FranklinCh04ff

4 BasicPropertiesofFeedback

A Perspective on the Properties of FeedbackA major goal of control design is to use the tools available to keep the errorsmall for any input and in the face of expected parameter changes. Al-though in this book we will focus on the selection of the controller transferfunction, the control engineer must be aware that changes to the plant maybe possible that will greatly help control of the process. It is also the casethat the selection and location of a sensor can be very important. Theseconsiderations illustrate the fact that control is a collaborative enterpriseand control objectives need to be considered at every step of the way fromconcept to finished product. However, in this book, we consider mainlythe case of control of dynamic processes and begin with models thatcan be approximated as linear, time-invariant, and described by transferfunctions. Discussion of the theoretical justification of this assumption isdeferred until Chapter 9, in which the theories of Lyapunov are introduced.

Given a model, the next step in the design is formulation of specifi-cations of what it is that the control is required to do. While maintainingthe essential property of stability, the control specifications include bothstatic and dynamic requirements such as the following:

• The permissible steady-state error in the presence of a constant or“bias” disturbance signal.• The permissible steady-state error while tracking a polynomial refer-

ence signal such as a step or a ramp.• The sensitivity of the system transfer function to changes in model

parameters.

PreTEX, Inc., Technical Typesetters Tel. (902)454-8111 FAX (902)454-2894 Franklin, Feedback Control of Dynamic Systems, 5e

166

Feedback Control of Dynamic Systems, Fifth Edition, by Gene F. Franklin, J. David Powell, and Abbas Emami-Naeini.

ISBN 0-13-149930-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

• The permissible transient error in response to a step in either the ref-erence or the disturbance input.

The two fundamental structures for realizing controls are open-loopcontrol (Fig. 4.1) and closed-loop control, also known as feedback con-Open-loop and closed-loop

control trol (Fig. 4.3). Open-loop control is generally simpler, does not require asensor to measure the output, and does not, of itself, introduce stabil-ity problems. Feedback control is more complex and may cause stabil-ity problems but also has the potential to give much better performancethan is possible with open-loop control. If the process is naturally (open-loop) unstable, feedback control is the only possibility to obtain a stablesystem and meet any performance specifications at all. Before specificdesign techniques such as the root locus are described, it is useful todevelop the equations of systems in general terms and to derive expres-sions for the several specifications in order to have a language describingthe objectives toward which the design is directed. As part of this activity,a comparison of open-loop to closed-loop control will expose both theadvantages and the challenges of feedback control.

Chapter OverviewThis chapter begins with consideration of the basic equations of feed-back and the comparison of a feedback structure with open-loop control.In Section 4.1 the equations are presented first in general form and usedto discuss the effects of feedback on disturbance rejection, parame-ter sensitivity, and command tracking. In Section 4.2 the steady-stateerrors in response to polynomial inputs are analyzed in more detail. Aspart of the language of steady-state performance, control systems arefrequently classified by type according to the maximum degree of theinput polynomial for which the steady-state error is a finite constant. InSection 4.3 the issue of dynamic tracking errors is introduced by con-sidering a modification of the closed-loop characteristic equation using aclassical structure of proportional, integral, and differential control, thePID controller. This study will illustrate the interaction of steady-state withtransient performance and will set the tone for the more sophisticated de-sign techniques to be described in later chapters. Finally, in Section 4.4,several extensions of the material of the chapter are presented that areinteresting and important, but something of a distraction from the mainissues of the chapter. Issues discussed there are digital controllers, tuningPID controllers, Truxel’s formula for error constants, and time-domain sen-sitivity. The most important of these is the implementation of controllersin digital form, introduced in Section 4.4.1. If time permits, considerationof this section is highly recommended because almost all modern con-trollers are realized by digital logic. A more complete discussion of thisimportant issue is given in Chapter 8.


167



168 Chapter 4 Basic Properties of Feedback

Figure 4.1Open-loop control system

yControllerDol

PlantG

Input shapingHr

U

W

R

4.1 The Basic Equations of Control

We begin by collecting the basic equations and transfer functions that will beused throughout the rest of the text. For the open-loop system of Fig. 4.1, if wetake the disturbance to be at the input of the plant, the output is given by

Yol = HrDolGR +GW (4.1)

and the error, the difference between reference input and system output, isgiven by

Eol = R − Yol (4.2)

= R − [HrDolGR +GW ] (4.3)

= [1−HrDolG]R −GW (4.4)

= [1− Tol]R −GW. (4.5)

The open-loop transfer function in this case is HrDolG, for which we will usethe generic notation Tol(s).

For feedback control, Figure 4.2 gives the basic structure of interest, butwith the disturbance and the sensor noise entering in unspecific ways. We willtake these signals to be at the inputs of the process and the sensor, respectively,as shown in Figure 4.3. The sensor transfer function is Hy and may show im-portant dynamics. However, the sensor can often be selected to be fast andaccurate. If this is the case, its transfer function can be taken to be a constantHy , with units of volts/unit-of-output. The reference input r has the same units

Figure 4.2Feedback control system

��

�

YController

Dcl

PlantG

SensorHy

Input shapingHr

W

V

R




Section 4.1 The Basic Equations of Control 169

Figure 4.3Basic feedback controlblock diagram

��

�

Y

V

ControllerDcl

PlantG

SensorHy

Input shapingHr

��

R

W

�

��

�

u

as the output, of course, and the input filter’s transfer function is Hr , also withunits of volts/unit-of-output. An equivalent block diagram is drawn in Fig. 4.4,with controller transfer function D(s) = HrDcl and with the feedback transferfunction as the ratio H = Hy

Hr. It is standard practice, especially if Hy is con-

stant, to select equal scale factors so that Hr = Hy and the block diagram can bedrawn as a unity feedback structure as shown in Figure 4.5. We will develop theequations and transfer functions for this standard structure. When we use theseequations later, it will be important to be sure that the preceding assumptionsactually apply. If the sensor has dynamics that cannot be ignored, for example,then the equations will need to be modified accordingly.

For the feedback block diagram of Figure 4.5, the equations for the outputand the control are

Ycl = DG

1+DGR + G

1+DGW − DG

1+DGV, (4.6)

U = D

1+DGR − DG

1+DGW − D

1+DGV. (4.7)

Perhaps more important than these is the equation of the error, Ecl = R − Ycl :

Ecl = R −[

DG

1+DGR + G

1+DGW − DG

1+DGV

](4.8)

= 11+DG

R − G

1+DGW + DG

1+DGV. (4.9)

Figure 4.4Equivalent feedback blockdiagram with Hr includedinside the loop �

�

�

Hr Dcl � D G

W

R ��

�

Y

V��

�

� HHr

Hy





Figure 4.5Unity feedback systemwhen Hr = Hy andletting D = HrDcl

��

�

W

R ��

�

Y

V��

�

uControllerD

PlantG

This equation is simplified by the definition of the sensitivity function � 1 andthe complementary sensitivity function T as

� = 11+DG

(4.10)

and

T = 1− � = DG

1+DG. (4.11)

In terms of these definitions, the equation for the closed-loop error is

Ecl = �R − �GW + TV. (4.12)

For future reference, it is standard to define the transfer function around a loopas the loop gain, L(s). In the case of Fig. 4.4, we have L = DGH , for example.

4.1.1 Watt’s Problem of Disturbance Rejection

One of the early uses of the steam engine in Britain was in mining, to pumpwater out of mines and to haul wagons loaded with coal. In carrying out thesetasks, the steady-state speed of early engines would change substantially whenpresented with added torque caused by a new load. To correct the problem,Watt’s company introduced the flying ball governor shown in Fig. 1.11, wherebythe speed of the engine was fed back to the steam chest to change the torqueof the engine. We will illustrate the principles of operation of this feedbackinnovation through study of the simple equations of motion of an engine withspeed ωe and external load torque τ� .

Equation (4.13) describes the dynamics of an engine with inertia J , viscousfriction b , control u, and load torque τ�(t):

J ωe + bωe = A1u+ A2τ�. (4.13)

1 The reason for the name, coined by H. W. Bode, will be given shortly.





If we take the Laplace transform of Eq. (4.13), let the velocity transform be�e(s) and the transform of the load torque be T�(s), we obtain the transformedequations of open-loop speed control as

sJ�e(s)+ b�e(s) = A1U(s)+ A2T�(s), (4.14)

sJ�e(s)+ b�e(s) = A1[U(s)+ A2

A1T�(s)], (4.15)

sτ�e(s)+�e(s) = A[U +W ]. (4.16)

In deriving Eq. (4.16), we have defined the parameters τ = J/b, A = A1/b ,and the disturbance variable to be W = A2

A1T� . In transfer function form the

equation is

�e(s) = A

(τs + 1)U(s)+ A

(τs + 1)W(s) (4.17)

= G(s)[U(s)+W(s)] (4.18)

= G(s)W(s) if U(s) = 0. (4.19)

In the feedback case, with no reference input and with control proportional toerror as U = −Kcl�e , the equations of proportional feedback control are

sτ�e(s)+�e(s) = A[−Kcl�e +W ], (4.20)

�e(s) = −G(s)Kcl�e(s)+G(s)W, (4.21)

[1+GKcl]�e(s) = GW, (4.22)

�e(s) = G

1+GKclW. (4.23)

In the open-loop case, if the control input is U(s) = 0 and W = wos , the final

value theorem gives2

ωss = G(0)wo = Awo. (4.24)

To make the comparison with the closed-loop case, suppose that G(0) = 1,wo = 1, and just for fun, we take the controller gain to be Kcl = 99. The steady-state output in the open-loop case is ωss = 1, and in the closed-loop case it is

ωss = 11+GKcl

∣∣∣s→0= 1

1+99 = 0.01. Thus the feedback system will have an error

to disturbance that is 100 times smaller than in the open-loop case. No wonderWatt’s engine was a success!

2 We assume for the moment that G(0) is finite.





This result is a particular case of application of the error equations. FromEq. (4.4) the error in the open-loop case is Eol = −GW , and from Eq. (4.12)in the feedback case the error is Ecl = −�GW = �Eol .Thus, in every case,the error due to disturbances is smaller by a factor � in the closed-loop casecompared with the open-loop case.

Major advantage of feedback System errors to constant disturbances can be made smaller with feedbackthan they are in open-loop systems by a factor of � = 1

1+DG(0), where DG(0)

is the loop gain at s = 0.

4.1.2 Black’s Problem: Sensitivity of SystemGain to Parameter Changes

During the 1920s, H. S. Black was working at Bell Laboratories to find a designfor an electronic amplifier suitable for use as a repeater on the long lines of thetelephone company. The basic problem was that electronic components driftedand he needed a design that maintained a gain with great precision in the face ofthese drifts. His solution was a feedback amplifier. To illustrate the advantageshe found, we compare the sensitivity of open-loop control with that of closed-loop control when a parameter changes. The change might come about becauseof external effects such as temperature changes, because of aging, or simplyfrom an error in the value used for the parameter from the start. Suppose thatthe plant gain in operation differs from its original design value of A to beA+ δA, which represents a fractional change of δA

A. The open-loop controller

gain is taken to be fixed at Dol(0) = Kol . In the open-loop case the nominaloverall gain is Tol = KolA,3 and the perturbed gain would be

Tol + δTol = Kol(A+ δA) = KolA+KolδA = Tol +KolδA.

Thus, δTol = KolδA. To give a fair comparison, we compute the fractionalchange in Tol , defined as δTol/Tol for a given fractional change in A. Substitutingthe values, we find that

δTol

Tol= KolδA

KolA= δA

A. (4.25)

This means that a 10% error in A would yield a 10% error in Tol . H. W. BodeSensitivitycalled the ratio of δT /T to δA/A the sensitivity � of the gain with respect tothe parameter A. In the open-loop case, therefore, � = 1.

The same change in A in the feedback case (Eq. (4.23)) yields the newsteady-state feedback gain

Tcl + δTcl = (A+ δA)Kcl

1+ (A+ δA)Kcl,

3 We use Tol and Tcl for the open-loop and closed-loop transfer functions, respectively. Theseare not to be confused with the transform of the disturbance torque Tol used earlier.





where Tcl is the closed-loop gain. We can compute the sensitivity of this closed-loop gain directly using differential calculus. The closed-loop steady-state gain is

Tcl = AKcl

1+ AKcl.

The first-order variation is proportional to the derivative and is given by

δTcl = dTcl

dAδA.

The general expression for sensitivity of a transfer function T to a parameterA is thus given by

δTcl

Tcl=(

A

Tcl

dTcl

dA

)δA

A

= (sensitivity)δA

A.

From this formula the sensitivity is seen to be

�TclA

�= sensitivity of Tcl with respect to A

�= A

Tcl

dTcl

dA,

so

�TclA =

A

AKcl/(1+ AKcl)

(1+ AKcl)Kcl −Kcl(AKcl)

(1+ AKcl)2

= 11+ AKcl

. (4.26)

This result, which explains our use of the name sensitivity earlier, exhibits an-other major advantage of feedback:

Advantage of feedback In feedback control, the error in the overall transfer function gain is less sen-sitive to variations in the plant gain by a factor of � = 1

1+DGcompared with

errors in open-loop control.

