Control Systems Module

8/8/2019 Control Systems Module

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Control Systems Module

Table of Contents

1. Introduction2. Mathematical model of systems3. State variable models4. Feedback control systems characteristics5. The performance of Feedback Control Systems6. The stability of Linear Feedback Systems7. Root Locus method8. Frequency Response Method9. Stability in the Frequency Domain10. The design of Feedback Control Systems11. The Design of State Variable Feedback Systems

12. Robust Control Systems13. Digital Control Systems

Appendix B: LabV IEW MathScript Basics.Appendix C: Symbols, Units, and Conversion Factors.Appendix D: Laplace Transform Pairs.Appendix E: An Introduction to Matrix Algebra.Appendix F: Decibel Conversion.Appendix G: Complex Numbers.Appendix H: z-Transfer Pairs Preface.

http://timeandspacetoday.blogspot.com/2008/12/free-control-systems-books.html

http://homepage.mac.com/sami_ashhab/courses/control/lectures/lecture_notes.htm

http://www.esr.ruhr-uni-bochum.de/rt1/syscontrol/node4.html

http://www.inf.ethz.ch/personal/fcellier/Lect/DC/Lect_dc_index.html

http://books.google.co.za/books?id=V-FpzJP5b EIC&dq=modern+control+systems,+google+book&printsec=frontcover&source=in&hl=en&ei=0ey ITIefH4SC4Abf9LzSBA&sa=X&oi=book_result&ct=result&resnum=12&ved=0CDwQ6A EwCw#v=onepage&q&f=false

Control Systems/Print version - Wikibooks, collection of open-content textbooks Page 7 of 209http://en.wikibooks.org/w/index.php?title=Control_Systems/Print_version&printable=yes10/30/2006


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1 Introduction to Control Systems

The study and design of automatic Control Systems , is a large and expansive area of study.Control systems, and control engineering techniques have become a pervasive part of modern

technical society. From devices as simple as a toaster, to complex machines like space shuttlesand rockets, control engineering is a part of our everyday life. This module will introduce thefield of control engineering, and will build upon those foundations to explore some of the moreadvanced topics in the field. Control systems are components that are added to other components, to increase functionality, or to meet a set of design criteria. Let's start off with animmediate example: We have a particular electric motor that is supposed to turn at a rate of 40RPM. To achieve this speed, we must supply 10 Volts to the motor terminals. However, with 10volts supplied to the motor at rest, it takes 30 seconds for our motor to get up to speed. This isvaluable time lost. Now, we have a little bit of a problem that, while simplistic, can be a point of concern to people who are both designing this motor system, and to the people who might

potentially buy it. It would seem obvious that we should increase the power to the motor at the

beginning, so that the motor gets up to speed faster, and then we can turn the power back downto 10 volts once it reaches speed.

Now this is clearly a simplistic example, but it illustrates one important point: That we can addspecial "Controller units" to preexisting systems, to increase performance, and to meet newsystem specifications. There are essentially two methods to approach the problem of designing anew control system: the Classical Approach , and the Modern Approach . It will do us good toformally define the term "Control System", and some other terms that are used throughout thismodule:

Control System

A Control System is a device, or a collection of devices that manage the behavior of other devices. Some devices are not controllable. A control system is an interconnection of components connected or related in such a manner as to command, direct, or regulate itself or another system.

Controller A controller is a control system that manages the behavior of another device or system.

Compensator A Compensator is a control system that regulates another system, usually by conditioning theinput or the output to that system. Compensators are typically employed to correct a single

design flaw, with the intention of affecting other aspects of the design in a minimal manner.

Classical and Modern


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Classical and Modern control methodologies are named in a misleading way, because the groupof techniques called "Classical" were actually developed later than the techniques labeled"Modern". However, in terms of developing control systems, Modern methods have been used togreat effect more recently, while the Classical methods have been gradually falling out of favor.Most recently, it has been shown that Classical and Modern methods can be combined to

highlight their respective strengths and weaknesses. Classical Methods are methods involving theLaplace Transform domain .

