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Studies in Systems, Decision and Control 16
Edward LayerKrzysztof Tomczyk
Signal Transforms in Dynamic Measurements
Studies in Systems, Decision and Control
Volume 16
Series editor
Janusz Kacprzyk, Polish Academy of Sciences, Warsaw, Polande-mail: [email protected]
About this Series
The series “Studies in Systems, Decision and Control” (SSDC) covers both newdevelopments and advances, as well as the state of the art, in the various areas ofbroadly perceived systems, decision-making and control-quickly, up to date and witha high quality. The intent is to cover the theory, applications, and perspectives on thestate of the art and future developments relevant to systems, decision-making, con-trol, complex processes and related areas, as embedded in the fields of engineering,computer science, physics, economics, social and life sciences, as well as the para-digms and methodologies behind them. The series contains monographs, textbooks,lecture notes and edited volumes in systems, decision making and control spanningthe areas of Cyber-Physical Systems, Autonomous Systems, Sensor Networks,Control Systems, Energy Systems, Automotive Systems, Biological Systems,Vehicular Networking and Connected Vehicles, Aerospace Systems, Automation,Manufacturing, Smart Grids, Nonlinear Systems, Power Systems, Robotics, SocialSystems, Economic Systems, and other. Of particular value to both the contributorsand the readership are the short publication time frame and the world-wide distri-bution and exposure which enable both a wide and rapid dissemination of researchoutput.
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Edward Layer • Krzysztof Tomczyk
Signal Transformsin Dynamic Measurements
123
Edward LayerFaculty of Electrical and ComputerEngineering
Cracow University of TechnologyCracowPoland
Krzysztof TomczykFaculty of Electrical and ComputerEngineering
Cracow University of TechnologyCracowPoland
ISSN 2198-4182 ISSN 2198-4190 (electronic)Studies in Systems, Decision and ControlISBN 978-3-319-13208-2 ISBN 978-3-319-13209-9 (eBook)DOI 10.1007/978-3-319-13209-9
Library of Congress Control Number: 2014955797
Springer Cham Heidelberg New York Dordrecht London© Springer International Publishing Switzerland 2015This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar ordissimilar methodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exemptfrom the relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material containedherein or for any errors or omissions that may have been made.
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Preface
The essence of every measurement is the determination of an unknown quantity,being a signal of various kinds generated by physical objects, the properties ofwhich are the subject matter of our research. For this purpose, we use differentsystems and measurement procedures, enabling the determination of that quantitywith the least error possible. The world surrounding us is analog, thus for obviousreasons measured signals also have an analogous and continuous form. For mea-surement purposes, these signals are usually converted into voltages and then, bymeans of a data acquisition card in cooperation with a computer, processed intodigital form. Measurements are taken mainly for purposes of documentation,control, and optimization, as well as for cognitive reasons, comprising widelyunderstood identification.
Depending on the purpose and destination of a measurement result, in eachof the cases above, one deals with a different set of input and output signals, whichconstitute the excitation and response of the object of interest, respectively.
The set of input signals may contain deterministic, periodic, and nonperiodicsignals, as well as random, stationary, and nonstationary ones. Within periodicdeterministic signals, one can include sinusoidal signals, as well as complex peri-odic ones, whereas in the case of nonperiod deterministic signals, they are almostperiodic and transient ones. The latter includes so-called standard pulse signals,often applied in measurement techniques. They are most often used for the purposeof object identification, both on the basis of the knowledge of dynamic properties ofthose objects and the measurement of the response signal.
In many cases, input signals are dynamically changing nonstationary signalswith characteristics that cannot be foreseen a priori. These are signals that are non-repetitive and occur most often in the reality surrounding us. In geology, forexample, they are earthquake vibrations; in meteorology—wind intensity anddirection in case of sudden weather changes; in mechanics—stroke of force,pressure, and moments; in electrical engineering—surges of voltages in powersystems; in biology and medicine—bioelectric ECG, EEG, and EMG signals, aswell as bioacoustic or biomagnetic signals, etc.
v
A substantial variety of phenomena in the research areas listed above, having inmind the relatively limited amount of measurement methods and ways of signalprocessing, requires the development of measurement systems that are character-ized by being pretty versatile. Unfortunately, such versatility of systems automat-ically generates many problems concerning their matching to the measured signals.Among the main factors that are subject to matching are:
• solutions used in dedicated measurement systems,• processing algorithms,• signal variability range,• system errors,• system sensitivity to external disturbances.
Solving the problem of versatility thus requires the development of ever morecomplex and computerized measurement systems, which invariably causes anincrease in their production costs. Unfortunately, these costs increase in a definitelynonlinear manner; additionally, very sophisticated equipment can only be operatedby well-trained specialists, which additionally increases the cost of measurement inexperimental studies of every kind. Model studies are an alternative to experimentalstudies which, although less reliable and less accurate, are generally much cheaper.For this purpose, we use more or less accurate mathematical models of a givenobject, obtained in the identification process, models of which describe theirbehavior in the definite moment of study and assigned time range. If the modelobtained fails to meet our requirements, we verify its adequacy, and when this is notsatisfactory, we repeat the identification procedure for another, more complexstructure of the model and estimate its new parameters, etc. Finding an optimalstructure of the model may be difficult in many cases, as its parameters do not havedirect physical interpretation, but are only a reflection of the conformity of suchmodel with experimental data. Additionally, difficulties may arise with correctestimation of model parameters, as the data used for this purpose are usuallyburdened with errors of various kinds. Despite the above difficulties, in a situationwhere computers of a higher class are commonly available, while existing softwareis continuously updated and completely new software developed, the use of modelsof various objects is gaining in popularity. This is caused mainly by their commonfeature, namely the expectation that such models will meet the prediction justifi-cation principle. This means that on the basis of their analysis, it will be possibleboth to predict the phenomena that may occur in the object modeled, as well aspredict their future responses to various external influences.
Signals are described by functions usually of time or frequency. The parametersof these functions result from certain mathematical relations and are scalar quan-tities. For example, one may include amplitude, mean value, rms value, shapefactor, peak factor, filling factor, standard deviation, etc. Functions of time orfrequency describe single signals or their mutual relations. These are, among others,autocorrelation function, cross-correlation function, distribution functions, andspectral characteristics.
vi Preface
In measurements, specific properties of signals are often used, which refer totheir orthogonal or orthonormal features. Properties of such type are used, amongothers, in digital measurements of electrical quantities, as well as for example, in thesynthesis of optimal mathematical models.
In the engineering practice, one often deals with the necessity of performingcertain mathematical operations on signals, among which the most frequent areconvolution transforms, Laplace, Fourier, Hilbert, wavelet transforms, and Ztransform. Convolution is most often applied for the determination of the outputsignal, knowing the form of the input and kernel of the object. The Laplacetransform may be of use in solving linear differential equations, state equations indetermining the exp(At) series, presenting models in the form of transfer function,solving transient states and checking stability, as well as the simplification of ordermathematical models. Similar to the former, and equally frequently used, theFourier transform is applied in the frequency analysis of signals. The Hilberttransform enables easy creation of analytic signals, commonly used in signalmodulation theory. Wavelet analysis of the signals, similar to Short Time FourierTransform, enables their decomposition, which is useful in such cases, for which inaddition to information about the frequency spectrum of the signal, informationabout their location over time is also needed. The Z transform is used for solvinglinear difference equations, analyzing linear systems with discrete data, and fordesigning digital filters.
Besides the transforms listed above, properties of signals are often described bymeans of systems of equations containing variables, their derivatives, and integrals.For systems with one variable, their differentiation allows elimination of integralsand differential equations to be obtained due to just one independent variable.Often, time is this variable, in which case it is convenient to present the equations inthe form of the state of equations, which is very popular, especially in technicaldomains. In engineering practice, linear equations are used, as in the great majorityof cases the linearity of the modeled objects is assumed or, alternatively, theassumption made is that the nonlinearities occurring in them are minor enough to beapproximated by means of linear equations. Such approximation is justified inmany practical cases, and the accuracy of the description is sufficient.
The measured signals most often are disturbed, in the majority of cases, by anadditive disturbance. The problem of disturbance reduction is an important partof the measurement process, and is particularly important in mathematical pro-cessing of signals, mainly differentiation, which causes amplification of such dis-turbance. Various methods of disturbance reduction are applied, among themKalman filtration and the time window method seem to be particularly attractive.The time window method is characterized by moving the differentiation operationto the window, and hence disturbances are not amplified.
Measurement data are obtained from experiments carried out using computer-aided measurement systems, and then processed using software dedicated for thepurpose. In such case, the computer must be equipped with a data acquisition card,which is a basic part of such a system. For measurement data processing, we usevarious procedures and mathematical operations realized by measurement and
Preface vii
control programs as well as software for numerical computation. The problem liesin the fact that the available and commonly used mathematical software is dedicatedto solving problems for continuous time variables, whereas the measurement dataare received at sampling moments, and have a discrete form. Hence, direct use ofsuch software is not possible. The development of special dedicated algorithms isthus required, which enable their application for discrete data. Several examples ofsolutions of such type can be found in the book.
The book is divided into ten chapters of which, following the introduction,Chap. 1 is devoted to classification and parameters of signals, Chaps. 2 and 3 toLaplace and Fourier transforms, Chap. 4 to the Z transform, and Chaps. 5 and 6 towavelet and Hilbert transforms, respectively. Besides the theoretical foundations,each of these chapters contains several examples of practical applications of thetransforms discussed. Chapter 7 discusses orthogonal signals and their applicationin the measurement of electrical quantities, while Chap. 8 is devoted to problems ofanalog and digital modulation. The two final Chaps. 9 and 10 discuss problemsconcerning convolutions and deconvolutions, as well as disturbance and itsreduction.
The authors hope that the book may be of interest to a wide group of engineersand specialists dealing with problems of measurement and signal processing, aswell as to students of various engineering disciplines.
viii Preface
Contents
1 Classification and Parameters of Signals . . . . . . . . . . . . . . . . . . . 11.1 Characteristics of Deterministic Signals . . . . . . . . . . . . . . . . . 11.2 Characteristics of Random Signals . . . . . . . . . . . . . . . . . . . . 31.3 Parameters of Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Examples of Analog Signals with Limited Energy. . . . . . . . . . 61.5 Examples of Analog Signals with Limited Power . . . . . . . . . . 101.6 Examples of Distributive Signals . . . . . . . . . . . . . . . . . . . . . 121.7 Discrete Signals with Limited Energy . . . . . . . . . . . . . . . . . . 131.8 Discrete Signals with Limited Power . . . . . . . . . . . . . . . . . . . 151.9 Examples of Analog Signals in MathCad . . . . . . . . . . . . . . . . 161.10 Examples of Discrete Signals in MathCad . . . . . . . . . . . . . . . 18
2 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1 Initial and Final Value. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Surface and Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3 Examples of Laplace Transforms . . . . . . . . . . . . . . . . . . . . . 242.4 Properties of Laplace Transform . . . . . . . . . . . . . . . . . . . . . . 252.5 Laplace Transform in Solving Differential Equation . . . . . . . . 262.6 Laplace Transform in Solving State Equation . . . . . . . . . . . . . 292.7 Simplification of Model Order . . . . . . . . . . . . . . . . . . . . . . . 342.8 Discretization of State Equation . . . . . . . . . . . . . . . . . . . . . . 362.9 Example in MathCad. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1 Continuous Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 433.2 Properties of Fourier Transform . . . . . . . . . . . . . . . . . . . . . . 453.3 Example of Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . 463.4 Discrete Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.4.1 Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 523.5 Short-time Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 553.6 Time Windows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
ix
3.7 Properties of Time Windows . . . . . . . . . . . . . . . . . . . . . . . . 573.8 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.9 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4 Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.1 Properties of Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 854.2 Determination of Z Transform . . . . . . . . . . . . . . . . . . . . . . . 864.3 Changing Sampling Interval . . . . . . . . . . . . . . . . . . . . . . . . . 894.4 Inverse Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.5 Digital Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.6 Example in MathCad. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5 Wavelet Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.1 Continuous Wavelet Transform. . . . . . . . . . . . . . . . . . . . . . . 975.2 Wavelet Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985.3 Discrete Wavelet Transform . . . . . . . . . . . . . . . . . . . . . . . . . 1005.4 Discrete Wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.5 Example of Three-Stage Wavelet Transform in LabVIEW . . . . 105
6 Hilbert Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.1 Examples of Hilbert Transform. . . . . . . . . . . . . . . . . . . . . . . 1106.2 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
7 Orthogonal Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177.1 Orthonormal Polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.2 Digital Measurement of Electrical Quantities . . . . . . . . . . . . . 124
7.2.1 Measurement of Active Power . . . . . . . . . . . . . . . . . . 1267.2.2 Measurement of Reactive Power . . . . . . . . . . . . . . . . . 1277.2.3 Digital Form of Current, Voltage, and Power . . . . . . . . 127
7.3 Measurement of Frequency . . . . . . . . . . . . . . . . . . . . . . . . . 1287.4 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.5 Examples in LabVIEW . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
8 Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.1 Analog Modulations (AM) . . . . . . . . . . . . . . . . . . . . . . . . . . 142
8.1.1 Double-Sideband Large Carrier Modulation(DSBLC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
8.1.2 Double Sideband with SuppressedCarrier Modulation (DSBSC) . . . . . . . . . . . . . . . . . . . 146
8.1.3 Single-Sideband (SSB) . . . . . . . . . . . . . . . . . . . . . . . 1478.1.4 Single Sideband with Suppressed Carrier (SSBSC)
Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.1.5 Vestigial Sideband (VSB) Modulation . . . . . . . . . . . . . 148
x Contents
8.2 Angle Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1498.2.1 Phase Modulation (PM) . . . . . . . . . . . . . . . . . . . . . . . 1508.2.2 Frequency Modulation (FM) . . . . . . . . . . . . . . . . . . . 151
8.3 Impulse Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1518.3.1 Pulse Width Modulation (PWM). . . . . . . . . . . . . . . . . 1518.3.2 Pulse Amplitude Modulation (PAM) . . . . . . . . . . . . . . 1528.3.3 PAM with Ideal Sampling . . . . . . . . . . . . . . . . . . . . . 1538.3.4 PAM with Real Sampling . . . . . . . . . . . . . . . . . . . . . 1538.3.5 PAM with Instantaneous Sampling . . . . . . . . . . . . . . . 1558.3.6 Pulse Duration Modulation (PDM) . . . . . . . . . . . . . . . 1568.3.7 Pulse Position Modulation (PPM) . . . . . . . . . . . . . . . . 1568.3.8 Pulse Code Modulation (PCM). . . . . . . . . . . . . . . . . . 1568.3.9 Differential Pulse Code Modulation (DPCM) . . . . . . . . 158
8.4 Digital Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1588.4.1 Modulation with Amplitude Shift Keying (ASK) . . . . . 1588.4.2 Modulation with Frequency Shift Keying (FSK) . . . . . . 1588.4.3 Phase Shift Keying (PSK) Modulation. . . . . . . . . . . . . 1598.4.4 Quadrature Amplitude Modulation (QAM). . . . . . . . . . 162
8.5 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
9 Convolution and Deconvolution . . . . . . . . . . . . . . . . . . . . . . . . . . 1699.1 Analog and Digital Convolution . . . . . . . . . . . . . . . . . . . . . . 1699.2 Properties of Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . 1709.3 Continuous and Digital Deconvolution . . . . . . . . . . . . . . . . . 1769.4 Deconvolution for Low-Pass System . . . . . . . . . . . . . . . . . . . 1789.5 Conjugate Operator and Maximum Integral Square
Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1799.6 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
10 Reduction of Signal Disturbance . . . . . . . . . . . . . . . . . . . . . . . . . 18910.1 Time Windows in Reduction of Disturbance . . . . . . . . . . . . . 18910.2 Signal Reconstruction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.3 Kalman Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.4 Examples in MathCad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19610.5 Kalman Filter in LabVIEW . . . . . . . . . . . . . . . . . . . . . . . . . 202
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
Contents xi
Chapter 1Classification and Parameters of Signals
We encounter signals in many fields of science, in particular in experimentalsciences, which deal with examination of the reality that surrounds us. The infor-mation carried by the signals enables description and analysis of that reality, if oneknows the mathematical relations concerning them. On the one hand, those relationsshould be general enough, to comprise the wide class of physical signals, while onthe other hand, they should enable easy analysis of the reality represented. Moreover,they should reflect certain characteristic properties, common for a given class ofsignals, and differentiate them from others. Deterministic signals are those that arerepeatable; that is, the measure of their value in a given time interval, determined in agiven moment, may be repeated at any moment at a later date. Such signals may bedescribed by means of strict mathematical relations, which can be real or complexfunctions of time. If signals do not repeat their values later, they are classified asrandom signals, which cannot be described by means of exact mathematical rela-tions, due to their uniqueness. The classification of deterministic signals is presentedin Fig. 1.1, while that of random signals is presented in Fig. 1.2.
1.1 Characteristics of Deterministic Signals
Periodic signals fulfill the condition
xðtÞ ¼ xðt � nTÞ for n ¼ �1; 2; 3; . . . ð1:1Þ
where A—amplitude and T—period.Monoharmonic signals consist of a single harmonic. They are
xðtÞ ¼ A sinðxt þ /Þ ð1:2Þ
where / is the initial phase.Polyharmonic signals have at least two harmonics, which have different
amplitudes and initial phases. An example of a polyharmonic signal is the sum oftwo harmonics.
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_1
1
Polyharmonic signals meet the following condition
xðtÞ ¼X1n¼1
An sinðxnt þ /nÞ ð1:3Þ
Polyharmonic signals become periodic signals, if all the frequencies contained inthem are integral multiples of the basic frequency.
Non-periodic signals are signals which fail to meet the condition of periodicity.Almost periodic signals are generated by summing two or more harmonic sig-
nals, for which the quotient of all possible pairs of frequencies is expressed by anirrational number. These signals are not periodic, despite the fact that they con-stitute periodic signals.
Transient signals may be described by means of time functions that are neitherperiodic nor almost periodic. They do not have a discrete spectrum, but a contin-uous spectrum, determined by a Fourier transform—Eq. (3.2).
Fig. 1.1 Classification of deterministic signals
Fig. 1.2 Classification of random signals
2 1 Classification and Parameters of Signals
1.2 Characteristics of Random Signals
Random signals cannot be presented by means of mathematical functions, becauseit is not possible to predict their values on the basis of previous values. Statisticalparameters are applied for the description of such signals, i.e., probability distri-bution, expected value, and variance.
Statistical parameters of stationary signals do not change in time, whereas suchchanges are possible in the case of non-stationary signals.
For ergodic signals, it is possible to determine statistical parameters, which is notpossible in the case of non-ergodic signals, even with long observation times.
1.3 Parameters of Signals
• Deterministic SignalsMean value
�x ¼ 1T
Zt0þT
t0
xðtÞdt
�x ¼ 1N
Xn 0þðN�1Þ
n¼n0
x½n�ð1:4Þ
where xðtÞ—signal, t0—initial time, x½n�—discrete signal, N—number ofsamples, and n0—initial sample.Mean value in the set interval
�x ¼ 1tn � t0
Ztnt0
xðtÞdt
�x ¼ 1nn � n0
Xnnn0
x½n�ð1:5Þ
RMS value
xrms ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1T
Ztnt0
x2ðtÞdt
vuuut
xrms ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1N
Xnnn0
x2½n�vuut
ð1:6Þ
1.2 Characteristics of Random Signals 3
Peak value
xpeak ¼ maxt0\t� t0þT
xðtÞj jxpeak ¼ max
n0\n� n0þðN�1Þx½n�j j ð1:7Þ
Shape factor
ks ¼ xrms
�xð1:8Þ
Peak factor
kp ¼ xpeakxrms
ð1:9Þ
Filling factor
kf ¼ �xxpeak
ð1:10Þ
Distortion factor
THD ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiP1
n¼2 x2rmsn
pxrms1
100 % ð1:11Þ
where xrmsn and xrms1 are the RMS of the nth harmonic and fundamentalcomponent of the signal.Nonlinear distortion factor
THDn ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiP1
n¼2 x2rmsn
pxrms
100 % ð1:12Þ
Signal power
P ¼ 1t2 � t1
Zt2t1
x2ðtÞdt
Px ¼ 1n2 � n1
Xn 2
n¼ n1
x2½n�ð1:13Þ
4 1 Classification and Parameters of Signals
Signal energy
E ¼Zt2t1
x2ðtÞdt
E ¼Xn 2
n¼ n1
x2½n�ð1:14Þ
Autocorrelation
RxxðsÞ ¼ limT!1
1T
ZT0
xðtÞxðt � sÞdt
RxxðmÞ ¼Xn¼1
n¼0
x½n�x½n� m�ð1:15Þ
Substituting into Eq. (1.15), the signals xðt � sÞ and x½n� m� by the signalsyðt � sÞ and y½n� m� gives the cross-correlation
RxyðsÞ ¼ limT!1
1T
ZT0
xðtÞyðt � sÞdt
RxyðmÞ ¼Xn¼1
n¼0
x½n�y½n� m�ð1:16Þ
If RxyðsÞ ¼ 0, then signals xðtÞ; x½n� and yðtÞ; y½n� are not correlated.The cross-correlation RxyðsÞ may have either a positive or a negative value; itdoes not have to have a maximum for s ¼ 0 and does not have to be an evenfunction.