As with the case of disturbance rejection, if the gain is such that 1+DG = 100,a 10% change in plant gain A will cause only a 0.1% change in the steady-stategain. The open-loop controller is 100 times more sensitive to gain changes thanthe closed-loop system with loop gain of 100.





The results in this section so far have been computed for the steady-stateerror in the presence of constant inputs, either reference or disturbance. Verysimilar results can be obtained for the steady-state behavior in the presence ofsinusoidal reference and disturbance signals. This is important because thereare times when such signals naturally occur, as with a disturbance of 60 Hz dueto power-line interference in an electronic system, for example. The concept isalso important because more complex signals can be described as containingsinusoidal components over a band of frequencies and analyzed using super-position of one frequency at a time. For example, it is well known that humanhearing is restricted to signals in the frequency range of about 60 to 15,000 Hz.A feedback amplifier and loudspeaker system designed for high-fidelity soundmust accurately track any sinusoidal (pure tone) signal in this range. If we takethe controller in the feedback system shown in Fig. 4.5 to have the transferfunction D(s) and we take the process to have the transfer function G(s), thenthe steady-state open-loop gain at the sinusoidal signal of frequency ωo will be|D(jωo)G(jωo)| and the error of the feedback system will be

|E(jωo)| = |R(jωo)|∣∣∣∣ 11+D(jωo)G(jωo)

∣∣∣∣ .Thus, to reduce errors to 1% of the input at the frequency ωo , we must make|1+DG| ≥ 100 or |D(jωo)G(jωo)|>∼100, and a good audio amplifier must havethis loop gain over the range 2π60 ≤ ω ≤ 2π15,000. We will revisit this conceptin Chapter 6 as part of the design based on frequency response techniques.

4.1.3 The Conflict with Sensor Noise

Finally, it must be noticed that the feedback system error has a term that ismissing from the open-loop case. This is due to the sensor, which is not neededin the open-loop case. The error due to this term is Ecl = TV and will be smallif T is small. Unfortunately, keeping both error due to W and error due to V

small requires that in the one case � be small and in the other case T be small.However, Eq. (4.11) shows that this is not possible. The standard solution to thisdilemma is frequency separation. The reference and the disturbance energiesare typically concentrated in a band of frequencies below some limit—let’s callit ωc . On the other hand, the sensor can usually be carefully designed so thatthe sensor noise V is held small in the low-frequency band below ωc , where theenergy in R and W are substantial.4 Thus the design should have � small whereR and W are large and where V is small, and should then make T be small (and� necessarily larger) for higher frequencies, where sensor noise is unavoidable.It is compromises such as this that will occupy most of our attention in thedesign of controllers in later chapters.

4 The moral of this is that money spent on a good sensor is usually money well spent.





4.1.4 The RADAR Problem: Tracking a TimeVarying Reference

In addition to rejecting disturbances, many systems are required to track a mov-ing reference, for which the generic problem is that of a tracking RADAR. Ina typical system, electric pulses are sent from a parabolic antenna, the echoesfrom the target airplane are received, and an error between the axis of the an-tenna and the vector pointing to the target is computed. The control is requiredto command the antenna pointing angles in such a way as to keep these vectorsaligned. The dynamics of the system are of central importance. A constant-gainopen-loop controller has no effect on the dynamics of the system for eitherreference or disturbance inputs. Only if an open-loop controller includes a dy-namic input filter, Hr(s), can the dynamic response to the reference signal bechanged, but the plant dynamics will still determine the system’s response todisturbances. On the other hand, feedback of any kind changes the dynamics ofthe system for both reference and disturbance inputs. In the case of open-loopspeed control, Eq. (4.17) shows that the plant dynamics are described by the(open-loop) time constant τ . The dynamics with proportional feedback controlare described by Eq. (4.23), and the characteristic equation of this system is

1+GKcl = 0, (4.27)

1+ AKcl

τs + 1= 0, (4.28)

τs + 1+ AKcl = 0, (4.29)

s = −1+ AKcl

τ. (4.30)

Therefore, the closed-loop time constant, a function of the feedback gain Kcl ,is given by τcl = τ

1+ AKcl, and is decreased as compared with the open-loop

value. It is typically the case that closed-loop systems have a faster responseas the feedback gain is increased and, if there were no other effects, this isgenerally desirable. As we will see, however, the responses of higher ordersystems typically become less well damped and eventually will become unstableas the gain is steadily increased. Thus a definite limit exists on how large wecan make the gain in our efforts to reduce the effects of disturbances andthe sensitivity to changes in plant parameters. Attempts to resolve the conflictbetween small steady-state errors and good dynamic response will characterizea large fraction of control design problems. The conclusion is as follows:

Property of feedback Feedback changes dynamic response and often makes a system both faster andless stable.





4.2 Control of Steady-State Error: System Type

In the speed-control case study in Section 4.1 we assumed both reference anddisturbances to be constants and also took D(0) and G(0) to be finite constants.In this section we will consider the possibility that either or both of D(s) andG(s) have poles at s = 0. For example, a well-known structure for the controlequation of the form

u(t) = kp + kI

∫ t

e(τ ) dτ + kD

de(t)

dt(4.31)

is called proportional plus integral plus derivative (PID) control, and the cor-responding transfer function is

D(s) = kp + kI

s+ kDs. (4.32)

In a number of important cases, the reference input will not be constant butcan be approximated as a polynomial in time long enough for the system toeffectively reach steady-state. For example, when an antenna is tracking the el-evation angle to a satellite, the time history as the satellite approaches overheadis an S-shaped curve as sketched in Fig. 4.6. This signal may be approximatedby a linear function of time (called a ramp function or velocity input) for asignificant time relative to the speed of response of the servomechanism. Inthe position control of an elevator, a ramp function reference input will directthe elevator to move with constant speed until it comes near the next floor. Inrare cases, the input can be approximated over a substantial period as having aconstant acceleration. In this section we consider steady-state errors in stablesystems with such polynomial inputs.

The general method is to represent the input as a polynomial in time and toconsider the resulting steady-state tracking errors for polynomials of differentdegrees. As we will see, the error will be zero for input polynomials below acertain degree, and will be unbounded for inputs of higher degrees. A stablesystem can be classified as a system type, defined to be the degree of the poly-Definition of system typenomial for which the steady-state system error is a nonzero finite constant. Inthe speed-control example, proportional control was used and the system had aconstant finite error to a step input, which is an input polynomial of zero degree;therefore this system is called a type zero (type 0) system. If the error to a rampor first-degree polynomial is a finite nonzero constant, such a system is calledtype one (type 1), and so on. System types can be defined with regard to eitherreference inputs or disturbance inputs, and in this section we will consider both

Figure 4.6Signal for satellite tracking

Time (sec)

us




Section 4.2 Control of Steady-State Error: System Type 177

classifications. Determining the system type involves calculating the transformof the system error and then applying the Final Value Theorem. As we will see,a determination of system type is easiest for the case of unity feedback, so wewill begin with that case.

4.2.1 System Type for Reference Tracking:The Unity Feedback Case

In the unity feedback case drawn in Fig. 4.5, the system error is given byEq. (4.9). If we consider only the reference input alone and set W = V = 0,then, using the symbol for loop gain, the equation is simply

E = 11+ L

R = �R. (4.33)

To consider polynomial inputs, we let r(t) = tk1(t), for which the transformis R = 1

sk+1 . As a generic reference nomenclature, step inputs for which k = 0are called “position” inputs, ramp inputs for which k = 1 are called “velocity”inputs, and if k = 2, the inputs are called “acceleration” inputs, regardless ofthe units of the actual signals. Application of the Final Value Theorem to theerror gives the formula

limt→∞ e(t) = ess = lim

s→0E(s) (4.34)

= lims→0

s1

1+ LR(s) (4.35)

= lims→0

s1

1+ L

1

sk+1. (4.36)

We consider first a system for which L has no pole at the origin and a step inputfor which R(s) = 1

s. In this case, Eq. (4.36) reduces to

ess = lims→0

s1

1+ L

1s

(4.37)

= 11+ L(0)

. (4.38)

We define such a system to be type 0 and we define the constant L(0)�= Kp as

the “position error constant.” If L has one pole at the origin, we could considerboth step and ramp inputs, but it is quite straightforward to evaluate Eq. (4.36)in a general setting. For this case, it is useful to be able to describe the behaviorof the controller and plant as s approaches 0. For this purpose, we collect allthe terms except the pole(s) at the origin into a function L(s), which is thus





finite at s = 0 so that we can define the constant Lo(0) = Kn and write the looptransfer function as

L(s) = Lo(s)

sn . (4.39)

For example, if L has no integrator, then n = 0. If the system has one integrator,then n = 1, and so forth. Substituting this expression into Eq. (4.36), we have

ess = lims→0

s1

1+ Lo(s)

sn

1

sk+1(4.40)

= lims→0

sn

sn +Kn

1

sk. (4.41)

From this equation we can see at once that if n > k , then e = 0, and if n < k ,then e→∞. If n = k = 0, then ess = 1

1+K0, and if n = k �= 0, then ess = 1

Kn. If

n = k = 0, the input is a zero-degree polynomial otherwise known as a step orposition, the constant Ko is called the “position constant,” written as Kp , andthe system is classified as “type 0,” as we saw before. If n = k = 1, the input isa first-degree polynomial, otherwise known as a ramp or velocity, the constantK1 is called the “velocity constant,” written as Kv , and the system is classified“type 1.” In a similar way, systems of “type 2” and higher types may be defined.The type information can be usefully gathered in a table of errors as follows:

TABLE 4.1 Errors as a Function of System Type

Input

Type Step (Position) Ramp (Velocity) Parabola (Acceleration)

Type 0 11+Kp

∞ ∞Type 1 0 1

Kv∞

Type 2 0 0 1Ka

The most common case is that of simple integral control leading to a “type1” system. In this case, the relationship between Kv and the steady-state error“Type 2” systemsto a ramp input is shown in Fig. 4.7. Looking back at the expression givenfor DcG in Eq. (4.39), we can readily see that the several error constants canbe calculated by counting the degree n of the poles of L at the origin (thenumber of integrators in the loop with unity gain feedback) and applying theappropriate one of the following simple formulas

Kp = lims→0

L(s), n = 0, (4.42)

Kv = lims→0

sL(s), n = 1, (4.43)

Ka = lims→0

s2L(s), n = 2. (4.44)





Figure 4.7Relationship between rampresponse and Kv

0 1 2 3 4 5 6 7 8 9 10

10

9

8

7

6

5

4

3

2

1

0

Time (sec)

r, yr y

1K�

ess �

EXAMPLE 4.1 System Type for Speed Control

Determine the system type and the relevant error constant for the speed-control exam-ple shown in Fig. 4.4, with proportional feedback given by D(s) = kp . The plant transferfunction is G = A

τs+1 .

Solution. In this case, L = kpA

τs+1 , and applying Eq. (4.42), we see that n = 0, as there isno pole at s = 0. Thus the system is type 0, and the error constant is a position constantgiven by Kp = kpA.

EXAMPLE 4.2 System Type Using Integral Control

Determine the system type and the relevant error constant for the speed-control exam-ple shown in Fig. 4.4, with PI feedback. The plant transfer function is G = A

τs+1 , and inthis case the controller transfer function is Dc = kp + kI

s.

Solution. In this case, the transfer function is L(s) = A(kps+kI )s(τs+1)

, and as a unity feedbacksystem with a single pole at s = 0, the system is immediately seen as type 1. The velocityconstant is given by Eq. (4.43) to be Kv = lim

s→0sL(s) = AkI .

The definition of system type helps us to identify quickly the ability ofa system to track polynomials. In the unity feedback structure, if the processparameters change without removing the pole at the origin in a type 1 system,the velocity constant will change, but the system will still have zero steady-stateerror in response to a constant input and will still be type 1. Similar statementscan be made for systems of type 2 or higher. Thus, we can say that system typeis a robust property with respect to parameter changes in the unity feedbackRobustness of system typestructure. Robustness is the major reason for preferring unity feedback overother kinds of control structure.





Figure 4.8Block diagram reductionto an equivalent unityfeedback system

��

�

YD

H � 1

GUE

��

R

4.2.2 System Type for Reference Tracking:The General Case

If the feedback H = Hy

Hrin Fig. 4.4 is different from unity, the formulas given in

the unity feedback case do not apply, and a more general approach is needed.There are two immediate possibilities. In the first instance, if one adds andsubtracts 1.0 from H , as shown by block diagram manipulation in Fig. 4.8, thegeneral case is reduced to the unity feedback case and the formulas can beapplied to the redefined loop transfer function L = DG

1+(H−1)DG, for which the

error equation is again E = 11+L

R = �R .Another possibility is to develop formulas directly in terms of the closed-

loop transfer function, which we call the complementary sensitivity functionT(s). From Fig. 4.4, the transfer function is

Y (s)

R(s)= T(s) = DG

1+HDG, (4.45)

and therefore the error is

E(s) = R(s)− Y (s) = R(s)− T(s)R(s).

