MCT Book - Chapter 1

7/30/2019 MCT Book - Chapter 1

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This version: 22/10/2004

Chapter 1

An introduction to linear control theory

With this book we will introduce you to the basics ideas of control theory, and the settingwill be that of single-input, single-output (SISO), finite-dimensional, time-invariant, linearsystems. In this section we will begin to explore the meaning of this lingo, and look atsome simple physical systems which fit into this category. Traditional introductory texts incontrol may contain some of this material in more detail [see, for example Dorf and Bishop2001, Franklin, Powell, and Emani-Naeini 1994]. However, our presentation here is intendedto be more motivational than technical. For proper background in physics, one should look

to suitable references. A very good summary reference for the various methods of derivingequations for physical systems is [Cannon, Jr. 1967].

Contents

1.1 Some control theoretic terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 An introductory example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Linear differential equations for physical devices . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.1 Mechanical gadgets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.2 Electrical gadgets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.3 Electro-mechanical gadgets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Linearisation at equilibrium points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 What you are expected to know . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1 Some control theoretic terminology

For this book, there should be from the outset a picture you have in mind of what youare trying to accomplish. The picture is essentially given in Figure 1.1. The idea is thatyou are given a plant, which is the basic system, which has an output y(t) that youdlike to do something with. For example, you may wish to track a reference trajectoryr(t). One way to do this would be to use an open-loop control design. In this case, onewould omit that part of the diagram in Figure 1.1 which is dashed, and use a controllerto read the reference signal r(t) and use this to specify an input u(t) to the plant whichshould give the desired output. This open-loop control design may well work, but it hassome inherent problems. If there is a disturbance d(t) which you do not know about,then this may well cause the output of the plant to deviate significantly from the reference
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2 1 An introduction to linear control theory 22/10/2004

r(t) controller plant y(t)

sensor

d(t)

u(t)

e(t)

s(t)

Figure 1.1 A basic control system schematic

trajectory r(t). Another problem arises with plant uncertainties. One models the plant,typically via differential equations, but these are always an idealisation of the plants actualbehaviour. The reason for the problems is that the open-loop control law has no idea whatthe output is doing, and it marches on as if everything is working according to an idealised

model, a model which just might not be realistic. A good way to overcome these difficultiesis to use feedback. Here the output is read by sensors , which may themselves be modelledby differential equations, which produce a signal s(t) which is subtracted from the referencetrajectory to produce the error e(t). The controller then make its decisions based on theerror signal, rather than just blindly considering the reference signal.

1.2 An introductory example

Lets see how this all plays out in a simple example. Suppose we have a DC servomotor whose output is its angular velocity (t), the input is a voltage E(t), and there is a

disturbance torque T(t) resulting from, for example, an unknown external load being appliedto the output shaft of the motor. This external torque is something we cannot alter. A littlelater in this section we will see some justification for the governing differential equations tobe given by

d(t)

dt+ 1

(t) = kEE(t) + kTT(t).

The schematic for the situation is shown in Figure 1.2. This schematic representation we

E(t) kE

kT

T(t)

u(t) ddt

+1

= u (t)

Figure 1.2 DC motor open-loop control schematic
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22/10/2004 1.2 An introductory example 3

give here is one we shall use frequently, and it is called a block diagram.1

Let us just try something nave and open-loop. The objective is to be able to drive themotor at a specified constant velocity 0. This constant desired output is then our referencetrajectory. You decide to see what you might do by giving the motor some constant torquesto see what happens. Let us provide a constant input torque E(t) = E0 and suppose thatthe disturbance torque T(t) = 0. We then have the differential equation

ddt

+ 1 = kEE0.