• Random SignalsMean value
�x ¼ limT!1
1T
ZT0
xðtÞdt ¼Zþ1
�1xðtÞdðxÞdx
�x ¼ 1N
XNn¼1
x½n� ¼Xn¼N
n¼�N
x½n�p½n�ð1:17Þ
where dðxÞ—density function of the random variable and p½n�—probability thatthe random variable will assume a given value.
1.3 Parameters of Signals 5
Variance
r2 ¼ limT!1
1T
ZT0
xðtÞ � �x½ �2dt
r2 ¼ 1N
XNn¼1
x½n� � �xf g2ð1:18Þ
where r—standard deviation.
1.4 Examples of Analog Signals with Limited Energy
• Rectangular signal (Fig. 1.3)
xðtÞ ¼ PðtÞ ¼ a for tj j\p
0 for tj j[ p
��x ¼ a; Ex ¼ 2a2p
ð1:19Þ
• Rectangular signal shifted in time (Fig. 1.4)
xðtÞ ¼ aPt � cb
� ��x ¼ a; Ex ¼ a2b
ð1:20Þ
Fig. 1.3 Rectangular signal
Fig. 1.4 Rectangular signal shifted in time
6 1 Classification and Parameters of Signals
• Triangular signal (Fig. 1.5)
xðtÞ ¼ KðtÞ ¼ a� tj j for tj j � a
0 for tj j[ a
�
�x ¼ a2; Ex ¼ 2
3a3
ð1:21Þ
• Cosinusoidal signal (Fig. 1.6)
xðtÞ ¼ X0 cosðx0tÞPðtÞ
�x ¼ 2X0
p; Ex ¼ pX2
0
2x0
ð1:22Þ
• Exponentially decreasing signal (Fig. 1.7)
xðtÞ ¼ X0e�a t for t� 0; a[ 0
0 for t\0
�
�x ¼ 0; Ex ¼ X20
2a
ð1:23Þ
Fig. 1.5 Triangular signal
Fig. 1.6 Cosinusoidal signal
1.4 Examples of Analog Signals with Limited Energy 7
• Exponentially decreasing sinusoidal signal (Fig. 1.8)
xðtÞ ¼ X0e�a t sinðx0tÞ for t � 0; a[ 0
0 for t [ 0
�
�x ¼ 0; Ex ¼ X20
4ax2
0
a2 þ x20
aðtÞ ¼ �X0e�a t
ð1:24Þ
• Sa signal (Fig. 1.9)
xðtÞ ¼sinðx0tÞx0t
for t 6¼ 0
0 for t ¼ 0
(
�x ¼ 0; Ex ¼ px0
ð1:25Þ
Fig. 1.7 Exponentially decreasing signal
Fig. 1.8 Exponentially decreasing sinusoidal signal
8 1 Classification and Parameters of Signals
• Gaussian signal (Fig. 1.10)
xðtÞ ¼ e�p t2
�x ¼ 0; Ex ¼ 1ffiffiffi2
p ð1:26Þ
• Unit step signal (Fig. 1.11)
xðtÞ ¼ 1ðtÞ ¼ 1 for t� 0
0 for t\0
�
�x ¼ 12; Px ¼ 1
ð1:27Þ
Fig. 1.10 Gaussian signal
Fig. 1.9 Sa signal
1.4 Examples of Analog Signals with Limited Energy 9
• Exponentially increasing signal (Fig. 1.12)
xðtÞ ¼ ð1� e�a tÞ1ðtÞ; a[ 0
�x ¼ 12; Px ¼ 1
2
ð1:28Þ
1.5 Examples of Analog Signals with Limited Power
• Harmonic signal (Fig. 1.13)
xðtÞ ¼ X0 sinðx0t þ uÞ; �1\t\þ1�x ¼ 0; Px ¼ 1
2X20
ð1:29Þ
Fig. 1.12 Exponentially increasing signal
Fig. 1.11 Unit step signal
10 1 Classification and Parameters of Signals
• Bipolar rectangular signal (Fig. 1.14)
�x ¼ 0; Px ¼ X20 ð1:30Þ
• Unipolar rectangular signal (Fig. 1.15)
�x ¼ TT0
X0; Px ¼ TT0
X20 ð1:31Þ
Fig. 1.14 Bipolar rectangular signal
Fig. 1.15 Unipolar rectangular signal
Fig. 1.13 Harmonic signal
1.5 Examples of Analog Signals with Limited Power 11
1.6 Examples of Distributive Signals
• Dirac delta (Fig. 1.16)
dðtÞ ¼ 0 for t 6¼ 0
1 for t ¼ 0
�Z1�1
dðtÞdt ¼ 1ð1:32Þ
• Comb signal (Fig. 1.17)
dTðtÞ ¼X1n¼�1
d t � nTð Þ ð1:33Þ
Fig. 1.16 Dirac delta
Fig. 1.17 Comb signal
12 1 Classification and Parameters of Signals
1.7 Discrete Signals with Limited Energy
Kronecker delta (Fig. 1.18)
x½n� ¼ d½n� ¼ 1 for n ¼ 0
0 for n 6¼ 0
��x ¼ 1; Ex ¼ 1
ð1:34Þ
Rectangular signal (Fig. 1.19)
x½n� ¼ 1 for n� Nj j0 for n[ Nj j
��x ¼ 1; Ex ¼ 2N þ 1
ð1:35Þ
Triangular signal (Fig. 1.20)
x½n� ¼ 1� nj jN for n� Nj j
0 for n[ Nj j
(
�x ¼ N2N þ 1
; Ex ¼ 2N2 þ 13N
ð1:36Þ
Fig. 1.18 Kronecker delta
Fig. 1.19 Rectangular signal
1.7 Discrete Signals with Limited Energy 13
Exponential signal (Fig. 1.21)
x½n� ¼ an; n� 0; 0\a\1
�x ¼ 0; Ex ¼ 11� a2
ð1:37Þ
Sa signal (Fig. 1.22)
x½n� ¼ Sa x0n½ � ¼sinðx0nÞx0n
for n 6¼ 0
1 for n ¼ 0
(
�x ¼ 0; Ex ¼ xtx
p
ð1:38Þ
Fig. 1.21 Exponential signal
Fig. 1.20 Triangular signal
Fig. 1.22 Sa signal
14 1 Classification and Parameters of Signals
1.8 Discrete Signals with Limited Power
Unit signal (Fig. 1.23)
x½n� ¼ 1½n� ¼ 1 for n� 0
0 for n\0
�
�x ¼ 12; Px ¼ 1
ð1:39Þ
Harmonic signal (Fig. 1.24)
x½n� ¼ X0 sin nxttþ u
� �; �1\n\1
�x ¼ 0; Px ¼ X20
2
ð1:40Þ
Fig. 1.23 Unit signal
Fig. 1.24 Harmonic signal
1.8 Discrete Signals with Limited Power 15
1.9 Examples of Analog Signals in MathCad
Exponentially decreasing signal
t :¼ �2;�1:99. . .8
X0 :¼ 1 a :¼ 0:5
xðtÞ :¼ X0 � e�a�t if t� 0
0 if t\ 0
����
−5 0 5 100
0.2
0.4
0.6
0.8
1
x(t)
t
t1 :¼ 0 t2 :¼ 1
xav :¼ 1t2 � t1
�Zt2t1
xðtÞdt E :¼Z10
xðtÞ2dt
xav ¼ 0 E ¼ 1
Exponentially decreasing sinusoidal signal
t :¼ �1; �0:99. . .8
X0 :¼ 1 a :¼ 0:5 x0 :¼ 3
xðtÞ :¼ X0 � e�a�t � sinðx0 � tÞ if t� 0
0 if t\0
����a1ðtÞ :¼ X0 � e�a�t
2 ðtÞ :¼ �X0 � e�a�t
−2 0 2 4 6 8 10−2
−1
0
1
2
)(tx
a1 )(t
a2 )(t
t
16 1 Classification and Parameters of Signals
t1 :¼ 0 t2 :¼ 1
xav :¼ 1t2 � t1
�Zt2t1
xðtÞdt E :¼Z10
xðtÞ2dt
xav ¼ 0 E ¼ 0:486
E1 :¼ X20
4 � a �x2
0
a2 þ x20
E1 ¼ 0:486
Sa signal
t :¼ �6;�5:99. . .6
x0 :¼ 3
xðtÞ :¼sinðx0�tÞðx0�tÞ if t 6¼ 0
1 if t ¼ 0
�����
t1 :¼ �3000 t2 :¼ 3000
xav :¼ 1t2 � t1
�Zt2t1
xðtÞdt E :¼Z1�1
xðtÞ2dt
xav ¼ 1:745� 10�4 E ¼ 1:047
Gaussian signal
t :¼ �1:5;�1:49. . .1:5
xðtÞ :¼ e�p�t2
−10 −5 0 5 10−0.5
0
0.5
1
(t)x
t
1.9 Examples of Analog Signals in MathCad 17
t1 :¼ �1 t2 :¼ 1
xav :¼ 1t2 � t1
�Zt2t1
xðtÞdt E :¼Z1�1
xðtÞ2dt
xav ¼ 0 E ¼ 0:707
1.10 Examples of Discrete Signals in MathCad
nstart :¼ �3 nend :¼ 3
n :¼ nstart. . .nend
Dirac delta signal
dðnÞ :¼ 1 if n ¼ 0
0 otherwise
����k :¼ 1
−2 −1 0 1 20
0.2
0.4
0.6
0.8
1
(t)x
t
−4 −2 0 2 40
0.5
1
n
18 1 Classification and Parameters of Signals
Unit signal
xðnÞ :¼ 1 if n� 00 if n\0
����
−4 −2 0 2 40
0.5
1
(n)x
n
Sinusoidal signal
x0 :¼ 1
xðnÞ :¼ sinðx0 � nÞ
−4 −2 0 2 4−1
0
1
(n)x
n
1.10 Examples of Discrete Signals in MathCad 19
Chapter 2Laplace Transform
The comprehensive use of information contained in signals requires performing onthem various mathematical operations, transforms, or conversions. One of the mostuseful transforms, commonly used in various fields of technical sciences andmathematics, is the Laplace transform. It has several practical applications, ofwhich some of the most noteworthy are the solution of ordinary linear differentialequations having constant coefficients, the examination of dynamic properties ofsystems, the synthesis of mathematical models, the simplification of their order, orthe determination of the expðAtÞ matrix, which is indispensable for solving the stateequation presented in the matrix form.
In case of linear differential equations, the Laplace transform algebraizes thoseequations, transforming them into algebraic equations. In consequence, the nth deriv-ative of a differential equation gets replaced by the nth power of an algebraic equation.The final solution of the differential equation is obtained by applying an inverse Laplacetransform, in which the pools of algebraic equation previously obtained are used.
The Laplace transform is
XðsÞ ¼Z1�1
xðtÞ e�stdt ð2:1Þ
which, for real signals that start at the time of t ¼ 0, is reduced to the form
XðsÞ ¼Z10
xðtÞ e�stdt for 0� t\1 ð2:2Þ
The Laplace integral Eq. (2.2) assigns to signal xðtÞ its transform, being afunction of the complex variable s ¼ rþ jx, while it is assumed that
Z10
xðtÞe�r tdt\1; r 2 Re ð2:3Þ
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_2
21
The range of ðr; xÞ values, for which the integral is convergent, is defined as theconvergence area.
If the Laplace integral of the function xðtÞ is convergent for s0 ¼ r0 þ jx, it isalso convergent in all the points in which r[ r0. The r0 is referred to as con-vergence abscissa (Fig. 2.1).
2.1 Initial and Final Value
From Eq. (2.2), we can easily determine the initial value for t ¼ 0 as well as thefinal value for t!1. Calculating the Laplace transform of derivative xðtÞ, we get
Z10
_xðtÞ e�stdt ¼ xðtÞe�stj10 þsZ10
xðtÞ e�stdt ¼ �xð0Þ þ sXðsÞ ð2:4Þ
where xð0Þ is the right-hand limit of xðtÞ for t ¼ 0.For the s! 0, left-hand side of Eq. (2.4) is
lims!0
Z10
_xðtÞ e�stdt ¼ limt!1 xðtÞ � xð0Þ½ � ð2:5Þ
Comparing for s! 0, the right-hand sides of Eqs. (2.4) and (2.5) give
limt!1 xðtÞ � xð0Þ½ � ¼ lim
s!0sXðsÞ � xð0Þ½ � ð2:6Þ
hence,
limt!1 xðtÞ ¼ lim
s!0sXðsÞ ð2:7Þ
and it is assumed that the limt!1 xðtÞ exists.
Fig. 2.1 Convergence area for Laplace integral
22 2 Laplace Transform
For s!1, the left-hand side of Eq. (2.4) equals zero
Z10
_xðtÞ e�stdt ¼ 0 ð2:8Þ
thus,
xð0Þ ¼ lims!1 sXðsÞ: ð2:9Þ
2.2 Surface and Moments
In order to calculate the surface under the signal xðtÞ, let us write the integralR10 xðtÞdt as a limit of Eq. (2.2) for s! 0
Z10
xðtÞdt ¼ lims!0
Z10
xðtÞ e�stdt ¼ Xð0Þ ð2:10Þ
The higher-order derivatives of Eq. (2.2) for s ¼ 0 give successive moments
Z10
tkxðtÞdt; k ¼ 1; 2; . . .; n ð2:11Þ
as we have
dds
Z10
xðtÞ e�stdt24
35 ¼ � Z1
0
t xðtÞ e�stdt ð2:12Þ
d2
ds2
Z10
xðtÞ e�stdt24
35 ¼ Z1
0
t2xðtÞ e�stdt ð2:13Þ
and
dk
dsk
Z10
xðtÞ e�stdt24
35 ¼ ð�1Þk Z
1
0
tkxðtÞ e�stdt ð2:14Þ
Substituting s ¼ 0 into Eqs. (2.12)–(2.14) gives
2.1 Initial and Final Value 23
• the moment of the first order
dds
XðsÞ½ �����s¼0¼ �
Z10
t xðtÞ dt ð2:15Þ
• the moment of the second order
d2
ds2XðsÞ½ �
����s¼0¼
Z10
t2xðtÞ dt ð2:16Þ
• the moment of the kth order
dk
dsk½XðsÞ�
����s¼0¼ ð�1Þk
Z10
tkxðtÞ dt: ð2:17Þ
2.3 Examples of Laplace Transforms
1. Dirac delta dðtÞ
L½dðtÞ� ¼Z10
dðtÞ e�stdt ¼ 1 ð2:18Þ
2. Unit step signal 1ðtÞ
L½1ðtÞ� ¼Z10
1 � e�stdt ¼ e�st
�s����1
0¼ 1
sfor Re s[ 0 ð2:19Þ
3. Signal e�a t
L½e�a t� ¼Z10
e�a t e�stdt ¼ e�ðs�aÞ t
�ðs� aÞ����1
0¼ 1
s� afor Re s[ � a ð2:20Þ
24 2 Laplace Transform
4. Signal e�jxt
L½e�jx t� ¼ 1s� jx
ð2:21Þ
5. Signal at
L½t� ¼Z10
at e�stdt ¼ a �eð�sÞ t� � s t þ 1
s2
����1
0¼ a
s2for Re s[ 0 ð2:22Þ
6. Signals sinx t and cosx t
L½cosx t þ j sinx t� ¼Z10
ejx t e�stdt ¼ e�ð s�jxÞ
�ðs� jxÞ����1
0
¼ 1s� jx
¼ sþ jxs2 þ x2 for Re s [ 0
ð2:23Þ
Comparing the real and imaginary parts of the last equation, we have
L½cosx t� ¼ ss2 þ x2 ð2:24Þ
and
L½sinx t� ¼ xs2 þ x2 : ð2:25Þ
2.4 Properties of Laplace Transform
1. Linearity
L½ax1ðtÞ þ bx2ðtÞ� ¼ a X1ðsÞ þ b X2ðsÞ ð2:26Þ
2. Shift in the s domain—multiplication by e�at
L½e�atxðtÞ� ¼ Xðsþ aÞ ð2:27Þ
3. Shift in the time domain
L½hðtÞ xðt � sÞ� ¼ XðsÞ e�s s ð2:28Þ
where hðtÞ is the unit step
2.3 Examples of Laplace Transforms 25
4. Integration in the time domain—division by s
L
Z t
0
xðsÞds24
35 ¼ 1
sXðsÞ ð2:29Þ
5. Change of time scale
L xðatÞ½ � ¼ 1aX
sa
� �ð2:30Þ
6. Differentiation in s domain—multiplication by t in time domain
L½t xðtÞ� ¼ � dds
XðsÞ ð2:31Þ
7. Transform of the first derivative
L x0ðtÞ½ � ¼ sXðsÞ � xð0þÞ ð2:32Þ
8. Transform of the second derivative
L x00ðtÞ½ � ¼ s2XðsÞ � s xð0þÞ � x0ð0þÞ ð2:33Þ
9. Transform of the nth derivative
L xnðtÞ½ � ¼ sn XðsÞ � sn�1 xð0þÞ � sn�2 x0ð0þÞ � � � � � xðn�1Þð0þÞ: ð2:34Þ
2.5 Laplace Transform in Solving Differential Equation
Applying the Laplace transform to both sides of linear differential equation
dnyðtÞdtn
þ an�1dn�1yðtÞdtn�1
þ � � � þ a1dyðtÞdtþ a0yðtÞ
¼ b0xðtÞ þ b1dxðtÞdtþ � � � þ bm�1
dm�1xðtÞdtm�1
þ bmdmxðtÞdtm
ð2:35Þ
with initial conditions that equal zero
yð0þÞ ¼ 0; y0ð0þÞ ¼ 0; . . .; yðn�1Þð0þÞ ¼ 0 for m\n; ak; bk 2 < ð2:36Þ
26 2 Laplace Transform
gives
YðsÞ sn þ an�1sn�1 þ � � � þ a1sþ a0� �
¼ XðsÞ bmsm þ bm�1sm�1 þ � � � þ b1sþ b0� � ð2:37Þ
The ratio of output YðsÞ to input XðsÞ in Eq. (2.37) is defined as the Laplacetransfer function KðsÞ
KðsÞ ¼ YðsÞXðsÞ ¼
bmsm þ bm�1sm�1 þ � � � þ b1sþ b0sn þ an�1sn�1 þ � � � þ a1sþ a0
ð2:38Þ
If the input XðsÞ is given, the solution of yðtÞ can be obtained by the inverseLaplace transform
yðtÞ ¼L�1½YðsÞ� ¼ 12p j
Zaþj1a�j1
YðsÞestds ð2:39Þ
In Eq. (2.39), the constant a must be selected in such a way that the integrationrange is contained within the convergence range. Because of calculation difficulties,the formula (2.39) is rarely used. Instead, the most commonly used method ofcalculating the inverse transform is the residue method. Depending on the form ofthe denominator poles, two cases can occur here:
• If YðsÞ has n single poles si, then
yðtÞ ¼Xni¼1
resYðsÞesit s ¼ 1; 2; . . .; n ð2:40Þ
where
res YðsÞ ¼ lims!siðs� siÞYðsÞ ð2:41Þ
• If YðsÞ has m multiple poles sm, then
yðtÞ ¼Xr
k¼1res YðsÞ tðk�1Þ
ðk � 1Þ!esmt ð2:42Þ
where r is the order of multiple pole sm, while
resYðsÞ ¼ 1ðr � kÞ! lims!sm
dðr�kÞ
dsðr�kÞ½ðs� smÞrYðsÞ� ð2:43Þ
Equation (2.41) is a specific case of Eq. (2.43) for r = k = 1. If in YðsÞ, bothsingle and multiple poles are present, then the solution yðtÞ is the sum of solutions
2.5 Laplace Transform in Solving Differential Equation 27
Eqs. (2.40) and (2.42). Poles occurring in YðsÞ may be real or complex conjugate.For t!1, poles with a negative real part give solutions for yðtÞ that approach aconstant value, whereas poles with a positive real part give solutions for yðtÞ thattend to infinity exponentially. Imaginary poles cause the generation of oscillations,which in the case of a negative real part decrease exponentially and in the case of apositive real part increase exponentially. The occurrence of complex conjugatedpoles without real part results in the generation of sustained oscillations.
Solving linear differential equations with constant coefficients consists oftransforming them into Laplace equations and then using the inverse transform toobtain the required form of time response.
Example 2.1 Solve the equation
d3
dt3yðtÞ þ 9
d2
dt2yðtÞ þ 26
ddtyðtÞ þ 24yðtÞ ¼ sin t ð2:44Þ
Writing Eq. (2.44) in Laplace transform form, we have
YðsÞ s3 þ 9s2 þ 26sþ 24� � ¼ 1
s2 þ 1ð2:45Þ
thus,
YðsÞ ¼ 1s2 þ 1ð Þ s3 þ 9s2 þ 26sþ 24ð Þ ð2:46Þ
or
YðsÞ ¼ 1ðs2 þ 1Þðsþ 2Þðsþ 3Þðsþ 4Þ ð2:47Þ
The transform YðsÞ has in the poles s1 ¼ �2, s2 ¼ �3, s3 ¼ �4, s4 ¼ þj, ands5 ¼ �j the following residues:
ResY s1ð Þ ¼ 110
; ResY s2ð Þ ¼ � 110
; ResY s3ð Þ ¼ 134
ResY s4ð Þ ¼ 1340
�3j� 5ð Þ; ResY s5ð Þ ¼ 1340
þ3j� 5ð Þð2:48Þ
The solution of the equation thus has the form
yðtÞ ¼ 110
expð�2tÞ � 110
expð�3tÞ þ 134
expð�4tÞ � 134
cosðtÞ þ 3170
sinðtÞ:ð2:49Þ
28 2 Laplace Transform
2.6 Laplace Transform in Solving State Equation
Differential equations with constant coefficients of the nth order Eq. (2.35) may bewritten in the form of state equations, that is, a system of n equations of the firstorder
_xðtÞ ¼ AxðtÞ þ BuðtÞyðtÞ ¼ CxðtÞ þ DuðtÞ ð2:50Þ
where x(t), y(t), and u(t) are state, output, and input vectors, while A, B, C, andD are state, input, output, and feedthrough matrices, respectively.