The reference-to-error transfer function is thus

E(s)

R(s)= 1− T(s),

and the system error transform is

E(s) = [1− T(s)]R(s) = �R.

We assume that the conditions of the Final Value Theorem are satisfied, namelythat all poles of sE(s) are in the left half plane. In that case the steady-stateerror is given by applying the Final Value Theorem to get

ess = limt→∞ e(t) = lim

s→0sE(s) = lim

s→0s[1− T(s)]R(s). (4.46)





With a polynomial test input, the error transform becomes

E(s) = 1sk+1

[1− T(s)],

and the steady-state error is given again by the Final Value Theorem:

ess = lims→0

s1− T(s)

sk+1= lim

s→0

1− T(s)

sk. (4.47)

The result of evaluating the limit in Eq. (4.47) can be zero, a nonzero constant,or infinite. If the solution to Eq. (4.47) is a nonzero constant, the system isreferred to as type k . For example, if k = 0 and the solution to Eq. (4.47) isa nonzero constant equal, by definition, to 1

1+Kp, then the system is type 0.

Similarly, if k = 1 and the solution to Eq. (4.47) is a nonzero constant, thenthe system is type 1 and has a zero steady-state error to a position input anda constant steady-state error equal, by definition, to 1/Kv to a unit velocityreference input. Type 1 systems are by far the most common in practice. Asystem of type 1 or higher has a closed-loop DC gain of 1.0, which means thatT (0) = 1.

EXAMPLE 4.3 System Type for a Servo with Tachometer Feedback

Consider an electric motor position-control problem, including a nonunity feedbacksystem caused by having a tachometer fixed to the motor shaft and its voltage (whichis proportional to shaft speed) is fed back as part of the control. The parameters corre-sponding to Fig. 4.4 are

G(s) = 1s(τ s + 1)

,

D(s) = kp,

H(s) = 1+ kt s.

Determine the system type and relevant error constant with respect to reference inputs.

Solution. The system error is

E(s) = R(s)− Y (s)

= R(s)− T(s)R(s)

= R(s)− DG(s)

1+HDG(s)R(s)

= 1+ (H(s)− 1)DG(s)

1+HDG(s)R(s).





The steady-state system error from Eq. (4.47) is

ess = lims→0

sR(s)[1− T(s)].

For a polynomial reference input, R(s) = 1/sk+1 , and hence

ess = lims→0

[1− T(s)]

sk= lim

s→0

1

sk

s(τ s + 1)+ (1+ kt s − 1)kp

s(τs + 1)+ (1+ kt s)kp

= 0 , k = 0,

= 1+ ktkp

kp

, k = 1.

Therefore the system is type 1 and the velocity constant is Kv = kp

1+ ktkp. Notice that

if kt > 0, this velocity constant is smaller than the unity feedback value of kp . Theconclusion is that if tachometer feedback is used to improve dynamic response, thesteady-state error is increased.

4.2.3 System Type with Respect to DisturbanceInputs

In most control systems, disturbances of one type or another exist. In practice,these disturbances can sometimes be usefully approximated by polynomial timefunctions such as steps or ramps. This would suggest that systems also be clas-sified with respect to the system’s ability to reject disturbance inputs in a wayanalogous to the classification scheme based on reference inputs. System typewith regard to disturbance inputs specifies the degree of the polynomial ex-pressing those input disturbances that the system can reject in the steady state.Knowing the system type, we know the qualitative steady-state response of thesystem to polynomial disturbance inputs such as step or ramp signals. Becausetype depends on the transfer function from disturbance to error, the systemtype depends on exactly where the disturbance enters into the control system.

The transfer function from the disturbance input W(s) to the error E(s) is

E(s)

W(s)= −Y (s)

W(s)= Tw(s), (4.48)

because, if the reference is equal to zero, the output is the error. In a similar wayas for reference inputs, the system is type 0 if a step disturbance input resultsin a nonzero constant steady-state error and is type 1 if a ramp disturbanceinput results in a steady-state value of the error that is a nonzero constant. Ingeneral, following the same approach used in developing Eq. (4.41), we assumethat a constant n and a function To,w(s) can be defined with the properties





that To,w(0) = 1Kn,w

and that the disturbance-to-error transfer function can bewritten as

Tw(s) = snTo,w(s). (4.49)

Then the steady-state error to a disturbance input that is a polynomial of degreek is

yss = lims→0

[sTw(s)

1sk+1

]

= lims→0

[To,w(s)

sn

sk

]. (4.50)

From Eq. (4.50), if n > k , then the error is zero, and if n < k , the error isunbounded. If n = k , the system is type k and the error is given by 1

Kn,w.

EXAMPLE 4.4 Satellite Attitude Control

Consider the model of a satellite attitude control system shown in Fig. 4.9(a), where

J = moment of inertia,

W = disturbance torque,

Hy = sensor gain, and

Dc(s) = the compensator.

With equal input filter and sensor scale factors, the system with PD control can beredrawn with unity feedback as in Fig. 4.9(b) and with PID control drawn as in Fig. 4.9(c).Assume that the control results in a stable system and determine the system types anderror responses to disturbances of the control system for

(a) System Fig. (4.9)(b) PD control

(b) System Fig. (4.9)(c) PID control

Solution.

(a) We see from inspection of Fig. 4.9(b) that, with two poles at the origin in the plant,the system is type 2 with respect to reference inputs. The transfer function fromdisturbance to error is

Tw(s) = 1

J s2 + kDs + kp

(4.51)

= To,w(s), (4.52)





Figure 4.9Model of a satellite attitudecontrol: (a) basic system;(b) PD control; (c) PIDcontrol

(a)

R ��

�

��

D (s) �

W

K

K

UJs1

s1

u � Yu

(b)

R ��

�

��

�

W

Js21

Y

�1.0

kp � kDs

(c)

R ��

�

��

�

W

Js21

Y

�1.0

kp � � kDsskI

for which n = 0 and Ko,w = kp . The system is type 0 and the error constant is kp ,so the error to a unit disturbance step is 1

kp.

(b) With PID control, the forward gain has three poles at the origin, so this system istype 3 for reference inputs, but the disturbance transfer function is

Tw(s) = s

J s3 + kDs2 + kps + kI

, (4.53)

n = 1, (4.54)

To,w(s) = 1

J s3 + kDs2 + kps + kI

, (4.55)

from which it follows that the system is type 1 and the error constant is kI , so theerror to a disturbance ramp of unit slope will be 1

kI.





EXAMPLE 4.5 System Type for a DC Motor Position Control

Consider the simplified model of a DC motor in unity feedback as shown in Fig. 4.10,where the disturbance torque is labeled W(s).

(a) Use the proportional controller

D(s) = kp, (4.56)

and determine the system type and steady-state error properties with respect todisturbance inputs.

(b) Let the control be PI, as given by

D(s) = kp + kI

s, (4.57)

and determine the system type and the steady-state error properties for disturbanceinputs.

Solution.

(a) The closed-loop transfer function from W to E (where R = 0) is

Tw(s) = −B

s(τs + 1)+ Akp

= s0To,w,

n = 0,

Ko,w = −Akp

B.

Applying Eq. (4.50), we see that the system is type 0 and the steady-state error toa unit step torque input is ess = −B

Akp. From the earlier section, this system is seen to

be type 1 for reference inputs and illustrates that system type can be different fordifferent inputs to the same system.

Figure 4.10DC motor with unityfeedback

�

W (s)

AB

��

�

�1.0

D (s)s(ts � 1)

AYR





(b) If the controller is PI, the disturbance error transfer function is

Tw(s) = −Bs

s2(τ s + 1)+ (kps + kI )A, (4.58)

n = 1, (4.59)

Kn,w = AkI

−B, (4.60)

and therefore the system is type 1 and the error to a unit ramp disturbance inputwill be

ess = −B

AkI

. (4.61)

4.3 Control of Dynamic Error: PID Control

We have seen in Section 4.1 basic properties of feedback control, and in Sec-tion 4.2 we examined the steady state response of systems to polynomial refer-ence and disturbance input. At the end of Section 4.1 we observed that propor-tional control changed the time constant of the simple speed-control system.In this section the impact of more sophisticated controls on system character-istic equations is examined in the context of a standard controller structure.The most basic feedback is a constant Proportional to error. As we saw in Sec-tion 4.2, addition of a term proportional to the Integral of error has a majorinfluence on the system type and steady-state error to polynomials. The finalterm in the classical structure term proportional to the Derivative of error.Combined, these three terms form the classical PID controller, which is widelyThe PID (proportional-integral-

derivative) controller used in the process and robotics industries.

4.3.1 Proportional Control (P)

When the feedback control signal is linearly proportional to the system error,we call the result Proportional feedback. This was the case for the feedbackused in the controller of speed in Section 4.1, for which the controller transferfunction is

U(s)

E(s)= Dc(s) = kp. (4.62)

As we saw in Section 4.1.4, the time constant of the feedback system was re-duced by a factor 1+Akp by proportional control. If the plant is second order,as, for example, is a DC motor with nonnegligible inductance, then the transferfunction can be written as

G(s) = A

s2 + a1s + a2. (4.63)




Section 4.3 Control of Dynamic Error: PID Control 187

In this case, the characteristic equation with proportional control is

1+ kpG(s) = 0, (4.64)

s2 + a1s + a2 + kp = 0. (4.65)

The designer can control the constant term and the natural frequency, but notthe damping of this equation. If kp is made large to get adequate steady-stateerror, the damping may be much too low for satisfactory transient response.

4.3.2 Proportional plus Integral Control (PI)

Adding an integral term to the controller results in the Proportional plus Inte-gral (PI) control equationProportional plus Integral

control

u(t) = kpe + kI

∫ t

t0

e(τ ) dτ, (4.66)

for which the Dc(s) in Fig. 4.5 becomes

U(s)

E(s)= Dc(s) = kp + kI

s. (4.67)

This feedback has the primary virtue that, in the steady-state, its control outputcan be a nonzero constant value even when the error signal at its input is zero.This comes about because the integral term in the control signal is a summationof all past values of e(t). In fact, the integral term will not stop changing untilits input is zero, and therefore if the system reaches a stable steady state, theinput signal to the integrator will of necessity be zero. This feature means that aconstant disturbance w (see Fig. 4.4) can be canceled by the integrator’s outputeven while the system error is zero.

If PI control is used in the speed example, the transform equation for thecontroller is

U = kp(�ref −�m)+ kI

�ref −�m

s, (4.68)

and the system transform equation with this controller is

(τ s + 1)�m = A(kp + kI

s)(�ref −�m)+ T W. (4.69)

If we now multiply by s and collect terms, we obtain

(τ s2 + (Akp + 1)s + AkI )�m = A(kps + kI )�ref + AsW. (4.70)

Because the PI controller includes dynamics, use of this controller will changethe dynamic response in more complicated ways than the simple speed-upwe saw with proportional control. We can understand this by considering the





characteristic equation of the speed control with PI control, as seen in Eq. (4.70).The characteristic equation is

τs2 + (Akp + 1)s + AkI = 0. (4.71)

The two roots of this equation may be complex and, if so, the natural frequency

is ωn =√

AkIτ

, and the damping ratio is ζ = Akp+12τωn

. These parameters areboth determined by the controller gains. If the plant is second order, then thecharacteristic equation is

1+ kps + kI

s

A

s2 + a1s + a2= 0, (4.72)

s3 + a1s2 + a2s + Akps + AkI = 0. (4.73)

In this case, the controller parameters can be used to set two of the coefficients,but not the third. For this we need derivative control.

4.3.3 Proportional-Integral-Derivative Control (PID)

The final term in the classical controller is derivative control, D, and the com-plete three-term controller is described by the transform equation we will use,namely,

Dc(s) = U(s)

E(s)= kp + kI

s+ kDs, (4.74)

or, equivalently, by the equation often used in the process industries, or

Dc(s) = kp[1+ 1TI s+ TDs], (4.75)

where the “reset rate” TI in seconds, and the “derivative rate,” TD , also inseconds, can be given physical meaning to the operator who must select valuesfor them to “tune” the controller. For our purposes, Eq. (4.74) is simpler to use.The effect of the derivative control term depends on the rate of change of theerror. As a result, a controller with derivative control exhibits an anticipatoryresponse, as illustrated by the fact that the output of a PD controller having aramp error e(t) = t1(t) input would lead the output of a proportional controller

having the same input by kDkp

�= TD seconds, as shown in Fig. 4.11.