Physical systems are modeled in the so-called "time domain", where the response of a givensystem is a function of the various inputs, the previous system values, and time. As time

progresses, the state of the system, and it's response change. However, time-domain models for systems are frequently modeled using high-order differential equations, which can becomeimpossibly difficult for humans to solve, and some of which can even become impossible for modern computer systems to solve efficiently. To counteract this problem, integral transforms,such as the Laplace Transform , and the Fourier Transform can be employed to change anOrdinary Differential Equation (OD E) in the time domain into a regular algebraic polynomial in

the transform domain.Once a given system has been converted into the transform domain, it can be manipulated withgreater ease, and analyzed quickly and simply, by humans and computers alike.Modern Control Methods, instead of changing domains to avoid the complexities of time-domainODE mathematics, converts the differential equations into a system of lower-order time domainequations called State Equations , which can then be manipulated using techniques from linear algebra (matrices ).

A third distinction that is frequently made in the realm of control systems is to divide analogmethods (classical and modern, described above ) from digital methods. Digital Control Methodswere designed to try and incorporate the emerging power of computer systems into previouscontrol methodologies. A special transform, known as the Z-Transform , was developed that canadequately describe digital systems, but at the same time can be converted (with some effort ) into the Laplace domain. Once in the Laplace domain, the digital system can be manipulated andanalyzed in a very similar manner to Classical analog systems. For this reason, this module willnot make a hard and fast distinction between Analog and Digital systems, and instead willattempt to study both paradigms in parallel.

In this section we demonstrate some control concepts through a detailed example. Nomathematical relations are used in explaining the example. The system (or process ) that we willconsider consists of a tank that contains oil. The tank sits on a heater which supplies energy tothe oil. The energy supplied by the heater depends on the input voltage to the heater which iscontrolled by a knob. Please refer to Figure 1.1 for a sketch of this thermal system.


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Figure 1.1: The oil, tank and heater system .

In this application we are interested in regulating the oil temperature. The system can berepresented by a block diagram as shown in Figure 1.2. The input to the system is the voltage tothe heater or knob position, whereas the output of the system is the oil temperature.

Figure 1.2: Block diagram of the oil, tank and heater system .

Let us suppose that this process is in a factory and a worker adjusts the voltage knob to set the oiltemperature at a desired value determined by his boss. The boss decides the desired temperature

based on the factory requirements. The worker uses a lookup table to position the knob. Thetable contains knob positions and the corresponding values of steady state temperature (we mean by steady state the final value since it takes some time to heat the oil to the desired temperature ).An experiment was performed in the past to obtain the lookup table. The knob was adjusted atthe different positions and the corresponding values of temperature were measured by atemperature sensor such as a thermometer or a thermocouple. See Figure 1.3.


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Figure 1.3: Open loop control of the oil temperature . Note that there is no device to measure the oiltemperature .

The situation in Figure 1.3 can be represented by a block diagram as shown in Figure 1.4. The boss decides the desired temperature which he tells the worker to adjust. Therefore, the input tothe worker is the desired temperature. The worker uses the lookup table to select the knob

position. If everything is OK the oil temperature should go to a value close to the desired value.The actual oil temperature will usually be different from the desired value due environmental andoperating condition changes. The worker does not know the final temperature since there is nosensor to provide the temperature measurement. The worker just adjusts the knob once and hehopes that the temperature will go to the desired value. This is called an open loop system. In thiskind of system the output (temperature here ) is controlled directly using an actuating device(human worker here ) without measuring this output.

Figure 1.4: Block diagram representation of the open loop system .

