6(b), and connecting the four form of damping must be introduced The elements in the appropriate order, we obtain the response would then take the form of either Fig system block diagram shown in Fig 10.7 From 10.11(a) or (b), depending on the amount of this figure we note that a control system is a damping

closed-loop system One of the variables (0,) is

subtracted from a variable (0,) which precedes it;

this is known as negative feedback

Using the techniques of Fig 10.5, Fig 10.7 can

be reduced to Fig 10.8

10.4 System response

Returning to equations 10.11 7 we can determine

the response of the system to particular inputs 0,

Suppose we want the load suddenly to rotate

through an angle a at time t = 0 This

corresponds to the step input 0, = 0, t < O ; 0, = a o ,

t 2 O shown in Fig 10.9 It is left as an exercise for

the reader to show that the response to this input

is given by

e, = a o ( i -coSw,t) (10.12)

-

One way of providing damping is to attach a damper to the load If the damper provides a torque Td which opposes the motion of the load and is proportional to the velocity (viscous damping), the constant of proportionality being

C , then equation 10.9(a) is replaced by

(10.13)

T, - CDO, = ID28, (10.14)

T, - Td = ld20,dt2

T = (ID2 + CD)O, T,l(ZD2 + CD) = 8,

The block diagram for the damped load is shown in Fig 10.12 We note that the effect of

160 Introduction to automatic control

tachogenerator is proportional to its angular velocity, so that

and the block-diagram form is shown in Fig lo.14

adding the damper is to replace I D 2 in the

undamped system by I D 2 + C D Hence, for the

damped system of Fig.lO.13 (cf equation

10.11 (a) ) 7

Consider the case of the undamped load with a tachogenerator attached The tachogenerator is a relatively small device and applies a negligible torque to the load so that equations 10.9 are applicable Assume that the voltage V, is subtracted from the voltage V by an operational- amplifier system so that the voltage V, applied to the motor is

( 10.19) The system block diagram for this case is shown in Fig 10.15 and we observe that the tachogenerator component parts of the system are listed below

(ID2 + C D + K)Oo = Kei (10.15)

where K = kl k2 Dividing by I to obtain the

standard form (see equation 9.22) we have

(10.16) where w: = KII and 5 = C/2d(KI)

equation 10.16 are (see also equation 9.33)

(D* + 2 c U , , ~ + w,,2)eo = w,,%i

For the Same input7 Fig lo.9> the so1utions to

appears in an inner loop The equations for the

(10.6)

e, = ao{i-e-~wnr[cosUdt

e, = ei - e,

+([/d12- l)sinhw,t]} [>1 T , = kl V,

Eliminating T,, V , , V, V, and 0, 7 we obtain

[<I

+ (l/d/l - 12) sin wd t ] }

= a o { l - e - c w n r [ l + w n t ] } [=1 (10.17)

= a { 1 - e- c w n r [cosh w e t

w h e r e c o d = q , d l - [ 2 a n d w , = w n d 1 2 - l

The output 0, does not settle to the required

value of a until (theoretically) an infinite time

has elapsed In practice, small amounts of

(ID2+klk4D+klk2)8, = klk2Bi (10.20) reasonably quickly

The viscous damper wastes power and cannot

readily be constructed to give a precise amount of The amount of damping in the system can be damping There are other methods of introducing altered by regulating the techogenerator voltage the first-derivative term (CDO,) into the system by a potentiometer circuit This method of equation 10.15, and one of these makes use of a introducing damping is known as output velocity d.c device known as a tachogenerator, driven by feedback Another common way of introducing the load The voltage Vt produced by the damping is to use proportional-plus-derivative

action (see problem 10.5)

Coulomb (dry) friction ensure that motion ceases ID28, = kl [k2(@ - 0,) - k4D8,]

Figure 10.15

10.5 System errors 161

10.5 System errors 0, = 0 What would be the steady-state error

A system equation relates one of the loop following the application of a constant external variables to the input(s) It is conventional to torque To to the load?

have the loop variable on the left-hand side of the

equation and the input(s) on the right For

example, in Fig 10.13, e,, V , T, and 0, are loop

variables and 0, is the input; equation 10.15 is an

output-input system equation To obtain the

error-input system equation we can replace 0, in

this equation by 0,- 0, from equation 10.6, to

obtain

Equation 10.14 is replaced by

(10.24)

or T, + To = (ID2 + CD)Oo (10.25)

We could equally well have written -To, since the direction was unspecified Putting

T = T,+ To and T/(ZD2+CD) = eo, we can draw the system block diagram (Fig 10.16) Notice that the external torque To appears as an extra input to the system Combining equations 10.25, 10.8, and 10.6 and putting 0, = 0, we have

(10.26) This system equation is identical in form with equation 10.15 with 0, and KOi replaced by 0, and -TO respectively and SO the solutions can be written down immediately from equation 10.17 The steady-state error can be obtained by letting

t-+ 00 and is found to be

T, - CDO, + To = ID2O0

( I D 2 + CD + K ) ( @ - ee) = KO,

(ZD2 + C D + K ) e e = ( I D 2 + C D ) @

(10.21)