Supposing, as is reasonable, that the motor starts with zero initial velocity, i.e., (0) = 0,the solution to the initial value problem is

(t) = kEE0

1 et/.We give a numerical plot for kE = 2, E0 = 3, and

1

= 12

in Figure 1.3. Well, we say, this all

2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.3 Open-loop response of DC motor

looks too easy. To get the desired output velocity 0 after a sufficiently long time, we needonly provide the input voltage E0 =

0kE

.However, there are decidedly problems lurking beneath the surface. For example, what

if there is a disturbance torque? Let us suppose this to be constant for the moment soT(t) = T0 for some T0 > 0. The differential equation is then

d

dt +1 = kEE0 kTT0,

and if we again suppose that (0) = 0 the initial value problem has solution

(t) = (kEE0 kTT0)

1 et/.If we follow our simple rule of letting the input voltage E0 be determined by the desired finalangular velocity by our rule E0 =

0kE

, then we will undershoot our desired final velocityby error = kTT0. In this event, the larger is the disturbance torque, the worse we doin
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2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.4 Open-loop response of DC motor with disturbance

fact, we can do pretty darn bad if the disturbance torque is large. The effect is illustrated

in Figure 1.4 with kT = 1 and T0 = 2.Another problem arises when we have imperfect knowledge of the motors physical char-

acteristics. For example, we may not know the time-constant as accurately as wed like.While we estimate it to be , it might be some other value . In this case, the actualdifferential equation governing behaviour in the absence of disturbances will be

d

dt+ 1

= kEE0,

which gives the solution to the initial value problem as

(t) = kEE01 et/.

The final value will then be in error by the factor

. This situation is shown in Figure 1.5for 1 =

58 .

Okay, I hope now that you can see the problem with our open-loop control strategy. Itsimply does not account for the inevitable imperfections we will have in our knowledge ofthe system and of the environment in which it works. To take all this into account, let usmeasure the output velocity of the motors shaft with a tachometer. The tachometer takesthe angular velocity and returns a voltage. This voltage, after being appropriately scaled bya factor ks, is then compared to the voltage needed to generate the desired velocity by feedingit back to our reference signal by subtracting it to get the error. The error we multiply by

some constant K, called the gain for the controller, to get the actual voltage input to thesystem. The schematic now becomes that depicted in Figure 1.6. The differential equationsgoverning this system are

d

dt+ ( 1

+ kEkSK) = kEKref+ kTT.

We shall see how to systematically obtain equations such as this, so do not worry if youthink it in nontrivial to get these equations. Note that the input to the system is, in some

1When we come to use block diagrams for real, you will see that the thing in the blocks are not differentialequations in the time-domain, but in the Laplace transform domain.
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22/10/2004 1.2 An introductory example 5

2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.5 Open-loop response of DC motor with actual motortime-constant

ref KE

kE

kT

T(t)

u(t) ddt

+1

= u (t)

kS

controller

plant

sensor

Figure 1.6 DC motor closed-loop control schematic

sense, no longer the voltage, but the reference signal ref. Let us suppose again a constantdisturbance torque T(t) = T0 and a constant reference voltage ref = 0. The solution tothe differential equation, again supposing (0) = 0, is then

(t) =kEK0 kTT0

1

+ kEkSK

1 e( 1+kEkSK)t.

Let us now investigate this closed-loop control law. As previously, let us first look at thecase when T0 = 0 and where we suppose perfect knowledge of our physical constants andour model. In this case, we wish to achieve a final velocity of0 = E0 kE as t , i.e., thesame velocity as we had attained with our open-loop strategy. We see the results of thisin Figure 1.7 where we have chosen kS = 1 and K = 5. Notice that the motor no longerachieves the desired final speed! However, we have improved the response time for the system
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2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.7 Closed-loop response of DC motor

significantly from the open-loop controller (cf. Figure 1.3). It is possible to remove the final

error by doing something more sophisticated with the error than multiplying it by K, butwe will get to that only as the course progresses. Now lets see what happens when weadd a constant disturbance by setting T0 = 2. The result is displayed in Figure 1.8. We

2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.8 Closed-loop response of DC motor with disturbance

see that the closed-loop controller reacts much better to the disturbance (cf. Figure 1.4),

although we still (unsurprisingly) cannot reach the desired final velocity. Finally we look atthe situation when we have imperfect knowledge of the physical constants for the plant. Weagain consider having 1 =

58 rather than the guessed value of

12 . In this case the closed-loop

response is shown in Figure 1.9. Again, the performance is somewhat better than that ofthe open-loop controller (cf. Figure 1.5), although we have incurred a largish final error inthe final velocity.