The boss would like to have the oil temperature more accurate, that is, the oil temperature should be very close to the desired value. He buys a thermometer (temperature sensor ) so that theworker could read the oil temperature and thus adjusts the knob appropriately to make thetemperature as close as possible to the desired value. If the measured temperature is less than thedesired one he will raise the knob, otherwise he will lower it. He keeps doing this until thedesired value of temperature is achieved. The worker is able to do a better job because he has asensor or feedback information about the oil temperature. In this case we have a closed loopsystem.


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Figure 1.5: Closed loop control of the oil temperature . Note that there is a thermometer to measure the oil

temperature .

The situation in Figure 1.5 can be represented by a block diagram as shown in Figure 1.6. The boss decides the desired temperature which he tells the worker to adjust. The worker uses thedesired and measured temperatures to calculate the error = desired - measured. If the error is

positive he raises the knob, otherwise he lowers it. He repeats this procedure until the error iszero, i.e., desired temp. = measured temp. Note that the lookup table is unnecessary in the closedloop case. Nevertheless, the lookup table gives the worker the initial guess for the knob positiondepending on the desired temperature and then he makes the appropriate changes to take theerror to zero.


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Figure 1.6: Block diagram representation of the closed loop (feedback) system .

In the above example the controller was a human (worker ). The controller is the part of thesystem that makes decisions based on an objective. The objective in the thermal system was toachieve the desired value of oil temperature. This was done by making the difference betweenthe desired and measured values of temperature equal to zero. The human could be replaced byan electronic circuit, computer or a mechanical controller. We used a thermometer for thetemperature measurement. A thermocouple which converts temperature to voltage could be usedfor the electronic circuit or computer controller. No sensor is needed for the mechanicalcontroller. The relationship between pressure and temperature could be used to design amechanism for moving the knob up and down to achieve the desired value of temperature.

EXA MPL ES OF CONTROL SYS T E M S

In this section we shall present two more examples of control systems.

Speed Control System:

The fundamental concept of a Watt's speed governor for an engine is shown in the schematicdiagram of Figure 1.7. The quantity of fuel entering the engine is tuned as per the difference

between the desired and the actual engine speeds.

The sequence of actions is stated as follows: The speed governor is tuned such that, at therequired speed, no pressured oil will flow into either side of the power cylinder. If the speeddrops below the required value due to any disturbance, then the centrifugal force of the speedgovernor decreases which causes the control valve to move downwards, supplying more fuel,and the speed of the engine increases until the required value is attained. Alternatively, if thespeed of the engine crosses the required value, then the increase in the centrifugal force of thegovernor makes the control valve to move upward. This curbs the supply of fuel, and the speedof the engine decreases until the required value is reached.


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In this speed control system, the engine is the plant (controlled system ) and the speed of theengine is the controlled variable. The error signal is the difference between the desired speed andthe actual speed. The control signal (the quantity of fuel ) to be applied to the plant (engine ) is theactuating signal. The disturbance is the external input to disturb the controlled variable. For example any unexpected change in the load is a disturbance.

Tem perature Control System:

A schematic diagram is shown in Figure 1.8 depicting the temperature control of an electricfurnace. A thermometer (analog device ) measures the temperature in the electric furnace. AnA/D converter converts the analog temperature to a digital temperature. An interface feeds thedigital temperature to a controller. It is then compared with the programmed input temperature,and if there is any inconsistency (error ), the controller sends out a signal to be heater, through aninterface, amplifier, and relay, to bring the furnace temperature to a desired value.


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This module is intended to accompany a course of study in under-graduate engineering.