Section 4.3 Control of Dynamic Error: PID Control 189

Figure 4.11Anticipatory nature ofderivative control

TD

u(t)

0 1 2 3 4 5

Time (sec)

PD

Proportional

Because of the sharp effect of derivative control on suddenly changingsignals, the “D” term is sometimes introduced into the feedback path as shownin Fig. 4.12(a), which would describe, for example, a tachometer on the shaftof a motor. The closed-loop characteristic equation is the same as if the termwere in the forward path, as given by Eq. (4.74) and drawn in Fig. 4.12(b), ifthe derivative gain is kD = kpkt but the zeros from the reference to the outputare different in the two cases. With the derivative in the feedback path, thereference is not differentiated, which may be a desirable result if the referenceis subject to sudden changes. With the derivative in the forward path, a stepchange in the reference input will, in theory, cause an intense initial pulse inthe control signal, which may be very undesirable.

To illustrate the effect of a derivative term on PID control, consider speedcontrol, but with the second-order plant. In that case, the characteristic equa-tion is

s2 + a1s + a2 + A(kp + kI

s+ kDs) = 0,

s3 + a1s2 + a2s + A(kps + kI + kDs2) = 0. (4.76)

Collecting terms results in

s3 + (a1 + AkD)s2 + (a2 + Akp)s + AkI = 0. (4.77)

Figure 4.12Alternative ways ofconfiguring rate feedback

(a)

(b)

��

�

1 � kt,Ds

kpR G(s) Y

��

�

kp � kDsR G Y





The point here is that this equation, whose three roots determine the natureof the dynamic response of the system, has three free parameters in kP , kI ,and kD , and by selection of these parameters, the roots can be uniquely and,in theory, arbitrarily determined. Without the derivative term, there would beonly two free parameters, but with three roots, the choice of roots of the char-acteristic equation would be restricted. To illustrate the effect more concretely,a numerical example is useful.

EXAMPLE 4.6 PID Control of Motor Speed

Consider the DC motor speed control with parameters5

Jm=1.13× 10−2 N-m- sec2 /rad, b=0.028 N-m-sec/rad, La=10−1 henry,

Ra=0.45 ohms, Kt=0.067 N-m/amp, Ke=0.067 V-sec/rad. (4.78)

Use the controller parameters

kp = 3, kI = 15 sec−1, kD = 0.3 sec. (4.79)

Discuss the effects of P, PI, and PID control on the responses of this system to stepsin the disturbance and steps in the reference input. Let the unused controller parametersbe zero.

Solution. Figure 4.13(a) illustrates the effects of P, PI, and PID feedback on the stepdisturbance response of the system. Note that adding the integral term increases theoscillatory behavior but eliminates the steady-state error, and that adding the derivativeterm reduces the oscillation while maintaining zero steady-state error. Figure 4.13(b)illustrates the effects of P, PI, and PID feedback on the step reference response, withsimilar results. The step responses can be computed by forming the numerator anddenominator coefficient vectors (in descending powers of s ) and using the step functionin MATLAB. For example, after the values for the parameters are entered, the followingcommands produce a plot of the response of PID control to a disturbance step:

numG = [La Ra 0];denG = [Jm*La Ra*b + Ke*Ke + Ke*kD Ra*Ke*Ke + Ke*kp Ke*ki];sysG = tf(numG,denG);y = step(sysG).

5 These values have been scaled to measure time in milliseconds by multiplying the true La andJm by 1000 each.




Section 4.4 Extensions to the Basic Feedback Concepts 191

�6

8

6

4

2

0

�2

�4

Am

plitu

de

0 1 2 3 4 5 6

Time (msec)

(a)

0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

Am

plitu

de

0 1 2 3 4 5 6

Time (msec)

(b)

P

PI

PID

P

PI

PID

Figure 4.13 Responses of P, PI, and PID control to (a) step disturbance input and (b) step reference input

4.4 Extensions to the Basic Feedback Concepts

4.4.1 Digital Implementation of Controllers

As a result of the revolution in the cost-effectiveness of digital computers, therehas been an increasing use of digital logic in embedded applications, such as con-trollers in feedback systems. With the formula for calculating the control signalin software rather than hardware, a digital controller gives the designer muchmore flexibility in making modifications to the control law after the hardwaredesign is fixed. In many instances, this means that the hardware and softwaredesigns can proceed almost independently, saving a great deal of time. Also, it iseasy to include binary logic and nonlinear operations as part of the function ofa digital controller. Special processors designed for real-time signal processingand known as digital signal processors, or DSPs, are particularly well suited foruse as real-time controllers. While, in general, the design of systems to use adigital processor requires sophisticated use of new concepts to be introducedin Chapter 8, such as the z-transform, it is quite straightforward to translate alinear continuous analog design into a discrete equivalent. A digital controllerdiffers from an analog controller in that the signals must be sampled and quan-tized.6 A signal to be used in digital logic needs to be sampled first, and thenthe samples need to be converted by an analog-to-digital converter, or A/Dconverter,7 into a quantized digital number. Once the digital computer has cal-culated the proper next control signal value, this value needs to be convertedback into a voltage and held constant or otherwise extrapolated by a digital-to-

6 A controller that operates on signals that are sampled but not quantized is called discrete, whileone that operates on signals that are both sampled and quantized is called digital.

7 Pronounced “A to D.”





analog converter, or D/A, in order to be applied to the actuator of the process.The control signal is not changed until the next sampling period. As a resultof the sampling, there are more strict limits on the speed or bandwidth of adigital controller than on analog devices. Discrete design methods that tendto minimize these limitations are described in Chapter 8 . A reasonable ruleof thumb for selecting the sampling period is that during the rise time of theresponse to a step, the input to the discrete controller should be sampled ap-proximately six times. By adjusting the controller for the effects of sampling,the sampling can be adjusted to 2 to 3 times per rise time. This corresponds to asampling frequency that is 10 to 20 times the system’s closed-loop bandwidth.The quantization of the controller signals introduces an equivalent extra noiseinto the system, and to keep this interference at an acceptable level, the A/Dconverter usually has an accuracy of 10 to 12 bits. For a first analysis, the effectsof the quantization are usually ignored. A simplified block diagram of a systemwith a digital controller is shown in Figure 4.14.

For this introduction to digital control, we will describe a simplified tech-nique for finding a discrete (sampled, but not quantized) equivalent to a givencontinuous controller. The method depends on the sampling period Ts beingshort enough that the reconstructed control signal is close to the signal that theoriginal analog controller would have produced. We also assume that the num-bers used in the digital logic have enough accurate bits so that the quantizationimplied in the A/D and D/A processes can be ignored. While there are goodanalysis tools to determine how well these requirements are met, here we willtest our results by simulation, following the well known advice that “The proofof the pudding is in the eating.”

Finding a discrete equivalent to a given analog controller is equivalent tofinding a recurrence equation for the samples of the control which will approx-imate the differential equation of the controller. The assumption is that wehave the transfer function of an analog controller and wish to replace it with adiscrete controller that will accept samples of the controller input, e(kTs), froma sampler and, using past values of the control signal, u(kTs), and present andpast samples of the input, e(kTs), will compute the next control signal to besent to the actuator. As an example, consider a PID controller with the transferfunction

U(s) = (kp + kI

s+ kDs)E(s), (4.80)

Figure 4.14Block diagram of a digitalcontroller

SensorH

U

T

��

�

YR A/De(kT ) u(kT )

D/ADigital controllerD(z)

Clock

PlantG





which is equivalent to the three terms of the time-domain expression

u(t) = kpe(t)+ kI

∫ t

0e(τ ) dτ + kD e(t) (4.81)

= uP + uI + uD. (4.82)

Using the fact that the system is linear, the next control sample can be computedterm by term. The proportional term is immediate:

uP (kTs + Ts) = kpe(kTs + Ts). (4.83)

The integral term can be computed by breaking the integral into two parts andapproximating the second part, which is the integral over one sample period,as follows:

uI (kTs + Ts) = kI

∫ kTs+Ts

0e(τ ) dτ (4.84)

= kI

∫ kTs

0e(τ ) dτ + kI

∫ kTs+Ts

kTs

e(τ ) dτ (4.85)

= uI (kTs)+ {area under e(τ ) over one period} (4.86)

∼= uI (kTs)+ kI

Ts

2{e(kTs + Ts)+ e(kTs)}. (4.87)

In Eq. (4.87) the area in question has been approximated by that of the trapezoidformed by the base Ts and vertices e(kTs + Ts) and e(kTs), as shown by thedashed line in Fig. 4.15.

The area can also be approximated by the rectangle of amplitude e(kTs)

and width Ts , shown by the solid blue in Fig. 4.15, to give uI (kTs + Ts) =uI (kTs)+kITse(kTs). These and other possibilities are considered in Chapter 8.

In the derivative term, the roles of u and e are reversed from integration,and the consistent approximation can be written down at once from Eq. (4.87)and Eq. (4.81) as

Ts

2{uD(kTs + Ts)+ uD(kTs)} = kD{e(kTs + Ts)− e(kTs)}. (4.88)

Figure 4.15Graphical interpretation ofnumerical integration

t

x

x � f (x, u)

� x dt

0 ti ti�1

x (ti)

0

t





As with linear analog transfer functions, these relations are greatly simplifiedand generalized by the use of transform ideas. At this time, the discrete trans-form will be introduced simply as a prediction operator z, much as if we de-scribed the Laplace transform variable s as a differential operator. Here wedefine the operator z as the forward shift operator in the sense that if U(z) isthe transform of u(kTs), then zU(z) will be the transform of u(kTs + Ts). Withthis definition, the integral term can be written as

zUI (z) = UI (z)+ kI

Ts

2[zE(z)+ E(z)] , (4.89)

UI (z) = kI

Ts

2z+ 1z− 1

E(z), (4.90)

and from Eq. (4.88) the derivative term becomes the inverse

UD(z) = kD

2Ts

z− 1z+ 1

E(z). (4.91)

The complete discrete PID controller is thus described by

U(z) =(

kp + kI

Ts

2z+ 1z− 1

+ kD

2Ts

z− 1z+ 1

)E(z). (4.92)

Comparing the two discrete equivalents of integration and differentiation withthe corresponding analog terms, it is seen that the effect of the discrete ap-proximation in the z-domain is as if everywhere in the analog transfer functionthe operator s has been replaced by the composite operator 2

Ts

z−1z+1 . This is the

trapezoid rule8 of discrete equivalents:

Trapezoid RuleThe discrete equivalent to Da(s) is Dd(z) = Da

(2Ts

z− 1z+ 1

). (4.93)

EXAMPLE 4.7 Discrete Equivalent

Find the discrete equivalent of the analog controller with transfer function

D(s) = U(s)

E(s)= 11s + 1

3s + 1, (4.94)

using the sample period Ts = 1.

8 The formula is also called Tustin’s Method after the English engineer who used the techniqueto study the responses of nonlinear circuits.





Solution. The discrete operator is 2(z− 1)z+ 1 , and thus the discrete transfer function is

Dd(z) = U(z)

E(z)= D(s)

∣∣∣∣s= 2

Tsz−1z+1

(4.95)

=11[

2(z− 1)

z+ 1

]+ 1

3[

2(z− 1)

z+ 1

]+ 1

. (4.96)

Clearing fractions, we get the discrete transfer function

Dd(z) = U(z)

E(z)= 23z− 21

7z− 5. (4.97)

Converting the discrete transfer function to a discrete difference equation by using thedefinition of z as the forward shift operator is done as follows: First we cross-multiplyin Eq. (4.97) to obtain

(7z− 5)U(z) = (23z− 21)E(z), (4.98)

and interpreting z as a shift operator, we find that this is equivalent to the differenceequation9

7u(k + 1)− 5u(k) = 23e(k + 1)− 21e(k), (4.99)

where we have replaced kTs + Ts with k + 1 to simplify the notation. To compute thenext control at time kTs + Ts , therefore, we solve the difference equation

u(k + 1) = 57u(k)+ 23

7e(k + 1)− 21

7e(k). (4.100)

Now let’s apply these results to a control problem. Fortunately, MATLABprovides us with the Simulink capability to simulate both continuous and dis-crete systems, allowing us to compare the responses of the systems with con-tinuous and discrete controllers.

EXAMPLE 4.8 Equivalent Discrete Controller for Speed Control

A motor speed control is found to have the plant transfer function

Y

U= 45

(s + 9)(s + 5). (4.101)

9 The process is similar to that used in Chapter 3 to find the ordinary differential equation towhich a rational Laplace transform corresponds.





A PI controller designed for this system has the transfer function

D(s) = U

E= 1.4

s + 6s

. (4.102)

The closed-loop system has a rise time of about 0.2 sec and an overshoot of about20%. Design a discrete equivalent of this controller, and compare the step responsesand control signals of the two systems. (a) Compare the responses if the sample periodis 0.07, which is about three samples per rise time. (b) Compare the responses with asample period of Ts = 0.035, which corresponds to about six samples per rise time.