I hope this helps to convince you that feedback is a good thing! As mentioned above, weshall see that it is possible to design a controller so that the steady-state error is zero, as thisis the major deficiency of our very basic controller designed above. This simple example,however, does demonstrate that one can achieve improvements in some areas (response time
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22/10/2004 1.3 Linear differential equations for physical devices 7

2 4 6 8 10

2

4

6

8

10

12

14

t

(t

Figure 1.9 Closed-loop response of DC motor with actual motortime-constant

in this case), although sometimes at the expense of deterioration in others (steady-stateerror in this case).

1.3 Linear differential equations for physical devices

We will be considering control systems where the plant, the controller, and the sensorsare all modelled by linear differential equations. For this reason it makes sense to providesome examples of devices whose behaviour is reasonably well-governed by such equations.The problem of how to assemble such devices to, say, build a controller for a given plant issomething we will not be giving terribly much consideration to.

1.3.1 Mechanical gadgets In Figure 1.10 is a really feeble idealisation of a car sus-pension system. We suppose that at y = 0 the mass m is in equilibrium. The spring, as we

m

k d

y(t)

F(t)

Figure 1.10 Simplified automobile suspension

know, then supplies a restoring force Fk = ky and the dashpot supplies a force Fd = dy,where means d

dt. We also suppose there to be an externally applied force F(t). If we

ask Isaac Newton, Isaac, what are the equations governing the behaviour of this system?
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he would reply, Well, F = ma, now go think on it.2 After doing so youd arrive at

my(t) = F(t) ky(t) dy(t) = my(t) + dy(t) + ky(t) = F(t).

This is a second-order linear differential equation with constant coefficients and with inho-mogeneous term F(t).

The same sort of thing happens with rotary devices. In Figure 1.11 is a rotor fixed to

a shaft moving with angular velocity . Viscous dissipation may be modelled with a force

(t)

Figure 1.11 Rotor on a shaft

proportional to the angular velocity: Fd = d. In the presence of an external torque (t),the governing differential equation is

J(t) + d(t) = (t)

where J is the moment of inertia of the rotor about its point of rotation. If one wishesto include a rotary spring, then one must consider not the angular velocity (t) as thedependent variable, but rather the angular displacement (t) = 0 + t. In either case, thegoverning equations are linear differential equations.

Lets look at a simple pendulum (see Figure 1.12). If we sum moments about the pivotwe get

m2 = mg sin = + g sin = 0.Now this equation, you will notice, is nonlinear. However, we are often interested in thebehaviour of the system near the equilibrium points which are (, ) = (0, 0) where 0

{0, }. So, let us linearise the equations near these points, and see what we get. We writethe solution near the equilibrium as (t) = 0 + (t) where (t) is small. We then have

+ g

sin = + g

sin(0 + ) = +g

sin 0 cos +g

cos 0 sin .

Now note that sin 0 = 0 if0 {0, }, and cos 0 = 1 if0 = 0 and cos 0 = 1 if 0 = .We also use the Taylor expansion for sin x around x = 0: sin x = x x3

3!+ x

5

5!+ . . . . Keeping

2This is in reference to the story, be it true or false, that when Newton was asked how hed arrived atthe inverse square law for gravity, he replied, I thought on it.


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Figure 1.12 A simple pendulum

only the lowest order terms gives the following equations which should approximate thebehaviour of the system near the equilibrium (0, 0):

+ g

= 0, 0 = 0

g

= 0, 0 = .

In each case, we have a linear differential equation which governs the behaviour near the equi-librium. This technique of linearisation is ubiquitous since there really are no linear physicaldevices, but linear approximations seem to work well, and often very well, particularly incontrol. We discuss linearisation properly in Section 1.4.