PrerequisitesUnderstanding of the material in this module will require a solid mathematical foundation. Thismodule does not currently explain, nor will it ever try to fully explain most of the necessarymathematical tools used in this text.For that reason, the reader is expected to have read the following books, or have backgroundknowledge comparable to them:

CalculusAlgebraLinear AlgebraDifferential Equations

Engineering Analysis

Module StructureThis module is organized as follows:-

y Chapter 1 Introduction to Control Systems general approach to designing and

building a control system

y Chapter 2 Mathematical Models of Systems this discusses the mathematicalmodels of such systems as mechanical, electrical and fluid

y Chapter 3 State Variable Models Mathematical models of systems in statevariable form are developed


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y Chapter 4 Feedback Control System Characteristics

y Chapter 5 Performance of Feedback Control Systems performance of a controlsystem correlated with the s-plane location of the poles and zeros of the transfer functionof the system.

y Chapter 6 The Stability of Linear Feedback System introduction to the Routh-Hurwitz criterion

y Chapter 7 Root Locus method the locus of roots in the s-plane is determined bygraphical method

y Chapter 8 Frequency Response steady state variable response of a system as the

frequency of the sinusoid is varied. Bode plot is considered

y Chapter 9 Stability in the Frequency Domain relative stability and the Nyquistcriterion are discussed in this chapter

Exercises


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2 Mathematical Models of Systems

The first step in control design is the development of a mathematical model for the process or system under consideration. In the modeling of systems, we assume a cause and effectrelationship described by the simple input/output diagram below, fig 2.1. An input is applied to a

system, and the system processes it to produce an output. In general, a system has the three basiccomponents listed below.

1. Inputs : These represent the variables under the designer's disposal. The designer produces these signals directly and applies them to the system under consideration. For example, the voltage source to a motor and the external torque input to a roboticmanipulator both represent inputs. Systems may have single or multiple inputs.

2. Outputs : These represent the variables which the designer ultimately wants to controland that can be measured by the designer. For example, in a flight control application anoutput may be the altitude of the aircraft, and in automobile cruise control the output isthe speed of the vehicle.

3. System or Plant : This represents the dynamics of a physical process which relate theinput and output signals. For example, in automobile cruise control, the output is thevehicle speed, the input is the supply of gasoline, and the system itself is the automobile.Similarly, an air conditioning system regulates the temperature in a room. The outputfrom the system is the air temperature in the room. The input is cool air added to theroom. The system itself is the room full of air with its air flow characteristics. Note thatthe system could be mechanical, thermal, fluid, electrical, electro-mechanical, etc.

Figure 2.1: Block diagram representation of a system.

In this section we will show by way of examples how mathematical models for simpleengineering systems can be developed. Notice that in each example four steps are taken.First, a diagram of all system components and externally applied inputs is drawn. Fromthis diagram, the inputs and outputs are identified. Then a diagram is made of the systemcomponents in which all internal signals are shown. Finally, the differential equationsgoverning the system dynamics are obtained. These equations form the mathematicalmodel of the system.

Example 1 Thermal system: Oil, tank and heater described in chapter 1.

The oil, tank and heater system was described in the previous lecture. Now, we develop amathematical model (differential equation ) for this system which describes its dynamics.The different signals and components needed to derive a mathematical model for thethermal system are shown in Figure 2.2.


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We first identify the input and output.Input: voltage to heater, v , output: oil temperature, T .

Figure 2.2: Diagram of the thermal system signals and components.

We apply the heat balance equation to obtain the differential equation:

Energy supplied by the heater = Energy stored by oil + Energy lost to the surroundingenvironment by conduction

where, k is a constant provided by the heater manufacturer, v is the voltage to the heater,c is thermal capacitance of the oil, T is the oil temperature, T a is the ambient temperature(environment ), R is the thermal resistance of the tank wall (heat conduction ) and t is thetime. Equation (1 ) is the differential equation that describes the dynamics of our thermalsystem. Note that the input and output appear in this equation. If we know the input, v then we can solve equation (1 ) for the output, T .

Example 2 Mechanical system: Spring-mass-damper system.

In this example we model the spring-mass-system shown in Figure 2.3(a ). The mass, m issubjected to an external force f . Let's suppose that we are interested in controlling the

position of m. The way to control the position of the mass is by chossing f .

We first identify the input and output.Input: external force, f , output: mass position, x.


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Figure 2.3: (a ) Diagram of the mechanical system components. (b ) Free body diagram of the mechanical system.

