kau1kor rtes pt09 embedded control systems engineering

8/3/2019 Kau1kor RTES PT09 Embedded Control Systems Engineering

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Name of the Student : Kaushik Kariappa D.VStudent Registration No : CHB0409015

Module Leader at MSRSAS : Dr. Chirag Jagadish

POSTGRADUATEENGINEERINGAN

DMANAGEMENTP

ROGRAMME(PEMP)

M.S.Ramaiah School of Advanced StudiesNew BEL Road, Gnanagangothri Campus, MSR Nagar, Bangalore-560 054Tel/Fax: 23605539/23601983; website: http://www.msrsas.org

Embedded Control Systems Engineering

Centre Name:Real Time Embedded SystemsCourse Name:M.Sc (Engg) in Real T ime Embedded Systems

PART TIME 2009 BATCH

ASSIGNMENT


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Declaration Sheet

Delegates Name Kaushik Kariappa D.V

Reg No CHB0409015

Course RTES Batch PT09

Module Code ESD 527

Module Title Embedded Control Systems Engineering

Module Start Date 14th November 2009 Submission Date 30th

January 2010

Module Leader Dr. Chirag JagadishSubmission ArrangementsThis assignment must be submitted to Academic Records Office (ARO) by the submission date before 1730 hours

for both Full-Time and Part-Time students.

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Up to one week late: Penalty of one grade (5 marks) One-Two weeks late: Penalty of two grades (10 marks) More than Two weeks late: Fail - 0% recorded (F2)All late assignments must be submitted to Academic Records Office (ARO). It is your responsibility to ensure that

the receipt of a late assignment is recorded in the ARO. If an extension was agreed, the authorization should besubmitted to ARO during the submission of assignment.

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Declaration

The assignment submitted herewith is a result of my own investigations and that I have conformed to the guidelines

against plagiarism as laid out in the PEMP Student Handbook. All sections of the text and results, which have been

obtained from other sources, are fully referenced. I understand that cheating and plagiarism constitute a breach of

University regulations and will be dealt with accordingly.

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DelegateDate

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Contents

Title Page i

Declaration sheet ii

Contents iii

Abstract iv

List of Tables vList of Figures vi

List of Symbols vii

Brief Introduction to control systems 1

1 CHAPTER -1

1.1 Introduction 21.2 Digital implementation of analog PID controller 21.3 Advantages of Analog controller design 41.4 Drawbacks of Analog controller design 41.5 Design and implementation in discrete domain 41.6 Advantages of Digital controller design 51.7 Drawbacks of Digital controller design 51.8 Conclusion between analog and digital PID controller Algorithms 51.9 Comparison of Performance of Discrete and Analog PID Controllers 6

2 CHAPTER -22.1 Introduction 72.2 Transfer Function of the passenger aircraft for open loop system 82.3 Transfer Function of the passenger aircraft for closed loop system 132.4 Discrete Model of the System 152.5 Design of PI controller 17

3 CHAPTER -33.1 Introduction 223.2 Design Driver of loudspeaker 22

3.2.1 Voice coil and magnet 223.2.2 Diaphragm 233.2.3 Surround 233.2.4 Spider 233.2.5 Basket 233.2.6 Dust Cap and Air Vent 23

3.3 Free body diagram of the loudspeaker driver 233.4 Voltage and force relations 24

3.4.1 Lorentz force 243.4.2 Back Emf 25


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3.5 Electrical model of the loudspeaker 253.6 Mechanical Governing Equation 253.7 Transfer Function 263.8 Bode Plot 27

3.8.1 The observations made from bode plot are 273.9 Impulse Response 28

3.9.1 Following observation is made from the impulse response plot 283.10State Space Model 293.11Bode Plot using State Space model 303.12Impulse response using state space model 313.13Simulink Mathematical model 32

3.13.1 sinusoidal input with frequency 350Hz and peak amplitude 1 V 323.13.2 Output response with input frequency 350Hz and input peak amplitude 1 V 333.13.3 Output response with input frequency 756Hz and input peak amplitude 1V 343.13.4 Output response with input frequency 10KHz and input peak amplitude 1V 343.13.5 Output response with input frequency 100KHz and input peak amplitude 1V 34

3.14Comparison of results of Simulink and Bode plot transfer function 36

4 References5 Appendix

5.1.1 Appendix-A Matlab code of aircraft model system 385.1.2 Appendix-B Matlab code of loudspeaker model system 39


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Abstract________________________________________________________________________________

This report presents an overview with examples for the design of embedded control systems

involving an Aircraft model system and then a Loudspeaker electromechanical system and explains

the strategies involved in designing these systems. Chapter 1 of this assignment deals with a digital

controller, designing in analog form and then converting to its discrete equivalent compared to a

controller converted to discrete first and then design of a digital controller. The design of an aircraft

model in chapter 2 for stable takeoff involves a MIMO model being converted to SISO. The open

loop analysis for the system is done for stability analysis. System behaviour is analysed for step

input with unity feedback and root locus and frequency analysis is carried. The above open loop

system is converted to its discrete equivalent with root locus plot and analysing frequency response.

A PI controller is also designed and analysis on system stability is done. Chapter 3 consists of an

electromechanical loudspeaker system whose schematic diagram is provided. Circuit diagram of the

electrical portion and mechanical portion governing equations were derived. Bode plot analysis is

made on the impulse response of the system. Impulse response is studied with State space model.

Implementation and design of digital controllers are made using Matlab and simulink tools. After

completion of this module, one will be able to determine the stability of the system, analyze its

performance, and control the system using a P, PI, PID controller to achieve a certain specification

requirement. Finally we propose an inexpensive and efficient manner of control system design.