Let us recall the basic rules for deriving the equations of motion for a mechanical system.

1.1 Deriving equations for mechanical systems Given: an interconnection of point masses andrigid bodies.

1. Define a reference frame from which to measure distances.2. Choose a set of coordinates that determine the configuration of the system.

3. Separate the system into its mechanical components. Thus each component should beeither a single point mass or a single rigid body.

4. For each component determine all external forces and moments acting on it.

5. For each component, express the position of the centre of mass in terms of the chosencoordinates.

6. The sum of forces in any direction on a component should equal the mass of the com-ponent times the component of acceleration of the component along the direction of the

force.7. For each component, the sum of moments about a point that is either (a) the centre of

mass of the component or (b) a point in the component that is stationary should equalthe moment of inertia of the component about that point multiplied by the angularacceleration.

This methodology has been applied to the examples above, although they are too simpleto be really representative. We refer to the exercises for examples that are somewhat moreinteresting. Also, see [Cannon, Jr. 1967] for details on the method we outline, and othermethods.
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1.3.2 Electrical gadgets A resistor is a device across which the voltage drop isproportional to the current through the device. A capacitor is a device across which thevoltage drop is proportional to the charge in the device. An inductor is a device acrosswhich the voltage drop is proportional to the time rate of change of current through thedevice. The three devices are typically given the symbols as in Figure 1.13. The quantity

I(t) E= RI

Resistor

q(t) E= 1Cq

Capacitor

I(t) E= LdIdt

Inductor

Figure 1.13 Electrical devices

R is called the resistance of the resistor, the quantity C is called the capacitance ofthe capacitor, and the quantity L is called the inductance of the inductor. Note thatthe proportionality constant for the capacitor is not C but 1C. The current I is related to

the charge q by I = dqdt

. We can then imagine assembling these electrical components insome configuration and using Kirchhoffs laws3 to derive governing differential equations. InFigure 1.14 we have a particularly simple configuration. The voltage drop around the circuit

E

+

R

L

C

Figure 1.14 A series RLC circuit

must be zero which gives the governing equations

E(t) = RI(t) + LI(t) + 1Cq(t) = Lq(t) + Rq(t) + 1Cq(t) = E(t)where E(t) is an external voltage source. This may also be written as a current equation bymerely differentiating:

LI(t) + RI(t) + 1C

I(t) = E(t).

In either case, we have a linear equation, and again one with constant coefficients.Let us present a methodology for determining the differential equations for electric cir-

cuits. The methodology relies on the notion of a tree which is a connected collection ofbranches containing no loops. For a given tree, a tree branch is a branch in the tree, anda link is a branch not in the tree.

3Kirchhoffs voltage law states that the sum of voltage drops around a closed loop must be zero andKirchhoffs current law states that the sum of the currents entering a node must be zero.


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1.2 Deriving equations for electric circuits Given: an interconnection of ideal resistors, capac-itors, and inductors, along with voltage and current sources.

1. Define a tree by collecting together a maximal number of branches to form a tree. Addelements in the following order of preference: voltage sources, capacitors, resistors, in-ductors, and current sources. That is to say, one adds these elements in sequence untilone gets the largest possible tree.

2. The states of the system are taken to be the voltages across capacitors in the tree branchesfor the tree of part 1 and the currents through inductors in the links for the tree frompart 1.

3. Use Kirchhoffs Laws to derive equations for the voltage and current in every tree branchin terms of the state variables.

4. Write the Kirchhoff Voltage Law and the Kirchhoff Current Law for every loop and everynode corresponding to a branch assigned a state variable.

The exercises contain a few examples that can be used to test ones understanding of theabove method. We also refer to [Cannon, Jr. 1967] for further discussion of the equationsgoverning electrical networks.

1.3.3 Electro-mechanical gadgets If you really want to learn how electric motorswork, then read a book on the subject. For example, see [Cannon, Jr. 1967].