We apply Newton's second law to obtain the differential equation of this mechanicalsystem. Using the free-body-diagram shown in Figure 2.3(b ), we have

Where, b is the damping coefficient and k is the spring stiffness. Equation (2 ) is thedifferential equation that describes the dynamics of the spring-mass-damper system. Notethat the input and output appear in this equation. If we know the input, f then we cansolve equation (2 ) for the output, x.

The thermal and mechanical systems described in this lecture can be represented by thefollowing block diagram:

Figure 2.4: Block diagram representation of the thermal or mechanical system.


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Example 3: Fluid System: Water Level Control

The system is a storage tank of cross-sectional area A whose liquid level or height is h. Theliquid enters the tank from the top and leaves the tank at the bottom through the valve, whosefluid resistance is R. The volume flow rate in and the volume flow rate out are qi and qo,

respectively. The fluid density p is constant. Please refer to Figure 2.5 for a schematic diagram of the system. In such a system it is desired to regulate the water level in the tank. Assume that thevariable the we can change to control the water level is qi.

Figure 2.5: Diagram of the fluid system components and signals.

We first identify the input and output of the systemInput: volume flow rate in, qi, output: water level, h.

In order to obtain the differential equation of the system we use the conservation of mass principle which states that

The time rate of change of fluid mass inside the tank = the mass flow rate in - mass flow rate out

where, A is the cross sectional area of the tank, g is the acceleration due gravity and R is the fluidresistance through the valve. Equation (1 ) is the differential equation that describes the dynamicsof our thermal system. Note that the input and output appear in this equation. If we know the


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input, qi then we can solve equation (1 ) for the output, h. A block diagram representation of thefluid system is shown in Figure 2.6.

Figure 2.6: Block diagram representation of the fluid system.

Example 4: Mechanical System: (2 mass )-spring-damper system

This system contains two masses, a spring, and a damper (refer to Figure 2.7 ). An external forceis applied to the first mass and we would like to control the position of the second mass. Theexternal force, f indirectly affects the motion of the second mass.

Figure 2.7: (a ) Diagram of the mechanical system components. (b ) Free body diagram of thesystem.

We first identify the input and output of the systeminput: external force, f , output: displacement of the second mass, z.

Next we find a set of differential equations describing this system. Newton's second law isapplied to each mass to obtain these differential equations.


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Notice how the system dynamics relate the input and output. Remember that f is the input and z is the output. A block diagram representation of the mechanical system is shown in Figure 2.8.

Figure 2.8: Block diagram representation of the mechanical system.

Example 5: Electrical System: RLC Circuit (Parallel Connection )

The system below, figure 2.9, shows an electrical circuit with a current source i, resistor R,inductor L, and capacitor C. All of these parts are connected in parallel. It is required to regulate

the capacitor voltage V.


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Figure 2.9: RLC circuit (parallel connection ).

In the next section we show how a state space representation of this mechanical system can beobtained.

We first identify the input and output of the systeminput: current, i, output: voltage, V .

Next we find a set of differential equation that describes this system. Kirchhoffs current law isapplied. The sum of the three currents ( R, L, and C ) is equal to the overall source current i.

In the next section we show how a state space representation of this mechanical system can beobtained.

State Space RepresentationIn deriving a mathematical model for a physical system one usually begins with a system of differential equations. It is often convenient to rewrite these equations as a system of first order differential equations. We will call this system of differential equations a state spacerepresentation. The solution to this system is a vector that depends on time and which contains

enough information to completely determine the trajectory of the dynamical system. This vector is referred to as the state of the system, and the components of this vector are called the statevariables. In order to illustrate these concepts consider the following two examples.