By applying the Laplace transform to both sides of Eq. (2.50), we have
sXðsÞ � Xð0Þ ¼ AXðsÞ þ BUðsÞYðsÞ ¼ CXðsÞ þ DUðsÞ ð2:51Þ
Simple transformations of Eq. (2.51) give
XðsÞ ¼ Is� Að Þ�1Xð0Þ þ Is� Að Þ�1BUðsÞYðsÞ ¼ C Is� Að Þ�1Xð0Þ þ C Is� Að Þ�1Bþ D
h iUðsÞ ð2:52Þ
Equation (2.52) in many practical cases is simplified, due to zeroing of thematrix D. This happens if in Eq. (2.38) the order of the numerator m is less than theorder of the denominator n. As a result, we have
XðsÞ ¼ Is� Að Þ�1Xð0Þ þ Is� Að Þ�1BUðsÞYðsÞ ¼ C Is� Að Þ�1Xð0Þ þ C Is� Að Þ�1B
h iUðsÞ ð2:53Þ
The solution of Eq. (2.53) is
xðtÞ ¼ eAtxð0Þ þZ t
0
eAðt�sÞBuðsÞds
yðtÞ ¼ CeAtxð0Þ þ CZ t
0
eAðt�sÞBuðsÞdsð2:54Þ
2.6 Laplace Transform in Solving State Equation 29
For zero initial conditions, Eq. (2.54) in the equivalent form is
xðtÞ ¼ eAðt�t0Þxðt0Þ þZ t
t0
eAðt�sÞBuðsÞds
yðtÞ ¼ CeAðt�t0Þxðt0Þ þ CZ t
t0
eAðt�sÞBuðsÞdsð2:55Þ
For Xð0Þ ¼ 0 and on the base of Eq. (2.53), we have
KðsÞ ¼ YðsÞUðsÞ ¼ C Is� A½ � �1B ð2:56Þ
For a single input UðsÞ and a single output YðsÞ, if the state equation is given inphase-variable canonical form, then matrices A, B, C, and D are
A ¼
0 1 0 . . . 0
0 0 1 . . . 0
..
. ... ..
. ... ..
.
0 0 0 . . . 1
�a0 �a1 . . . . . . �an�1
26666664
37777775
B ¼
0
0
..
.
0
1
26666664
37777775
C ¼ b0 b1 . . . bm½ � D ¼ lims!1KðsÞ ¼ 0 for m\n
ð2:57Þ
and the transfer function (2.56) equals (2.38).The expression expðAtÞ in solutions of Eq. (2.55) represents an infinite series
eAt ¼ Iþ At þ 12!A2t2 þ 1
3!A3t3 þ � � � ð2:58Þ
in which I is the unit matrix. This series may be determined by the inverse Laplacetransform. We thus have
L½eAt� ¼ ½Is� A��1 ð2:59Þ
from that
eAt ¼L�1½ðIs� AÞ�1� ð2:60Þ
Example 2.2 Calculate the state equations, where the state variables are: currenti1ðtÞ and voltages across the capacitors uC1ðtÞ; uC2ðtÞ and the output is voltage uðtÞacross the resistor R3—Fig. 2.2.
30 2 Laplace Transform
By assumption, the vector of state variables has the form
xðtÞ ¼uC1ðtÞuC2ðtÞi1ðtÞ
264
375 ð2:61Þ
from that
_xðtÞ ¼_uC1ðtÞ_uC2ðtÞ_i1ðtÞ
264
375 ð2:62Þ
and the output vector yðtÞ
yðtÞ ¼ uðtÞ ¼ i3ðtÞR3 ð2:63Þ
Voltage across the capacitor C1 is
_uC1 ¼1Ci1ðtÞ ð2:64Þ
The Kirchhoff’s current law for the node gives
_uC2 ¼i1ðtÞC2� i3ðtÞ
C2ð2:65Þ
We have to eliminate the current i3ðtÞ from the last equation, as it does not appearin the state equation. From Kirchhoff’s voltage law for the second loop, we have
i3ðtÞ ¼ 1R3
uC2ðtÞ þR2C2
R3_uC2 �
e2ðtÞR3
ð2:66Þ
R1
L
e1(t)
uC2(t)
e2(t)
R3 R2
i1(t)
i3(t)
i2(t)
u(t)uC1(t)
C1 C2
Fig. 2.2 Electrical circuit, for which the state variables are uC1 ðtÞ; uC2 ðtÞ; i1ðtÞ
2.6 Laplace Transform in Solving State Equation 31
which, after substituting into Eq. (2.65) and simplifying, gives
_uC2 ¼i1ðtÞR3
C2 R2 þ R3ð Þ �uC2ðtÞ
C2 R2 þ R3ð Þ þe2ðtÞ
C2 R2 þ R3ð Þ ð2:67Þ
From Kirchhoff’s voltage law for the first loop, we have
i1ðtÞR1 þ L_i1ðtÞ þ uC1ðtÞ þ uðtÞ ¼ e1ðtÞ ð2:68Þ
Voltage uðtÞ does not appear in the state equation; thus, we have to eliminate it.From Kirchhoff’s voltage law for the second loop, we have
i3ðtÞ ¼ uC2ðtÞR2 þ R3
þ i1ðtÞR2
R2 þ R3� e2ðtÞR2 þ R3
ð2:69Þ
then
uðtÞ ¼ i3ðtÞR3 ¼ uC2ðtÞR3
R2 þ R3þ i1ðtÞR2R3
R2 þ R3� e2ðtÞR3
R2 þ R3ð2:70Þ
Substituting Eq. (2.70) into Eq.(2.68), we get the state variable _i1ðtÞ in the form
_i1ðtÞ ¼ � uC1ðtÞL� uC2ðtÞR3
LðR2 þ R3Þ �i1ðtÞðR1R2 þ R1R3 þ R2R3Þ
LðR2 þ R3Þþ e2ðtÞR3
LðR2 þ R3Þ þe1ðtÞL
ð2:71Þ
Equations (2.64), (2.67) and (2.71) expressed in the matrix form give therequired state equation:
_uC1ðtÞ_uC2ðtÞ_i1ðtÞ
264
375 ¼
0 0 1C
0 � 1C2ðR2 þ R3Þ
R3C2ðR2 þ R3Þ
� 1L � R3
LðR2 þ R3Þ � R1R2 þ R1R3 þ R2R3LðR2 þ R3Þ
2664
3775
uC1ðtÞuC2ðtÞiðtÞ
264
375
þ0 0
0 1C2ðR2 þ R3Þ
1L
R3LðR2 þ R3Þ
2664
3775 e1ðtÞ
e2ðtÞ
" # ð2:72Þ
The output equation is obtained directly from Eq. (2.70) and is
uðtÞ ¼ 0 R3R2 þ R3
R2R3R2 þ R3
h i uc1ðtÞuc2ðtÞi1ðtÞ
24
35 þ 0 �R3
R2 þ R3
h i e1ðtÞe2ðtÞ
� �ð2:73Þ
32 2 Laplace Transform
Example 2.3 Solve the state equation Eq. (2.54) for zero initial conditions, ifmatrices A, B, and C have the form
A ¼0 1 00 0 1�4 �8 �5
24
35; B ¼
001
24
35; C ¼ 1 0 0½ �; and uðtÞ ¼ 1ðtÞ
ð2:74Þ
Matrix Is� A½ � equals
Is� A½ � ¼s �1 00 s �14 8 sþ 5
24
35 ð2:75Þ
From that
Is� A½ ��1¼s2þ5sþ8
s3þ5s2þ8sþ4sþ5
s3þ5s2þ8sþ41
s3þ5s2þ8sþ4�4
s3þ5s2þ8sþ4sðsþ5Þ
s3þ5s2þ8sþ4s
s3þ5s2þ8sþ4�4s
s3þ5s2þ8sþ4�4ð2sþ1Þ
s3þ5s2þ8sþ4s2
s3þ5s2þ8sþ4
264
375 ð2:76Þ
Applying inverse Laplace transform to Eq. (2.76) gives
eAt ¼4e�t � 2te�2t � 3e�2t 4e�t � 3te�2t � 4e�2t e�t � te�2t � e�2t
�4e�t þ 4te�2t þ 4e�2t �4e�t þ 6te�2t þ 5e�2t �e�t þ 2te�2t þ e�2t
4e�t � 8te�2t � 4e�2t 4e�t � 12te�2t � 4e�2t e�t � 4te�2t
24
35
ð2:77Þ
thus, Eq. (2.54) is
yðtÞ ¼Z t
0
1 0 0½ �4e�ðt�sÞ � 2ðt � sÞe�2ðt�sÞ � 3e�2ðt�sÞ
�4e�ðt�sÞ þ 4ðt � sÞe�2ðt�sÞ þ 4e�2ðt�sÞ
4e�ðt�sÞ � 8ðt � sÞe�2ðt�sÞ � 4e�2ðt�sÞ
264
4e�ðt�sÞ � 3ðt � sÞe�2ðt�sÞ � 4e�2ðt�sÞ e�ðt�sÞ � ðt � sÞe�2ðt�sÞ � e�2ðt�sÞ
�4e�ðt�sÞ þ 6ðt � sÞe�2ðt�sÞ þ 5e�2ðt�sÞ �e�ðt�sÞ þ 2ðt � sÞe�2ðt�sÞ þ e�2ðt�sÞ
4e�ðt�sÞ � 12ðt � sÞe�2ðt�sÞ � 4e�2ðt�sÞ e�ðt�sÞ � 4ðt � sÞe�2ðt�sÞ
375
0
0
1
264
375dsð2:78Þ
which, after simple calculations, gives
yðtÞ ¼ 14� e�t þ 3
4e�2t þ 1
2te�2t: ð2:79Þ
2.6 Laplace Transform in Solving State Equation 33
2.7 Simplification of Model Order
Let the Laplace transform of n-order model (Eq. (2.38)) be expressed by numeratorand denominator in the form of Ruth tables, which are as follows:
• for the numerator
b1;1 b1;2 b1;3 b1;4 . . .b2;1 b2;2 b2;3 b2;4 . . .b3;1 b3;2 b3;3. . .bm;1bmþ1;1
ð2:80Þ
• for the denominator
a1;1 a1;2 a1;3 a1;4 . . .a2;1 a2;2 a2;3 a2;4 . . .a3;1 a3;2 a3;3. . .an;1anþ1; 1
ð2:81Þ
where in (2.80)
b1;1 ¼ bm b1;2 ¼ bm�2 b1;3 ¼ bm�4 b1;4 ¼ bm�6b2;1 ¼ bm�1 b2;2 ¼ bm�3 b2;3 ¼ bm�5 b2;4 ¼ bm�7
bi;j ¼ � 1bi�1;1
bi�2;1 bi�2;jþ1bi�1;1 bi�1;jþ1
�������� i ¼ 3; 4; . . .; n j ¼ 1; 2; . . .
ð2:82Þ
and in (2.81)
a1;1 ¼ an ¼ 1 a1;2 ¼ an�2 a1;3 ¼ an�4 a1;4 ¼ an�6a2;1 ¼ an�1 a2;2 ¼ an�3 a2;3 ¼ an�5 a2;4 ¼ an�7
ai;j ¼ � 1ai�1;1
ai�2;1 ai�2;jþ1ai�1;1 ai�1;jþ1
�������� i ¼ 3; 4; . . .; n j ¼ 1; 2; . . .
ð2:83Þ
The model of Eq. (2.38) of the nth order is described by the first two lines oftables (2.82) for the numerator and (2.83) for the denominator. Subsequent lines, (2)and (3), (3) and (4), (4) and (5), etc., allow us to reduce this model, respectively, toorders ðn� 1Þ; ðn� 2Þ; ðn� 3Þ, etc. Thus, the model of order ðn� 1Þ is
34 2 Laplace Transform
Kn�1ðsÞ ¼ bm�1sm�1 þ b3;1sm�2 þ bm�3sm�3 þ b3;2sm�4 þ � � �an�1sn�1 þ a3;1sn�2 þ an�3sn�3 þ a3;2sn�4 þ � � � ð2:84Þ
and the model of order ðn� 2Þ is
Kn�2ðsÞ ¼ b3;1sm�2 þ b4;1sm�3 þ b3;2sm�4 þ b4;2sm�5 þ � � �a3;1sn�2 þ a4;1sn�3 þ a3;2sn�4 þ a4;2sn�5 þ � � � ð2:85Þ
In a similar way, one can determine further models of lower orders.
Example 2.4 Obtain, using the Ruth’s method, a third-order model of a seventh-order pitch rate of a supersonic aircraft (given by Sinha and de Bruin 1973)described by the transfer function
K7ðsÞ ¼ 375;000 ðsþ 0:08333Þs7 þ 83:64 s6 þ 4;097 s5 þ 70;342 s4 þ 853;703 s3
� � � þ2;814;271 s2 þ 3;310;875 sþ 281;250
ð2:86Þ
Ruth table (2.83) of the model denominator has the form
1 4;097 853;703 3;310;87583:64 70;342 2;814;271 281;2503:256� 103 8:201� 105 3:308� 106
4:928� 104 2:729� 106 281;2506:398� 105 3:289� 106
2:476� 106 281;2503:216� 106
ð2:87Þ
Fig. 2.3 Impulse responses of models K7ðsÞ (2.86) and K3ðsÞ (2.88)
2.7 Simplification of Model Order 35
It can be easily checked that the third-order model is generated by the fifth andsixth lines of the table (2.87). One thus obtains
K3ðsÞ ¼ 375;000 ðsþ 0:08333Þ6:398� 105s3 þ 2:476� 106s2 þ 3:289� 106sþ 281;250
ð2:88Þ
The plots in the Fig. 2.3 show the impulse responses of the models k7ðtÞ and k3ðtÞ.
2.8 Discretization of State Equation
In order to discretize the solution yðtÞ of the state equation (2.55), let us assume thatthe signal yðtÞ will be sampled with the D step.
t ¼ ½nþ 1�D; t0 ¼ nD for n ¼ 0; 1; 2; . . . ð2:89Þ
Substituting Eq. (2.89) into Eq. (2.55), we have
yf½nþ 1�Dg ¼ CeAf½nþ1�D�nDgxðnDÞ þ CZðnþ1ÞDnD
eAf½nþ1�D�sgBuðsÞds ð2:90Þ
Equation (2.90) can be simplified to the form
y½ðnþ 1ÞD� ¼ CeADxðnDÞ þ CZðnþ1ÞDnD
eA½ðnþ1ÞD�s�BuðsÞds ð2:91Þ
Assuming that uðsÞ is constant between consecutive sampling moments
uðsÞ ¼ uðnDÞ for nD\s\½nþ 1�D ð2:92Þ
and substituting Eq. (2.92) into Eq. (2.91), we have
yf½nþ 1�Dg ¼ CeADxðnDÞ
þ CZðnþ1ÞDnD
eAf½nþ1�D�sgdsBuðnDÞ; s 2 nD; ½nþ 1�Df g ð2:93Þ
Let
k ¼ ½nþ 1�D� s ð2:94Þ
36 2 Laplace Transform
then,
y ½nþ 1�Df g ¼ eADyðnDÞ �Z0
D
eAkdkBuðnDÞ; k 2 D; 0½ � ð2:95Þ
Changing the limits of integration in Eq. (2.95), we have
y ½nþ 1�Df g ¼ eADyðnDÞ þZD0
eAkdkBuðnDÞ; k 2 D; 0½ � ð2:96Þ
Equation (2.96) may be written in the simple form
yf½nþ 1�Dg ¼ UyðnDÞ þWuðnDÞ ð2:97Þ
where
U ¼ eAD and W ¼ZD0
eAkdkB ð2:98Þ
Due to difficulties connected with the determination of the series expðADÞ;matrix U may be presented in the equivalent form
U ¼ IþX1k¼1
ADð Þkk!
ð2:99Þ
Rewriting Eq. (2.97) in the matrix form
y1½nþ 1�...
yk½nþ 1�
264
375 ¼
u1;1 � � � u1;k
..
. ...
uk;1 � � � uk;k
264
375
y1½n�...
yk½n�
264
375þ
w1
..
.
wk
264
375u½n� ð2:100Þ
and taking into account that the state variable y1½n� is measured directly, we canwrite
y1½n� ¼ y½n�; y1½nþ 1� ¼ y½nþ 1� ð2:101Þ
and then,
2.8 Discretization of State Equation 37
y½nþ 1�y2½nþ 1�...
yk½nþ 1�
26664
37775 ¼
u1;1 u1;2 . . . u1;nu2;1 u2;2 . . . u2;n
..
. ... ..
. ...
uk;1 uk;2 . . . uk;k
26664
37775
y½n�y2½n�...
yk½n�
26664
37775þ
w1w2
..
.
wk
26664
37775u½n� ð2:102Þ
where yð0Þ; y2ð0Þ; . . .; ykð0Þ ¼ 0:Equation (2.102) provides an easy way to perform recurrent calculations of the
signal yðtÞ using appropriate mathematical software, e.g., MathCad, MATLAB, etc.
Example 2.5 Determine the output response of the model
KðsÞ ¼ 1s2 þ 0:8sþ 4
ð2:103Þ
to the input
uðtÞ ¼ sinð0:3p tÞ þ cosð0:5p tÞ ð2:104Þ
Model Eq. (2.103) has the impulse response which is
kðtÞ ¼ 0:51 expð�0:4 tÞ sinð1:96 tÞ ð2:105Þ
From Eqs. (2.103) and (2.57), we have
A ¼ 0 1�4 �0:8
� �; B ¼ 0
1
� �ð2:106Þ
Fig. 2.4 Signals uðtÞ (2.104) and kðtÞ (2.105)
38 2 Laplace Transform
and
U ¼ 1 0�0:04 0:992
� �; W ¼ 0
0
� �ð2:107Þ
Figures 2.4 and 2.5 present the input uðtÞ; the response kðtÞ in [0, 10s.], and theoutput y½n� for 104 samples.
2.9 Example in MathCad
T :¼ 10 D :¼ 0:01
t :¼ 0;D;. . .; T
KðsÞ :¼ 1s2 þ 0:8sþ 4
KðtÞ :¼ 0:510eð�0:400Þ�t � sinð1:96tÞ
Fig. 2.5 Output y½n�
0 2.5 5 7.5 10
0
0.25
0.5
k(t)
t
− 0.25
− 0.5
2.8 Discretization of State Equation 39
uðtÞ :¼ sin 3 � p � f � tð Þ þ cos 5 � p � f � tð Þ
yðtÞ :¼Z t
0
kðt � sÞ � uðsÞds
A :¼ 0 1�4 �0:8
B :¼ 0
1
U :¼for k 2 0; . . .; TD� 1Uk uðk � DÞU
������ TD :¼for i 2 0; . . .; TD� 1TDi i � DTD
������eA�D :¼ 0:99980053891365841134 0:009959442460565568843
�0:039837769842262275375 0:99183298494520595626
U :¼ 0:99980053891365841134 0:009959442460565568843�0:039837769842262275375 0:99183298494520595626
0 2 4 6 8 10− 2
− 1
0
1
2
u(t)
t
0 2.5 5 7.5 10
0
0.5
1
y(t)
t
− 0.5
− 1
40 2 Laplace Transform
ZD0
e
0 1�4 �0:8
�kdk � 0
1
! 0
0
W :¼ 00
Y :¼
Y0;0 0Y10;0 0for k 2 0; . . .; TD� 1
Ykþ1;0 U0;0 � Yk;0 þ U0;1 � Y1k;0 þW0;0 � Uk;0
Y1kþ1;0 U1;0 � Yk;0 þ U1;1 � Y1k;0 þW1;0 � Uk;0
�����Y
�������������
0 2 4 6 8 10− 1
− 0.5
0
0.5
1
Y
TΔ
2.9 Example in MathCad 41
Chapter 3Fourier Transform
The Fourier transform converts the signal x(t) from the time domain to thefrequency domain, showing the way in which particular frequencies create theoriginal signal. The Fourier transform X(ω) of the signal x(t) presents a specific caseof the Laplace transform, for which s = jω, and for which the x(t) signal it assumedto meet the Dirichlet condition that it is periodic, monotonic in every finite sub-interval, absolutely convergent on the whole axis, which means that the integral ofits absolute value is finite
Z1�1
xðtÞj jdt\1; ð3:1Þ
in the interval of one period it has a finite number of local maxima and minima, italso has in it a finite number of discontinuity points, in which it has its left-hand andright-hand limit.