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List of Tables________________________________________________________________________________

Table 3.1 Comparison chart of result of Transfer function and simulink 44


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List of Figures________________________________________________________________________________

Figure 1.1 Feedback Control System Block DiagramFigure 2.1: Aircraft model

Figure 2.2: Root locus of open loop SISO system

Figure 2.3: Open loop simulink modelFigure 2.4: Step response of Open loop system using simulink

Figure 2.5: Impulse response of Open loop system using simulink

Figure 2.6: Step response of Open loop system using matlab

Figure 2.7: Impulse response of Open loop system using matlab

Figure 2.8: closed loop simulink model with unity negative feedback

Figure 2.9: Step response of closed loop negative unity feedback system

Figure 2.10: Root Locus of closed loop system with negative unity feedback

Figure 2.11: Bode plot of closed loop system with negative feedback and unity gain

Figure 2.12: SISO model with zero order hold

Figure 2.13: Discrete open loop system simulink model

Figure 2.14: Step response of discrete open loop system

Figure 2.15: Root locus of discrete closed loop system

Figure 2.16: Discrete closed loop system simulink modelFigure 2.17: Step response of discrete closed loop system

Figure 3.1: Components of loud speaker driverFigure 3.2: Electrical and mechanical schematic of a loudspeaker driver

Figure 3.3: Electrical model of a loudspeakerFigure 3.4: Mechanical model of loud speaker driver.

Figure 3.5: Bode plot for the Transfer functionFigure 3.6: Impulse response for the Transfer function

Figure 3.7: Bode plot for the State Space model

Figure 3.8: Impulse response for the State Space model

Figure 3.9: Simulink mathematical model of loudspeaker driverFigure 3.10: Sinusoidal input with frequency 350Hz and peak amplitude 1 V

Figure 3.11: Output response with input frequency 350 Hz and input peak amplitude 1 V

Figure 3.12: Output response with input frequency 756Hz and input peak amplitude 1 V

Figure 3.13: Output response with input frequency 10 KHz and input peak amplitude 1 V

Figure 3.14: Output response with input frequency 100 KHz and input peak amplitude 1 V


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List of Symbols________________________________________________________________________________

SYMBOLS MEANING

fs Sampling frequency

Ts Sampling Time

Wn Natural frequency of the system

Magnetic field strength in Tesla

pi

Damping ratio of the system

Pitch angle of the airplane


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_______________________________________________________________________________

Brief Introduction to control systems

The study and design of automatic Control Systems, a field known as control engineering, 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 or a toilet, to

complex machines like space shuttles and power steering, control engineering is a part of our

everyday life. 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 an immediate example:

We have a particular electric motor that is supposed to turn at a rate of 40 RPM. To achieve this

speed, we must supply 10 Volts to the motor terminals. However, with 10 volts supplied to the

motor at rest, it takes 30 seconds for our motor to get up to speed. This is valuable time lost.

Now, we have a little bit of a problem that, while simplistic, can be a point of concern both to the

people designing this motor system, and to the people who might potentially buy it. It would seem

obvious that we should start the motor at a higher voltage, so that the motor accelerates faster, and

then we can reduce the supply back down to 10 volts once it reaches speed.

Now this is clearly a simplistic example, but it illustrates one important point: That we can add

special "Controller units" to pre-existing systems, to increase performance, and to meet new system

specifications. There are essentially two methods to approach the problem of designing a new

control system: the Classical Approach, and the Modern Approach.

Definition of the term "Control System", goes as, A Control System is a device, or a collection of

devices that manage the behaviour 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.


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CHAPTER 1

________________________________________________________________________________

1.1 Introduction

Discrete equivalents of analog controllers includes electronic controllers, which although discrete in

nature, implements control by emulating the continuous nature of analog control strategies. A typical

example is the electronic PI/PID algorithm. There are a number of ways by which this common and

versatile controller can be implemented in discretised form ultimately.

1) Design the controller for the continuous domain, analyze the system and then convert the

designed controller into its discrete equivalent.

2) Before designing convert the analog system to its discrete equivalent and then design the digital

controller followed with the analysis of the closed loop system

Figure 1.1 shows block diagram of a general control system with feedback.

Figure 1.1 Feedback Control System Block Diagram

1.2 Digital implementation of analog PID controller

The procedure for designing the controller is given as below

1) Controller for the analog system is designed to meet the desired design specifications.

2) Convert the analog controller to a digital controller by using a suitable transformation.


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3) The gain of the transfer function is tuned with proportional Z- Domain design to meet the

requirements

4) Make sure that the design specifications have been met by checking the responses of the digital

system

Analog controller mapped to a digital controller using the transformation should satisfy the

following requirements:

1) A stable analog controller must map to a stable digital controller.

2) The frequency response of the digital controller must closely resemble the analog controller

frequency response in the frequency range of 0 to fs/2 where fs being the sampling frequency.

For large change in the output for a given change in the error, for accelerating the movement of the

process towards set point and eliminates the residual steady-state error and to reduce the magnitude

of the overshoot produced by the integral component and improve the combined controller-process

stability a PID controller can be designed. Consider the ideal PID controller equation written in the

continuous time domain form:

(1)

To discretize the controller, we need to approximate the integral and the derivative terms to forms

suitable for computation by a computer. From a purely numerical point of view, we can use:

And (2)

The discretized PID algorithm is therefore:

(3)

which is now in the form of a difference equation, suitable for coding in an appropriate

programming language. This particular form of the PID algorithm is known as the 'positional' PID

controller, because the control signal is calculated with reference to a base level, uo.


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1.3 Advantages of Analog controller design

The advantage of such a design method is that the systems which need to be controlled can be easily

represented and visualized in continuous time domain and hence designing the controller in

continuous time domain becomes easy.

1.4 Drawbacks of Analog controller design

The designing involves approximation of design in continuous domain to discrete domain. This

approximation results in distortion in design. The approximation is valid only when the sampling

time is very small compared to system response time. The location of poles and zeros are often

restricted to subsets of unit circle. This restriction in poles and zeros limit designers ability to

reshape the system root locus. Two or more poles in S domain will be mapped to same point in Z

domain making the designing of controller difficult. Another complication in digital approximation

of analog controller is the need to have a pole at 0 in place of the pole at 1, as obtained by direct

digital transformation, to avoid an unbounded frequency response at the folding frequency. This may

result in a significant difference between the digital and analog controllers and may complicate the

design process considerably Once the design of the controller is converted from analog to digital re-

tuning of controller gain is required.