Solution.

(a) Using the substitution given by Eq. (4.93), the discrete equivalent for Ts = 0.07 is

given by replacing s by s ← 20.07

z− 1z+ 1 in D(s) as follows:

Dd(z) = 1.4

2.07

z− 1z+ 1

+ 6

2.07

z− 1z+ 1

, (4.103)

= 1.42(z− 1)+ 6 ∗ 0.07(z+ 1)

2(z− 1), (4.104)

= 1.41.21z− 0.79

(z− 1). (4.105)

On the basis of this expression, the equation for the control is (the sample periodis suppressed)

u(k + 1) = u(k)+ 1.4 ∗ [1.21e(k + 1)− 0.79e(k)]. (4.106)

(b) For Ts = 0.035, the discrete transfer function is

Dd = 1.41.105z− 0.895

z− 1, (4.107)

for which the difference equation is

u(k + 1) = u(k)+ 1.4[1.105 e(k + 1)− 0.895 e(k)].

A Simulink block diagram for simulating the two systems is given in Fig. 4.16, andplots of the step responses are given in Fig. 4.17(a). The respective control signals areplotted in Fig. 4.17(b). Notice that the discrete controller for Ts = 0.07 results in asubstantial increase in the overshoot in the step response, while with Ts = 0.035, thedigital controller matches the performance of the analog controller fairly well.

For controllers with many poles and zeros, making the continuous-to-discrete sub-stitution called for in Eq. (4.93) can be very tedious. Fortunately, MATLAB providesa command that does all the work. If one has a continuous transfer function given by





��

Step

MuxControl

ss�6PI Control

1.4Slider Kc

s�99

Tau 1

Mux1Output

s�55

Tau 2

��

z�11.21z�0.79Discrete

PI control 1.4Slider Kd

s�99

Tau 1

s�55

Tau 2

Figure 4.16 Simulink block diagram to compare continuous and discrete controllers

Dc(s) = numDdenD

represented in MATLAB as sysDa=tf(numD,denD), then the discreteequivalent with sampling period Ts is given by

sysDd = c2d (sysDa, Ts , 't'). (4.108)

In this expression, of course, the polynomials are represented in MATLAB form. Thelast parameter in the c2d function given by 't' calls for the conversion to be done usingthe trapezoid method. The alternatives can be found by asking MATLAB for help c2d.For example, to compute the polynomials for Ts = 0.07 for the preceding example, thecommands would be

numDa = [1 6];denDa = [1 −1];sysDa = tf(numD,denD)sysDd = c2d( sysDa,0.07,'t')

0

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (sec)

(a)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (sec)

(b)

0

0.5

1.0

1.5

2.0

2.5

Continuous controller

Digital controller (T � 0.07 sec)

Discrete controller (T � 0.035 sec)Continuous controller

Digital controller (T � 0.07 sec)

Discrete controller (T � 0.035 sec)

Figure 4.17 Comparison plots of a speed-control system with continuous and discrete controllers: (a) outputresponses; (b) control signals





▲ 4.4.2 Ziegler–Nichols Tuning of PID Regulators

As we will see in later chapters, sophisticated methods are available to developa controller that will meet steady-state and transient specifications for bothtracking input references and rejecting disturbances. These methods requirethat the designer have either a dynamic model of the process in the form ofequations of motion or a detailed frequency response over a substantial rangeof frequencies. Either of these data can be quite difficult to obtain, and the dif-ficulty has led to the development of sophisticated techniques of system modelidentification. Engineers early on explored ways to avoid these requirements.

Callender et al. (1936) proposed a design for the widely used PID controllerby specifying satisfactory values for the controller settings based on estimatesof the plant parameters that an operating engineer could make from experi-ments on the process itself. The approach was extended by J. G. Ziegler andN. B. Nichols (1942, 1943) who recognized that the step responses of a largenumber of process control systems exhibit a process reaction curve like thatshown in Fig. 4.18, which can be generated from experimental step responsedata. The S-shape of the curve is characteristic of many systems and can beapproximated by the step response ofTransfer function for a

high-order system with acharacteristic processreaction curve Y (s)

U(s)= Ae−std

τ s + 1, (4.109)

which is a first-order system with a time delay of td seconds. The constants inEq. (4.109) can be determined from the unit step response of the process. If atangent is drawn at the inflection point of the reaction curve, then the slope ofthe line is R = A/τ and the intersection of the tangent line with the time axisidentifies the time delay L = td .

Ziegler and Nichols gave two methods for tuning the PID controller forsuch a model. In the first method the choice of controller parameters is designedTuning by decay ratio of 0.25to result in a closed-loop step response transient with a decay ratio of approxi-

Figure 4.18Process reaction curve

t

y(t)

L � tdLag

t

A

tA

Slope R �At � reaction rate





Figure 4.19Quarter decay ratio

0.25

1 Period

t

y(t)

mately 0.25. This means that the transient decays to a quarter of its value afterone period of oscillation, as shown in Fig. 4.19. A quarter decay correspondsto ζ = 0.21 and is a reasonable compromise between quick response and ade-quate stability margins. The authors simulated the equations for the system onan analog computer and adjusted the controller parameters until the transientsshowed the decay of 25% in one period. The regulator parameters suggestedby Ziegler and Nichols for the controller terms, defined by

Dc(s) = kp(1+ 1TI s+ TDs), (4.110)

are given in Table 4.2.

TABLE 4.2 Ziegler–Nichols Tuning for the RegulatorD(s) = kp(1+ 1/TI s + TDs), for a decay ratio of 0.25

Type of Controller Optimum Gain

Proportional kp = 1/RL

PI

{kp = 0.9/RL,

TI = L/0.3

PID

kp = 1.2/RL,

TI = 2L,

TD = 0.5L

In the ultimate sensitivity method, the criteria for adjusting the parametersTuning by evaluation at limit ofstability (ultimate sensitivitymethod)

are based on evaluating the amplitude and frequency of the oscillations ofthe system at the limit of stability rather than on taking a step response. Touse the method, the proportional gain is increased until the system becomesmarginally stable and continuous oscillations just begin, with amplitude limitedby the saturation of the actuator. The corresponding gain is defined as Ku





Figure 4.20Determination of theultimate gain and period

��

�

Processry

Kue

(called the ultimate gain) and the period of oscillation is Pu (called the ultimateperiod). These are determined as shown in Figs. 4.20 and 4.21. Pu should bemeasured when the amplitude of oscillation is as small as possible. Then thetuning parameters are selected as shown in Table 4.3.

TABLE 4.3 Ziegler–Nichols Tuning for the RegulatorDc(s) = kp(1+ 1/TI s + TDs), Based on the UltimateSensitivity Method

Type of Controller Optimum Gain

Proportional kp = 0.5Ku

PI

{kp = 0.45Ku,

TI = Pu

1.2

PID

kp = 0.6Ku,

TI = 12Pu,

TD = 18Pu

Experience has shown that the controller settings according to Ziegler–Nichols rules provide acceptable closed-loop response for many systems. Theprocess operator will often do final tuning of the controller iteratively on theactual process to yield satisfactory control.10

Figure 4.21Neutrally stable system

Pu

t

y(t)

10 Tuning of PID controllers has been the subject of continuing study since 1936. A modernpublication on the topic is H. Panagopoulous, K. J. Åström, and T. Hagglund, Proceedings ofthe American Control Conference, San Diego, CA, June 1999.





Figure 4.22A measured processreaction curve

0.0 100.0 200.0 300.0 400.0Time (sec)

1.2

1.0

0.8

0.6

0.4

0.2

0

y

EXAMPLE 4.9 Tuning of a Heat Exchanger: Quarter Decay Ratio

Consider the heat exchanger of Example 2.13. The process reaction curve of this systemis shown in Fig. 4.22. Determine proportional and PI regulator gains for the system usingthe Zeigler–Nichols rules to achieve a quarter decay ratio. Plot the corresponding stepresponses.

Solution. From the process reaction curve, we measure the maximum slope to beR ∼= 1

90 and the time delay to be L ∼= 13 sec. According to the Zeigler–Nichols rules ofTable 4.2 the gains are as follows:

y

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

y

400.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0300.0200.0100.00.0

Time (sec)

(b)

400.0300.0200.0100.00.0

Time (sec)

(a)

PI PI

Proportional

Proportional

Figure 4.23 Closed-loop step responses





Proportional : kp = 1RL= 90

13= 6.92,

PI : kp = 0.9RL= 6.22 and TI = L

0.3= 13

0.3= 43.3

Figure 4.23(a) shows the step responses of the closed-loop system to these two regula-tors. Note that the proportional regulator results in a steady-state offset, while the PIregulator tracks the step exactly in the steady-state. Both regulators are rather oscilla-tory and have considerable overshoot. If we arbitrarily reduce the gain kp by a factorof 2 in each case, the overshoot and oscillatory behaviors are substantially reduced, asshown in Fig. 4.23(b).

EXAMPLE 4.10 Tuning of a Heat Exchanger: Oscillatory Behavior

Proportional feedback was applied to the heat exchanger in the previous example untilthe system showed nondecaying oscillations in response to a short pulse (impulse) input,as shown in Fig. 4.24. The ultimate gain was Ku = 15.3, and the period was measuredat Pu = 42 sec. Determine the proportional and PI regulators according to the Zeigler–Nichols rules based on the ultimate sensitivity method. Plot the corresponding stepresponses.

Solution. The regulators from Table 4.3 are

Proportional : kp = 0.5Ku = 7.65,

PI : kp = 0.45 Ku = 6.885 and TI = 11.2

Pu = 35.

The step responses of the closed-loop system are shown in Fig. 4.25(a). Note that theresponses are similar to those in Example 4.9. If we reduce kp by 50%, then the overshootis substantially reduced, as shown in Fig. 4.25(b).

Figure 4.24Ultimate period of heatexchanger

Impu

lse

resp

onse

0 20 40 60 80 100 120

0.010

0.008

0.006

0.004

0.002

0.00

�0.002

�0.006

�0.004

�0.008

�0.010

Time (sec)





y

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

y

400.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0300.0200.0100.00.0

Time (sec)

(b)

400.0300.0200.0100.00.0

Time (sec)

(a)

PI

PI

Proportional Proportional

Figure 4.25 Closed-loop step response

▲ 4.4.3 Truxal’s Formula for the Error Constants

In this chapter we have derived formulas for the error constants in terms of thesystem transfer function. The most common case is the type 1 system, whoseerror constant is Kv , the velocity error constant. Truxal (1955) derived a formulafor the velocity constant in terms of the closed-loop poles and zeros, a formulathat connects the steady-state error to the dynamic response. Since controldesign often requires a trade-off between these two characteristics, Truxal’sformula can be useful to know. Its derivation is quite direct. Suppose the closed-loop transfer function T(s) of a type 1 system is

T(s) = K(s − z1)(s − z2) · · · (s − zm)

(s − p1)(s − p2) · · · (s − pn). (4.111)

Since the steady-state error in response to a step input in a type 1 system iszero, the DC gain is unity. Thus,

T(0) = 1. (4.112)

The system error is given by

E(s)�= R(s)− Y (s) = R(s)

[1− Y (s)

R(s)

]= R(s)[1− T(s)]. (4.113)





The system error due to a unit ramp input is given by

E(s) = 1− T(s)

s2 . (4.114)

Using the Final Value Theorem, we get

ess = lims→0

1− T(s)

s. (4.115)

Using L’Hôpital’s rule, we rewrite Eq. (4.115) as

ess = − lims→0

dT

ds(4.116)

or

ess = − lims→0

dT

ds= 1

Kv

. (4.117)

Equation (4.117) implies that 1/Kv is related to the slope of the transfer func-tion at the origin, a result that will also be shown in Section 6.1.2. UsingEq. (4.112), we can rewrite Eq. (4.117) as

ess = − lims→0

dT

ds

1T

(4.118)

or

ess = − lims→0

d

ds[ln T(s)]. (4.119)

Substituting Eq. (4.111) into Eq. (4.119), we get

ess = − lims→0

d

ds

{ln[K

∏mi=1(s − zi)∏ni=1(s − pi)

]}(4.120)

= − lims→0

d

ds

[K +

m∑i=1

ln(s − zi)−m∑

i=1

ln(s − pi)

], (4.121)

or

1Kv

=∣∣∣∣−d ln T

ds

∣∣∣∣s=0=

n∑i=1

− 1pi

+m∑

i=1

1zi

. (4.122)

We observe from Eq. (4.122) that Kv increases as the closed-loop polesTruxal’s formulamove away from the origin. Similar relationships exist for other error coeffi-cients, and these are explored in the problems.





EXAMPLE 4.11 Truxal’s Formula

A third order type 1 system has closed-loop poles at −2± 2j and −0.1. The system hasonly one closed-loop zero. Where should the zero be if a Kv = 10 is desired?