3.1 Continuous Fourier Transform
The continuous Fourier transform (CFT) is
XðxÞ ¼Z1�1
xðtÞe�jx tdt ð3:2Þ
which can be presented by
XðxÞ ¼Z1�1
xðtÞ cosðxtÞdt � jZ1�1
xðtÞ sinðxtÞdt ð3:3Þ
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_3
43
The module and phase of the spectrum equal
XðxÞj j ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZ1�1
xðtÞ cosðx tÞdt24
352
þZ1�1
xðtÞ sinðx tÞdt24
352
vuuut ð3:4Þ
and
uðxÞ ¼ arctg
R1�1 xðtÞ sinðx tÞdtR1�1 xðtÞ cosðx tÞdt ð3:5Þ
The transform Eq. (3.2) is lossless, which means that the signal x(t) may becompletely reconstructed. For that purpose, we use the inverse transform
xðtÞ ¼ F�1½XðxÞ� ¼ 12p
Z1�1
XðxÞejx tdx ð3:6Þ
If the signal x(t) does not meet the Dirichlet conditions, generally one of thefollowing two cases occurs:
• The signal x(t) is not periodic, but we can find the range [−T/2, T/2], over whichit will overlap with a periodic signal, and the Fourier transform may be per-formed on it—Fig. 3.1.
• The signal x(t) is not absolutely integrable, and in order to apply the Fouriertransform, we multiply it by e�r t; and then
Z1�1
xðtÞj j e�r tdt\1 ð3:7Þ
Fig. 3.1 Non-periodic and periodic signals
44 3 Fourier Transform
and
xðtÞe�r t ¼ 12p
Z1�1
Z10
xðtÞe�rte�jx tdt ejx tdx ð3:8Þ
while x(t) in Eq. (3.8) is calculated as limit
xðtÞ ¼ limr!0
xðtÞe�r t ¼ limr!0
12p
Z1�1
ejx tdxZ10
e�r txðtÞ e�jx tdt ð3:9Þ
3.2 Properties of Fourier Transform
1. Linearity
xðtÞ $ XðxÞ and yðtÞ $ YðxÞ;axðtÞ þ byðtÞa$ a XðxÞ þ bYðxÞ ð3:10Þ
2. Symmetry
XðxÞ $ 2pxð�xÞ ð3:11Þ
3. Change of scale
xta
� �$ aj jX x
a
� �ð3:12Þ
4. Shift
– in the time domain
xðt � t0Þ $ X ðxÞ e�jxt0 ð3:13Þ
– in the frequency domain
xðtÞ e�jx0 t $ X ðx� x0Þ ð3:14Þ
3.1 Continuous Fourier Transform 45
5. Convolution
xðtÞ � yðtÞ $ XðxÞ YðxÞ ð3:15Þ
6. Multiplication
xðtÞ � yðtÞ $ 12p
XðxÞ � YðxÞ ð3:16Þ
7. Differentiation
– in the time domain
dnxðtÞdtn
$ ðjxÞnXðxÞ ð3:17Þ
– in the frequency domain
ð�jÞntn xðtÞ $ dnXðxÞdxn
ð3:18Þ
8. Integration
Z t
�1xðsÞ ds$ 1
jxXðxÞ ð3:19Þ
9. Parseval equality
Z1�1
xðtÞj j2dt$ 12p
Z1�1
XðxÞj j2dx ð3:20Þ
3.3 Example of Fourier Transforms
1. Dirac delta
xðtÞ ¼ dðtÞXðxÞ ¼ 1
ð3:21Þ
2. Unit signal
xðtÞ ¼ 1
XðxÞ ¼ 2pdðxÞ ð3:22Þ
46 3 Fourier Transform
3. Dirac impulse series
xðtÞ ¼X1k¼�1
dðt � kTÞ
X xð Þ ¼ x0
X1k¼�1
dðx� kx0Þ; x0 ¼ 2pT
ð3:23Þ
4. Harmonic signal
xðtÞ ¼ a ejx0 t
XðxÞ ¼ 2apd ðx� x0Þð3:24Þ
5. Cosinusoidal signal
xðtÞ ¼ cosðx0tÞ ¼ 12ðejx0 t þ e�jx0tÞ
XðxÞ ¼ p dðx� x0Þ þ dðxþ x0Þ½ �ð3:25Þ
6. Sinusoidal signal
xðtÞ ¼ sinðx0tÞ ¼ 12j
ejx0t þ e�jx0t� �
XðxÞ ¼ �jp dðx� x0Þ � dðxþ x0Þ½ �ð3:26Þ
7. Sign signal
xðtÞ ¼ sgnðtÞ ¼�1 for t\0
0 for t ¼ 0
1 for t[ 0
8><>:
XðxÞ ¼ 2jx
ð3:27Þ
8. Unit step signal
xðtÞ ¼ �1 for t\0
1 for t[ 0
�
XðxÞ ¼ p dðxÞ þ 1jx
ð3:28Þ
9. Sa signal
xðtÞ ¼ sinðXtÞXt
X xð Þ ¼ p2Sx0ð�xÞ; Sx0ðxÞ ¼ Sx0ðxÞ ¼
0 for xj j[X
1 for xj j �X
� ð3:29Þ
3.3 Example of Fourier Transforms 47
10. Gaussian signal
xðtÞ ¼ e�at2
X xð Þ ¼ffiffiffipa
re�x24a
ð3:30Þ
11. Exponential signal
xðtÞ ¼ 0 for t\0
e�at for t 0
�
X xð Þ ¼ 1aþ jx
; a[ 0ð3:31Þ
Example 3.1 Determine the spectrum of a rectangular signal of magnitude one over[0, T]. From Eq. (3.2), we have
XðxÞ ¼ZT0
e�jxTdt ¼ 1jxð1� e�jxTÞ ¼ 1
jxð1� e�j
xT2 � e�jxT2 Þ
¼ 2xsin
xT2
� e�j
xT2 ¼ T
xT2
sinxT2
� e�j
xT2 ¼ T � SaxT
2
� e�j
xT2 ð3:32Þ
where
KðxÞj j ¼ T � SaxT2
ð3:33Þ
is the module of the spectrum and
uðxÞ ¼ x T2
for 0\x\2pT
ð3:34Þ
is its phase. It is easy to see that
KðxÞj j ¼ 0 forx T2¼ np; n ¼ 1; 2; 3; . . . ð3:35Þ
and
KðxÞj j ¼ 2Tpð2nþ 1Þ for
x T2¼ p
2ð2nþ 1Þ ð3:36Þ
Figure 3.2 presents this signal and its characteristics.
48 3 Fourier Transform
while
uðxÞ ¼ xT2� p for
2npT
\x\ð2nþ 2Þp
T; n ¼ 1; 2; 3; . . . ð3:37Þ
Example 3.2 Determine the output of an ideal filter with zero attenuation over[−ω0, ω0] when a rectangular input signal over [0, T] and unit step input signal areapplied.
From Eqs. (3.6) and (3.31), we have
yðtÞ ¼ 12p
Zx0
�x0
1jxð1� e�jxTÞ ejxtdx ¼ 1
2p
Zx0
�x0
ejxT � e�jx ðt�TÞ
jxdx
¼ 12p j
Zx0
�x0
cosx t � cosx ðt � TÞx
dxþ 12p
Zx0
�x0
sinxt � sinx ðt � TÞx
dx
ð3:38Þ
Fig. 3.2 Rectangular signal x(t), frequency distribution K(ω) and phase uðxÞ
3.3 Example of Fourier Transforms 49
The first integral of the last equation is equal to zero, hence
yðtÞ ¼ 1p
Zx0
0
sinx tx� sinxðt � TÞ
x
� dx
¼ 1p
Zx0t
0
sin xx
dx�Zx0ðt�TÞ
0
sin xx
dx
264
375
¼ 1pSiðx0tÞ � Siðx0t � TÞ½ �
ð3:39Þ
Figure 3.3 presents the solution of Eq. (3.39).
For the unit step, we have directly
yðtÞ ¼ limr!0
12p
Z1�1
ejxtdxZ10
e�ðrþjxÞtdt ¼ limr!0
12p
Z1�1
ejxt
rþ jxdx
¼ limr!0
12p
Z1�1
cosx trþ jx
dxþ jZ1�1
sinx trþ jx
dx
24
35
ð3:40Þ
After calculations, Eq. (3.40), we have
yðtÞ ¼ limr!0
12p
p e�rt þ jZ1�1
sinx trþ jx
dx
24
35 ¼ 1
2þ 1p
Z10
sinx tx
dx ð3:41Þ
Fig. 3.3 Response of an ideal low-pass filter to rectangular signal
50 3 Fourier Transform
A similar result can be obtained by extending the window from [0, T] to [0, ∞]in Eq. (3.39). We then have
yðtÞ ¼ 1p
Zx0t
0
sin xx
dx�Zx0ðt�1Þ
0
sin xx
dx
264
375 ð3:42Þ
Substituting
Zx0ðt�1Þ
0
sin xx
dx ¼Z�10
sin xx
dx ¼ � p2
ð3:43Þ
into Eq. (3.42), we have
yðtÞ ¼ 12þ 1p
Zx0t
0
sin xx
dx ¼ 12þ 1p
Zx0
0
sinxx
dx ð3:44Þ
Let ω0 → ∞, then, we finally obtain
yðtÞ ¼ 12þ 1p
Z10
sinxx
dx ð3:45Þ
.
3.4 Discrete Fourier Transform
The discrete Fourier transform DFT and its inverse transform are
X½k� ¼XN�1n¼0
x½n� e�j 2 p k nN ; k ¼ 0; 1; . . .; N � 1 ð3:46Þ
and
x½n� ¼ 1N
XN�1k¼0
X½k� ej 2p k nN ; n ¼ 0; 1; . . .; N � 1 ð3:47Þ
.
3.3 Example of Fourier Transforms 51
3.4.1 Fast Fourier Transform
Fast Fourier Transform FFT is an algorithm for determining the DFT and the IDFTwith fewer arithmetic operations. FFT allows to reduce the N2 multiplications andthe N additions required in the DFT to approximately N
2 log2 N:Reducing the number of calculations for the DFT can be realized in many ways.
We will present two of them as an example:
1. Decomposition of N point DFT in two N/2 point DFTThis is a very effective procedure for determining the DFT, provided that thesize of the DFT is the total power of two. The method is based on the division ofthe number of samples of the input signal
xðnÞ ¼ xð0Þ; xð1Þ; . . .; xðN � 1Þ ð3:48Þ
into two parts
xð0Þ; xð1Þ; . . .; x N2� 1
� �ð3:49Þ
and
xN2
� �; x
N2þ 1
� �; . . .; x N � 1ð Þ ð3:50Þ
The DFT Eq. (3.46) of both sequence of samples (3.48) and (3.49) is
X½k� ¼XðN=2Þ�1n¼0
x½n�e�j 2 p k nN þXN�1n¼N=2
x½n�e�j 2 p k nN ð3:51Þ
Substituting
n ¼ nþ N=2 ð3:52Þ
into the second component of the sum (3.51) gives
X½k� ¼XðN=2Þ�1n¼0
x½n�e�j 2p k nN þXðN=2Þ�1n¼0
x nþ N2
� e�j 2pðnþN=2Þ k
N ð3:53Þ
52 3 Fourier Transform
Recalculation of Eq. (3.53) gives
X½k� ¼XðN=2Þ�1n¼0
x½n�e�j 2 p k nN þ e�j kpXðN=2Þ�1n¼0
x nþ N2
� e�j 2p k n
N ð3:54Þ
Taking into account that
e�j kp ¼ ðe�j pÞk ¼ cos p� j sinpð Þk¼ ð�1Þk ð3:55Þ
equation (3.54) takes the form
X½k� ¼XðN=2Þ�1n¼0
x½n� þ ð�1Þkx nþ N2
� � e�j 2p k n
N ð3:56Þ
For even values of k, Eq. (3.56) is
X½k� ¼XðN=2Þ�1n¼0
x½n� þ x nþ N2
� � e�j 2p k n
N ð3:57Þ
while for odd values of k, it is
X½k� ¼XðN=2Þ�1n¼0
x½n� � x nþ N2
� � e�j 2p k n
N ð3:58Þ
Substituting k = 2k for even k and k = 2k + 1 for odd k into Eqs. (3.57) and(3.58) gives
X½2k� ¼XðN=2Þ�1n¼0
x½n� þ x nþ N2
� � e�j 4p k n
N ð3:59Þ
and
X½2k þ 1� ¼XðN=2Þ�1n¼0
x½n� � x nþ N2
� � e�j 2p n
N e�j 4p k n
N ð3:60Þ
Let us introduce in Eqs. (3.59) and (3.60) the following notations
a½n� ¼ x½n� þ x nþ N2
� ; b½n� ¼ x½n� � x nþ N
2
� ð3:61Þ
3.4 Discrete Fourier Transform 53
and
e�j 4p k n
N ¼ e�j 2 p k n
N=2 ð3:62Þ
Then, we have
X½2k� ¼XðN=2Þ�1n¼0
a½n�WnkN=2 ð3:63Þ
and
X½2k þ 1� ¼XðN=2Þ�1n¼0
b½n�WnNW
nkN=2 ð3:64Þ
where (Fig 3.4)
WmN ¼ exp
�j 2 pmN
� �ð3:65Þ
2. Decomposition of two N/2 point DFT into four N/4 point DFT
The method is based on the division of the sequences að0Þ; að1Þ; . . .; a N2 � 1� �
and bð0ÞW0N ; bð1ÞW1
N ; . . .; bN2 � 1� �
WN2�1N
obtained in the first step into four N/4point sequences (Fig 3.5).
Fig. 3.4 Example of the decomposition of an N point DFT into two N/2 point DFT for N = 8
54 3 Fourier Transform
3.5 Short-time Fourier Transform
The short-time Fourier transform (STFT) in the time domain is
STFTTx ðt; f Þ ¼
Zþ1�1
xðsÞwðs� tÞe�j 2 p f sds ð3:66Þ
while in the frequency domain it is
STFTFx ðt; f Þ ¼ e�j 2 pft
Zþ1�1
XðvÞWðv� f Þe�j 2 p v tdv ð3:67Þ
where W(f) is the Fourier spectrum of the time window w(t).The inverse Fourier transform normalized by the window w(t) in t = 0 is
xðtÞ ¼ 1wð0Þ
Zþ1�1
STFTFx ðt; f Þej2 p f tdf ð3:68Þ
The signal spectrum is represented by the second power of STFTFx ðt; f Þ
SSPECx ðt; f Þ ¼ STFTFx ðt; f Þ
2 ð3:69Þ
Fig. 3.5 Example of the decomposition of two N/2 point DFT into four N/4 point DFT for N = 8
3.5 Short-time Fourier Transform 55
In the STFT, a narrow window w(t) gives good time resolution but poor fre-quency resolution. A wide window gives the reverse result. It is thus impossible toattain high resolution in the time and the frequency domain at the same time.
The discrete form STFT is expressed as
DSTFTðn; kÞ ¼Xþ1
m¼�1x½m�w½n� m� ej 2p
N kð Þm; k ¼ 0; 1; . . .; N � 1 ð3:70Þ
In Eq. (3.70), N should be greater than or equal to the number of samples M ofthe window w(n).
3.6 Time Windows
The time windows occurring in Eqs. (3.65) and (3.66) are used for “cutting out” onthe time axis a sector of the signal, in order to perform its spectral analysis. Theapplication of the inverse transform IDFT enables the reproduction of sample seriesfor the signal analyzed.
For a periodic signal, the part used for analysis is a multiple of its period, and theanalysis is referred to as synchronous. In such a case, the best results are obtainedwhen a rectangular time window is applied. That is due to the fact that frequenciesof the signal considered are located exactly at the points, for which the Fourierspectrum is calculated. The spectrum values, except for the fundamental compo-nent, occur at zero points of the Fourier spectrum. The application of windows otherthan rectangular worsens the frequency resolution.
In the case of non-periodic signals, the application of a rectangular window inthe IDFT generates side lobes after transformation. For this reason, the window of
Fig. 3.6 Synchronous spectral analysis
56 3 Fourier Transform
the shape different than the rectangular is applied. Such an analysis is referred to asasynchronous.
Figures 3.6 and 3.7 present synchronous and asynchronous spectral analysiswith the application of a rectangular window and a Hanning window as an example.
3.7 Properties of Time Windows
The time windows used in practice have a specific shape which is symmetricalabout the peak of the curve, which occurs at the midpoint of its width. Due to thesymmetrical shape of the window, its phase spectrum is linear. The limited lengthof a window causes an infinite amplitude spectrum which starts at zero from a valueequal to the sum of the samples. The window spectrum consists of the main lobeand side lobes. An ideal time window should have the main lobe as narrow aspossible, and its side lobes should be as small as possible. This is a contradictoryrequirement, as narrowing of the main lobe causes widening of side lobes. Theexisting windows are a compromise between the two above requirements. How-ever, it is worth underlining that even a small, insignificant change to the windowshape may cause substantial change in the spectrum distribution (Fig 3.8).
The two most important parameters which influence the quality of a time win-dow are as follows:
1. Width of the main lobe Sg, which is the distance between the point of charac-teristics occurring for f = 0 and the point for which the amplitude spectrumachieves the nearest minimum value. Because the width of the main lobe isinversely proportional to the width of the window N, it is convenient to use theproduct Sg N as a window comparative measure.
Fig. 3.7 Asynchronous spectral analysis
3.6 Time Windows 57
2. Attenuation of the side lobe Ps expressed in decibel as the difference betweenthe maximum of the main lobe and maximum of the highest side lobe.Eqs. (3.70)–(3.79) present formulae for the discrete and continuous timewindows.
– Rectangular window
Rectang ðnÞ ¼ 1 for n ¼ 1; 2; . . .;N
Rectang ðtÞ ¼ 1 for t 2 0; Tð Þ ð3:71Þ
– Triangular window
TriangðnÞ ¼ 1� 2n� NN
for n ¼ 1; 2; . . .;N
TriangðtÞ ¼ 1� 2t � TT
for t 2 0; Tð Þ
ð3:72Þ
– Barlett window
BarlettðnÞ ¼ N � 12� n� N � 1
2
for n ¼ 1; 2; . . .; N
BarlettðtÞ ¼ T2� t � T
2
for t 2 0; Tð Þ
ð3:73Þ
– Hanning window
HanðnÞ ¼ 0:5� 0:5 cos2p nN
� �for n ¼ 1; 2; . . .; N
HanðtÞ ¼ 0:5� 0:5 cos2p tT
� �for t 2 0; Tð Þ
ð3:74Þ
Fig. 3.8 Shape of a typical time window and its amplitude spectrum
58 3 Fourier Transform
– Hamming window
HamðnÞ ¼ 0:54� 0:46 cos2p nN
� �for n ¼ 1; 2; . . .;N
HamðtÞ ¼ 0:54� 0:46 cos2p tT
� �for t 2 0; Tð Þ
ð3:75Þ
– Blackman window
BlackðnÞ ¼ 0:42� 0:5 cos2p nN � 1
� �þ 0:08 cos
4p nN � 1
� �for n ¼ 1; 2; . . .;N
BlackðtÞ ¼ 0:42� 0:5 cos2p tT
� �þ 0:08 cos
4p tT
� �for t 2 ð0; TÞ
ð3:76Þ
– Gaussian window
GaussðnÞ ¼ exp �0:5 n N�12
r N�12
!224
35 for n ¼ 1; 2; . . .;N; r� 0:5
GaussðtÞ ¼ exp �0:5 n T2
r T2
� �2" #
for t 2 ð0; TÞ; r� 0:5
ð3:77Þ
– Flat Top window
FTðnÞ ¼ 0:28� 0:52 cos2pnN � 1
� �� 0:2 cos
4pnN � 1
� �for n ¼ 1; 2; . . .;N
FTðtÞ ¼ 0:28� 0:52 cos2ptT
� �� 0:2 cos
4ptT
� �for t 2 ð0; TÞ
ð3:78Þ
– Exponential window
ExpðnÞ ¼ fn
N�1ð Þ for n ¼ 1; 2; . . .;N; f 2 ð0; 1ÞExpðtÞ ¼ f
tTð Þ for t 2 ð0; TÞ; f 2 ð0; 1Þ
ð3:79Þ
3.7 Properties of Time Windows 59
– Kaiser window
KaðnÞ ¼I0 pa
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 2n
N�1� 1� �2q�
I0ðpaÞ for n ¼ 1; 2; . . .;N; a ¼ 3
where I0ðmÞ ¼ 1þX1k¼1
m2
� �kk!� zero order Bessel function
KaðtÞ ¼I0 pa
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 2t
T � 1� �2q�
I0ðpaÞ for t 2 ð0; TÞ; a ¼ 3
where I0ðxÞ ¼ 1þX1k¼1
x2
� �kk!