1.5 Design and implementation in discrete domain

The controller design can be done directly in Z-plane. Because of similarity in Z-domain and S-

domain root locus the methods used to design controller from S-domain root locus can be applied for

Z-domain root locus. We can also formulate discrete PID controllers directly from the Laplace

domain. Here, the ideal PID algorithm is written as:

(4)

Now we can apply either the backward difference or bilinear transformation methods to get an

equivalent discrete PID controller. Say we apply the backward difference method. Then,

(5)

Simplification yields:


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(6)

This PID controller is different in structure to that obtained from time-domain considerations and is

known as the 'velocity' PID algorithm.

As opposed to the fixed control reference used in the positional algorithm, here, the calculation of

current control uses the previous control value as reference. In essence, the control is calculated as a

change, hence the term 'velocity form'. Application of the Bilinear Transform will yield a similar

velocity form algorithm, due to the presence of the term in the approximation.

1.6 Advantages of Digital controller design

The advantage of such a design procedure is that there will not be any errors due to approximation ofmodels from one domain (Continuous) to another domain (Discrete), however the designing is not easy

as it is difficult to visualize a system which is inherently analog in nature as a digital system. However it

is possible to select any point within the unit circle in Z domain in controller design.

1.7 Drawbacks of Digital controller design

1) The stable region is now the unit circle in Z domain as opposed to the much larger left half of S-

Plane.

The selection of pole locations in z domain directly is very difficult compared to the selection ofPole in S domain. However the selection of pole is simplified by the use of Matlab tools.

2) No integrity check: It is very difficult to remove noise waves from the desired signals. Digital

control design eliminates this analog problem because value has status and error checking.

3) High power consumption: Analog designs need more power due to its inherent ampere consuming

devices. Digital controllers consume less power compared to its analog counter parts.

4) Range limited: For an analog transmitter the range is usually set much narrower than the

minimum and maximum sensor limits.

1.8 Conclusion between analog and digital PID controller Algorithms

Although the structures of the positional and velocity PID algorithms appear very different, they are

in fact related. The positional algorithm derived is obtained from equation (1). Time shifting back

one sampling interval, we obtain


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(7)

Subtracting this from the original, we end up with the velocity form, i.e.

(8)

1.9 Comparison of Performance of Discrete and Analog PID Controllers

1) Commissioning discrete PID ControllersTo calculate control, the analog PID requires knowledge ofuo, which is some steady-state control

output level. This may have implications during commissioning. With the digital form however, the

previous control can be set to any reasonable arbitrary level, and hence commissioning is simpler.

2) Integral Windup:A more important aspect is the use of a summation to calculate the contribution of the integral term.

This can lead to problems leading to a phenomenon known as integral windup, causing long periods

of overshoots in the controlled response. This can be caused by a poorly tuned controller or the

controller output is overly constrained. With the digital PID algorithm, because it does not make use

of a sum of errors to generate integral action, the problem of integral windup will not occur, even if

the control signal is constrained.

3) Controller tuning:All the tuning procedures for continuous time PID controllers apply. These range from the empirical

Ziegler-Nichol ultimate gain method, to the recipe based integral of error criteria. With recipe based

methods, a factor of 0.5Ts is usually added to the process dead-time to account for the delay caused

by the sampler.


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CHAPTER 2

________________________________________________________________________________

2.1 Introduction

There are three primary ways for an aircraft to change its orientation relative to the passing air. Pitch

(movement of the nose up or down),Roll (rotation around the longitudinal axis, that is, the axis

which runs along the length of the aircraft) and Yaw (movement of the nose to left or right.) Turning

the aircraft (change of heading) requires the aircraft firstly to roll to achieve an angle of bank; when

the desired change of heading has been accomplished the aircraft must again be rolled in the

opposite direction to reduce the angle of bank to zero. Pitch is controlled by the rear part of the tail

planes horizontal stabilizer being hinged to create an elevator. By moving the elevator control

backwards the pilot moves the elevator up (a position of negative camber) and the downwards force

on the horizontal tail is increased. The angle of attack on the wings increased so the nose is pitched

up and lift is generally increased. The elevator angle needs to be controlled for the aircraft pitch to

be within permissible limits. The state space mathematical model of a passenger aircraft is given by,

.q .

.

HLJ

IMK =

@0.313 56.7 0

@0.0139 @ 0.426 00 56.7 0

HLJ

IMK

q

HLJ

IMK+

0.2320.0203

0

HLJ

IMK [ ] (1)

Figure 2.1: Aircraft model


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2.2 Transfer Function of the passenger aircraft for open loop system

The state space model is represented as shown below

X = A X + B U

Y = C X + D U

Where X is state vector, U is input vector and Y is output vector.

Therefore from equation (1)

[email protected] 56.7 0

@0.0139 @ 0.426 00 56.7 0

HLJ

IMK B=

0.2320.0203

0

HLJ

IMK

The pitch angle is dependent on elevator angle . The pitch angle of the aircraft needs to be

controlled here. The output Y dependent on.

Y =

Since we are interested in only, the above equation can be re-written as

Y = [0 0 1]

q

HLJ

IMK+ [0] [ ] (2)

From equation (2)

C= [0 0 1] D = [0]

The transfer function of the above system can be obtained using matlab. The above operation is

performed by ss2tfcommand in matlab. The matlab commands used are mentioned below:

>> A=[ -0.313 56.7 0; -0.0139 -0.426 0; 0 56.7 0]

>> B=[0.232; 0.0203; 0]

>> C=[0 0 1]

>> D=0

>> [NUM, DEN]=ss2tf(A,B,C,D)

>> aircraft_pitch =tf(NUM,DEN)

The transfer function obtained by matlab is as given below

H(s) = -7.772e-016 s2

+ 1.151 s + 0.1774

------------------------------------------ (3)

s3 + 0.739 s2 + 0.9215 s

Here since the coefficients of s2 for

Zeros is very small (-7.772e-016). This coefficient can be

removed from the transfer function.

>> aircraft_pitch =tf([1.151 0.1774],[1 0.739 0.9215 0])


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I.e. H(s) = 1.151 s + 0.1774

------------------------------------------ (4)

s3+ 0.739 s

2+ 0.9215 s

The Zeros of the system can be obtained by equating the numerator of transfer function to Zero.