Example 6: spring-mass-damper system


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Example 7: (2 mass )-spring-damper system

This is the same example discussed in the previous lecture where we derived the mathematicalmodel of the system (set of differential equations ). The equations describing the evolution of thissystem can be written as (see equations (2 ) and (3 ) from the previous section ):

Suppose that the input to the system is f and the output is z. In this problem we have two secondorder differential equations which means that we need four initial conditions to solve themuniquely. For example, specifying the value of w(to), dw/dt(to), z(to), and dz/dt(to) will allow for solution of w(t) and z(t) for t>to . Therefore, we will also require four state variables to describethis system. We will label each state variable xi where i = 1, ...,4 and make the assignments

Rewriting the two differential equations above in terms of the state variables xi yields thefollowing state-space description for the system:


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We can use matrix notation to rewrite the equations in the above state space representation.

So for the above example, the state vector takes the form:

In general, the linear state space models assume the form:

where x(0) is given.If the number of states is equal to n, number of inputs is equal to p and the number of outputs is


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equal to q, then A is an nxn square matrix, B is an nx p matrix, C is an qxn matrix, and D is anqx p matrix.

The previous examples illustrate how to go from a physical problem to a mathematicaldescription in state space form. Pictorially, figure 2.10:

Figure 2.10 Pictorial illustration

We will usually want to choose the smallest set of state variables which will still fully describethe system. A system description in terms of the minimum number of state variables is called aminimal realization of the system. We will want next to compute the solution of the state spaceequation. This leads us to the study of Laplace transforms.

Laplace TransformsLaplace transforms are mathematical operations that transform functions of a single real variableinto functions of a complex variable. They are extremely useful in the study of linear systems for many reasons. The most important reason is that by using Laplace transforms one can tranform alinear differential equation into an algebraic equation. Another important reason for Laplacetransforms is that they supply a different representation for a linear system. Such a representaionwill be the basis for the classical control techniques. Given a function of time, x(t) for t>0 , wedefine the Laplace transform of x(t) as follows:

A natural question to ask is: what are the conditions that the signal x(t) must satisfy in order for the Laplace transform to exist? The Laplace transform exists for signals for which the abovetransformation integral converges. However, the following fact supplies an easy way to check sufficient condition for the existence of a Laplace transform.

Fact Suppose that

1. x(t) is piecewise continuous2. | x(t) |< K exp( at ) when t>0 for some positive real numbers K and a

then the Laplace transform L(x(t)) = X(s) exists for Re(s)>a . Note that Re(s) is the real part of s.


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Since any physically realizable signal is continuous and will not blow up so fast that noexponential can bound it, the previous fact assures us that the Laplace transform can be definedfor any physically realizable signal. Let us now consider a few examples of how to calculateLaplace transforms. Note: For this course, we will use the letter 'j' to represent the complexconstant square root of -1.

Example 8: x(t)=Step Function

Example E xponential Fun c tion

We will now start studying the properties of the Laplace transform.

Linearity of Laplace Transforms : If then for any constants

.

This property can be derived in the following way:


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Laplace Transform of a Derivative : If then


This property shows that the process of differentiation in the time-domain corresponds tomultiplication by s in the Laplace domain plus the addition of the constant - x(0) .

Laplace Transforms of Higher Derivatives : If then


In general we have


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Laplace Transform of a Signal Multiplied by an Exponential : If then

.

This property can be derived as follows.

This property shows that multiplication by an exponential in time corresponds to shifting in s.

Example : Find the Laplace transform for the following functions

We can find the Laplace transform for t and sin2 t using the Laplace transform table. In part (a ) we need to shift s by 3 whereas we shift s by -1 in part (b ). Therefore the Laplace transforms of the above two functions are given as

Inverse Laplace TransformsIn this lecture we discuss how to calculate the inverse Laplace transform. The simplest way toinvert a Laplace transform is to convert the function to be inverted into a sum of functions whichare Laplace transforms that we have already calculated. By the linearity of Laplace transforms,the given function can be inverted directly and its inverse will be a sum of the correspondingtime domain functions. The table given under the supplementary material in this web site will bevery useful in this. It lists some common time domain functions and their Laplace transforms.Some of these relationships were derived in previous section.