ð3:80Þ
3.8 Fourier Series
In measurement practice, we usually deal with two types of signal: periodic signalsand undetermined signals. For periodic signals, the share of particular frequenciesin the original signal is determined by decomposition of the x(t) signal into aFourier series. As a result, we obtain the constant component a0 of this signal,as well as sinusoidal and cosinusoidal components, having frequencies x1; 2x1;3x1; . . .; nx1—Eq. (3.81)
xðtÞ ¼ a0 þX1n¼1
an cosðnx1tÞ þ bn sinðnx1tÞ½ � ¼ a0 þX1n¼1
cn cosðnx1t þ /nÞ½ �
ð3:81Þ
In order to determine coefficients a0, an, bn of the series in Eq. (3.81), let usintegrate it in the limits of T=2; T=2½ �: We obtain then
ZT2
�T2
xðtÞdt ¼ZT
2
�T2
a0dtþX1n¼1
an
ZT2
�T2
cosðnx1tÞdt þ bn
ZT2
�T2
sinðnx1tÞdt
264
375 ð3:82Þ
and, as the right side of Eq. (3.82) is equal to zero, we have
60 3 Fourier Transform
ZT2
�T2
xðtÞdt ¼ZT
2
�T2
a0dt ð3:83Þ
from that
a0 ¼ 1T
ZT2
�T2
xðtÞdt ð3:84Þ
The constant component a0 thus represents the mean value of the signal x(t) in�T=2; T=2½ �: Let us now multiply the series (3.81) by cos(mω1t) and integrate it in�T=2; T=2½ �: We thus obtain
ZT2
�T2
cosðmx1tÞxðtÞdt¼ a0
ZT2
�T2
cosðmx1tÞdt
þX1n¼1
an
ZT2
�T2
cosðnx1tÞ cosðmx1tÞ dt þ bn
ZT2
�T2
sinðnx1tÞ cosðmx1tÞdt
264
375
ð3:85Þ
As we have
ZT2
�T2
cosðmx1tÞdt ¼ 0
ZT2
�T2
sinðnx1tÞ cosðmx1tÞdt ¼ 0
ð3:86Þ
and for m = n
ZT2
�T2
cosðnx1tÞ cosðnx1tÞdt ¼ T2
ð3:87Þ
3.8 Fourier Series 61
equation (3.87) becomes
ZT2
�T2
cosðnx1tÞxðtÞdt ¼ anT2
ð3:88Þ
from that
an ¼ 2T
ZT2
�T2
xðtÞ cosðnx1tÞdt ð3:89Þ
Multiplication of the series (3.81) by sin(mω1t) and integration in the limits of�T=2; T=2½ �; gives
ZT2
�T2
sinðmx1tÞxðtÞdt¼ a0
ZT2
�T2
sinðmx1tÞdt
þX1n¼1
an
ZT2
�T2
cosðnx1tÞ sinðmx1tÞdt þ bn
ZT2
�T2
sinðnx1tÞ sinðmx1tÞdt
264
375ð3:90Þ
As we have
ZT2
�T2
sinðmx1tÞdt ¼ 0;ZT
2
�T2
cosðnx1tÞ sinðmx1tÞdt ¼ 0 ð3:91Þ
and for m = n
ZT2
�T2
sinðnx1tÞ sinðmx1tÞdt ¼ T2
ð3:92Þ
so Eq. (3.85) becomes
ZT2
�T2
sinðnx1tÞxðtÞdt ¼ bnT2
ð3:93Þ
62 3 Fourier Transform
from that
bn ¼ 2T
ZT2
�T2
xðtÞ sinðnx1tÞdt ð3:94Þ
Depending on the shape of the signal x(t), the trigonometric series contain onlyparticular components:
For odd x(t), we have
a0 ¼ 1T
ZT2
�T2
xðtÞdt ¼ 1T
Z0�T2
xðtÞdt þZT
2
0
xðtÞdt
264
375 ¼ 0 ð3:95Þ
an ¼ 2T
ZT2
�T2
xðtÞ cosðnx1tÞdt¼ 2T
Z0�T2
xðtÞ cosðnx1tÞdt
þ 2T
ZT2
0
xðtÞ cosðnx1tÞdt ¼ 0
ð3:96Þ
bn ¼ 2T
ZT2
�T2
xðtÞ sinðnx1tÞdt¼ 2T
Z0�T2
xðtÞ sinðnx1tÞdt
þ 2T
ZT2
0
xðtÞ sinðnx1tÞdt 6¼ 0
ð3:97Þ
and the series (3.81) contain only sinusoidal components.For even x(t), we have
a0 ¼ 1T
Z�T2�T2
xðtÞdt ¼ 2T
ZT2
0
xðtÞdt ð3:98Þ
an ¼ 2T
ZT2
�T2
xðtÞ cosðnx1tÞdt ¼ 4T
ZT2
0
xðtÞ cosðnx1tÞdt ð3:99Þ
3.8 Fourier Series 63
bn ¼ 2T
ZT2
�T2
xðtÞ sinðnx1tÞdt¼ 2T
Z0�T2
xðtÞ sinðnx1tÞdt
þ 2T
ZT2
0
xðtÞ sinðnx1tÞdt ¼ 0
ð3:100Þ
and the series (3.81) contains only cosinusoidal components and the constantcomponent.
Fourier series may be presented in exponential or trigonometric form. Theexponential form of the Fourier series is
xðtÞ ¼X1n¼�1
Anej nx1t ¼ A0 þ A1ej x1t þ A2ej 2x1t þ � � � þ A�1e�j x1t
þ A�2e�j 2x1t þ � � � þ A�ne�j nx1t þ � � �ð3:101Þ
where
A0 ¼ 1T
ZT2
�T2
f ðtÞdt; An ¼ 1T
ZT2
�T2
xðtÞe�jnx1tdt;
A�n ¼ 1T
ZT2
�T2
xðtÞejnx1tdt
ð3:102Þ
Example 3.3 Determine the first 5 components of the Fourier series for the signalx(t) in [−1, 3]—Fig. 3.9
Fig. 3.9 Signal x(t)
64 3 Fourier Transform
The constant component equals
a0 ¼ 14
Z3�1
xðtÞ cos 02pT
t
� �� dt ¼ 1 ð3:103Þ
The particular components an of the Fourier series are
an ¼ 24
Z3�1
xðtÞ cos n2pT
t
� �� dt ð3:104Þ
From (3.103), we have a1 = 0.637, a2 = 0, a3 = −0.212, a4 = 0, and a5 = 0.127,and the Fourier series is the sum
xðtÞ ¼ a0 þX5n¼1
an cos n2pT
t
� �ð3:105Þ
Figures 3.10, 3.11 and 3.12 present particular components of these sums andsignal corresponding to them.
Fig. 3.10 Components x1(t) and x3(t) of the signal x(t)
3.8 Fourier Series 65
3.9 Examples in MathCad
Discrete windows
N :¼ 100
n :¼ 0; 1; . . .;N � 1
Rectangular
Rect nð Þ :¼ 1
Fig. 3.12 Sum of components x013 ðtÞ ¼ a0 þ x1ðtÞ þ x3ðtÞ and x0135ðtÞ ¼ a0 þ x1ðtÞ þ x3ðtÞþx5ðtÞ
Fig. 3.11 Components x5(t) and x01ðtÞ ¼ a0 þ x1ðtÞ
66 3 Fourier Transform
0 20 40 60 80 1000.999
0.9995
1
1.0005
1.001
Rect(n)
nTriangular
TriangðnÞ :¼ 1� 2 � n� NN
0 20 40 60 80 1000
0.2
0.4
0.6
0.8
1
Triang(n)
nBartlett
BartðnÞ :¼ N � 12� n� N � 1
2
0 20 40 60 80 1000
10
20
30
40
50
Bart(n)
nHanning
HanðnÞ :¼ 0:5� 0:5 � cos 2 � p � nN
� �
3.9 Examples in MathCad 67
0 20 40 60 80 1000
0.2
0.4
0.6
0.8
1
Han(n)
nHamming
HamðnÞ :¼ 0:54� 0:46 � cos 2 � p � nN
� �
0 20 40 60 80 1000
0.2
0.4
0.6
0.8
1
Ham(n)
n
Gaussianσ := 0.4
�0:5 � n� N�12
r � N�12
!2
Gaussa(n) := e
0 20 40 60 80 1000
0.2
0.4
0.6
0.8
1
Gaussa(n)
n
68 3 Fourier Transform
Flat top
a0 :¼ 0:28 a1 :¼ 0:52 a2 :¼ 0:2
FTðnÞ :¼ a0 � a1 � cos 2 � p � nN
� �þ a2 � cos 4 � p � n
N
� �� �
0 20 40 60 80 1000.5−
0
0.5
1
FT(n)
nExponential
f :¼ 0:2
ExpðnÞ :¼ fn
N�1ð Þ
0 20 40 60 80 1000.2
0.4
0.6
0.8
1
Exp(n)
nBlackman
a0 :¼ 0:41 a1 :¼ 0:5 a2 :¼ 0:08
FTðnÞ :¼ a0 � a1 � cos 2 � p � nN � 1
� �þ a2 � cos 4 � p � n
N � 1
� �� �
3.9 Examples in MathCad 69
0 20 40 60 80 1000.2−0
0.2
0.4
0.6
0.8
1
FT(n)
nSynchronous discrete analysis of signal x(t) for rectangular and Hanning
windows
D :¼ 0:001 T :¼ 1 f :¼ 5
t :¼ 0;D; . . .; T
xðtÞ :¼ sinð2 � p � f � tÞ
0 0.2 0.4 0.6 0.8 11−
0.5−
0
0.5
1
x(t)
tDiscretization of x(t)
xn :¼for i 2 0; . . .;
TD
xi xði � DÞX
n :¼
for i 2 0; . . .;TD
ni i
n
N ¼ TD
70 3 Fourier Transform
0 6001−
0.5−
0
0.5
1
xn
Spectral analysis of the product of signal x[n] and rectangular window
DFTSPEC Rec x :¼
for k 2 0; . . .;N � 1
DFTSPEC Rec xk 1N
PN�1n¼0
1 � xnn � e�i�2�p�k�n
N
� �
DFTSPEC Rec x
k :¼ 0; . . .; 20
0 5 10 15 200
0.1
0.2
0.3
0.4
0.5
DFTSPEC_Rec_xk
kHanning window
HanðnÞ :¼ 0:5� 0:5 � cos 2 � p � nN
� �
0 400 8000
0.2
0.4
0.6
0.8
1
Han(n)
n
3.9 Examples in MathCad 71
Han x :¼for k 2 0; . . .;N
Han xk HanðkÞ � xnkHanx
0 400 8001−
0.5−
0
0.5
1
Han_x
nSpectral analysis of the product of the signal x[n] and Hanning window
DFTSPEC Han xk :¼
for k 2 0; . . .;N � 1
DFTSPEC Han xk 1N
XN�1n¼0
Han xn � e�i�2�p�k�nN
� �
DFTSPEC Han x
0 5 10 15 200
0.1
0.2
0.3
DFTSPEC_Han_xk
kAsynchronous discrete analysis of the signal for rectangular and Hanning
windowsSignal shifted by 100 samples to the left. Analysis refers to 900 samples.
D :¼ 0:001 T :¼ 1 f :¼ 5
t :¼ 0; D; . . .; T
xðtÞ :¼ sinð2 � p � f � tÞshift :¼ 100
72 3 Fourier Transform
xsn :¼for i 2 0; . . .;
TD� shift
xsn xði � DÞxsn
n1 :¼
for i 2 0; . . .;TD� shift
n1 i
n1
N1 :¼ T
D� shift
Spectral analysis of the product of the signal x[n] and rectangular window
DFTSPEC Rec x :¼
for k 2 0; . . .;N1
DFTSPEC Rec xk 1N1
XN1�1
n¼0
1 � p � k � nN1
� �
DFTSPEC Rec x
k :¼ 0; . . .; 20
0 5 10 15 200
0.1
0.2
0.3
0.4
DFTSPEC_Rec_xk
k
Spectral analysis of the product of the signal x[n] and Hanning window
DFTSPEC Han x :¼
for k 2 0; . . .;N1 � 1
DFTSPEC Han xk 1N1
PN1�1
n¼0Han xn � e
�1�p�k�nN1
� �
DFTSPEC Han xk
0 5 10 15 200
0.1
0.2
0.3
DFTSPEC_Han_xk
k
3.9 Examples in MathCad 73
STFT in MathCad for rectangular window
D :¼ 0:05 T :¼ 10 F :¼ 8
t :¼ 0; D; . . .; T
f 1 :¼ 3 f 2 :¼ 5 f 3 :¼ 1 f 4 :¼ 7
f :¼ 0; D. . .F
xðtÞ :¼
0 if 0� t \ 1sin 2 � p � f1 � tð Þ if 1� t \ 20 if 2� t \ 4sin 2 � p � f2 � tð Þ if 4� t \ 50 if 5� t \ 7sin 2 � p � f3 � tð Þ þ sin 2 � p � f4 � tð Þ if 7� t \ 90 if 9� t \ 10
0 2 4 6 8 102−
1−
0
1
2
x(t)
tSTFT for good time resolution
OtðtÞ :¼ 2 if 0� t \ 0:40 otherwise
����
0 2 4 6 8 100
0.5
1
1.5
2
Ot )(t
t
74 3 Fourier Transform
STETt�SPECðt; f Þ :¼ZT0
xðvÞ � Otðv� tÞ � e�i�2�p�f �vdv
0@
1A
2
Atðt; f Þ :¼STETt�SPECðt; f Þtf
0@
1A
ta :¼ 0 tb :¼ 10
f a :¼ 0 f b :¼ 8
grida :¼ 30 gridb :¼ 30
St :¼ CreateMesh ðAt; ta; tb; f a; f b; grida; gridb
StSTFT for good frequency resolution
Of ðtÞ :¼ 2 if 0� t \ 60 otherwise
0 2 4 6 8 100
0.5
1
1.5
2
Of )(t
t
3.9 Examples in MathCad 75
STFTt�SPFCðt; f Þ :¼ZT0
xðvÞ � O2ðv� tÞ � e�i�2�p�f �vdv
0@
1A
2
Atðt; f Þ :¼STFTt�SPECðt; f Þtf
0@
1A
Sf :¼ CreateMesh ðAf ; ta; tb; f a; f b; grida; gridbÞ
Sf
STFT for resolution being a compromise between time and frequency
Otf ðtÞ :¼ 2 if 0� t \ 0:20 otherwise
0 2 4 6 8 100
0.5
1
1.5
2
Otf )(t
t
76 3 Fourier Transform
STFTtf �SPECðt; f Þ :¼ZT0
xðvÞ � Otf ðv� tÞ � e�i�2�p�f �vdv
0@
1A
2
Atf ðt; f Þ :¼STFTtf �SPECðt; f Þtf
0@
1A
Stf :¼ CreateMesh ðAtf ; ta; tb ; f a ; f b; grida; gridbÞ
StfFourier seriesEven function
t :¼ �2;�0:99; . . .; 2
xðtÞ :¼
x �1 if � 2� t \ � 1
x t if � 1� t \ 0
x �t if 0� t \ 1
x �1 if 1� t \ 2
x
2− 1− 0 1 21−
0.8−
0.6−
0.4−
0.2−
x t( )
t
3.9 Examples in MathCad 77
N :¼ 5
T :¼ 4 n :¼ 1; . . .;N
a0 :¼ 1T�Z2�2
xðtÞ � cos 0 � 2 � pT� t
� �� �dt
an :¼ 2T�Z2�2
xðtÞ � cos n � 2 � pT� t
� �� �dt
x1ðtÞ :¼ a0 þXNn¼1
an � cos n � 2 � pT� t
� �
2− 1− 0 1 21.5−
1−
0.5−
0
)(tx
x1 )(t
tOdd function
xðtÞ :¼
x �1 if � 8� t \ � 6x 1 if � 6� t \ � 4x 0 if � 4� t \ � 2x �1 if 0� t \ 2x 1 if 2� t \ 4x 0 if 4� t \ 6x �1 if 6� t \ 8x
5− 0 51−
0
1
)(tx
t
78 3 Fourier Transform
N :¼ 20
T :¼ 16 n :¼ 1; . . .;N
bn :¼ 2T�Z8�8
xðtÞ � sin n � 2 � pT� t
� �� �dt
x1ðtÞ :¼XNn¼1
bn � sin n � 2 � pT� t
� �
5− 0 5
1−
0
1
)(tx
x1 )(t
t
3.9 Examples in MathCad 79
Chapter 4Z Transform
In Chap. 2, we discussed the Laplace transform, which is widely used in theanalysis of linear systems described by linear differential equations with constantcoefficients. However, many systems are described by means of difference equa-tions referring to discrete moments of time. They occur wherever we deal with A/Dconverters, digital transmission and signal processing, digital filters, etc. For suchsystems, the Z transform realizes a similar mathematical operation to that of theLaplace transform for systems with continuous time. In systems with continuoustime, in which inputs and outputs are represented by means of differential equa-tions, the Laplace transform enables solving them, and transfer function enablestheir description. Difference equations describing systems in discrete moments aresolved by means of the Z transform, while the transfer function in Z space is used torepresent them. This chapter is devoted to the methods of determining and appli-cation of the Z transform in the description of systems with discrete data.
Let us consider the x[n] series of samples of the x(t) analog signal. This series,with amplitudes proportional to x(t) is obtained by the process of sampling—Fig. 4.1.
Series of sampling impulses is
dTðtÞ ¼ dðtÞ þ dðt � TÞ þ dðt � 2TÞ þ � � � ð4:1Þ
which can be represented in the simple form
dTðtÞ ¼X1n¼0
dðt � nTÞ ð4:2Þ
Because the values x(t) in the sampling process are read only for t = nT, theoutput signal from the sampling system, if x(t) = 0 for t < 0, is
x nTð Þ ¼X1n¼0
x nTð Þd t � nTð Þ ð4:3Þ
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_4
81
The Laplace transform of Eq. (4.3) gives
LX1n¼0
xðnTÞdðt � nTÞ ¼X1n¼0
xðnTÞe�nTs ð4:4Þ
Let us define Z transform as
z ¼ eTs; ð4:5Þ
and then, the Eq. (4.4) may be expressed as
LX1n¼0
xðnTÞdðt � nTÞ ¼ Z½xðnTÞ� ¼ X½z� ¼X1n¼0
xðnTÞz�n ð4:6Þ
where z is a complex number. The domain of the Z transform is a set of complexnumbers, for which the series (4.6) is convergent.
In order to simplify the notation, the series x[nT] is usually defined by x[n].Then, the formula (4.6) is
Z x n½ �gf ¼ X z½ � ¼X1n¼0
x nð Þz�n ð4:7Þ
Let us consider the transform Z x n½ �gf of sample series x n½ � ¼ an
X z½ � ¼X1n¼0
anz�n ¼X1n¼0
az
� �n
¼ 11� a
z
¼ zz� a
ð4:8Þ
Fig. 4.1 Sampling principle for analog signals
82 4 Z Transform
Transform X[z] Eq. (4.8) has the pole at z = a and zero at z = 0. It represents aninfinite geometric series tending to zero if 0\a\1 and to infinity if a > 1. Thesetwo behaviors are similar to the behavior of the inverse transform X(s) for realnegative poles and for real positive poles, respectively. For a negative value of a,oscillations are generated for which �1\a\0 tends toward zero and for a < − 1tends toward infinity. This time, the two behaviors are similar to the behavior of theinverse transform of X(s) for complex conjugate poles with negative real parts andpositive real parts, respectively (Fig. 4.2).
The significant difference between the systems defined by means of the trans-form X(s) and X[z] lies in the fact that for the generation of oscillations for X(s), atleast two complex conjugate poles are necessary, whereas for oscillations for X[z],one negative pole is enough. In the general case, the stability of the s plane isdetermined by the abscissa of convergence σ, and of the z plane by a circle of radiusexp rTð Þ. For signals that are absolutely integrable, for which σ = 0, we are dealingwith an imaginary axis for plane s and a unitary circle for plane z. The mapping of
Fig. 4.2 Transform X½z� for n ¼ 20 sample series Eq. (4.8)
4 Z Transform 83
the plane s into the plane z results directly from the definition of the transformZ—Eq. (4.5). Substitution s ¼ rþ jx into Eq. (4.5) gives
z ¼ eTs ¼ er T cosxT þ j sinxTð Þ ð4:9Þ
Dividing the imaginary axis of Eq. (4.9) into sections with a width of xs=4;where xs ¼ 2p=T , we get
for x ¼ xs=4 ! e jx T ¼ e jp=2
for x ¼ xs=2 ! e jx T ¼ e jp
for x ¼ 3xs=4 ! e jx T ¼ e j3p=2
and for x ¼ xs ! e jxT ¼ e j2p
In this way, the Z transform maps the ordinate rþ j1 into a circle of radius erT .The situation is identical for the ordinate r� j1. We have now
for x ¼ �xs=4 ! e jx T ¼ e�jp=2;
for x ¼ �xs=2 ! e jxT ¼ e�jp etc. (Fig. 4.3).
In the analysis of properties of systems with continuous time, two importantsignals are used. These are the Dirac delta dðtÞ and the unit step 1ðtÞ. Let usdetermine the Z transforms for these signals. The Z transform of Dirac delta dðtÞEq. (1.34) is (Fig. 4.4)
Z d n½ �gf ¼X1�1
d nð Þ z�n ¼ z�0 ¼ 1 ð4:10Þ
Fig. 4.3 Mapping of s plane into z plane
84 4 Z Transform
The Z transform of Dirac delta dðtÞ shifted by nT of samples is
Z d t � nTð Þ½ � ¼X1n¼0
dðt � nTÞ z�n ¼ z�n; t� 0 ð4:11Þ
The Z transform of unit step (1.41) is (Fig. 4.5)
Z 1 n½ �gf ¼X1n¼0
1 n½ �z�n ¼ 1z0
þ 1z1
þ 1z2
þ � � � ¼ 11� 1
z
¼ zz� 1
for1z
��������\1
ð4:12Þ
4.1 Properties of Z Transform
1. Linearity
Z ax1ðtÞ þ bx2ðtÞ½ � ¼ aX1ðsÞ þ bX2ðsÞ ð4:13Þ
2. Shift in time domain
Z xðt � sÞ½ � ¼ XðzÞ z�s ð4:14Þ
Fig. 4.4 Z transform of Dirac delta d½n�
Fig. 4.5 Z transform of unit step 1ðnÞ
4 Z Transform 85
3. Change of frequency scale
Z e jxtxðtÞ� � ¼ Xðe�jxzÞ ð4:15Þ
4. Convolution
Z x1ðtÞ � x2ðtÞ½ � ¼ X1ðzÞ � X2ðzÞ ð4:16Þ
5. Time reversal
Z xð�tÞ½ � ¼ Xðz�1Þ ð4:17Þ
6. Derivative
Z t xðtÞ½ � ¼ �zdXðzÞdz
ð4:18Þ
7. Transform of sum
ZXm�1
n¼0
x nT½ � ¼ zz� 1
FðzÞ ð4:19Þ
8. Transform of difference
Z x k þ 1½ �T � x kT½ �gf ¼ z� 1ð ÞXðzÞ � zxð0Þ ð4:20Þ
9. Initial value
limt!0
xðtÞ ¼ limz!1XðzÞ ð4:21Þ
10. Finale value
limt!1 xðtÞ ¼ lim
z!11� z�1XðzÞ� � ð4:22Þ
4.2 Determination of Z Transform
For single poles of X(s) the transform X[z] is
X½z� ¼Xmk
resXðskÞ 11� eskTz�1
¼Xmk
resXðskÞ zz� eskT
; k ¼ 1; 2; . . .;m ð4:23Þ
86 4 Z Transform
where sk is the kth pole of the transform X(s), m is the number of single poles, T isthe sampling interval.