Likewise, the poles of the system can be obtained by equating the denominator of transfer function

to Zero.

s3+ 0.739 s

2+ 0.9215 s =0

Solving above equation the poles of the system are found to be at s=0, s = -0.3695 + 0.8860i and

s = -0.3695 - 0.8860i. This system consists of pair of complex conjugate poles in the left half of S-

Domain and a pole at the origin of S-Domain.

A system is stable if the poles of this transfer function lie strictly in the closed left half of the

complex plane (i.e. the real part of all the poles is less than zero). When the appropriate conditions

above are satisfied a system is said to be asymptotically stable. The variables of an asymptotically

stable control system always decrease from their initial value and do not show permanent

oscillations. Permanent oscillations appear when a pole has a real part exactly equal to zero (in the

continuous time case) or a modulus equal to one (in the discrete time case). If a simply stable system

response neither decays nor grows over time, and has no oscillations, it is marginally stable as in

this case. In this case the system transfer function has non-repeated poles at complex plane origin

(i.e. their real and complex component is zero in the continuous time case). Oscillations are present

when poles with real part equal to zero have an imaginary part not equal to zero.

Figure 2.2: Root locus of open loop SISO system


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The root locus of closed loop system can be obtained using the matlab command rlocus.

>> rlocus(aircraft_pitch)

The root locus of open loop system is as shown in figure 2.2

The simulink model of the system(step and impulse input) is as shown in figure 2.3

Figure 2.3: Open loop simulink model

For step response keep the initial value 0 and final value 1. In contrary for impulse response keep the

initial value 1 and final value 0.

For a step input of 1v the output appears to be increasing without bound, hence the system is

unstable. The output of the system for step input is as shown in figure 2.4

Figure 2.4: Step response of Open loop system using simulinkFor an impulse input, a constant output is seen after initial impulse response. The output of the

system for impulse input is as shown in figure 2.5


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Figure 2.5: Impulse response of Open loop system using simulink

The step response of the open loop system can also be obtained by using the matlab command step.

>> step (aircraft_pitch)

The impulse response of the open loop system can also be obtained by using the matlab command

impulse.

>> impulse(aircraft_pitch)

Figure 2.6: Step response of Open loop system using matlab


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Figure 2.7: Impulse response of Open loop system using matlab

Even though the system has a pair of complex conjugate poles in left half of S Plane, the pole at the

origin of S plane place a very significant in the instability of the system. Complex conjugate poles

are associated with an oscillatory response that decays exponentially for LHP poles and increases

exponentially for RHP poles. A pole at the origin of the s-plane implies an infinite gain at zero

frequency. The pole at origin affects the transient response causing instability and can be negated by

introducing zero close to the pole.

2.3 Transfer Function of the passenger aircraft for closed loop system

The formula for transfer function for the closed loop system with unity proportional controller is

given by:

H(s)closed = H(s)open

-------------------- (5)

1 + H(s) open

Where H(s) closed corresponds to transfer function of closed loop system with unity proportional

controller. H(s)open corresponds to transfer function of open loop system and is given by equation (4).

The transfer function of closed loop system with unity proportional controller can also be obtained

using the matlab function feedback

>> aircraft_closedloop_tf=feedback (aircraft_pitch, 1)

Transfer function of closed loop system with unity proportional controller is given by


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H(s)closed = 1.151 s + 0.1774

------------------------------------------ (6)

s3

+ 0.739 s2

+ 2.072 s + 0.1774

Equating the denominator of closed loop transfer function to Zero, the poles of the transfer function

can be found.

s3 + 0.739 s2 + 2.072 s + 0.1774 = 0

Poles of the closed loop system are situated at s= -0.0881, s = -0.3255 + 1.3816i and s = -0.3255 -

1.3816i.

The closed loop system rigged up in the simulink is as shown in figure 2.8

Figure 2.8: closed loop simulink model with unity negative feedback

For a step input of 1v the output of the closed loop unity negative feedback system is as shown in

figure 2.9.

Figure 2.9: Step response of closed loop negative unity feedback system

The system is stable as all the poles of the system lie on the left half of S Plane (Pole at origin is

shifted to Left half of S plane due to the feedback). From the step response it is clear that the system

is stable and has a settling time of around 36 seconds to reach 98% of maximum amplitude. The first


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peak is observed at 3.3sec and magnitude is around 0.9v. The second peak is observed at 8sec and a

magnitude of 0.82..

The root locus of closed loop system can be obtained using the matlab command rlocus.

>> rlocus(aircraft_closedloop_tf)

The root locus plot obtained is as shown in fig 2.10

Figure 2.10: Root Locus of closed loop system with negative unity feedback

From the root locus it is observed that the system is stable for any value of the proportional gain.

The frequency response of closed loop system i.e. bode plot can be obtained by using the matlab

command bode.

>>bode(aircraft_closedloop_tf)

The bode plot obtained is as shown in figure 2.11

The observations made from bode plot are listed below

a) The bandwidth of the closed loop system is around 0.0267 Hz.b) Maximum response is obtained at 0.214Hz and gain is 2.13db.c) The system is stable as absolute gain is less than one when absolute phase is greater than or

equal to 180 o.


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Figure 2.11: Bode plot of closed loop system with negative feedback and unity gain

2.4 Discrete Model of the System

The controller (PD, PI, and PID) which is used to control the system is implemented on a digital

microcontroller which operates at a certain frequency. In order to design and implement the

controller and also to understand the behavior, it is required to discretize the continuous time domainsystem and analyze the response of the discretized model to various inputs. The discrete model of

the system is represented by transfer function in Z-Domain.

Zero order hold is commonly used data reconstruction device. It can be mathematically described as

en(t) = e(nT), nT


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Figure 2.12: SISO model with zero order hold

Where G(s) = transfer function of the Zero-Order Hold and SISO model = (1 - -Ts) Gp(s) (8)

-----------------

s

Gp(s) = transfer function of SISO model represented by equation (4)

The output can be now represented as

C(s) = E*(s) G(s) (9)

Taking z transform the above equation is represented as

C(z) = E(z) G(z) (10)

The transfer function of the discrete model is given by

H(z) = C(z)/E(z) = G(z) (11)

The Z-transform of G(s) i.e. G (z) can be obtained using residue theorem or using matlab command

c2d with zoh as parameter for conversion.