A DC servo motor works by running current through a rotary toroidal coil which sits ina stationary magnetic field. As current is run through the coil, the induced magnetic fieldinduces the rotor to turn. The torque developed is proportional to the current through thecoil: T = KtI where T is the torque supplied to the shaft, I is the current through thecoil, and Kt is the torque constant. The voltage drop across the motor is proportionalto the motors velocity; Em = Ke where Em is the voltage drop across the motor, Ke is aconstant, and is the angular position of the shaft. If one is using a set of consistent units

with velocity measured in rads/sec, then apparently Ke = Kt.Now we suppose that the rotor has inertia J and that shaft friction is viscous and so thefriction force is given by d. Thus the motor will be governed by Newtons equations:

J = d + KtI = J + d = KtI.To complete the equations, one need to know the relationship between current and . Thisis provided by the relation Em = Ke and the dynamics of the circuit which supplies currentto the motor. For example, if the circuit has resistance R and inductance L then we have

LdI

dt+ RI = E Ke

with E being the voltage supplied to the circuit. This gives us coupled equations

J + d = KtI

LdI

dt+ RI = E Ke

which we can write in first-order system form as v

I

=

0 1 00 d

JKtJ

0 KeL RL

v

I

+

00

1L

E


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where we define the dependent variable v = . If the response of the circuit is much fasterthan that of the motor, e.g., if the inductance is small, then this gives E = Ke + RI andso the equations reduce to

J +

d + KtKeR

= KtR E.

Thus the dynamics of a DC motor can be roughly described by a first-order linear differentialequation in the angular velocity. This is what we saw in our introductory example.

Hopefully this gives you a feeling that there are a large number of physical systems whichare modelled by linear differential equations, and it is these to which we will be restrictingour attention.

1.4 Linearisation at equilibrium points

When we derived the equations of motion for the pendulum, the equations we obtainedwere nonlinear. We then decided that if we were only interested in looking at what is goingon near an equilibrium point, then we could content ourselves with linearising the equations.We then did this in a sort of hacky way. Lets see how to do this methodically.

We suppose that we have vector differential equations of the formx1 = f1(x1, . . . , xn)

x2 = f2(x1, . . . , xn)

...

xn = fn(x1, . . . , xn).

The n functions (f1, . . . , f n) of the n variables (x1, . . . , xn) are known smooth functions. Letus denote x = (x1, . . . , xn) and f(x) = (f1(x), . . . , f n(x)). The differential equation canthen be written as

x

=f

(x

).

(1.1)It really only makes sense to linearise about an equilibrium point. An equilibrium pointis a point x0 Rn for which f(x0) = 0. Note that the constant function x(t) = x0 is asolution to the differential equation ifx0 is an equilibrium point. For an equilibrium pointx0 define an n n matrix Df(x0) by

Df(x0) =

f1x1

(x0)f1x2

(x0) f1xn (x0)f2x1

(x0)f2x2

(x0) f2xn (x0)...

.... . .

...fnx1

(x0)fnx2

(x0) fnxn (x0)

.

This matrix is often called the Jacobian off at x0. The linearisation of (1.1) about anequilibrium point x0 is then the linear differential equation

= Df(x0).

Lets see how this goes with our pendulum example.

1.3 Example The nonlinear differential equation we derived was

+ g sin = 0.
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22/10/2004 1.5 What you are expected to know 13

This is not in the form of (1.1) since it is a second-order equation. But we can put this intofirst-order form by introducing the variables x1 = and x2 = . The equations can then bewritten

x1 = = x2

x2 = = g sin = g sin x1.

Thusf1(x1, x2) = x2, f2(x1, x2) = g

sin x1.

Note that at an equilibrium point we must have x2 = 0. This makes sense as it meansthat the pendulum should not be moving. We must also have sin x1 = 0 which means thatx1 {0, }. This is what we determined previously.

Now let us linearise about each of these equilibrium points. For an arbitrary pointx = (x1, x2) we compute

Df(x) =

0 1

g

cos x1 0

.