For multiple poles, the transform X[z] of X(s) is
X½z� ¼ Prk¼1
resXðsÞ � ð�1Þk�1
ðk�1Þ!dk�1
dsðk�1Þz
z�esT
� ����s¼sw
; k ¼ 1; 2; . . .; r ð4:24Þ
where r is the order of the multiple pole sw, and the residuum of X(s) is given byEq. (2.43).
In the case where single and multiple poles appear simultaneously, the Ztransform is given by the sum of (4.23) and (4.24).
Example 4.1 Determine X[z] of the signal x(t)
xðtÞ ¼ e�at � e�bt
b� að4:25Þ
The Laplace transform of x(t) is
XðsÞ ¼ 1ðsþ aÞðsþ bÞ ð4:26Þ
which at the pole s ¼ �a has the
res1
ðsþ aÞðsþ bÞ����s¼�a
¼ 1b� a
ð4:27Þ
and at the pole, s ¼ �b has the residuum
res1
ðsþ aÞðsþ bÞ����s¼�b
¼ 1a� b
ð4:28Þ
Using the notation of Eq. (4.23) gives
X½z� ¼ 1b� a
zz� e�aT
� zz� e�bT
ð4:29Þ
Example 4.2 Determine X½z� of the XðsÞ transform
XðsÞ ¼ 1
s3ðsþ aÞ2 ð4:30Þ
4.2 Determination of Z Transform 87
The XðsÞ transform has one triple pole at zero
s1 ¼ 0
s2 ¼ 0
s3 ¼ 0
9>=>; k ¼ 1; 2; 3; r ¼ 3 ð4:31Þ
and one double pole at �a
s4 ¼ �a
s5 ¼ �a
)k ¼ 1; 2; 3; r ¼ 2 ð4:32Þ
Residua corresponding to those poles Eq. (2.48) equal
resXðsÞjs1¼0¼12!
d2
ds2ðs� 0Þ3 1
s3ðsþ aÞ2" #
s¼0
¼ 12
6
ðsþ aÞ4�����s¼0
¼ 3a4
for k ¼ 1
ð4:33Þ
resXðsÞjs2¼0¼11!
dds
ðs� 0Þ3 1
s3ðsþ aÞ2" #
s¼0
¼ � 2
ðsþ aÞ3�����s¼0
¼ � 2a3
for k ¼ 2
ð4:34Þ
resXðsÞjs3¼0¼10!
ðs� 0Þ3 1
s3ðsþ aÞ2" #
s¼0
¼ 1
ðsþ aÞ2�����s¼0
¼ 1a2
for k ¼ 3
ð4:35Þ
resXðsÞjs4¼�a¼11!
dds
ðsþ aÞ2 1
s3ðsþ aÞ2" #
s¼�a
¼ � 3s4
����s¼�a
¼ � 3a4
for k ¼ 1
ð4:36Þ
resXðsÞjs5¼�a¼10!
ðsþ aÞ2 1
s3ðsþ aÞ2" #
s¼�a
¼ 1s3
����s¼�a
¼ � 1a3
for k ¼ 2
ð4:37Þ
The components of the transform X[z] corresponding to these residua are
X½z�js1¼0¼3a4
ð�1Þ01ð1� 1Þ!
zz� esT
� �����s¼s1
¼ 3a4
zz� 1
; k ¼ 1 ð4:38Þ
88 4 Z Transform
X½z�j s2¼0 ¼ � 2a3
ð�1Þ11ð2� 1Þ!
dds
zz� esT
� �����s¼s2
¼ � 2a3
TzeTs
ðz� eTsÞ2�����s¼0
¼ 12a2
T2zeTs
ðz� eTsÞ2 þ2T2ze2Ts
ðz� eTsÞ3 !
s¼0
¼ � 2a3
Tz
ðz� 1Þ2 ; k ¼ 2
ð4:39Þ
X½z�j s3¼0 ¼1a2
ð�1Þ2ð3� 1Þ!
d2
ds2z
z� esT
� �����s¼s3
¼ 1a2
T2z
ðz� 1Þ2 þ2T2z
ðz� 1Þ3 !
¼ 12a2
T2zðzþ 1Þðz� 1Þ3 ; k ¼ 3
ð4:40Þ
X½z�js4¼�a¼ � 3a4
ð�1Þ01ð1� 1Þ!
zz� esT
� �����s¼s4
¼ � 3a4
zðz� e�aTÞ ; k ¼ 1 ð4:41Þ
X½z�j s5¼�a ¼ � 1a3
ð�1Þ11ð2� 1Þ!
dds
zz� esT
� �����s¼s5
¼ � 1a3
TzeTs
ðz� eTsÞ2�����s¼�a
� 1a3
Tze�Ta
ðz� e�TaÞ2 ; k ¼ 2
ð4:42Þ
The transform X[z] is the sum of the components
X½z� ¼ 3a4
zz� 1
� 2a3
Tz
ðz� 1Þ2 þ12a2
T2zðzþ 1Þðz� 1Þ3
� 3a4
zðz� e�aTÞ �
1a3
Tze�Ta
ðz� e�TaÞ2 ð4:43Þ
4.3 Changing Sampling Interval
Let us consider a change of sampling interval from T to T1: Let us rewriteEq. (4.23) in the following form
X½z� ¼Xmk
resXðskÞ 11� akðTÞz�1 ; k ¼ 1; 2; . . .;m ð4:44Þ
4.2 Determination of Z Transform 89
where
akðTÞ ¼ eskT ð4:45Þ
Changing T into T1 in Eq. (4.45), we have
akðT1Þ ¼ eskT1 ð4:46Þ
Logs of both sides of Eq. (4.45) gives
sk ¼ 1Tln akðTÞ ð4:47Þ
Substitution (4.47) into (4.46) gives
akðT1Þ ¼ eT1T ln akðTÞ ð4:48Þ
The new X½z� is thus given by
X½z� ¼Xmk
resXðskÞ 11� akðT1Þz�1 ; k ¼ 1; 2; . . .;m ð4:49Þ
Example 4.3 For X ½z� ,
X½z� ¼ 101� 0:012z�1 ð4:50Þ
sampled every 1 s determine X½z� sampled every 0.5 s.From Eq. (4.48), we have
að0:5Þ ¼ e0:51 ln 0:012 ¼ 0:11 ð4:51Þ
Thus, the new model has the form
X½z� ¼ 101� 0:11z�1 ð4:52Þ
4.4 Inverse Z Transform
The transform x½n� inverse to X½z� is
x½n� ¼Xmk¼1
resfzn�1X½zk�g�����z¼zk
; k ¼ 1; 2; . . .;m ð4:53Þ
90 4 Z Transform
where for single poles,
resX½zk� ¼ ðz� zkÞX½z�jz¼zk ð4:54Þ
and for multiple poles,
resX½zk� ¼ 1ðr � 1Þ!
dðr�1Þ
dzðr�1Þ fðz� zwÞrX½z]g�����z¼zk
ð4:55Þ
where r is the order of the multiple pole zw:
Example 4.4 Determine the inverse transform x½n� for
X½z� ¼ zðz� 1Þðz� 2Þðz� 3Þ ð4:56Þ
The transform X[z] has three single poles
z1 ¼ 1; z2 ¼ 2; z3 ¼ 3 ð4:57Þ
Components of x½n� corresponding to those poles are
x½nz1� ¼ res fzn�1X½z�g��z1¼1¼ ðz� 1Þ zn
ðz� 1Þðz� 2Þðz� 3Þ
z¼1¼ 1
2ð4:58Þ
x½nz2� ¼ res fzn�1X½z�g��z2¼2¼ ðz� 2Þ zn
ðz� 1Þðz� 2Þðz� 3Þ
z¼2¼ �2n ð4:59Þ
x½nz3� ¼ res fzn�1X½z�g��z2¼3¼ ðz� 3Þ zn
ðz� 1Þðz� 2Þðz� 3Þ
z¼3¼ 3n
2ð4:60Þ
The transform x½n� has the final form
x½n� ¼ 12� 2n þ 3
2
n
ð4:61Þ
A method often used to determine the transform x½n� is decomposition of X½z�into partial fractions. If X½z� is given in the form of the quotient of two polynomials,we can decompose it into partial fractions
X½z� ¼ LðzÞMðzÞ ¼
Xnk¼1
resX½zk�z� zk
ð4:62Þ
4.4 Inverse Z Transform 91
and then, the transform X½z� is
X½z� ¼ z�mXnk¼1
resX½zk�1� zkz�1 ð4:63Þ
where resX½z� is given by Eq. (4.54).The poles of zk may be located inside or outside of the convergence area. For the
poles zk located in the convergence area, the component of sum (4.63) is
resX ½zk�1� zkz�1 $ resX½zk�ðzkÞn1 ½n� ð4:64Þ
whereas in case of poles zk located outside this area,
resX½zk�1� zkz�1 $ �resX½zk�ðzkÞn1 ½�n� 1� ð4:65Þ
The presentation of X½z� as a sum of fractions (4.63) requires exclusion of theultimate powers of z from polynomials in the numerator LðzÞ and the denominatorMðzÞ and, in consequence, introduction of the common multiplicands zm. Thiscauses shifting of the sum (4.63) forward or back, depending on the sign of m.
Example 4.5 Solve example (4.4) by the method of decomposition X½z� into partialfractions
X½z� ¼ zðz� 1Þðz� 2Þðz� 3Þ ¼
12
1z� 1
� 21
z� 2þ 32
1z� 3
ð4:66Þ
The transform X½z� has three poles 1; 2; 3 and corresponding residua:12 ; �2; 3
2 : Excluding z from denominators of (4.66) gives
X½z� ¼ 1z
12
11� 1z�1 � 2
11� 2z�1 þ
32
11� 3z�1
ð4:67Þ
Taking into account Eq. (4.64), we have
X½n� ¼ 1z
121n � 2� 2n þ 3
2� 3n
ð4:68Þ
If X½z� is multiplied by zk in solving the equations, we make use shift of argu-ment. They are as follows: for shifting the argument to the left
zfx½n� k�g ¼ z�kX½z� ð4:69Þ
92 4 Z Transform
and shifting the argument to the right
zfx½n� k�g ¼ zkX½z� �Xk�1
r¼0
xðrÞzk�r ð4:70Þ
where it is assumed that x½�n� ¼ 0For small values of k, the last formula is reduced to the form
z x½nþ 1� ¼ z X½z� � z x½0� for k ¼ 1 ð4:71Þ
z x½nþ 2� ¼ z2X½z� � z2x½0� � zx½1� for k ¼ 2 ð4:72Þ
z x½nþ 3� ¼ z3X½z� � z3x½0� � z2x½1� � zx½2� for k ¼ 3 ð4:73Þ
Utilizing Eq. (4.69), we evaluate Eq. (4.68) to get
x½n� ¼ 121n�1 � 2� 2n�1 þ 3
2� 3n�1 ¼ 1
2� 2n þ 3n
2ð4:74Þ
Example 4.6 Solve the equation
x½nþ 2� þ 5x½nþ 1� þ 6x½n� ¼ 0 ð4:75Þ
for the initial conditions: x½0� ¼ 2; x½1� ¼ �5Applying Eqs. (4.71) and (4.72), we get
z2X½z� � z2x½0� � z x½1� þ 5ðz X½z� � z x½0�Þ þ 6X½z� ¼ 0 ð4:76Þ
Inserting the initial conditions gives
X½z�ðz2 þ 5zþ 6Þ ¼ z½2z� 5� ð4:77Þ
and
X½z� ¼ z2z� 5
z2 þ 5zþ 6¼ z
2z� 5ðzþ 2Þðzþ 3Þ ð4:78Þ
Hence, the inverse transform of X½z� is
x½n� ¼ zðn�1Þ zð2z� 5Þzþ 3
����z¼�2
þ zðn�1Þ zð2z� 5Þzþ 2
����z¼�3
¼ �9ð�2Þn þ 11ð�3Þn
ð4:79Þ
4.4 Inverse Z Transform 93
4.5 Digital Filters
The idea of digital filters design depends on the calculation of the transfer functionfor an analog filter that meets the assumed requirements and then determining adigital filter, corresponding to the analog one. A popular method of transforminganalog filters designed in the Laplace space into digital filters is to approximate theoperator s to the operator z by means of a bilinear transformation
s ¼ 2Tð1� z�1Þð1þ z�1Þ ð4:80Þ
Hence, we have
KðzÞ ¼ KðsÞjs¼2
Tð1�z�1Þð1þz�1Þ
ð4:81Þ
where T is the sampling interval.The transformation (4.81) represents a nonlinear relationship between the analog
frequency xa and the digital frequency xc. Substituting z ¼ ejxc T into Eq. (4.80)gives
s ¼ jx ¼ 2T1� e�jxcT
1þ e�jxcT¼ 2
Te jxcT=2ðe jxcT=2 � e�jxcT=2Þe jxcT=2ðe jxcT=2 þ e�jxcT=2Þ
¼ 2T
ðe jxcT=2 � e�jxcT=2Þ�j2ðe jxcT=2 þ e�jxcT=2Þ=2 ¼ j2
TsinðxcT=2ÞcosðxcT=2Þ ¼
j2TtanðxcT=2Þ ð4:82Þ
It can be easily confirmed that the last relation is almost linear for small xc\0:5.The frequency characteristic of the filter is
KðejxcÞ ¼ KðzÞjz¼ejxc T ð4:83Þ
where xc ¼ X Tp is the normalized frequency in relation to sampling rate.
Example 4.7 Design the transfer function for a digital filter with a 1-kHz samplingfrequency on the basis of Butterworth low-pass filter
KðsÞ ¼ 1
s2 þ ffiffiffiffi2
psþ 1
ð4:84Þ
94 4 Z Transform
Applying the transform (4.80), we have
KðzÞ ¼ KðsÞ js¼ 2
10�3ð1�z�1Þð1þz�1Þ
¼ 1
210�3
ð1�z�1Þð1þz�1Þ
� �2þ ffiffiffi
2p
210�3
ð1�z�1Þð1þz�1Þ þ 1
¼ 10�6 z2 þ 2zþ 14:003z2 � 7:999zþ 3:997
ð4:85Þ
Substituting z ¼ ejx�10�3
into Eq. (4.85), we get
KðejxcÞ ¼ 106 sinðxc�103Þ2 þ sinðxc
�2� 103Þ2 � 1
1013½0:4 sinðxc=103Þ2 � 1:6 sinðxc=2� 103Þ2 � 1�
þ j
ffiffiffi2
p103½2 sinðxc
�103Þ þ sinð2xc
�103�
1013½0:4 sinðxc=103Þ2 � 1:6 sinðxc=2� 103Þ2 � 1� ð4:86Þ
4.6 Example in MathCad
x :¼ ð1 2 3ÞT
Zðx; zÞ :¼XrowsðxÞ�1
n¼0
xnz�nð Þ
Zvðx; zÞ ! 2zþ 3z2
þ 1
2zþ 3z2
þ 1� �
invztrans ! dðn; 0Þ þ 2� dðn� 1; 0Þ þ 3� dðn� 2; 0Þ
n :¼ 0. . .2
dðn; 0Þ þ 2� dðn� 1; 0Þ þ 3� dðn� 2; 0Þ:
1
3
2
4.5 Digital Filters 95
Chapter 5Wavelet Transform
In Chap. 3, we discussed the Fourier transform, which converts the stationary signalx(t) from the time domain to the frequency domain XðxÞ and thus allows us toperform a frequency analysis. Thanks to this transform, we can determine theamplitudes and frequencies of the sine and cosine making up the signal x(t), but wecannot determine at what time the corresponding amplitude occurs. The STFtransform used in the analysis of non-stationary signals allows us to obtain thedistribution of frequency components in time, but we are faced with the problem ofselecting the appropriate window width. Selection of the wrong width blurs thetime–frequency data obtained as a result of applying the transform. In the wavelettransform, the problem of time–frequency resolution is solved by replacing the timewindow with a wavelet function.
5.1 Continuous Wavelet Transform
The continuous wavelet transform (CWT) is defined as
Wf ða; sÞ ¼Z10
xðtÞWa;sðtÞdt ð5:1Þ
in which
Wa;sðtÞ ¼ 1ffiffiffia
p wt � sa
� �ð5:2Þ
where w is the mother wavelet, s is the shift factor, and a is the scaling factor.When a < 1, the wavelet is narrowed, whereas when a[ 1, the wavelet is
stretched. The 1=ffiffiffia
pfactor normalizes the wavelet. For low values of a, the wavelet
has a maximum instantaneous value and the wavelet decreases as a increases. Themother wavelet W, depending on the form of xðtÞ and the requirements for itsanalysis, fulfills the following conditions:
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_5
97
• The mean value of the wavelet equals zero
Z10
wðtÞdt ¼ 0 ð5:3Þ
• The norm of the wavelet equals one
wðtÞk k ¼ 1 ð5:4Þ
and the integral must be finite
Zþ1
�1
jW xð Þ2jx
dx ¼ finite\1 ð5:5Þ
where W xð Þ is the Fourier transform of wðtÞ:Figure 5.1 presents an example of a wavelet translation along the signal xðtÞ:During the CWT, the wavelet is translated along the signal, and for each of its
translations, the value ofWf ða; sÞ is calculated. After reaching the end of signal, thewavelet is rescaled and shifted back to the beginning of the signal and the procedureis repeated. The data matrix thus obtained is the representation of the signal in thewavelet domain.
The inverse wavelet transform is
xðtÞ ¼ 1Cw
Zþ1
0
Zþ1
0
Wf ða; sÞ 1ffiffiffia
p wt � sa
� �ds
daa2
: ð5:6Þ
5.2 Wavelet Functions
The Morlet wavelet
wðt; rÞ ¼ e�t22 e�itr ð5:7Þ
is used for the estimation of the amplitude–frequency signal components. Parameterr is the acceptability condition.
The Marr wavelet (“Mexican hat”) is used to estimate the extremes of the signaldistribution
98 5 Wavelet Transform
wðtÞ ¼ ð1� t2Þe�t22 ð5:8Þ
The Meyer wavelet is defined only in the frequency domain.
WðxÞ ¼1 for jxj � 2
3 pcos p
2 t34p jxj � 1� �� �
for 23 p� jxj � 4
3 p0 for jxj � 4
3 p
8<: ð5:9Þ
where
tðtÞ ¼ 0 for t� 01 for t� 1
:
ð5:10Þ
Fig. 5.1 CWT procedure, k is a successive wavelet translation
5.2 Wavelet Functions 99
5.3 Discrete Wavelet Transform
The discrete wavelet transform DWT of the signal xn is determined by simultaneousfiltration using a FIR digital filter bank. Outputs of the low-pass filter and the high-pass filter have the following forms
yhn ¼XL�1
l¼0
xlhn�l ð5:11Þ
and
ygn ¼XL�1
l¼0
xlgn�l; n ¼ 0; 1; . . .; L� 1 ð5:12Þ
where L is the number of coefficients of impulse responses hn and gn:Equations (5.11) and (5.12) present the digital convolution. They decompose the
input signal in such a way that the low-pass filter transmits the constant componentand attenuates the component that has the period of p rad. The high-pass filterattenuates the constant component and transmits the component that has a fre-quency of p rad. To meet the above conditions, we have
gn ¼ �ð�1ÞnhL�n�1 ð5:13Þ
and
PL�1
n¼0hn ¼
ffiffiffi2
p
PL�1
n¼0hnhnþ2m ¼ dm; m ¼ 0; 1; . . .; L2 � 1
8>><>>: ð5:14Þ
where
dm ¼ 0 for m 6¼ 01 for m ¼ 0
ð5:15Þ
presents Kronecker delta.The values of the signal samples yhn and ygn denote the coefficients of discrete
wavelet transform. Their sum contains twice the number of samples, and as a result,their coding would require twice the number of memory cells compared to a singlecoding. The reduction to half the number of samples is realized by removing everysecond sample from the output filters.