>> c2d(aircraft_pitch,0.1,'zoh')

Output

Transfer function:

0.005641 z^2 - 2.268e-005 z - 0.005447

--------------------------------------

z^3 - 2.92 z^2 + 2.849 z - 0.9288

>> air_dis_num=[0.005641 -2.267e-005 -0.005447]

>> air_dis_den = [1 -2.92 2.849 -0.9288]

>> H= tf(air_dis_num,air_dis_den,0.1)

The transfer function of the discrete system is given by

Transfer function:

H(z) = 0.005641 z^2 - 2.267e-005 z - 0.005447 (12)

--------------------------------------


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z^3 - 2.92 z^2 + 2.849 z - 0.9288

The simulink model of the discrete system is as shown in figure 2.13

Figure 2.13: Discrete open loop system simulink model

The response of the system to a step input of 1 v is shown in figure 2.14

Figure 2.14: Step response of discrete open loop system

From the response it is clear that the system is stable however it have steady state error of around (1

0.8) = 0.2 = 20 %. This is unacceptable and should be reduced, this can be achieved by introducing

PI controller in feedback.

2.5 Design of PI controller

A PI controller is used to ensure that the system has a zero steady state error. This is achieved by

increases the type of the system by one (i.e. adding a pole). However the addition of a pole affects


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the transient response and might cause the system to become unstable. Hence the affect of this pole

must be negated by having a Zero very close to the pole (pole Zero cancellation).

D(z) = Ki z / (z -1) + Kp (13)

D(z) = [(Ki+Kp)z Kp] /[z-1] (14)

Where Ki corresponds to the gain of the integrator and Kp corresponds to the gain of the

proportional controller.

From equation (12), the transfer function of discrete open loop system is given by

H(z) = 0.005641 z^2 - 2.268e-005 z - 0.005447

------------------------------------------------

z^3 - 2.92 z^2 + 2.849 z - 0.9288

Representing the above equation in Zero-Pole form

H(z) = (z (0.0020 + 0.9827i)) (z (0.0020 - 0.9827i))

-------------------------------------------------------- (15)

(z-(0.9732 + 0.0829i)) (z-(0.9732 - 0.0829i)) (z 0.9736)

From the above equation it is clear that all poles lie with the region of stability i.e. unit circle in Z-

domain. Hence the system is stable.

The PI controller adds a pole at Zp = 1 and a zero at Zo = (Kp/ (Ki+Kp)). The overall gain K is

given by K=Ki+Kp

From equation (8) we find that there is already a pole at z = 0.97936. The poles at z=1 and z =

0.97936 can be negated by having a zero such that 1 > Zo > 0.96. As the zero approaches 1 the effect

of integrator reduces and the time taken to reach zero steady state error increases, However. When

zero is moved away from pole at 1 the time taken to reach zero steady state error decreases,

However the transient response appears to become unstable.

Assuming the zero to be at 0.96, the transfer function of controller is given by

D(z) = [K(z 0.96)]/[z-1] where K = (Ki + Kp) and 0.96 = Kp/(Ki + Kp) (16)


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The transfer function of the closed loop system with controller in forward path is given by

Hcl(z) = D(z) H(z)

--------------

1 + D(z) H(z)

The transfer function of closed loop system can be obtained using the matlab command feedback

>> aircraft_PI = tf( [1 -0.96],[1 -1],0.1)

>> aircraft_closed_H = feedback(H* aircraft_PI,1)

The closed loop transfer function obtained is

Transfer function:

Hcl(z) = 0.005641 z^3 - 0.005438 z^2 - 0.005425 z + 0.005229

---------------------------------------------------

z^4 - 3.914 z^3 + 5.764 z^2 - 3.783 z + 0.934

The root locus of the closed loop system is as shown in figure 2.15.

Figure 2.15: Root locus of discrete closed loop system

From the root locus it is clear that the system is stable only for a small range of gain. This range is as

given below

3 > K > 0.01

3 > (Ki + Kp) > 0.01


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But 0.96 = Kp/ (Ki + Kp)

3 > Kp/0.96 > 0.01

The value of K can be selected based on the required damping ratio and natural frequency.

Lets us assume that the controller needs to be designed for a damping ration of =0.00348 and wn =

2.23 rad/s.

Then the value of gain obtained from root locus for =0.078 and wn = 1.2 rad/s is 0.278

K = 0.909

(Ki + Kp) = 0.909

But 0.96 = Kp/ (Ki + Kp)

Kp = 0.87264

Ki = 0.03636

Hence the transfer function of controller is given by

D(z) = [0.909*(z -0.96)]/ [z-1]

The simulink model of the designed system is as shown below

Figure 2.16: Discrete closed loop system simulink model

The output of the model for a unit input of 1v is as shown below fig 20

From the figure it is clear that the steady state error has become zero and the system settling time is

around 45s which confirms to the theoretical value of Ts = 4/ wn = 42.73s.


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Figure 2.17: Step response of discrete closed loop system


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CHAPTER 3

________________________________________________________________________________

3.1 Introduction

A loudspeaker (or "speaker") is an electro acoustic transducer that converts an electrical signal into

sound. The speaker pulses in accordance with the variations of an electrical signal and causes sound

waves to propagate through a medium such as air or water. The main components in a loudspeaker

are voice coil, magnet, diaphragm, surround, spider basket, dust and air vent. When a time varying

current is applied to the voice coil, which is placed in a permanent magnetic field, Lorentz force is

produced. Due to this Lorentz force the cone and coil vibrate at the frequency of the applied

electrical signal. This is the principle behind generation of audio signals.

3.2 Design Driver of loudspeaker

Figure 3.1: Components of loud speaker driver

The various components of the loudspeaker are described below.

3.2.1 Voice coil and magnet


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The voice coil is a lightweight tube usually made of plastic or cardboard. A specified length of

copper or silver wire is wrapped tightly around the tube. The voice coil is suspended inside the

magnet. The voice coil is attached to the speaker diaphragm at one end. When a current is applied to

the voice coil, the resulting Lorentz force causes mechanical motion of the diaphragm.