At the equilibrium point x1 = (0, 0) we thus have

Df(x1) =

0 1

g

0

,

and at the equilibrium point x2 = (0, ) we thus have

Df(x1) =

0 1g

0

.

With these matrices at hand, we may write the linearised equations at each equilibriumpoint.

1.5 What you are expected to know

There are five essential areas of background that are assumed of a student using thistext. These are

1. linear algebra,

2. ordinary differential equations, including the matrix exponential,

3. basic facts about polynomials,

4. basic complex analysis, and

5. transform theory, especially Fourier and Laplace transforms.

Appendices review each of these in a cursory manner. Students are expected to have seenthis material in detail in previous courses, so there should be no need for anything but rapidreview in class.

Many of the systems we will look at in the exercises require in their analysis straight-forward, but tedious, calculations. It should not be the point of the book to make you gothrough such tedious calculations. You will be well served by learning to use a computerpackage for doing such routine calculations, although you should try to make sure you areasking the computer to do something which you in principle understand how to do yourself.
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I have used Mathematica to do all the plotting in the book since it is what I am familiarwith. Also available are Maple and Matlab. Matlab has a control toolbox, and is themost commonly used tool for control systems.4

You are encouraged to use symbolic manipulation packages for doing problems in thisbook. Just make sure you let us know that you are doing so, and make sure you know whatyou are doing and that you are not going too far into black box mode.

1.6 Summary

Our objective in this chapter has been to introduce you to some basic control theoreticideas, especially through the use of feedback in the DC motor example. In the remainder ofthese notes we look at linear systems, and to motivate such an investigation we presentedsome physical devices whose behaviour is representable by linear differential equations, per-haps after linearisation about a desired operating point. We wrapped up the chapter with aquick summary of the background required to proceed with reading these notes. Make sureyou are familiar with everything discussed here.

4Mathematica and Maple packages have been made available on the world wide web for doing thingssuch as are done in this book. See http://mast.queensu.ca/~math332/.
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Exercises for Chapter 1 15

Exercises

E1.1 Probe your life for occurrences of things which can be described, perhaps roughly, bya schematic like that of Figure 1.1. Identify the components in your system whichare the plant, output, input, sensor, controller, etc. Does your system have feedback?Are there disturbances?

E1.2 Consider the DC servo motor example which we worked with in Section 1.2. Determineconditions on the controller gain K so that the voltage E0 required to obtain a desiredsteady-state velocity is greater for the closed-loop system than it is for the open-loop system. You may neglect the disturbance torque, and assume that the motormodel is accurate. However, do not use the numerical values used in the notesleaveeverything general.

E1.3 An amplifier is to be designed with an overall amplification factor of 2500 50. Anumber of single amplifier stages is available and the gain of any single stage maydrift anywhere between 25 and 75. The configuration of the final amplifier is given inFigure E1.1. In each of the blocks we get to insert an amplifier stage with the large

Vin K1 K2 KN Vout

Figure E1.1 A multistage feedback amplifier

and unknown gain variation (in this case the gain variation is at most 50). Thus the

gain in the forward path is K1K2 KN where N is the number of amplifier stagesand where Ki [25, 75]. The element in the feedback path is a constant 0 < < 1.(a) For N amplifier stages and a given value for determine the relationship between

Vin and Vout.

The feedback gain is known precisely since it is much easier to design a circuitwhich provides accurate voltage division (as opposed to amplification). Thus, we canassume that can be exactly specified by the designer.

(b) Based on this information find a value of in the interval (0, 1) and, for thatvalue of , the minimal required number of amplifier stages, Nmin, so that thefinal amplifier design meets the specification noted above.

E1.4 Derive the differential equations governing the behaviour of the coupled masses inFigure E1.2. How do the equations change if viscous dissipation is added betweeneach mass and the ground? (Suppose that both masses are subject to the samedissipative force.)

The following two exercises will recur as exercises in succeeding chapters. Since the compu-tations can be a bit involvedcertainly they ought to be done with a symbolic manipulationpackageit is advisable to do the computations in an organised manner so that they maybe used for future calculations.