100 5 Wavelet Transform
Removing the samples is performed using decimators according to
an ¼XL�1
l¼0
xnhn�l
( )#2
ð5:16Þ
dn ¼XL�1
l¼0
xngn�l
( )#2
ð5:17Þ
In Eqs. (5.16) and (5.17), the notation #2 presents convolution of every secondsample. After substituting Eqs. (5.11) and (5.17) into Eqs. (5.11) and (5.12), wehave
an ¼ x2n ¼XL�1
l¼0
xn h2n�l ð5:18Þ
and
dn ¼ x2n ¼XL�1
l¼0
xn g2n�l ð5:19Þ
Equations (5.18) and (5.19) present the two-point decimation, which realize theMallat algorithm (Fig. 5.2).
The signal an at the output of low-pass filter is referred to as the approximation(trend), whereas the signal dn at the output of high-pass filter is referred to as thedetail (fluctuation).
Figure 5.3 presents the diagram of the multistage wavelet transform, for whichthe Eqs. (5.18) and (5.19) are
Fig. 5.2 DWT with decimators
5.3 Discrete Wavelet Transform 101
amn ¼XL�1
l¼0
amþ1lh2n�l ð5:20Þ
dmn ¼XL�1
l¼0
amþ1lg2n�l ð5:21Þ
whereM is the number of decomposition stages, and m ¼ 1; 2; . . .;M. This way, thesignal xn is presented as a sum of the approximations of the last level of a1n anddetails d1n; . . .dmn; dmþ1n from all stages of decomposition.
Reconstruction of the coefficient amþ1n is realized by means of an inverse dis-crete wavelet transform (IDWT). It is
amþ1n ¼XL�1
l¼0
faml~hn�2l þ dml~gn�2lg ð5:22Þ
where ~h and ~g represent the impulse responses of the filters.In order to provide ideal reconstruction of the signal, equating ~xn with xn, the
Z transforms of filters hn; gn; ~hn; ~gn must meet the following conditions
HðzÞ~HðzÞ þ GðzÞ~GðzÞ ¼ 2 ð5:23Þ
Hð�zÞ~HðzÞ þ Gð�zÞ~GðzÞ ¼ 0 ð5:24Þ
Figure 5.4 presents signal reconstruction by means of a filters bank withexpanders. These insert a zero value between every second sample.
Fig. 5.3 Diagram of multistage wavelet transform
102 5 Wavelet Transform
5.4 Discrete Wavelets
In the DWT analysis, the wavelets used are generated indirectly by determinationof coefficients of the filters gn and hn—Eqs. (5.13) and (5.14). In practical appli-cations, the most popular discrete wavelets are the Harr and the Daubechieswavelets.
For Haar wavelets of length L ¼ 2; we have
h0 þ h1 ¼ffiffiffi2
ph20 þ h21 ¼ 1
ð5:25Þ
hence,
h0 ¼ffiffiffi2
p
2; h1 ¼
ffiffiffi2
p
2ð5:26Þ
Substituting Eq. (5.26) into Eq. (5.13) gives
g0 ¼ffiffiffi2
p
2; g1 ¼ � ffiffiffi
2p
2ð5:27Þ
For Daubechies wavelets of order higher than 2, we have
PL�1
n¼0hn ¼
ffiffiffi2
p
PL�1
n¼2mhnhnþ2m ¼ dm for m ¼ 0; 1; 2; . . .; L2 � 1
PL�1
q¼0qkð�1ÞqhL�1�q ¼ 0 for k ¼ 1; 2; . . .; L2 � 1
8>>>>>>><>>>>>>>:
ð5:28Þ
Fig. 5.4 IDWT with expanders
5.4 Discrete Wavelets 103
A typical set of equations for a filter of the length L = 4 is a system of followingequations
h0 þ h1 þ h2 þ h3 ¼ffiffiffi2
ph20 þ h21 þ h22 þ h23 ¼ 1h0h2 þ h2h3 ¼ 00h3 � 1h2 þ 2h1 � 3h0 ¼ 0
8>><>>: ð5:29Þ
The solution of Eq. (5.29) gives
h0 ¼ 1þ ffiffiffi3
p
4ffiffiffi2
p ; h1 ¼ 3þ ffiffiffi3
p
4ffiffiffi2
p
h2 ¼ 3� ffiffiffi3
p
4ffiffiffi2
p ; h3 ¼ 1� ffiffiffi3
p
4ffiffiffi2
pð5:30Þ
Hence, by Eqs. (5.13) and (5.30), we have high-pass filter parameters
g0 ¼ h3; g1 ¼ �h2; g2 ¼ h1; g3 ¼ �h0 ð5:31Þ
The modifications of the Daubechies wavelets are Coiflet and Symplet wavelets.The values of the filter coefficients gn and hn; for these wavelets, are characterizedby a symmetry which is close to the ideal.
104 5 Wavelet Transform
5.5 Example of Three-Stage Wavelet Transformin LabVIEW
5.5 Example of Three Stage Wavelet Transform in LabVIEW 105
Chapter 6Hilbert Transform
The Hilbert transform H[x(t)] presents the integral convolution of the signalsx(t) and g(t)
H xðtÞ½ � ¼ ~xðtÞ ¼Z1�1
xðsÞgðt � sÞds ð6:1Þ
in which
gðtÞ ¼ 1p t
ð6:2Þ
The Hilbert transform thus has the form
~xðtÞ ¼ 1p
Z1�1
xðsÞt � s
ds ð6:3Þ
while for t ¼ s, the integralR1
�1xðsÞt�sds should be considered in the sense of the
Cauchy principal value, so
Z1�1
xðsÞt � s
ds ¼ lime!0
Zt�e
�1
xðsÞt � s
dsþZ1tþe
xðsÞt � s
ds
24
35 ð6:4Þ
The Inverse Hilbert transform is
H�1 xðtÞ½ � ¼ � 1p
Z1�1
~xðtÞt � s
ds ð6:5Þ
Let the Hilbert transform be given as
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_6
107
H xðtÞ½ � ¼ 1pt
� xðtÞ ð6:6Þ
or in the frequency domain as a product of spectra
HðxÞ ¼ KðxÞXðxÞ ¼ �j sgnðxÞXðxÞ ð6:7Þ
where �j sgnðxÞ is the spectrum of 1pt :
The spectrum KðxÞ in Eq. (6.7) has the form
KðxÞ ¼ �j sgnðxÞ ¼�j for x[ 0j for x\ 00 for x ¼ 0
8<: ð6:8Þ
or
KðxÞ ¼e�jðp=2Þ for x[ 0ejðp=2Þ for x\ 00 for x ¼ 0
8<: ð6:9Þ
Substituting Eq. (6.8) into Eq. (6.7), we have
HðxÞ ¼�jXðxÞ for x[ 0jXðxÞ for x\ 00 for x ¼ 0
8<: ð6:10Þ
From Eq. (6.10), we can see that the spectrum of the Hilbert transform HðxÞdiffers from the spectrum XðxÞ only in that the two halves of the spectrum XðxÞ aremultiplied, depending on the sign of x, by either j or �j, that is their phases areshifted by 90�. From Eq. (6.8), it is easy to see that KðxÞj j ¼ 1 for all values of xand that the argument equals
argKðxÞ ¼ �p=2 for x[ 0p=2 for x\ 0
�ð6:11Þ
For this reason, the Hilbert transform is often referred to as the phase shifter.Figure 6.1 presents the Hilbert transform characteristics KðxÞj j and argKðxÞ:The Hilbert transform is used to determine complex analytic signals xaðtÞ: The
real part of xaðtÞ is the original signal xðtÞ, and the imaginary part is its Hilberttransform ~xðtÞ
xaðtÞ ¼ xðtÞ þ j~xðtÞ ð6:12Þ
108 6 Hilbert Transform
Let us consider the analytic signal xa of two variables, real r and imaginary u
xa ¼ rðv; yÞ þ juðv; yÞ ð6:13Þ
The derivatives of the signal xa are calculated from the following relations
_xa ¼ drdv
þ jdudy
ð6:14Þ
or
_xa ¼ dudy
� jdrdv
ð6:15Þ
The analytic signal meets the conditions
drdv
¼ dudy
anddrdy
¼ � dudv
ð6:16Þ
For example
xa ¼ evþjy ¼ ev cos yþ jev sin y ¼ rðv; yÞ þ juðv; yÞ ð6:17Þ
is analytic, because the conditions (6.16) are satisfied
drdv
¼ ev cos y;dudy
¼ ev cos y ð6:18Þ
and
drdy
¼ �ev sin y; � dudv
¼ �ev sin y ð6:19Þ
Fig. 6.1 The Hilbert transform characteristics
6 Hilbert Transform 109
The signal described in Eq. (6.13) may be presented in the exponential form
xaðtÞ ¼ EðtÞejuðtÞ ¼ EðtÞ cosuðtÞ þ j sinuðtÞ½ � ð6:20Þ
where EðtÞ is the envelope of the signal
EðtÞ ¼ � xaðtÞj j ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2ðtÞ þ ~x2ðtÞ
pð6:21Þ
and uðtÞ is its phase
uðtÞ ¼ arctg~xðtÞxðtÞ ð6:22Þ
The derivative of uðtÞ presents the phase frequency
xðtÞ ¼ duðtÞdt
¼ x2ðtÞE2ðtÞ ð6:23Þ
Figure 6.2 presents signal xðtÞ ¼ A cosðxtÞ, its transform ~xðtÞ ¼ A sinðxtÞ, andenvelopes �EðtÞ.
6.1 Examples of Hilbert Transform
H sinðtÞ ¼ � cosðtÞ ð6:24Þ
H cosðtÞ ¼ sinðtÞ ð6:25Þ
Fig. 6.2 Signals xðtÞ;~xðtÞ and their envelopes �EðtÞ
110 6 Hilbert Transform
HsinðtÞt
¼ 1� cosðtÞt
ð6:26Þ
HdðtÞ ¼ 1pt
ð6:27Þ
H _dðtÞ ¼ � 1pt2
ð6:28Þ
H€dðtÞ ¼ 2pt3
ð6:29Þ
Hejt ¼ �jejt ð6:30Þ
He�jt ¼ je�jt ð6:31Þ
Hejbt ¼ j sgnðbÞejbt ð6:32Þ
x1ðtÞ; x2ðtÞh i ¼ ~x1ðtÞ;~x2ðtÞh i ð6:33Þ
xðtÞ; xðtÞh i ¼ ~xðtÞ;~xðtÞh i ð6:34Þ
xðtÞ;~xðtÞh i ¼ 0 ð6:35Þ
Hx1ðtÞ; x2ðtÞh i ¼ x1ðtÞ � Hx2ðtÞh i ð6:36Þ
H½x1ðtÞ � x2ðtÞ� ! ~x1ðtÞ � x2ðtÞ ¼ x1ðtÞ � ~x2ðtÞ ¼ �~x1ðtÞ � ~x2ðtÞ ð6:37Þ
H c1x1ðtÞ þ c2x2ðtÞ½ � ¼ c1~x1ðtÞ þ c2~x2ðtÞ ð6:38Þ
H xðtÞxðtÞ½ � ¼ xðtÞ~xðtÞ ð6:39Þ
H½c� ¼ 0 ð6:40Þ
H xðtÞ þ c½ � ¼ H xðtÞ½ � þ HðcÞ ¼ ~xðtÞ ð6:41Þ
H xðatÞ½ � ¼ sgn a~xðatÞ ð6:42Þ
H�1xðtÞ ¼ �HxðtÞ ð6:43Þ
H2 xðtÞ½ � ¼ �xðtÞ ð6:44Þ
H4 xðtÞ½ � ¼ xðtÞ ð6:45Þ
HdxðtÞdt
� �¼ d
dtH xðtÞ½ � ð6:46Þ
6.1 Examples of Hilbert Transform 111
HdkxðtÞdtk
� �¼ dk
dtkH xðtÞ½ � ð6:47Þ
Figures 6.3 and 6.4 show examples of analytic signals in the complex coordinatesystem. Figure 6.3 presents xðtÞ ¼ cosðxtÞ, ~xðtÞ ¼ sinðxtÞ, and xaðtÞ ¼ cosðxtÞþj sinðxtÞ. Figure 6.4 presents xðtÞ ¼ e�at cosðxtÞ, ~xðtÞ ¼ e�at sinðxtÞ, andxaðtÞ ¼ e�at cosðxtÞ þ j sinðxtÞ½ �.
Fig. 6.4 The signal xðtÞ (harmonic horizontal function), its transform ~xðtÞ (harmonic verticalfunction), and analytic signal xaðtÞ (conic helix)
Fig. 6.3 The signal xðtÞ (harmonic horizontal function), its transform ~xðtÞ (harmonic verticalfunction), and analytic signal xaðtÞ (circular helix)
112 6 Hilbert Transform
6.2 Examples in MathCad
Example referring to Fig. 6.2
i :¼ 0; . . .; 1000 A :¼ 2 x0 :¼ 0:025
xi :¼ A � cos x0 � ið Þx1 :¼ hilbertðxÞ
Ei :¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðxiÞ2 þ ðx1iÞ2
q
0 200 400 600 800 1 103×
−1
−0.5
0
0.5
1
xi
i
0 200 400 600 800 1 103×
−2
−1
0
1
2
x1i
Ei
Ei−
i
6.2 Examples in MathCad 113
Example referring to Fig. 6.3
t :¼ �10;�9:999; . . .; 10
x :¼ 1
X1ðtÞ :¼cosðx � tÞsinðx � tÞ
t
0@
1A X2ðtÞ :¼
0sinðx � tÞ
t
0@
1A X3ðtÞ :¼
cosðx � tÞ0t
0@
1A
t1 :¼ �10 t2 :¼ 10 grid :¼ 500
A1 :¼ CreateSpaceðF1; t1; t2; gridÞ
A2 :¼ CreateSpaceðF2; t1; t2; gridÞ
A3 :¼ CreateSpaceðF3; t1; t2; gridÞ
114 6 Hilbert Transform
Example referring to Fig. 6.4
t :¼ �2;�1:999; . . .; 2
a :¼ 0:6 x :¼ 10
x1ðtÞ :¼ e�a�t � cosðx � tÞ x2ðtÞ :¼ e�a�t � sinðx � tÞ
X2ðtÞ :¼0
x1ðtÞt
0@
1A X1ðtÞ :¼
x1ðtÞx2ðtÞt
0@
1A X3ðtÞ :¼
x2ðtÞ0t
0@
1A
t1 :¼ �2 t2 :¼ 2 grid :¼ 500
A1 :¼ CreateSpaceðX1; t1; t2; gridÞ
A2 :¼ CreateSpaceðX2; t1; t2; gridÞ
A3 :¼ CreateSpaceðX3; t1; t2; gridÞ
6.2 Examples in MathCad 115
116 6 Hilbert Transform
Chapter 7Orthogonal Signals
Orthogonal signals are commonly used in various practical and theoretical appli-cations, in particular in metrology, automatic control engineering, medicine, com-munication, approximation theory, theory of polynomials, and many other fields.For that reason they play an important role in the theory of signals.
The set of signals {x(t)} is named orthogonal over the interval [a, b] with respectto the weight function w(t), if
Zb
a
wðtÞxjðtÞ xkðtÞ dt ¼ 0 for j 6¼ kak [ 0 for j ¼ k
�ð7:1Þ
If additionally ak ¼ 1 for each k = 0, 1, …, n, and the energy of the signalsequals one
Ex ¼Zb
a
xkðtÞj j2 dt ¼ 1 ð7:2Þ
then these signals are orthonormal. It is easy to see that sets of signals
fxðtÞg ¼ x0ðtÞ; x1ðtÞ; . . .; xnðtÞf g; k ¼ 1; 2; . . .; n ð7:3Þ
in which
xkðtÞ ¼ Ak sinð2p kf0 tÞ T0 ¼ 1=f0 ð7:4Þ
and
fyðtÞg ¼ y0ðtÞ; y1ðtÞ; . . .; nnðtÞf g; k ¼ 1; 2; . . .; n ð7:5Þ
in which
ykðtÞ ¼ Bk cosð2p kf0 tÞ T0 ¼ 1=f0 ð7:6Þ
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_7
117
are orthogonal over [0, T0], because
ZT00
Ak sinð2pkf0tÞAm sinð2pmf0 tÞdt
¼ZT00
Ak cosð2pkf0tÞAm cosð2pmf0 tÞdt ¼ 0; k 6¼ m
ð7:7Þ
ZT00
Ak sinð2pkf0 tÞAm sinð2pmf0tÞdt ¼ �AkAm ½sinð4pkÞ � 4pk�8p f k
¼ ak [ 0; k ¼ m
ð7:8Þ
ZT00
Ak cosð2pkf0 tÞAm cosð2pmf0 tÞdt ¼ AkAm ½sinð4pkÞ þ 4pk�8pfk
¼ ak [ 0; k ¼ m
ð7:9Þ
Let the amplitude Ak and Bk be
Ak ¼ Bk ¼ffiffiffiffiffiffiffiffiffiffi2=T0
p; k ¼ 1; 2; . . .; n ð7:10Þ
then the signals of the sets {x(t)} and {y(t)} over [0, T0] are orthonormal, because
ZT00
2T0
sinð2pkf0 tÞ2dt ¼ZT00
2T0
cosð2pkf0 tÞ2dt ¼ 1; k ¼ 1; 2; . . .; n ð7:11Þ
The signals of sets {xk(t)} and {yk(t)}, k = 1, 2, …, n are also mutually ortho-normal, because
ZT00
2T0
sinð2pkf0tÞ cosð2pmf0tÞ dt
¼ZT00
1T0fsin½ðk � mÞ2pf0t� þ sin½ðk þ mÞ2pf0t�g dt
¼ ðk þ mÞ sin½pðk � mÞ�2 þ ðk � mÞ sin½pðk þ mÞ�2pðk2 � m2Þ ¼ 0; k 6¼ m
ð7:12Þ
118 7 Orthogonal Signals
If the signals in the sets {xk(t)} and {yk(t)} have different phase shifts, they arealso orthonormal, because for {xk(t)}, k = 1, 2, …, n, we have
ZT00
xkðtÞ xmðtÞdt ¼ZT00
ffiffiffiffiffi2T0
rsinð2p kf0 t þ ukÞ
ffiffiffiffiffi2T0
rsinð2pmf0 t þ umÞdt
¼ sin½ðk � mÞ2pþ ðuk � umÞ�ðk � mÞ 2p � sin½ðk þ mÞ2p þ ðuk þ umÞ�
ðk þ mÞ2p� �
� sinðuk � umÞðk � mÞ 2p �
sinðuk þ umÞðk þ mÞ2p
� �¼ 0; k 6¼ m ð7:13Þ
and
ZT00
xkðtÞj j2 dt ¼ 2T0
ZT00
sinð2p kf0 t þ ukÞ2dt
¼ � sinð4p k þ 2ukÞ þ sinð2ukÞ þ 4p k4p k
¼ 1 ð7:14Þ
Similarly we can check orthonormality for signals of the set {yk(t)},k = 1, 2, …, n for which we have
ZT00
ykðtÞ ymðtÞdt ¼ZT00
ffiffiffiffiffi2T0
rcosð2p kf0 t þ ukÞ
ffiffiffiffiffi2T0
rcosð2pmf0 t þ umÞdt
¼ sin½ðk � mÞ2pþ ðuk � umÞðk � mÞ2p þ sin½ðk þ mÞ2pþ ðuk þ umÞ
ðk þ mÞ2p� �
� sinðuk � umÞðk � mÞ2p þ
sinðuk þ umÞðk þ mÞ2p
� �¼ 0; k 6¼ m ð7:15Þ
and
ZT00
ykðtÞj j2 dt ¼ 2T0
ZT00
cosð2p kf0 t þ ukÞ2dt
¼ sinð4p k þ 2ukÞ � sinð2ukÞ þ 4p k4p k
¼ 1
ð7:16Þ
The common relation for signals from sets {xk(t)} and {yk(t)}, k = 1, 2, … ,n results in
7 Orthogonal Signals 119
ZT00
xkðtÞ ymðtÞdt ¼ZT00
ffiffiffiffiffi2T0
rsinð2p kf0 t þ ukÞ
ffiffiffiffiffi2T0
rcosð2pmf0 t þ umÞdt
¼ cosðuk � umÞ � cos½ðuk � umÞ þ 2p ðk � mÞ�ðk � mÞ2p
þ cosðuk þ umÞ � cos½ðuk þ umÞ þ 2p ðk þ mÞ�ðk þ mÞ2p ¼ 0; k 6¼ m
ð7:17Þ
For the case where the scalar product of the two signals xk(t) and yk(t) over theinterval [a, b] is significantly smaller than the energy of each of them that is
Zb
a
xkðtÞykðtÞdt� Ex � Ey ð7:18Þ
and additionally those energies are close to one
Ex � Ey � 1 ð7:19Þ
then such signals are referred to as quasi-orthogonal. Examples of quasi-orthogonalsignals may include signals xk(t) and yk(t), for which the relationship betweenfrequency f0 and period T0 is a real number, and not a natural number.
Let us assume that f0 = x/T0 where x is a real number. We then have
2T0
ZT00
sinð2pkf0tÞ cosð2pkf0tÞ dt ¼ 1T0
ZT00
sinð4pkf0tÞ dt
¼ 1T0� cosð4pkf0tÞ
4pkf0
� �T0
0
¼ 1� cosð4p xÞ4p x
� 0
ð7:20Þ
Ex ¼ 2T0
ZT00
sin2ð2pkf0tÞdt ¼ 1T0
ZT00
½1� cosð4pkf0tÞ�dt
¼ 1T0
t þ sinð4pkf0tÞ4pkf0
� �T0
0
¼ 1þ sinð4pxÞ4px
� 1
ð7:21Þ
120 7 Orthogonal Signals
and
Ey ¼ 2T0
ZT00
cos2ð2pkf0tÞ dt ¼ 1T0
ZT00
½1þ cosð4pkf0tÞ�dt
¼ 1T0
t � sinð4pkf0tÞ4pkf0
� �T0
0¼ 1� sinð4pxÞ
4px� 1
ð7:22Þ
and signals x(t) and y(t) are therefore quasi-orthogonal.