3.2.2 Diaphragm

The diaphragm, or cone, is used to improve the acoustic efficiency. It is an extension of the voice

coil. The cone is made of materials such as paper, plastic, aluminum, magnesium, or titanium have

very different mechanical resonance properties, and each has a unique "sound". The cone introduces

mass, stiffness and mechanical damping.

3.2.3 Surround

The surround connects the diaphragm to the frame, or basket, of the loudspeaker. It is a compliant

material, usually rubber or foam, which allows the cone to move a specified displacement. The

surround adds additional mechanical damping to the cone.

3.2.4 Spider

The spider is a compliant damping material connected to the diaphragm just above the voice coil. It

adds stiffness and damping to the diaphragm.

3.2.5 Basket

The basket is the frame, usually constructed of metal. It holds the diaphragm onto the magnet. The

design objective is to keep the mechanical resonance of the basket outside the bandwidth the speaker

drives.

3.2.6 Dust Cap and Air Vent

The dust cap is an inverted dome placed over the voice coil to keep the environment out of the

magnet chamber. To improve sound dispersion at high frequencies, cone-shaped dust caps, known as

"phase plugs" are employed. The air vent is located on the bottom of the magnet. It cools the voice

coil by allowing airflow in and out of the magnet chamber.

3.3 Free body diagram of the loudspeaker driver

The schematic diagram of the loudspeaker, indicating the electrical and mechanical portions is

shown in Fig 3.2


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Figure 3.2: Electrical and mechanical schematic of a loudspeaker driver

The loudspeaker driver is an consists of both an electrical portion and mechanical portion

a) The electrical portion consists of a voltage source and a coil of wire in a magnetic field. The coil

has a resistance,R and inductance,L. The motion of the coil in the magnetic field generates a back

electromotive force (back emf), which needs to be taken into consideration.

b) The mechanical portion consists of the spider and surround. They are rigidly attached to the

basket in a manner that permits movement of the cone only in the axial direction. The mechanical

suspension can be modeled as a nonlinear spring and a linear damper connected in parallel. Themechanical system therefore consists of the mass, m of the speaker cone connected to the rest of the

speaker by a spring constant k and damper with coefficient b.

3.4Voltage and force relations

3.4.1 Lorentz force

When an electric current passes through the voice coil placed in a magnetic field, Lorentz force is

produced. Due to the Lorentz force, the cone and the voice coil move.

Let the Lorentz force be represented by

Let Radius of voice coil = r

Number of coil turns =N

Current through the coil = i


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Magnetic field strength =

Length of the coil = circumference * number of turns = 2rN

The Lorentz force is equal to the product of the length of the wire coil, magnetic field strength, and

the current through the coil.

So the equation for Lorentz force would be, = (2rN)i (1)

3.4.2 Back Emf

The movement of voice coil in the magnetic field generates a voltage called Back Emf. Let the

velocity with which the coil moves be denoted by v. The Back Emf is then equal to product of

magnetic field strength (), length of the coil (2rN) and the velocity of the coil (v). Let the Back

Emf be represented by = (2rN) v (2)

Since velocity is the rate of change of distance v=dx/dt (where x is the displacement of the coil)

= (2rN) (dx/dt) (3)

3.5 Electrical model of the loudspeaker

The circuit diagram of the electrical portion is as shown below in fig 3.3.The electric part of the

loudspeaker driver would consists of the input voltage source, coil resistance (R), coil inductance (L)

and the back Emf created due to movement of voice coil.

Figure 3.3: Electrical model of a loudspeaker

From Kirchhoffs voltage law, the sum total of all voltage drops in a closed loop electrical circuit is

zero. Hence,

-ei + iR+ + L(di /dt)

Substituting for from equation (3) we get,

ei = iR+ (2rN) dx/dt)+ L(di /dt)

3.6 Mechanical Governing Equation


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The mechanical system consists of mass, spring, damper and Lorentz force that acts on the voice

coil. The Lorentz force and Back Emf provide the link between electric and mechanical parts of the

speaker.

From Newtons second law, the sum of all force acting on a body must be Zero.

Force applied = Mass * Acceleration + Spring Constant * Displacement of spring + Damping factor

* velocity. i.e. = mx + kx + bx (6)

The inertial force, spring force, and damping force act in opposition to the Lorentz force.

The Free body diagram is as shown below.

Figure 3.4: Mechanical model of loud speaker driver.

Substituting equation (1) in (6) we get, (2rN) .i = mx + kx + bx (7)Above equation represents the mechanical model.

3.7 Transfer Function

Transfer function of the speaker system refers to the ratio of Laplace transform of displacement

output to the Laplace transform of the voltage input, assuming all initial conditions to be zero.

Taking the Laplace transform of equation (5)

E(s) = RI(s) + (2rN) [sX(s) x(0)] + L[sI(s) i(0)] (8)

Assuming initial conditions to be zero i.e. i(0) = x(0) = 0.

E(s) = (R + Ls)I(s) + (2rN) sX(s) (9)

Taking the Laplace transform of equation (7)

(2rN) I(s) = m [ X(s)-sx(0)-x(0)] + b[sX(s) x(0)] + k X(s) (10)

Assuming Zero initial conditions i.e. x(0) = 0; x(0) = 0

(2rN) I(s) = m X(s)+b sX(s) + k X(s) (11)

I(s) = [m X(s)+b sX(s) + k X(s)] / (2rN) (12)

Substituting equation (12) in equation (9) we get,

E(s) = (R + Ls) [m X(s)+b sX(s) + k X(s)] / (2rN) + (2rN) sX(s) (13)

E(s)=X(s)[ (R+Ls) (m+bs+ k)/ (2rN) + (2rN) s] (14)

E(s) = X(s)[ (R+Ls) (m+bs+ k) + 2rNs] / (2rN) (15)


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Therefore transfer function is got by Laplace transform of displacement (x) to the Laplace transform

of input voltage (ei).