If 0, has the constant value a as shown in

Fig 10.9 then all its derivatives are zero and, for

this input, equation 10.21 becomes, for t > O ,

(1o-22)

We already have the solution for eo, equations

10.17 Subtracting these functions of 0, from 0, we

obtain

k,k20e+ To = ( I D 2 + CD)(-ee) ( I D 2 + C D + kl k2)Oe = - To ( I D 2 + C D + K ) e e = 0

e, = croe-conr {coswdt

= aoe- w n f { 1 + w , t }

= age- l w n r {coshw,t

[eel, = [eels, = -To/(klkz)

l= 1 (10.23) which is independent of the amount of damping

(Note that for zero damping the system oscillates indefinitely with a mean error value of -To/ The complete solution of equation 10.26

+ [l/V(J2 - l ) ] sinhwet} l> 1

where w 2 = KII and 5 = i C / V ( Z K ) Each of the

above three equations contains the negative consists of (a) the complementary function, which exponential term e-5wnr so that, as t+ 00, ee+O is the transient part of the solution and dies away and we say that the final or steady-state error is with time, provided some positive damping is zero and write present, and (b) the particular integral or

steady-state solution which remains after the transients have died away For a constant forcing

We do not need to solve equation 10.22 to find function, the steady-state solution must be a the steady-state value of 0, since this is merely the constant function

particular-integral part of the solution, which is Equation 10.26 describes the system for all time clearly zero That the steady-state error is not from 0 to 00 In the steady-state, therefore, always zero can be seen from the following

example

Consider the position-control system with

viscously damped load which has already been

described Assume that the system is at rest with

(kl k2 1)

I

+ [</V(I - <2)]sinwdt} l< 1

[eelr-m = [eelss = 0

Dee = D20e = 0 and equation 10.26 becomes

k ~ k z [ f ) ~ ] , , = -To and the steady-state error is

= - T o / ( ~ I ~ z )

162 introduction to automatic control

Consider once again Fig 10.13 Assume that

the system is initially at rest then, at time t = 0, it

is required that the load have a constant angular

velocity R i The desired position or input is

therefore

e, = o , t < o

and Oj=Rit, t r O

as shown in Fig 10.17 This is known as a ramp

input

The error-input equation for this particular

input is, from equation 10.21,

(ID2 + C D + K ) e e

= ( I D 2 + C D ) R i t = CRi (10.27)

The steady-state error is equal to C Q / K and

the error response will be of the same form as

equations 10.17 Since 0, = 0, - e,, the output

response can be obtained by subtracting the error

response from the input function The result is

illustrated in Figs 10.18(a) and (b)

in Fig 10.19

The error-input equation for this system can be written down directly from equation 10.27, with

K = k l k2 replaced by kl (kZ + k5/D) Thus [ID2 + C D + kl (k2 + ks/D)]ee = CRj

To convert this to a purely differential equation

we simply differentiate with respect to time by multiplying by D:

[ID3 + C D 2 + klk2D + kl k 5 ] e e

= DCRi = 0 (10.29) since CRj is constant The steady-state error is the particular integral of the above equation so that, for the ramp input,

[ e e 1 s = 0

10.6 Stability of control systems

The introduction of integral action in the above example had the effect of removing the steady- state error to a ramp input It also had the effect

of raising the order of the system The order is

defined as the highest power of D on the left-hand side of a system equation, and in the example it was raised from two to three

For any particular control system, the system- equation loop variable, whichever one is chosen, will be preceded by the same polynomial in operator D (see problem 10.2) Thus the complementary functions (transient responses) for the loop variables will have different initial

A control system with a residual error is conditions but will otherwise be of the same form normally unsatisfactory Certain steady-state Before the concept of integral action was errors can be overcome by using a controller introduced in the previous section, all the system

which incorporates integral action Suppose that, equations were of order two; that is, they were of

in the above example, the voltage V , applied to the form

to the error e,, is given by

The transient response, and thus the stability of

v, = k2ee+k,/'eedt such a system, depends only on the coefficients

provided that a,>O and a 2 > 0 , the com-

plementary function will not contain any positive time exponentials and the system will be stable If

a, = 0 (zero damping) the complementary func- tion will oscillate indefinitely with constant amplitude and, although not strictly unstable, this represents unsatisfactory control Such a system is

[a2D2+alD+ao]x = f ( D ) y

(10.28)

0 ao, a , , and a2 Assuming that ao>O, then,

In D-operator form this is written

v,,, = ( k 2 + 3 e e

and so the block diagram representation of this

proportional-plus-integral controller is as shown

described as marginally stable If either a l < O

(negative damping) o r a2 < 0 (negative mass), the

transient will contain positive exponentials and

the system will be unstable Figure 10.20

illustrates the various types of stability

iii)

a1 a2 > aOa3 (10.32)

Hurwitz conditions for stability of a control

We give below, without proof, the Routh-

10.6 Stability of control systems 163

Consider now the array

al a0 0 > O

a3 a2 a1

(10.46)