We will next give an example that demonstrates how the Laplace transform can be used insolving an ordinary differential equation. Note that the inverse Laplace transform will be neededto solve the problem.


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Example 1:

The differential equation of a system is given as

where u is the input to the system and x is the output of the system. Find the output signal x(t) if

the input

Take Laplace transform of both sides of the differential equation

In the following we will study by examples the techniques of inverting Laplace transforms. Eachexample will represent a particular form for a Laplace transform, and the solution can be viewedas a model on which to base the procedure for similar cases. Note that in the previous examplewe provided a method for inverting the Laplace transform.

Example 2: Calculate the inverse Laplace transform for the following function


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Using the partial fraction expansion, we have

Example 3: Calculate the inverse Laplace transform for the following function

Using the partial fraction expansion, we have


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Example 4: Calculate the inverse Laplace Transform for the following function. Note that thedenominator has real and complex roots.

The roots of the denominator of X(s) are s = -6, s = -3+4 j and s = -3-4 j . We first calculate the partial fraction expansion of X(s) without factoring the second order polynomial which has thecomplex roots.


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Solutions of Linear SystemsIn this lecture we will begin studying a few methods for the solution of linear differentialequations. In particular we will focus on those differential equations that arise from modelinglinear mechanical systems such as a mass-spring system. First consider the following differentialequation:

In order to solve this equation for a unique y(t) , we need the initial conditions:

Let us start by writing this equation in state space form. Define . This givesus:

With initial conditions: . In matrix notation we have


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This fits the general form:

with

Before we develop the complete solution for this general problem we will look at the solution of a very simple case. Suppose x(t) and u(t) are scalars. This would give us A = a and B = b alsoscalars. Consider the first order differential equation

We operate on this equation as follows

multiply both sides by

and integrate the expression with respect to time

This last equation is known as the scalar form of the variation of parameters formula. Thecomponents of this equation are as follows:


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Example : Calculating the solution to a specific case of the scalar problem given above.

With input

Using the variation of parameters formula above we find:

We now consider the more general matrix case:

where A is n x n , B is n x m, x(t) is n x 1 and u(t) is m x 1. Let us define the matrix J (t) whichsolves the following matrix differential equation:

where I is the n x n identity matrix. By using Laplace transforms we can transform the aboveequation into

and thus


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and

which is the inverse Laplace transform. One can also verify that J (t) is equal to the following

infinite sum by showing that the sum converges and satisfies the above matrix differentialequation.

Observe that we can use J (t) as an integrating factor in the following way:

Note that for one of the steps above we have used the fact that J (t) commutes with A; that isAJ X = J (t) A. Using the infinite sum expression for J (t), the reader can easily verify this fact.In a similar way to the scalar case we arrive at

In the general case where the initial condition is given at some point t 0, we have

where

is given. This is the matrix form of the variation of parameters formula. The matrix is calledthe state transition matrix.


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Example 1 Solve the following matrix differential equation.

Suppose u(t) = 0 .

We first find

Therefore,

and

Hence for the unforced case ( u(t) = 0 ), we have by the variation of parameters formula

Therefore, the solution is given by

Example 2 Solve the following differential equation with a non-zero input.

This time let u(t) = sin2 t .


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We may use the J (t) which we found previously since no part of its derivation has changed.Hence, we have

Now using the variation of parameters formula given above, we can see that our system satisfiesthe equation:

Therefore, the solution is given by


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Example 3 For the shown spring-mass-damper system, find the position and velocity of the massif the initial position is 1 m, the initial velocity is zero and the external force f (t ) = 5 N.

The state variables are the position and velocity of the mass. Recall that the state spacerepresentation for this system is written as

We need to solve this equation in order to find the state variables. The variation of parametersformula will be used to solve the problem. We first calculate the state transition matrix

The variation of parameters formula gives


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Control Systems Module

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