X(s) / E(s) = (2rN) / ((R+Ls) (ms2+bs+ k) + 2rN)s (16)

X(s) / E(s) = (2rN) / (mL + (mR + bL) + (bR + kL+2rN)) s + kR) (17)

Equation (17) represents the transfer function of the loudspeaker system.

The numerical values for the parameters are mentioned below:

Coil Resistance R = 5 ohms

Coil Impedance L = 0.05mH

Spring Constant k = 2e5N/m

Damper Co-efficient b = 50 Ns/m

Mass of the coil m = 4e-3kg

Radius of the coil r = 0.04 m

Number of turns in the coil N = 50

Magnetic field strength = 0.25 Tesla

Substituting the above parameter values in equation (17) we get

X(s)/E(s) = 3.1416 / (2*10+ 0.0225 + 269.869 s + 10)

The transfer function in Matlab is computed using tffunction.

>> Num = [ 3.1416];

>> Den = [ 2e-7 0.0225 269.869 1e6];

>> speaker_transf = tf (Num, Den);

Transfer function: 3.142

----------------------------------------------------- (18)

2e-007 s^3 + 0.0225 s^2 + 269.9 s + 1e006

3.8 Bode Plot

Bode plot gives Magnitude and phase versus the frequency. The Bode plot of the loudspeaker can be

obtained by using the Matlab bodecommand.

>> bode (speaker_transf)

The obtained Bode plot is as shown below in figure 3.5

3.8.1 The observations made from bode plot are:

a) It is observed that the high frequency response rolls off by -3b at approximately 756Hz.

b) Maximum response is obtained at Zero Hz and remains at the value till a frequency of

approximately 461 Hz is reached.


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c) From the magnitude plot, it is seen that the magnitude is 218dB at 1MHz, while it is 278 dB at

10MHz. The difference in magnitude is approximately 60dB/decade. Thus system is of the order 3.

d)The phase plot shows a drop from ~0 to ~270. This also indicates that this system is of third

order.

Figure 3.5: Bode plot for the Transfer function

3.9 Impulse Response

Impulse response is used to determine how the system behaves for an impulse.

Consider an impulse input Ei(t) governed by equation,

Ei(t) = 1 v for t=0 and 0 v for t>0 (19)

The Laplace transform of Ei(t) is Ei(s) = 1.

The matlab command used to obtain the impulse response is impulse function.

>> impulse(speaker_transf)

The impulse response obtained is as shown in figure 3.6

3.9.1 Following observation is made from the impulse response plot.

a) The maximum displacement of voice coil is x=8.62 mm for an impulse input of 1v.b) A delay of 0.15 ms is observed from the time when the impulse is applied to the system to

reach maximum displacement.

c) The setting time is around 0.95 ms


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Figure 3.6: Impulse response for the Transfer function

3.10 State Space Model

The state space model representation is as shown below

X = A X + B U

Y = C X + D U

Where X is state vector, U is input vector and Y is output vector.

In loudspeaker driver, the state variables are the current that flows through the voice coil i, and the

displacement of the voice coil due to Lorentz forcex and the rate of change of displacement is

velocity v.We know Rate of change of displacement is given by.

x = v (20)Substituting equation (20) in Equation (7) we get

(2rN) i = mv + kx + bv or (21)

v = [(2rN) i kx bv] / m (22)substituting (dx/dt) as x and (di/dt) as i in Equation (5) we get,

ei = iR+ (2rN) x+ Li (23)

Substituting equation (20) in the above equation (23) we get,ei = iR+ (2rN) v+ Li or (24)

i = [ei iR (2rN) v]/L (25)The output equation is Y = x (26)

The state vector X is defined as


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X =xiv

HJ

IK (27)

Equations (20), (22), (25) can be re-written in matrix form as shown below

x .i.v .

HLJ

IMK=

0 0 1

0 @RLfffff 2rN

` ab

Lfffffffffffffffffffffffffffff

@k

m

ffffff

@2rN

` ab

m

fffffffffffffffffffffffffffff

@b

m

ffffff

H

LLLLLLLJ

I

MMMMMMMK

xiv

HJ

IK+

01

L

ffff

0

HLLLLJIMMMMK ei t

` aB C(28)

Re-Writing Equation Y=x we get,

Y= 1 0 0@ A x

iv

HJ

IK+ 0@ A ei t` a

B C(29)

Therefore,

A=

0 0 1

0 @R

Lfffff 2rN

` ab

Lfffffffffffffffffffffffffffff

@k

m

ffffff

@2rN

` ab

m

fffffffffffffffffffffffffffff

@b

m

ffffff

HLLLLLLLJ

IMMMMMMMKB=

01

L

ffff

0

HLLLLJ

IMMMMK C= 1 0 0@ A

D= 0@ A

(30)

Matlab commands:

>> A= [0 0 1; 0 -R/L -(2rN)/L; -K/M (2rN)/M -b/M];

>>B= [0; 1/L ; 0];

>>C= [1 0 0];

>>D= [0];

>>speaker_transf_ss = ss(A,B,C,D)

3.11 Bode Plot using State Space model

The matlab command bode can be used to obtain the bode plot using the state space model.

>> bode (speaker_transf_ss)

The plot obtained is as shown in fig 3.7

The observations made in Bode plot for state space model are same as bode plot using transfer

function method.


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Figure 3.7: Bode plot for the State Space model

3.12 Impulse response using state space model

Figure 3.8: Impulse response for the State Space model

The Matlab command impulse can be used to obtain the impulse response using the state space

model.

The plot obtained is as shown in fig 3.8.

>> impulse (transfer_func_ss)

The observations made in impulse response plot for state space model are same as impulse response

plot using transfer function method.


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3.13 Simulink Mathematical model

From the state space mode we get the three governing equations: equation 20, equation 22 and

equation 25

Here since value of r, N and is fixed let (2rN) be a constant C; i.e. C=3.14

The simulink model representing the above equation is as given in figure 3.9:

Figure 3.9: Simulink mathematical model of loudspeaker driver

Using the above simulink mathematical model shown in figure 3.9, the response of the

electromechanical system to sinusoidal input voltages of different frequencies can be studied for

input signal of amplitude 1v. The resulting outputs can be compared with the bode plot obtained

using Transfer function/ state space model derived earlier. The bode plot obtained from transfer

function is as shown in figure 3.5. From the bode plot it is clear that till 350 Hz maximum gain is

observed. At a frequency of 756 Hz the gain reduces by 3db and this corresponds to bandwidth. At 5

KHz and 10 KHz, high frequency roll-off is observed. In order to compare the results of simulink

with the results obtained from transfer function, let us consider the two important input frequencies

350Hz, 756 Hz.