Solution. From Truxal’s formula we have

1Kv

= − 1−2+ 2j

− 1−2− 2j

− 1−0.1

+ 1z,

or

0.1 = 0.5+ 10+ 1z.

Therefore, the closed-loop zero should be at z = −0.1.

▲ 4.4.4 Sensitivity of Time Response to ParameterChange

We have considered the effects of errors on the steady-state gain of a dynamicsystem and have shown how feedback control can reduce these errors. Sincemany control specifications are in terms of the step response, the sensitivity ofthe time response to parameter changes is sometimes very useful to explore. Forexample, by looking at the sensitivity plot we can tell whether increasing a par-ticular parameter will increase or decrease the overshoot of the response.11 Theanalysis that follows is also a good exercise in small-signal linearization.

To consider the sensitivity of the output y(t, θ) of a system having a param-eter of interest, θ , we compute the effect of a perturbation in the parameter,δθ , on the nominal response by using the Taylor’s series expansion

y(t, θ + δθ) = y(t, θ)+ ∂y

∂θδθ + · · · . (4.123)

The first-order approximation of the parameter perturbation effect is the term

δy(t) = ∂y

∂θδθ. (4.124)

This function can be generated from the system itself as shown by Perkinset al, 1991. We assume that the response depends linearly on the parameterand therefore that the overall transfer function T(s, θ) is composed of com-ponent transfer functions that can be defined to bring out the dependence onthe parameter explicitly. A block diagram of the transfer function in terms of

11 As we shall see in the next chapter, the development of the MATLAB root locus interfacerltool gives the designer a computer aid to this result.





Figure 4.26Block diagram showing thedependence of output y onparameter θ

u��

�

�

��

YR T11

T12

T22

T21X Z V

the components Tij (s) can be expressed as shown in Fig. 4.26, where we havelabeled the parameter as θ and its input signal as Z . In terms of this blockdiagram, the equations relating Y and Z to the reference input can be writtenimmediately:

Y = T11R + T21θZ (4.125)

and

Z = T12R + T22θZ. (4.126)

The perturbed equations are

Y + δY = T11R + T21(θ + δθ)(Z + δZ) (4.127)

and

Z + δZ = T12R + T22(θ + δθ)(Z + δZ). (4.128)

Multiplying these out and ignoring the small term δθδZ , the expressions forthe perturbations in Y and Z are given by

δY = T21(Zδθ + θδZ) (4.129)

and

δZ = T22(Zδθ + θδZ). (4.130)

The solutions to these equations can be best presented as a block diagram,shown in Fig. 4.27(a). The output of this figure is δY = ∂y

∂θδθ , and we notice

that the input Z is multiplied by a gain of δθ . Therefore, if we drop the blockδθ , the output will be simply ∂y

∂θ, as shown in Fig. 4.27(b). Finally, to compute

the sensitivity as the variation to a percent change in the parameter, which is∂y

∂ ln θ=

∂y(t,θ)∂θ

∂ ln θ∂θ

= θ∂y

∂θ, we need only shift the input Z from the output side of

the θ block to its input as shown in Fig. 4.27(c). We are now in a position togive the final block diagram of the system as it is to be implemented, shown inFig. 4.28.





Figure 4.27Block diagrams showingthe generation of (a) δY

and δZ , (b) ∂y

∂θ, and

(c) θ∂y

∂θ

(a)

du dZ��

�

T22Z

T21

u

dY � du�u

�y

(b)

��

�

T22Z

T21

u

�u

�y

(c)

��

�

T22

Z T21u�u

u�y

In this figure, it is clear that to compute the sensitivity of the output to aparameter, one needs to simulate two copies of the system. The input to thefirst system is the reference input of interest, and the input to the second systemis at the input to the parameter of interest of the variable Z , taken from theinput to the parameter in the original system. The transfer function from thereference input to the output sensitivity is readily computed to be

T12θT21

(1− θT22)2 . (4.131)

Response sensitivity From this function it is clear that to keep the sensitivity of the output signal to aparameter change low, it is important to have feedback with high gain aroundthe parameter in question.

Figure 4.28Block diagram showingthe computation of θ

∂y

∂θ

from the original transferfunction

R Y

u u�

�

�ZV

u�y�u





Figure 4.29Block diagram showingthe computation of thesensitivity of the output ofthe speed-control example

R

K

�s � 1A

�

�

YKcl

�s � 1AKcl

�

��

Z

�K�Y

EXAMPLE 4.12 Time-Domain Sensitivity

Compute the sensitivity of the output of the speed-control example described byEq. (4.20) with respect to the control gain, Kcl . Take the nominal values to be Kcl = 9,τ = 0.01 sec, and A = 1 rad/volt-sec.

Solution. The required block diagram for the computation is given in Fig. 4.29, basedon Fig. 4.28. In MATLAB, we will construct the several transfer functions with Tij = nij

dij

and will implement Eq. (4.131). For comparison, we compute the nominal response fromFig. 4.26 and add 10% of the sensitivity to the nominal response. The instructions to dothe computation in MATLAB are

% script to compute sensitivity for Fig. 4.29

% First input the data for the component transfer functions Tij

% and the nominal parameter, Kcl for this problem

Kcl = 9; tau = .01;

n11 = 0; d11 = 1;

n12 = 1; d12 = 1;

n22 = [0 −1]; d22 = [tau 1];

n21 = 1; d21 = [tau 1];

% Now compute the numerator and denominator polynomials of the transfer

% functions using the convolution function conv to multiply the polynomials

% and put them into system transfer function forms with the MATLAB function tf.

% The overall transfer function is

% Y/R = n11/d11 + (n12*n21* d22)/(d12*d21* [d22-Kcl*n22]) = sysy

% The transfer function from the reference input to the sensitivity is

% Kcl*(dy/dKcl)/R = sysdy

% Now define the numerators and denominators of several intermediate

% intermediate transfer functions

n1 = Kcl*conv(n21,n12);

d1 = conv(d21,d12);

n2 = d22;





d2 = [d22-Kcl*n22];

ny = conv(n1,n2);

dy = conv(d1,d2);

% Now put these together to form two intermediate transfer functions

sysy1 = tf(ny,dy);

sysy2 = tf(n11,d11);

% Now construct the final transfer functions

% The overall transfer function Y/R

sysy = sysy1+sysy2;

% The sensitivity transfer function

ndy = conv(ny,n2);

ddy = conv(dy,d2);

sysdy = tf(ndy,ddy);

% Now use these to compute the step responses and

% plot the output, the sensitivity and a perturbed response

[y,t] = step(sysy);

[yd,t] = step(sysdy);

plot(t,[y yd y+.1*yd]);

These instructions are constructed to compute the sensitivity for any system, given theseveral transfer functions. The script input is for the specific example. Plots of the output,its sensitivity, and the result of a 10% change in the parameter value are given in Fig. 4.30.

Figure 4.30Plots of the output, thesensitivity, and the resultof a 10% change in theparameter value for thespeed-control example

1.0

0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6

Time (msec)

y

y � dy

kcl dkcl

dy





SUMMARY

• The system error in a feedback control can be compactly described bydefining the loop gain L, the sensitivity function �, and the complementarysensitivity function T.

• Compared with open-loop control, feedback can be used to reduce steady-state errors to disturbances, reduce the system’s transfer function sensitivityto parameter variations, and speed up the transient response.

• Sensor noise introduces a conflict between efforts to reduce error causedby plant disturbances and those caused by the sensor noise.

• Classifying a system as type k indicates the ability of the system to achievezero steady-state error to polynomials of degree less than but not equalto k . A stable unity feedback system is type k with respect to referenceinputs if the loop gain L has k poles at the origin, in which case we canwrite

L(s) = (s + z1)(s + z2) · · ·sk(s + p1)(s + p2) · · · ,

and the error constant is given by

Kk = lims→0

skL(s). (4.132)

• A table of steady-state errors for unity feedback systems of types 0, 1, and2 to reference inputs is given in Table 4.1.

• Systems can be classified as to type for rejecting disturbances by computingthe system error to polynomial disturbance inputs. The system is type k todisturbances if the error is zero to all disturbance polynomials of degreeless than k , but nonzero for a polynomial of degree k .

• The standard PID controller is described by the equation▲

U(s) =(

kp + kI

s+ kDs

)E(s)

or

U(s) = kp

(1+ 1

TI s+ TDs

)E(s) = D(s)E(s).

This latter form is ubiquitous in the process-control industry and is thebasic controller in many control systems.

• Increasing the proportional feedback gain reduces steady-state errors, buthigh gains almost always destabilize the system. Integral control providesrobust reduction in steady-state errors, but often makes the system lessstable. Derivative control usually increases damping and improves stability.These three kinds of control combined form the classical PID controller.




End-of-Chapter Questions 211

• Useful guidelines for tuning PID controllers were presented in Tables 4.2▲and 4.3.

• A difference equation describing a digital controller to be used to replacea given analog controller can be found by replacing s with 2

Ts

z−1z+1 in the

analog controller’s transfer function and using z as a forward shift operatorin the sense that if U(z) corresponds to u(kTs), then zU(z) corresponds tou(kTs + Ts).

• MATLAB can compute a discrete equivalent with the command c2d.

End-of-Chapter Questions

1. Give three advantages of feedback in control.

2. Give two disadvantages of feedback in control.

3. A temperature control system is found to have zero error to a constant trackinginput and an error of 0.5◦C to a tracking input that is linear in time, rising at the rateof 40◦C/sec. What is the system type of this control system and what is the relevanterror constant [Kp or Kv or etc.]?

4. What are the units of Kp , Kv , and Ka?

5. What is the definition of system type with respect to reference inputs?

6. What is the definition of system type with respect to disturbance inputs?

7. Why does system type depend on where the external signal enters the system?

8. What is the main objective of introducing integral control?

9. What is the major objective of adding derivative control?

10. Why might a designer wish to put the derivative term in the feedback path ratherthan in the error path?

11. Give two reasons to use a digital controller rather than an analog controller.

12. Give two disadvantages to using a digital controller.

13. Give the substitution in the discrete operator z for the Laplace operator s if theapproximation to the integral in Eq. (4.87) is taken to be the rectangle of heighte(kTs) and base Ts .

14. What is the advantage of having a “tuning rule” for PID controllers?▲





Problems

Problems for Section 4.1: The Basic Equations of Control

4.1. Consider a system with the configuration of Fig. 4.5, where D is the constant gainof the controller and G is that of the process. The nominal values of these gainsare D = 5 and G = 7. Suppose a constant disturbance w is added to the controlinput u before the signal goes to the process.

(a) Compute the gain from w to y in terms of D and G.

(b) Suppose the system designer knows that an increase by a factor of six in theloop gain DG can be tolerated before the system goes out of specification.Where should the designer place the extra gain if the objective is to minimizethe system error r − y due to the disturbance? For example, either D or G

could be increased by a factor of six, or D could be doubled and G tripled,and so on. Which choice is the best?

4.2. Bode defined the sensitivity function relating a transfer function G to one of itsparameters k as the ratio of percent change in k to percent change in G. We definethe reciprocal of Bode’s function as

SGk =

dG/G

dk/k= d ln G

d ln k= k

G

dG

dk.

Thus, when the parameter k changes by a certain percentage, S tells us whatpercent change to expect in G. In control systems design, we are almost alwaysinterested in the sensitivity at zero frequency, or when s = 0. The purpose of thisexercise is to examine the effect of feedback on sensitivity. In particular, we wouldlike to compare the topologies shown in Fig. 4.31 for connecting three amplifierstages with a gain of −K into a single amplifier with a gain of −10.

Figure 4.31Three-amplifier topologiesfor Problem 4.2

b1R Y�K �K �K

(a)

(b)

(c)

R Y�K �K �K��

�

b3

��

�

R �K ��

�

�K

b2

��

�

�K Y

b2b2




Problems 213

(a) For each topology in Fig. 4.31, compute βi so that, if K = 10, Y = −10R .

(b) For each topology, compute SGk when G = Y/R . [Use the respective βi values

found in part (a).] Which case is the least sensitive?

(c) Compute the sensitivities of the systems in Fig. 4.31(b, c) to β2 and β3 . Us-ing your results, comment on the relative need for precision in sensors andactuators.

4.3. Compare the two structures shown in Fig. 4.32 with respect to sensitivity to changesin the overall gain due to changes in the amplifier gain. Use the relation

S = d ln F

d ln K= K

F

dF

dK

as the measure. Select H1 and H2 so that the nominal system outputs satisfyF1 = F2 , and assume KH1 > 0.

��

�

R

H1

K ��

�

H1

K F1

(a)

��

�

R

H2

K K F2

(b)

Figure 4.32 Block diagrams for Problem 4.3

4.4. A unity feedback control system has the open-loop transfer function

G(s) = A

s(s + a).

(a) Compute the sensitivity of the closed-loop transfer function to changes in theparameter A.