164 Introduction to automatic control

where 0 , is the natural circular frequency of the

oscillation and A is the real amplitude

We will use as an illustration the third-order

system described by equation 10.31 With the

right-hand side set equal to zero we have, for the

complementary function,

(10.48) where we assume that a , al , a2 and a3 are all real

and positive

[a3 D3 + a2 D2al D + % ] A e'"nt = 0

Now

DA e i o n t = j w ~ ei"d (10.49)

(10.50)

D ~ A ei0.t = ( j w ) 2 ~ ei"d

and it follows that

D'Ael"n'= (jwn)'A e i o n f (10.51)

where r is any integer

Equation 10.48 becomes

[a3 (jwd3 + a2 ( i w J 2

+al(jw,)+ao]Ae'""'= 0 (10.52)

so that

a3 O'wrJ3 + a2 (jwn12

+al(jw,)+ao = 0 (10.53) sinceAei""#O

Hence

or

The real and imaginary parts must separately be

zero, hence

-a3 jw; - a2 wn2 + al jw, + a = 0

( -a2wn2 + a o ) + 0 , ( -a3wn2 + al ) j = o

( 10.54)

We conclude that if a l a 2 = ~ 0 then the ~ 3

third-order system will be marginally stable and

will oscillate at a circular frequency 0 , given by

equation 10.54 We learned above (inequality

10.32) that if a1a2>aoa3 the system will be stable

It is clear that if this inequality is reversed the

system will be unstable

A physical reason why this inequality deter-

mines the stability of the system can be found by

considering a small applied sinusoidal forcing

term, Fe' "', where F is a complex force amplitude

and w is close to w,

The Argand diagram without the forcing term

is as shown in Fig 10.21(a) and that with the

forcing term is shown in Fig 10.21(b) For energy

W , 2 = - a0 = - a1

a2 0 3

Y

to be fed info the system the force must have a

component which is in phase with the velocity (i.e the imaginary part of the force must be positive) It follows that if energy is required to keep the system oscillating then the system must

be stable So we see that

a1 W > U 3 W 3

or a1>a3w2

Now since w is close to w, we can write

a0 = U2W,2 = a 2 ( w + E)'

where E is a small quantity So as E+O then

w2+ ao/a2

Hence for a stable system

a1 > a3 (ao/a2 1

or ala2>a3ao

Note that as previously mentioned all the a's must

be positive because if any one a is negative the output will diverge for zero input

10.7 Frequency response methods

An assessment of the behaviour of a closed-loop control system can be made from an examination

of the frequency response of the open-loop system Graphical methods are often employed in this work

The main reasons for using open-loop system response methods are

(a) the overall open-loop system response can

be built up quickly using standard response curves

10.7 Frequency response methods 165

of the component parts of the system; In frequency response methods we are only (b) in practice most open-loop systems are concerned with the steady-state oscillations stable which is an advantage if experimental (particular integral) part of the solution and so we techniques are used! ignore the transient (complementary function) Consider again a simple position-control sys- response Since the system is linear, the particular tem with proportional control driving an inertia integral solution of equation 10.58 must be load with viscous damping The block diagram for sinusoidal and at the same frequency w as the the closed-loop system is shown in Fig 10.22 input The steady-state solution is therefore of the which corresponds to Fig 10.13 with K = k l k 2 form

0, = Be I w'

Substituting for 0, in equation 10.58,

D (1 + r D ) B e I"' = KOA e I"' (10.59)

or, from equation 10.51

jw(l+rjw)BeJ"'= K,AeJw' (10.60) The forward-pafh Operator (3 (Dl is given by We see that, for sinusoidal inputs of frequency w

to the open loop system, the ratio of output to

If we disconnect the feedback loop we have the - - - - KO = G(jw) (10.61) open-loop system of Fig 10.23 and it can be seen 8 AeJw' j w ( l + r j w )

K

00

0, ID2 + CD

G ( D ) = - = (10.55) input is

0, B el "'

- -

that G ( D ) is ako the open-loop transfer Operator-

Here ei is simp1y the input to the open-loop which corresponds to the transfer operator G ( D )

of equation 10.56 with D replaced by j w G ( j w ) is

We turn our attention now to the closed-loop system with unity feedback A unity feedback

system is, by definition, one for which the error

0, = 0, - 0, and therefore, since

eliminating 0, we obtain [ l + G ( D ) ] O o = G ( D ) e l (10.63) (For a system with non-unity feedback see example 10.7.)

Suppose now, as was discussed at the end of

G ( D ) = (10.56) section 10.6, that the closed-loop system is

marginally stable, i.e it oscillates continuously at frequency w, say, for no input In this case the particular integral part of the solution is zero, but

the complementary function, or 'transient' part is

We can write G ( D ) in standard form as

KO

D ( l + r D )

where K, = KIC and r = IIC (Note that r h a s the

dimensions of time.) So, for the open-loop

D ( i + a ) e o = = , e (10*57) sinusoidal

0, = Ce Jwn'

We wish to consider the frequency response of

the open-loop so the input must be sinusoidal and

right-hand side set equal to zero we have, for the complementary function

0 = A e J w r

[ l + G ( j w , ) ] C e J " n ' = 0 where A is complex Equation 10.57 now

becomes

(10.58) therefore

D (1 + r D ) 0, = KOA elw'