3.13.1 Sinusoidal input with frequency 350Hz and peak amplitude 1 V


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Figure 3.10: Sinusoidal input with frequency 350Hz and peak amplitude 1 V

3.13.2 Output response with input frequency 350Hz and input peak amplitude 1 V


It is clear from the output that the peak amplitude of the output is around 2.9*10-6. this

corresponds to a gain of 20log10[2.9*10-6] = -110.75db. Also it can be observed that the first cycle

of the output is completed in 3.13 ms, which corresponds to a time delay of 0.273 ms with respect to


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the input because 1/350s=2.857ms. i.e. phase lag = (0.273/2.857)*360 = 34.39. From the above both

the magnitude and phase lag is in concurrence with the result obtained from transfer function.

3.13.3 Output response with input frequency 756Hz and input peak amplitude 1V


It is clear from the output that the peak amplitude of the output is around 2.35*10-6. This

corresponds to a gain of 20log10[2.35*10-6] = -112.570db. Also it can be observed that the first

cycle of the output is completed in 1.469 ms, which corresponds to a time delay of 0.147 ms with

respect to the input. i.e. phase lag = (0.147/1.322)*360 = 69.6. From the above both the magnitude

and phase lag is in concurrence with the result obtained from transfer function.

3.13.4 Output response with input frequency 10KHz and input peak amplitude 1V

It is clear from the output that the peak amplitude of the output is around 3.35*10-8. This

corresponds to a gain of 20log10[3.35*10-8] = -149.49db. From the above both the magnitude and

phase lag is in concurrence with the result obtained from transfer function.


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Figure 3.13: Output response with input frequency 10KHz and input peak amplitude 1 V

3.13.5 Output response with input frequency 100KHz and input peak amplitude 1V

Figure 3.14: Output response with input frequency 100KHz and input peak amplitude 1 V

It is clear from the output that the peak amplitude of the output is around 1.4*10-10. this


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corresponds to a gain of 20log10[1.4*10-10] = -197.077db. From the above both the magnitude and

phase lag is in concurrence with the result obtained from transfer function.

3.14 Comparison of results of Simulink and Bode plot transfer function

Frequency in radian

(rad/sec)

Input Frequency

(HZ)

Result from TF / SS

model (Bode Plot)

Result from

SIMULINK

2198 350 Gain is -110 dBOutput lags input by

34 degrees

Gain is -110.75 dB

Output lags input by

34.39 degrees

4750 756 Gain is -113 dBOutput lags input by

66.8 degrees

Gain is -112.570 dB

Output lags input by

69.6 degrees

62800 10K Gain is -150 dB Gain is -149 dB628000 100K Gain is -205 dB Gain is -197 dB

Table 3.1 Comparison chart of result of Transfer function and simulink


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References

________________________________________________________________________________

http://lorien.ncl.ac.uk/ming/digicont/digimath/dpid1.htm

http://en.wikipedia.org/wiki/Control_theory

http://en.wikipedia.org/wiki/Control_Systems

http://en.wikibooks.org/wiki/Control_Systems/Introduction

http://www.tpub.com/content/doe/h1013v2/css/h1013v2_117.htm

http://en.wikipedia.org/wiki/PID_controller#Proportional_term

http://books.google.co.in/books?id=s0k9kGs5bHYC&pg=PA553&lpg=PA553&dq=design+continu

ous+time+controller+and+then+convert+it+discrete+equivalent+or+design+discrete+controller&source=bl&ots=kBbypkRWIH&sig=C1ox5rdkZHr9DmUAzyOlUZatbOo&hl=en&ei=LcRhS9Ef16CR

BeyorOcL&sa=X&oi=book_result&ct=result&resnum=8&ved=0CC0Q6AEwBw#v=onepage&q=d

esign%20continuous%20time%20controller%20and%20then%20convert%20it%20discrete%20equi

valent%20or%20design%20discrete%20controller&f=false

Benjamin C Kuo, Automatic Control Systems, Princeton Hall Co., Seventh Edition,

2003.


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Appendix-A

________________________________________________________________________________

Matlab code of chapter 2- Aeroplane model

A=[ -0.313 56.7 0; -0.0139 -0.426 0; 0 56.7 0]B=[0.232;0.0203;0]

C=[0 0 1]

D=0

[NUM,DEN]=ss2tf(A,B,C,D)

aircraft_pitch =tf(NUM,DEN)

aircraft_pitch =tf([1.151 0.1774],[1 0.739 0.9215 0])

rlocus(aircraft_pitch)

step (aircraft_pitch)

impulse(aircraft_pitch)

aircraft_closedloop_tf=feedback (aircraft_pitch, 1)

rlocus(aircraft_closedloop_tf)bode(aircraft_closedloop_tf)

c2d(aircraft_pitch,0.1,'zoh')

air_dis_num=[0.005641 -2.267e-005 -0.005447]air_dis_den = [1 -2.92 2.849 -0.9288]

H= tf(air_dis_num,air_dis_den,0.1)aircraft_PI = tf( [1 -0.96],[1 -1],0.1)

aircraft_closed_H = feedback(H* aircraft_PI,1)


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Appendix-B

________________________________________________________________________________

Matlab code of chapter 3- Loudspeaker model

Num = [ 3.1416];

Den = [ 2e-7 0.0225 269.869 1e6];

speaker_transf = tf (Num, Den);

bode (speaker_transf)

impulse(speaker_transf)

A= [0 0 1; 0 -R/L -(2rN)/L; -K/M (2rN)/M -b/M];B= [0; 1/L ; 0];

C= [1 0 0];D= [0];

speaker_transf_ss = ss(A,B,C,D)bode (speaker_transf_ss)

impulse (transfer_func_ss)

kau1kor rtes pt09 embedded control systems engineering

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