(b) Compute the sensitivity of the closed-loop transfer function to changes in theparameter a .

(c) If the unity gain in the feedback changes to a value of β �= 1, compute thesensitivity of the closed-loop transfer function with respect to β .

(d) Assuming that A = 1 and a = 1, plot the magnitude of each of the precedingsensitivity functions for s = jω using the semilogy command in MATLAB.Comment on the relative effect of parameter variations in A, a, and β atdifferent frequencies ω , paying particular attention to DC (when ω = 0).

Problems for Section 4.2: Control of Steady-State Error

4.5. Consider the second-order plant

G(s) = 1(s + 1)(5s + 1)

.





(a) Determine the system type and error constant with respect to tracking poly-nomial reference inputs of the system for P, PD, and PID controllers (asconfigured in Fig. 4.5). Let kp = 19, kI = 0.5, and kD = 4

19 .

(b) Determine the system type and error constant of the system with respect todisturbance inputs for each of the three regulators in part (a) with respect torejecting polynomial disturbances w(t) at the input to the plant.

(c) Is this system better at tracking references or rejecting disturbances? Explainyour response briefly.

(d) Verify your results for parts (a) and (b) using MATLAB by plotting unit stepand ramp responses for both tracking and disturbance rejection.

4.6. Consider a system with the plant transfer function G(s) = 1/s(s + 1). You wishto add a dynamic controller so that ωn = 2 rad/sec. and ζ ≥ 0.5. Several dynamiccontrollers have been proposed:

1. D(s) = (s + 2)/2

2. D(s) = 2 s + 2s + 4

3. D(s) = 5 (s + 2)s + 10

4. D(s) = 5 (s + 2)(s + 0.1)(s + 10)(s + 0.01)

(a) Using MATLAB, compare the resulting transient and steady-state responsesto reference step inputs for each controller choice. Which controller is bestfor the smallest rise time and smallest overshoot?

(b) Which system would have the smallest steady-state error to a ramp referenceinput?

(c) Compare each system for peak control effort; that is, measure the peak mag-nitude of the plant input u(t) for a unit reference step input.

(d) Based on your results from parts (a) to (c), recommend a dynamic controllerfor the system from the four candidate designs.

4.7. A certain control system has the following specifications: rise time tr ≤ 0.010 sec,overshoot Mp ≤ 16%, and steady-state error to unit ramp ess ≤ 0.005.

(a) Sketch the allowable region in the s -plane for the dominant second-orderpoles of an acceptable system.

(b) If Y/R = G/(1 + G), what condition must G(s) satisfy near s = 0 for theclosed-loop system to meet specifications; that is, what is the required asymp-totic low-frequency behavior of G(s)?




Problems 215

4.8. For the system in Problem 4.35, compute the following steady-state errors:

(a) for a unit-step reference input;

(b) for a unit-ramp reference input;

(c) for a unit-step disturbance input;

(d) for a unit-ramp disturbance input.

(e) Verify your answers to parts (a) to (d) using MATLAB. Note that a rampresponse can be generated as the step response of a system modified by anadded integrator at the reference input.

4.9. Consider the system shown in Fig. 4.33. Show that the system is type 1 and computethe Kv .

Figure 4.33Control system forProblem 4.9

K(as � b)s(s � 1)�

�

�YR

4.10. Consider the DC motor control system with rate (tachometer) feedback shownin Fig. 4.34(a).

� ��

uur

(a)

Kp ��

Kk1Km

s(1 � tms)

kts

��

uur

(b)

K�

s(1 � tms)

1 � k�ts

�

Figure 4.34 Control system for Problem 4.10

(a) Find values for K ′ and k′t so that the system of Fig. 4.34(b) has the sametransfer function as the system of Fig. 4.34(a).

(b) Determine the system type with respect to tracking θr and compute the systemKv in terms of parameters K ′ and k′t .

(c) Does the addition of tachometer feedback with positive kt increase or de-crease Kv?

4.11. Consider the system shown in Fig. 4.35, where

D(s) = K(s + α)2

s2 + ω2o

.

(a) Prove that if the system is stable, it is capable of tracking a sinusoidal referenceinput r = sin ωot with zero steady-state error. (Look at the transfer functionfrom R to E and consider the gain at ωo.)






��

�

D(s) YR1

s(s � 1)

(b) Use Routh’s criteria to find the range of K such that the closed-loop systemremains stable if ωo = 1 and α = 0.25.

4.12. Consider the system shown in Fig. 4.36, which represents control of the angle of apendulum that has no damping.


��

�

��

�

W

s21

D(s)R

K

Y��

�

(a) What condition must D(s) satisfy so that the system can track a ramp refer-ence input with constant steady-state error?

(b) For a transfer function D(s) that stabilizes the system and satisfies the condi-tion in part (a), find the class of disturbances w(t) that the system can rejectwith zero steady-state error.

(c) Show that, although a PI controller satisfies the condition derived in part (a),it will not yield a stable closed-loop system. Will a PID controller work—thatis, satisfy part (a) and stabilize the system? If so, what constraints must kp ,kI , and kD satisfy?

(d) Discuss qualitatively and briefly the effects of small variations on the con-troller parameters kp , kI , and kD on the system’s step response rise time andovershoot.

4.13. A unity feedback system has the overall transfer function

Y (s)

R(s)= T(s) = ω2

n

s2 + 2ζωns + ω2n

.

Give the system type and corresponding error constant for tracking polynomialreference inputs in terms of ζ and ωn .

4.14. Consider the second-order system

G(s) = 1s2 + 2ζ s + 1

.

We would like to add a transfer function of the form D(s) = K(s + a)/(s + b) inseries with G(s) in a unity-feedback structure.(a) Ignoring stability for the moment, what are the constraints on K , a , and b so

that system is type 1?(b) What are the constraints placed on K , a , and b so that the system is stable

and type 1?(c) What are the constraints on a and b so that the system is type 1 and remains

stable for every positive value for K?




Problems 217

4.15. The transfer function for the plant in a motor position control is given by

G(s) = A

s(s + a).

If we were able to select values for both A and a , what would they be to result ina system with Kv = 20 and ζ = 0.707?

4.16. Consider the system shown in Fig. 4.37(a).

(a) What is the system type? Compute the steady-state tracking error due to aramp input r(t) = rot1(t).

(b) For the modified system shown in Fig. 4.37(b), give the value of Hf so thatthe system is type 2 for reference inputs, and compute the Ka in this case.

(c) Is the resulting type 2 property of this system robust with respect to changesin Hf (i.e., will the system remain type 2 if Hf changes slightly)?


(a)

(b)

��

�

YRs(�s � 1)

A

��

�

Y��

�

Rs(�s � 1)

A

Hf s

Hr

4.17. A controller for a satellite attitude control with transfer function G = 1/s2 hasbeen designed with a unity feedback structure and has the transfer function D(s) =10(s + 2)

s + 5 .

(a) Find the system type for reference tracking and the corresponding error con-stant for this system.

(b) If a disturbance torque adds to the control so that the input to the process isu+w , what is the system type and corresponding error constant with respectto disturbance rejection?

4.18. A compensated motor position control system is shown in Fig. 4.38. Assume thatthe sensor dynamics are H(s) = 1.

(a) Can the system track a step reference input r with zero steady-state error?If yes, give the value of the velocity constant.

(b) Can the system reject a step disturbance w with zero steady-state error? Ifyes, give the value of the velocity constant.

(c) Compute the sensitivity of the closed-loop transfer function to changes in theplant pole at −2.






Sensor

��

�

YR�

�

�

W

1s(s � 2)

Plant

H (s)

s � 4s � 30

Compensator

160

(d) In some instances there are dynamics in the sensor. Repeat parts (a) to (c)for H(s) = 20/(s + 20) and compare the corresponding velocity constants.

4.19. Consider the system shown in Fig. 4.39 with PI control.


skPs � kI�

�

�

Y

W

UR �

�

�

s2 � s � 2010

(a) Determine the transfer function from R to Y .

(b) Determine the transfer function from W to Y .

(c) Use Routh’s criteria to find the range of (kp, kI ) for which the system is stable.

(d) What is the system type and error constant with respect to reference tracking?

(e) What is the system type and error constant with respect to disturbancerejection?

4.20. The general unity feedback system shown in Fig. 4.40 has disturbance inputs w1 ,w2 , and w3 and is asymptotically stable. Also,

G1(s) =K1∏m1

i=1(s + z1i )

sl1∏m1

i=1(s + p1i ), G2(s) =

K2∏m1

i=1(s + z2i )

sl2∏m1

i=1(s + p2i ).

Figure 4.40Single input–single outputunity feedback system withdisturbance inputs �

�

�

YR�

�

�

W1

G1(s)

W2

��

�

W3

G2(s)

(a) Show that the system is of type 0, type l1 , and type (l1 + l2) with respect todisturbance inputs w1 , w2 , and w3 .




Problems 219

4.21. One possible representation of an automobile speed-control system with integralcontrol is shown in Fig. 4.41.

Figure 4.41System using integralcontrol

��

�

�

W

smVc �

��

k1k2s

k1E

k3

k1

FV

(a) With a zero reference velocity input (vc = 0), find the transfer functionrelating the output speed v to the wind disturbance w .

(b) What is the steady-state response of v if w is a unit-ramp function?

(c) What type is this system in relation to reference inputs? What is the value ofthe corresponding error constant?

(d) What is the type and corresponding error constant of this system in relationto tracking the disturbance w?

4.22. For the feedback system shown in Fig. 4.42, find the value of α that will make thesystem type 1 for K = 5. Give the corresponding velocity constant. Show thatthe system is not robust by using this value of α and computing the tracking errore = r − y to a step reference for K = 4 and K = 6.


��

�

Rs � 2

KYa

4.23. A position control system has the closed-loop transfer function (meter/meter)given by

Y (s)

R(s)= b0s + b1

s2 + a1s + a2.

(a) Choose the parameters (a1, a2, b0, b1) so that the following specifications aresatisfied simultaneously:

i. The rise time tr < 0.1 sec.

ii. The overshoot Mp < 20%.

iii. The settling time ts < 0.5 sec.

iv. The steady-state error to a step reference is zero.

v. The steady-state error to a ramp reference input of 0.1 m/sec is not morethan 1 mm.

(b) Verify your answer via MATLAB simulation.





4.24. Suppose you are given the system depicted in Fig. 4.43(a), where the plant param-eter a is subject to variations.

��

�

R Ys � a

1s1

41

��

�4

��

�

U

�

x

(a)

��

�

R YG(s)

(b)

E(t)

Figure 4.43 Control system for Problem 4.24

(a) Find G(s) so that the system shown in Fig. 4.43(b) has the same transferfunction from r to y as the system in Fig. 4.43(a).

(b) Assume that a = 1 is the nominal value of the plant parameter. What is thesystem type and the error constant in this case?

(c) Now assume that a = 1 + δa , where δa is some perturbation to the plantparameter. What is the system type and the error constant for the perturbedsystem?

4.25. Two feedback systems are shown in Fig. 4.44.

��

�

YRK0

4s � 1

(a)

��

�

YK0

4s � 1

(b)

R K3K2K1

sU U

Figure 4.44 Two feedback systems for Problem 4.25

(a) Determine values for K1 , K2 , and K3 so that both systems

i. exhibit zero steady-state error to step inputs (that is, both are type 1)

ii. whose static velocity error constant Kv = 1 when K0 = 1.

(b) Suppose K0 undergoes a small perturbation: K0 → K0 + δK0 . What effectdoes this have on the system type in each case? Which system has a typewhich is robust? Which system do you think would be preferred?

(c) Estimate the transient response of both systems to a step reference input, andgive estimates for ts , tr , and Mp . In your opinion, which system has a bettertransient response at the nominal parameter values?




Problems 221

4.26. You are given the system shown in Fig. 4.45, where the feedback gain β is subjectto variations. You are to design a controller for this system so that the output y(t)

accurately tracks the reference input r(t).


��

�

YR Di(s) 10(s � 1)(s � 10)

b

a

(a) Let β = 1. You are given the following three options for the controller Di(s):

D1(s) = kp, D2(s) = kps + kI

s, D3(s) = kps2 + kI s + k2

s2.

Choose the controller (including particular values for the controller con-stants) that will result in a type 1 system with a steady-state error to a unitreference ramp of less than 1

10 .

(b) Next, suppose that there is some attenuation in the feedback path that ismodeled by β = 0.9. Find the steady-state error due to a ramp input for yourchoice of Di(s) in part (a).

(c) If β = 0.9, what is the system type for part (b)? What are the values of theappropriate error constant?

4.27. Consider the system shown in Fig. 4.46.


��

�

YR1

(s � p1)(s � p2) � � � (s � pq)E

(a) Find the transfer function from the reference input to the tracking error.

(b) For this system to respond to inputs of the form r(t) = tn1(t) (where n < q )with zero steady-state error, what constraint is placed on the open-loop polesp1, p2, · · · , pq ?