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m m

k k k

x1 x2

Figure E1.2 Coupled masses

x

Figure E1.3 Pendulum on a cart

E1.5 Consider the pendulum on a cart pictured in Figure E1.3. Derive the full equationswhich govern the motion of the system using coordinates (x, ) as in the figure. HereM is the mass of the cart and m is the mass of the pendulum. The length of thependulum arm is . You may assume that there is no friction in the system. Linearisethe equations about the points (x,, x, ) = (x0, 0, 0, 0) and (x,, x, ) = (x0, , 0, 0),where x0 is arbitrary.

E1.6 Determine the full equations of motion for the double pendulum depicted in Fig-ure E1.4. The first link (i.e., the one connected to ground) has length 1 and mass

1

2

Figure E1.4 Double pendulum

m1, and the second link has length 2 and mass m2. The links have a uniform massdensity, so their centres of mass are located at their midpoint. You may assume that
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there is no friction in the system. What are the equilibrium points for the doublependulum (there are four)? Linearise the equations about each of the equilibria.

E1.7 Consider the electric circuit of Figure E1.5. To write equations for this system we

E

+

R

C L

Figure E1.5 Electric circuit

need to select two system variables. Using IC, the current through the capacitor, and

IL, the current through the inductor, derive a first-order system of equations in twovariables governing the behaviour of the system.

E1.8 For the circuit of Figure E1.6, determine a differential equation for the current through

E

+

R1

R2 C

Figure E1.6 Another electric circuit

the resistor R1 in terms of the voltage E(t).

E1.9 For the circuit of Figure E1.7, As dependent variable for the circuit, use the volt-

E

+

RL

L C

RC

Figure E1.7 Yet another electric circuit

age across the capacitor and the current through the inductor. Derive differentialequations for the system as a first-order system with two variables.


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E1.10 The mass flow rate from a tank of water with a uniform cross-section can be roughlymodelled as being proportional to the height of water in the tank which lies abovethe exit nozzle. Suppose that two tanks are configured as in Figure E1.8 (the tanks

Figure E1.8 Coupled water tanks

are not necessarily identical). Determine the equations of motion which give the massflow rate from the bottom tank given the mass flow rate into the top tank. In thisproblem, you must define the necessary variables yourself.

In the next exercise we will consider a more complex and realistic model of flow in coupledtanks. Here we will use the Bernoulli equation for flow from small orifices. This says thatif a tank of uniform cross-section has fluid level h, then the velocity flowing from a small

nozzle at the bottom of the tank will be given by v = 2gh, where g is the gravitationalacceleration.E1.11 Consider the coupled tanks shown in Figure E1.9. In this scenario, the input is the

volume flow rate Fin which gets divided between the two tanks proportionally to theareas 1 and 2 of the two tubes. Let us denote =

11+2

.

(a) Give an expression for the volume flow rates Fin,1 and Fin,2 in terms ofFin and .

Now suppose that the areas of the output nozzles for the tanks are a1 and a2, andthat the cross-sectional areas of the tanks are A1 and A2. Denote the water levels inthe tanks by h1 and h2.

(b) Using the Bernoulli equation above, give an expression for the volume flow rates

Fout,1 and Fout,2 in terms of a1, a2, h1, and h2.(c) Using mass balance (assume that the fluid is incompressible so that mass and

volume balance are equivalent), provide two coupled differential equations for theheights h1 and h2 in the tanks. The equations should be in terms ofFin, , A1,A2, a1, a2, and g, as well as the dependent variables h1 and h2.

Suppose that the system is in equilibrium (i.e., the heights in the tanks are constant)with the equilibrium height in tank 1 being 1.

(d) What is the equilibrium input flow rate ?

(e) What is the height 2 of fluid in tank 2?
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Fin

1

Fin,1

2

Fin,2

a1

a2

A1

A2

Fout,1

Fout,2

Figure E1.9 Another coupled tank scenario


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MCT Book - Chapter 1

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