4.28. The feedback control system shown in Fig. 4.47 is to be designed to satisfy thefollowing specifications: (1) steady-state error of less than 10% to a ramp referenceinput, (2) maximum overshoot for a unit-step input of less than 5%, and (3) 1%settling time of less than 3 sec.

(a) Compute the closed-loop transfer function.

(b) Sketch the region in the complex plane where the closed-loop poles may lie.

(c) What does specification (1) imply about the possible values of A?






��

�

YRA

s (s � 2)

1 � kts

(d) What does specification (3) imply about the closed-loop poles?

(e) Find the error due to a unit-ramp input in terms of A and kt .

(f) Suppose A = 32. Find the value of kt that yields closed-loop poles on theright-hand boundary of the feasible region. Use MATLAB to check whetherthis choice for kt satisfies the desired specifications. If not, adjust kt until itdoes.

(g) Using A = 32 and the value for kt computed in part (f), estimate the settlingtime of the system. Use MATLAB to check your answer.

4.29. The transfer functions of speed control for a magnetic tape-drive system are shownin Fig. 4.48. The speed sensor is fast enough that its dynamics can be neglectedand the diagram shows the equivalent unity feedback system.

Figure 4.48Speed-control system for amagnetic tape drive

K��

�

100.5s � 1

Amplifier

Torquemotor

��

�Torque

Disturbancetorque

1Js � b

Tapedynamics

mReferencespeed, r

J � 0.10 kg�m2

b � 1.00 N�m�sec

(a) Assuming ωr = 0, what is the steady-state error due to a step disturbancetorque of 1 N·m? What must the amplifier gain K be in order to make thesteady-state error ess ≤ 0.001 rad/sec?

(b) Plot the roots of the closed-loop system in the complex plane, and accuratelysketch the time response ω(t) for a step input ωr using the gain K computedin part (a). Are these roots satisfactory? Why or why not?

(c) Plot the region in the complex plane of acceptable closed-loop poles corre-sponding to the specifications of a 1% settling time of ts ≤ 0.1 sec and anovershoot Mp ≤ 5%.

(d) Give values for kp and kD for a PD controller that will meet the specifications.

(e) How would the disturbance-induced steady-state error change with the newcontrol scheme in part (d)? How could the steady-state error to a disturbancetorque be eliminated entirely?




Problems 223

4.30. A linear ODE model of the DC motor with negligible armature inductance (La =0) and disturbance torque w was given earlier in the chapter; it is restated here,in slightly different form, as

JRa

Kt

θm +Keθm = va + Ra

Kt

w,

where θm is measured in radians. Dividing through by the coefficient of θm , weobtain

θm + a1θm = b0va + c0w,

where

a1 = KtKe

JRa

, b0 = Kt

JRa

, c0 = 1J

.

With rotating potentiometers, it is possible to measure the positioning error be-tween θ and the reference angle θr or e = θref − θm . With a tachometer we canmeasure the motor speed θm . Consider using feedback of the error e and themotor speed θm in the form

va = K(e − TDθm),

where K and TD are controller gains to be determined.

(a) Draw a block diagram of the resulting feedback system showing both θm andθm as variables in the diagram representing the motor.

(b) Suppose the numbers work out so that a1 = 65, b0 = 200, and c0 = 10. Ifthere is no load torque (w = 0), what speed (in rpm) results from va = 100 V?

(c) Using the parameter values given in part (b), find kp and kD so that a stepchange in θref with zero load torque results in a transient that has an approx-imately 17% overshoot and that settles to within 5% of steady-state in lessthan 0.05 sec.

(d) Derive an expression for the steady-state error to a reference angle input,and compute its value for your design in part (c), assuming θref = 1 rad.

(e) Derive an expression for the steady-state error to a constant disturbancetorque when θref = 0, and compute its value for your design in part (c), as-suming that w = 1.0.

4.31. We wish to design an automatic speed-control for an automobile. Assume that(1) the car has a mass m of 1000 kg, (2) the accelerator is the control U andsupplies a force on the automobile of 10 N per degree of accelerator motion, and(3) air drag provides a friction force proportional to velocity of 10 N·sec/m.

(a) Obtain the transfer function from control input U to the velocity of theautomobile.





(b) Assume the velocity changes are given by

V (s) = 1s + 0.002

U(s)+ 0.05s + 0.02

W(s),

where V is given in meters per second, U is in degrees, and W is the percentgrade of the road. Design a proportional-control law U = −kpV that willmaintain a velocity error of less than 1 m/sec in the presence of a constant2% grade.

(c) Discuss what advantage (if any) integral control would have for this problem.

(d) Assuming that pure integral control (that is, no proportional term) is ad-vantageous, select the feedback gain so that the roots have critical damping(ζ = 1).

4.32. Consider the automobile speed-control system depicted in Fig. 4.49.

Figure 4.49Automobile speed-controlsystem

YR ��

�

W

��

�

Hy

Hr kp s � aA

s � aB

R �Y �

W �

Desired speedActual speedRoad grade, %

(a) Find the transfer functions from W(s) and from R(s) to Y (s).

(b) Assume that the desired speed is a constant reference r , so that R(s) = ro/s .Assume that the road is level, so w(t) = 0. Compute values of the gains K ,Hr , and Hf to guarantee that

limt→∞ y(t) = ro.

Include both the open-loop (assuming Hy = 0) and feedback cases (Hy �= 0)

in your discussion.

(c) Repeat part (b) assuming that a constant grade disturbance W(s) = wo/s ispresent in addition to the reference input. In particular, find the variation inspeed due to the grade change for both the feed forward and feedback cases.Use your results to explain (1) why feedback control is necessary and (2) howthe gain kp should be chosen to reduce steady-state error.

(d) Assume that w(t) = 0 and that the gain A undergoes the perturbation A+δA.Determine the error in speed due to the gain change for both the feed forwardand feedback cases. How should the gains be chosen in this case to reducethe effects of δA?




Problems 225

4.33. For a system with impulse response h(t), prove that the velocity constant is givenby

1Kv

=∫ ∞

0th(t) dt,

and the acceleration constant is given by

1Ka

= −12

∫ ∞0

t2h(t) dt.

4.34. Consider the multivariable system shown in Fig. 4.50. Assume that the system isstable. Find the transfer functions from each disturbance input to each output, anddetermine the steady-state values of y1 and y2 for constant disturbances. We definea multivariable system to be type k with respect to polynomial inputs at wi if thesteady-state value of every output is zero for any combination of inputs of degreeless than k , and at least one input is a nonzero constant for an input of degree k .What is the system type with respect to disturbance rejection at w1 ? At w2 ?

Figure 4.50Multivariable system

��

�

��

�

W1

s1R1 Y1

s � 11

��

�

R2 ��

�

s � 11

s � 21

��

�

W2

Y2

Problems for Section 4.3: Control of Dynamic Error: PID Control

4.35. The DC motor speed control shown in Fig. 4.51 is described by the differentialequation

y + 60y = 600va − 1500w,

where y is the motor speed, va is the armature voltage, and w is the load torque.Assume that the armature voltage is computed by using the PI control law

va = −(

kpe + kI

∫ t

0e dt

),

where e = r − y .

Figure 4.51DC motor speed controlblock diagram forProblem 4.35

��

�

Y

W

e �aR �

�

�

D 600

1500

s � 601





(a) Compute the transfer function from W to Y as a function of kp and kI .

(b) Compute values for kp and kI so that the characteristic equation of the closed-loop system will have roots at −60± 60j .

4.36. Consider the system shown in Fig. 4.52, which consists of a prefilter and a unityfeedback system.

Figure 4.52Unity feedback system forProblem 4.36

Kr s � 1

s � a ��

�YR

s(ts � 1)A

(a) Determine the transfer function from R to Y .

(b) Determine the steady-state error due to a step input.

(c) Discuss the effect of different values of (Kr, a) on the system’s response.

(d) For each of the three cases,

(1) A = 1, τ = 1, (2) A = 10, τ = 1, (3) A = 1, τ = 2,

use MATLAB to find values for Kr and a so that (if possible)

i. the rise time is less than 1.5 sec,

ii. the overshoot is less than 20%,

iii. the settling time is less than 10 sec, and

iv. the steady-state error is less than 5%.

In cases in which the specifications are easily met, try to make the rise time assmall as possible. If the specifications cannot be met, find the design to meetas many of the specifications as possible, in the order given.

4.37. Consider the satellite attitude-control problem shown in Fig. 4.53, where the nor-malized parameters are

J = 10 spacecraft inertia, N-m-sec2/rad.

θr = reference satellite attitude, rad.

θ = actual satellite attitude, rad.

Hy = 1 sensor scale factor volts/rad.

Hr = 1 reference sensor scale factor, volts/rad.

w = disturbance torque N-m.




Problems 227

Figure 4.53Satellite attitude control

Hy

��

�

��

�Js

1

w

Hr s1D(s) uur

(a) Use proportional control, P, with D(s) = kp , and give the range of values forkp for which the system will be stable.

(b) Use PD control, let D(s) = (kp + kDs), and determine the system type anderror constant with respect to reference inputs.

(c) Use PD control, let D(s) = (kp + kDs), and determine the system type anderror constant with respect to disturbance inputs.

(d) Use PI control, let D(s) = (kp + kI /s), and determine the system type anderror constant with respect to reference inputs.

(e) Use PI control, let D(s) = (kp + kI /s), and determine the system type anderror constant with respect to disturbance inputs.

(f) Use PID control, let D(s) = (kp + kI /s + kDs), and determine the systemtype and error constant with respect to reference inputs.

(g) Use PID control, let D(s) = (kp + kI /s + kDs), and determine the systemtype and error constant with respect to disturbance inputs.

Problems for Section 4.4: Extensions to the Basic Feedback Concepts▲

4.38. Compute the discrete equivalents for the controllers of Problem 4.6 by using the▲trapezoid rule of Eq. (4.93). Let Ts = 0.05 in each case.

(a) D1(s) = (s + 2)/2

(b) D2(s) = 2 s + 2s + 4

(c) D3(s) = 5 (s + 2)s + 10

(d) D4(s) = 5 (s + 2)(s + 0.1)(s + 10)(s + 0.01)

4.39. Give the difference equations corresponding to each of the discrete controllers▲respectively found in Problem 4.38

(a) For D1(s).

(b) For D2(s).

(c) For D3(s).

(d) For D4(s).

4.40. The unit-step response of a paper machine is shown in Fig. 4.54(a), where the▲input into the system is stock flow onto the wire and the output is basis weight(thickness). The time delay and slope of the transient response may be determinedfrom the figure.

(a) Find the proportional, PI, and PID-controller parameters by using the Zeigler–Nichols transient-response method.





Figure 4.54Paper-machine responsedata for Problem 4.40

Step

res

pons

e

1.0

0.8

0.6

0.4

0.2

0.0

Uni

t im

puls

e re

spon

se

0.4

0.3

0.2

0.1

0.0

�0.1

�0.2

�0.3

�0.40

(b)

Time (sec)

1 2 3 4 5 6 7 8 9 100

(a)

Time (sec)

1 2 3 4 5 6 7 8 9 10

(b) Using proportional feedback control, control designers have obtained aclosed-loop system with the unit impulse response shown in Fig. 4.54(b).When the gain Ku = 8.556, the system is on the verge of instability. Deter-mine the proportional-, PI-, and PID-controller parameters according to theZeigler–Nichols ultimate sensitivity method.

4.41. A paper machine has the transfer function▲

G(s) = e−2s

3s + 1,

where the input is stock flow onto the wire and the output is basis weight orthickness.

(a) Find the PID-controller parameters using the Zeigler–Nichols tuning rules.

(b) The system becomes marginally stable for a proportional gain of Ku = 3.044,as shown by the unit-impulse response in Fig. 4.55. Find the optimal PID-controller parameters according to the Zeigler–Nichols tuning rules.

Figure 4.55Unit impulse responsefor paper machine inProblem 4.41

0.0 5.0 10.0 15.0 20.0 25.0

0.020

0.015

0.010

0.005

0.00

�0.005

�0.010

�0.015

Time (sec)

y




Problems 229

4.42. Prove that for a type 2 system, the acceleration error constant is given by▲

1Ka

= 12

(m∑

i=1

1

z2i

−n∑

i=1

1

p2i

),

where zi and pi are the closed-loop zeros and poles of the system.

4.43. For the unity feedback system with proportional control D = kp and process▲transfer function G(s) = A

s(τs + 1).

(a) Draw the block diagram from which to compute the sensitivity to changesin the parameter τ of the output response to a reference step input. Let theparameter be θ = 1/τ .

(b) Use MATLAB to compute and plot the sensitivity computed from the blockdiagram of part (a) if A = τ = kp = 1.




FranklinCh04ff

Documents

technical

upper saddle

zeiglernichols

kp ki

relevant error

loop transfer

forward shift

controller