7.1 Orthonormal Polynomials
Sets of orthogonal signals are often used in approximation theory. Let us assumethat {xn(t)} is a set of orthogonal signals over the interval [a, b] with weightfunction w(t) and P(t) is the polynomial
PðtÞ ¼Xnk¼ 0
ak xkðtÞ ð7:23Þ
approximating signal y(t) with a minimum integral square error. Coefficients ak inEq. (7.23) minimizing the error
Eða0; a1; . . .; anÞ ¼Zb
a
wðtÞ yðtÞ �Xnk¼0
akxkðtÞ" #2
dt ð7:24Þ
result from zeroing of derivatives
ddaj
Eða0; a1; . . .; anÞ
¼ 2Zb
a
wðtÞ yðtÞ �Xnk¼0
akxkðtÞ" #
xjðtÞdt ¼ 0; j ¼ 0; 1; . . .; nð7:25Þ
that is
Zb
a
wðtÞyðtÞxjðtÞdt ¼Xnk¼0
ak
Zb
a
wðtÞxkðtÞxjðtÞdt; j ¼ 0; 1; . . .; n ð7:26Þ
7 Orthogonal Signals 121
Substituting j = k into Eq. (7.26), we have
Zb
a
wðtÞyðtÞxkðtÞdt ¼ ak
Zb
a
wðtÞ½xkðtÞ�2dt ð7:27Þ
from which we finally get
ak ¼R ba wðtÞyðtÞxkðtÞdtR ba wðtÞ½xkðtÞ�2dt
ð7:28Þ
For a set of orthonormal signals, for which w(t) = 1 andR ba ½xðtÞ�2 ¼ 1, Eq. (7.28)
is reduced to the form
ak ¼Zb
a
xkðtÞyðtÞdt ð7:29Þ
If the set of signals {x(t)} is orthogonal, then it may be orthonormalized bydividing each signal xk(t) by ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZb
a
½xkðtÞ�2vuuut k ¼ 1; 2; . . .; n ð7:30Þ
Equation (7.31) presents example of the set of orthogonal Tchebyshev poly-nomials {xn(t)} over [−1, 1]
x0ðtÞ ¼ 1
x1ðtÞ ¼ t
x2ðtÞ ¼ t2 � 13
x3ðtÞ ¼ t3 � 35t
x4ðtÞ ¼ t4 � 67t2 þ 3
35
x5ðtÞ ¼ t5 � 109t3 þ 5
21t
ð7:31Þ
122 7 Orthogonal Signals
The orthonormal set corresponding to (7.31) is
x0ðtÞ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 dt
q ¼ffiffiffi2p
2
x1ðtÞ ¼ tffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 t
2 dtq ¼
ffiffiffi6p
2t
x2ðtÞ ¼t2 � 1
3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 ðt2 � 1
3Þ2 dtq ¼
ffiffiffiffiffi10p
4ð3t2 � 1Þ
x3ðtÞ ¼t3 � 3
5 tffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 ðt3 � 3
5 tÞ2 dtq ¼
ffiffiffiffiffi14p
4ð5t3 � 3tÞ
x4ðtÞ ¼t4 � 6
7 t2 þ 3
35ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 ðt4 � 6
7 t2 þ 3
35Þ2 dtq ¼ 3
ffiffiffi2p
16ð35 t4 � 30t2 þ 3Þ
x5ðtÞ ¼t5 � 10
9 t3 þ 5
21 tffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR 1�1 ðt5 � 10
9 t3 þ 5
21 tÞ2 dtq ¼
ffiffiffiffiffi22p
16ð63t5 � 70t3 þ 15tÞ
ð7:32Þ
Example 7.1 Reduce the seven-order polynomial
yðtÞ ¼ t3 þ t7 ð7:33Þ
to the third order (Fig. 7.1) using the orthonormal set (7.32).Using Eq. (7.29), we get
a0 ¼Z1
�1
ffiffiffi2p
2ðt3 þ t7Þdt ¼ 0
a1 ¼Z1
�1
ffiffiffi6p
2tðt3 þ t7Þdt ¼ 0:762
a2 ¼Z1
�1
ffiffiffiffiffi10p
4ð3t2 � 1Þðt3 þ t7Þdt ¼ 0
a3 ¼Z1
�1
ffiffiffiffiffi14p
4ð5t3 � 3tÞðt3 þ t7Þdt ¼ 0:441
ð7:34Þ
7.1 Orthonormal Polynomials 123
hence the approximating polynomial, with minimal integral square error is
PðtÞ ¼ 0:762
ffiffiffi6p
2t þ 0:441
ffiffiffiffiffi14p
4ð5t3 � 3tÞ ¼ 2:063t3 þ 0:304t ð7:35Þ
7.2 Digital Measurement of Electrical Quantities
Let us consider the signal
xðtÞ ¼ Xm cosðx tÞ ð7:36Þ
and the signal shifted by 2s
xðt � 2sÞ ¼ Xm cosðx t � 2sÞ ð7:37Þ
Taking into account the sum and difference of the signals (7.36) and (7.37), weget
xðtÞ þ xðt � 2sÞ ¼ 2Xm cosðx t � sÞ cosðsÞ ð7:38Þ
and
xðtÞ � xðt � 2sÞ ¼ �2Xm sinðx t � sÞ sinðsÞ ð7:39Þ
thus, the two signals are shifted relative to each other by p2 : The xaðtÞ signal
resulting from the sum (7.38)
xaðtÞ ¼ Xm cosðx t � sÞ ¼ 12xðtÞ þ xðt � 2sÞ
cosðxsÞ ð7:40Þ
Fig. 7.1 Polynomials y(t) and P(t)
124 7 Orthogonal Signals
and the signal xb(t) resulting from the difference (7.39)
xbðtÞ ¼ Xm sinðx t � sÞ ¼ � 12xðtÞ � xðt � 2sÞ
sinðxsÞ ð7:41Þ
are orthogonal signals. They can be used in the measurement of power, current,voltage, and frequency, using numerical methods. Let us denote the orthogonalcomponents of voltage by
uaðtÞ ¼ Um cosðx t � sÞ ¼ 12uðtÞ þ uðt � 2sÞ
cosðsÞ ð7:42Þ
and
ubðtÞ ¼ Um sinðx t � sÞ ¼ � 12uðtÞ � uðt � 2sÞ
sinðsÞ ð7:43Þ
and the orthogonal components of current shifted by u as
iaðtÞ ¼ Im cosðx t � sþ uÞ ¼ 12iðtÞ þ iðt � 2sÞ
cosðsÞ ð7:44Þ
and
ibðtÞ ¼ Im sinðx t � sþ uÞ ¼ � 12iðtÞ � iðt � 2sÞ
sinðsÞ ð7:45Þ
For sinusoid signals, we have
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðxa þ jxbÞðxa � jxbÞ
p¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2a þ x2b
q¼
ffiffiffiffiffiffiX2m
q¼ Xm ð7:46Þ
and
xrms ¼ Xmffiffiffi2p ¼ 1ffiffiffi
2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2a þ x2b
qð7:47Þ
Substituting Eqs. (7.42)–(7.43) and (7.44)–(7.45) into Eq. (7.47) gives for thevoltage
urms ¼ 1ffiffiffi2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12uðtÞ þ uðt � 2sÞ
cosðxsÞ� �2
þ 12uðtÞ � uðt � 2sÞ
sinðxsÞ� �2
sð7:48Þ
7.2 Digital Measurement of Electrical Quantities 125
and for the current
irms ¼ 1ffiffiffi2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12iðtÞ þ iðt � 2sÞ
cosðxsÞ� �2
þ 12iðtÞ � iðt � 2sÞ
sinðxsÞ� �2
sð7:49Þ
7.2.1 Measurement of Active Power
Active power is
P ¼ urmsirms cosu ¼ 12UmIm cosu ¼ 1
2UmIm cosð�uÞ ð7:50Þ
Adding and subtracting xðt � sÞ for u into (7.50), we obtain
P ¼ 12UmIm cos xðt � sÞ � xðt � sÞ � u½ �
¼ 12UmIm cos ðxt � sÞ � ðxt � sþ uÞ½ �
ð7:51Þ
After simple transformation, we get
P ¼ 12UmIm cosðxt � sÞ cosðxt � sþ uÞ½
þ sinðxt � sÞ sinðxt � sþ uÞ�ð7:52Þ
and
P ¼ 12½Um cosðx t � sÞ Im cosðx t � sþ uÞþ Um sinðx t � sÞIm sinðx t � sþ uÞ�
ð7:53Þ
Taking into account Eq. (7.53) in the Eqs. (7.42)–(7.45), we have
P ¼ 12uaðtÞiaðtÞ þ ubðtÞibðtÞ½ � ð7:54Þ
that is
P ¼ 12½uðtÞ þ uðt � 2sÞ� ½iðtÞ þ iðt � 2sÞ�
4 cos2ðxsÞ�
þ ½uðtÞ � uðt � 2sÞ� ½iðtÞ � iðt � 2sÞ�4 sin2ðxsÞ
� ð7:55Þ
126 7 Orthogonal Signals
7.2.2 Measurement of Reactive Power
Reactive power is
Q ¼ urms irms sinu ¼ � 12Um Im sinð�uÞ ð7:56Þ
Transforming Eq. (7.56) in a similar way to that for the case of active power, weget
Q ¼ � 12Um sin½x ðt � sÞ� Im cos½x ðt � sÞ þ u�
� 12Um cos½x ðt � sÞ� Im sin½x ðt � sÞ þ u�
ð7:57Þ
that is
Q ¼ � 12uaðtÞibðtÞ þ ubðtÞiaðtÞ½ � ð7:58Þ
and
Q ¼ iðtÞuðt � 2sÞ � uðtÞiðt � 2sÞ4 sinðxsÞ cosðxsÞ ð7:59Þ
7.2.3 Digital Form of Current, Voltage, and Power
Setting t = n and s ¼ k into Eqs. (7.48)–(7.49) and (7.55)–(7.59) we can write.
• for current
irms ¼ 1ffiffiffi2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12i ½n� þ i ½n� 2 k�
cosðx kÞ� �2
þ 12i ½n� � i ½n� 2 k�
sinðx kÞ� �2
sð7:60Þ
• for voltage
urms ¼ 1ffiffiffi2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12u ½n� þ u ½n� 2 k�
cosðx kÞ� �2
þ 12u ½n� � u ½n� 2 k�
sinðx kÞ� �2
sð7:61Þ
7.2 Digital Measurement of Electrical Quantities 127
• for active power
P ¼ 12fu½n� þ u½n� 2k�g fi½n� þ i½n� 2k�g
4 cos2ðx kÞ þ fu½n� � u½n� 2k�g fi½n� � i½n� 2k�g4 sin2ðx kÞ
� �ð7:62Þ
• and for reactive power
Q ¼ i ½n� u½n� 2k� � u ½n�i ½n� 2k�4 sinðx kÞ cosðx kÞ ð7:63Þ
where n is the number of sample, and k is the shift of samples.
7.3 Measurement of Frequency
Let us present the signals x(t) and xðt � sÞ as a sum of two orthogonal signalsxa(t) and xb(t) shifted relative to each other by s ¼ p=2
xðtÞ ¼ xaðtÞ þ jxbðtÞ ¼ Xm exp jxðt � sÞ½ � ð7:64Þ
xðt � sÞ ¼ xaðt � sÞ þ jxbðt � sÞ ¼ Xm exp jx ðt � 2sÞ½ � ð7:65Þ
Let us write the product of signals xðtÞ with xðt � sÞ
xðtÞxðt � sÞ ¼ Xm exp½jxðt � sÞ�Xm exp½�jxðt � 2sÞ� ¼ X2m expðjxsÞ ð7:66Þ
where xðtÞ is conjugate with x(t) and shifted by x(t).Substituting Eqs. (7.64)–(7.65) into Eq. (7.66) and comparing the real and
imaginary parts, we get
xaðtÞxaðt � sÞ þ xbðtÞxbðt � sÞ ¼ X2m cosðxsÞ ð7:67Þ
and
xbðtÞxaðt � sÞ � xaðtÞxbðt � sÞ ¼ X2m sinðxsÞ ð7:68Þ
The last equation, for the shift of 2s has the form
xbðtÞxaðt � 2sÞ � xaðtÞxbðt � 2sÞ ¼ X2m sinð2xsÞ ð7:69Þ
128 7 Orthogonal Signals
The quotient of Eqs. (7.69) and (7.68) gives
xbðtÞxaðt � 2sÞ � xaðtÞxbðt � 2sÞxbðtÞxaðt � sÞ � xaðtÞxbðt � sÞ ¼ 2 cosðxsÞ ð7:70Þ
hence
f ¼ 12ps
arc cosxbðtÞxaðt� 2sÞ � xaðtÞxbðt� 2sÞxbðtÞxaðt� sÞ � xaðtÞxbðt� sÞ ð7:71Þ
where the constant s occurring in Eq. (7.71) causes orthogonality of the signalsx(t) and xðt � 2sÞ:
The Eq. (7.71) in discrete form is
f ¼ 12ps
arc cosxb½n�xa½n� 2 k� � xa½n�xb½n� 2 k�xb½n�xa½n� k� � xa½n�xb½n� k� ð7:72Þ
where k determines the number of samples and causes orthogonality of the signalsx[n] and x½n� 2 k�:
7.4 Examples in MathCad
Determination of the root mean square of current, voltage, active and reactivepower.
1. Continuous signal.
T : ¼ 5 D :¼ 0:01
t : ¼ 0;D; . . .T x :¼ 5 u :¼ p12
Um : ¼ 10 Im :¼ 4
uðtÞ : ¼ Um � sinðx � tÞ iðtÞ :¼ Im � sinðx � t � uÞIrms : ¼ Imffiffiffi
2p Urms :¼ Umffiffiffi
2p s :¼ 1
7.3 Measurement of Frequency 129
Irms :¼ Imffiffiffi2p Urms :¼ Umffiffiffi
2p s :¼ 1
P :¼ Umffiffiffi2p � Imffiffiffi
2p � cosðuÞ Q :¼ Umffiffiffi
2p � Imffiffiffi
2p � sinðuÞ
Irms ¼ 2:828 Urms ¼ 7:071 P ¼ 19:319 Q ¼ 5:176
I1:rmsðtÞ :¼ 1ffiffiffi2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12� iðtÞ þ iðt � 2Þ � s
cosðx � sÞ� �2
þ 12� iðtÞ � iðt � 2Þ � s
sinðx � sÞ� �2
s
U1:rmsðtÞ :¼ 1ffiffiffi2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12� uðtÞ þ uðt � 2Þ � s
cosðx � sÞ� �2
þ 12� uðtÞ � uðt � 2Þ � s
sinðx � sÞ� �s
I1:rmsðTÞ ¼ 2:828 U1:rmsðTÞ ¼ 7:071
uaðtÞ :¼ 12� uðtÞ þ uðt � 2Þ � s
cosðx � sÞ ubðtÞ :¼ �12 �uðtÞ � uðt � 2Þ � s
sinðx � sÞiaðtÞ :¼ 1
2� iðtÞ þ iðt � 2Þ � s
cosðx � sÞ ibðtÞ :¼ �12 �iðtÞ � iðt � 2Þ � s
sinðx � sÞ
0 1 2 3 4 5−10
−5
0
5
10
)( tu
)( ti
t
0 1 2 3 4 5−10
−5
0
5
10
ua )(t
t0 1 2 3 4 5
−10
−5
0
5
10
ub )(t
t
0 1 2 3 4 5−4
−2
0
2
4
ia )(t
t0 1 2 3 4 5
−4
−2
0
2
4
ib )(t
t
130 7 Orthogonal Signals
P1ðtÞ :¼ 12� ðuaðtÞ � ibðtÞ þ ubðtÞ � iaðtÞ
Q1ðtÞ :¼ �12 � ðuaðtÞ � ibðtÞ � ubðtÞ � iaðtÞP1ðTÞ ¼ 19:319 Q1ðTÞ ¼ 5:176
P2ðtÞ :¼ 12:
uðtÞ � uðt � 2 � sÞð Þ � iðtÞ � iðt � 2 � sÞð Þ4: cosðx � sÞ2
" #
þ uðtÞ � uðt � 2 � sÞð Þ � iðtÞ � iðt � 2 � sÞð Þ4 � sinðx � sÞ2
" #
Q2ðtÞ :¼ iðtÞ � uðt � 2 � sÞ � uðtÞ � iðt � 2 � sÞ4 � sinðx � sÞ � cosðx � sÞ
P2ðTÞ ¼ 19:319 Q2ðTÞ ¼ 5:176
2. Discrete signal.
T :¼ 5 D :¼ 0:01
t :¼ 0; D::T x :¼ 5 u :¼ p12
Um :¼ 10 Im :¼ 4
uðtÞ :¼ Um � sinðx � tÞ
uðtÞ :¼ Um � sinðx � tÞ iðtÞ :¼ Im � sinðx � t � uÞ
id :¼form 2 0. . . TD
idm iðmÞid
��������ud :¼
form 2 0. . . TD
udm uðmÞud
��������n :¼ 2. . . TD k :¼ 1
IdrmsðnÞ : ¼ 1ffiffiffi2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12� idn þ idn�2�kcosðx � kÞ
� �2
þ 12� idn � idn�2�ksinðx � kÞ
� �2s
UdrmsðnÞ : ¼ 1ffiffiffi2p �
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12� udn þ idn�2�kcosðx � kÞ
� �2
þ 12� udn � udn�2�ksinðx � kÞ
� �2s
7.4 Examples in MathCad 131
Idrms TD
¼ 2:828 Udrms TD
¼ 7:071
PdðnÞ : ¼ 12:
uðnÞ þ uðn� 2 � kÞð Þ � iðnÞ þ iðn� 2 � kÞð Þ4: cosðx � kÞ2
" #
þ uðnÞ � uðn� 2 � kÞð Þ � iðnÞ � iðn� 2 � kÞð Þ4 � sinðx � kÞ2
" #
QdðnÞ :¼iðnÞ � uðn� 2 � kÞ � uðnÞ � iðn� 2 � kÞ
4 � sinðx � kÞ � cosðx � sÞ
PdTD
¼ 19:319 QdTD
¼ 5:176
Determination of frequency.
1. Continuous signal.
T :¼ 0:5 D :¼ 0:00001
t :¼ 0;D; . . .; T x :¼ 5 f 0 :¼ 3 Xm :¼ 1
xðtÞ :¼ Xm � sinð2 � p � f 0 � tÞ
0 0.1 0.2 0.3 0.4 0.5−1
−0.5
0
0.5
1
)(tx
t
132 7 Orthogonal Signals
Shift of signal
id :¼
j 0for i 2 D;Dþ D; . . .; T � Dif xði� DÞ\xðiÞ[ xðiþ DÞj 1break
����j
�����������s ¼ 0:083
f 0e :¼ 14�s
f 0e ¼ 3
xcðtÞ :¼ 12� xðtÞ þ xðt � 2 � sÞcosð2 � p � f 0e � sÞ
xsðtÞ :¼ �12 �xðtÞ þ xðt � 2 � sÞsinð2 � p � f 0e � sÞ
f ðtÞ :¼ 12 � p � s � a cos 0:5 � xsðtÞ � xcðt � 2 � sÞ � xcðtÞ � xSðt � 2 � sÞ
xsðtÞ � xcðt � sÞ � xcðtÞ � xsðt � sÞ� �
f ðTÞ ¼ 3
2. Discrete signals.
T :¼ 0:5 D :¼ 0:00001
t :¼ 0;D; . . .; T f 0 :¼ 3 Xm :¼ 1
x tð Þ :¼ Xm � sinð2 � p � f 0 � tÞ
7.4 Examples in MathCad 133
Discretization of signal and time
xd :¼form 2 0. . . TDxdm iðm � DÞ
xd
������ Td :¼form 2 0. . . TD
Tdm mTd
������
04 4
2×10 4×10−1
−0.5
0
0.5
1
xd
Td
k :¼
j 0for i 2 1; 2. . . TD� 1if xdi�1\xdi [ xdiþ1j ibreak
����j
�����������k ¼ 8:333� 103
f 0e :¼1
4 � k
f 0e ¼ 3� 10�5
xcdðnÞ :¼ 12� xðnÞ þ xðn� 2 � kÞcosð2 � p � f 0de � kÞ
xsdðnÞ :¼ �12 �xðnÞ þ xðn� 2 � kÞsinð2 � p � f 0de � kÞ
f dðnÞ :¼1
2 � p � k � D � a cos 0:5 � xsðnÞ � þ xcðn� 2 � kÞ � xcðnÞ � xSðn� 2 � kÞxsðnÞ � þxcðn� kÞ � xcðnÞ � xsðn� kÞ
� �
f dTD� 1
� �¼ 3
134 7 Orthogonal Signals
7.5 Examples in LabVIEW
Measurement of current
Block diagram
Front panel
7.5 Examples in LabVIEW 135
Measurement of voltage
Block diagram
Front panel
136 7 Orthogonal Signals
Measurement of active power
Block diagram
Front panel
7.5 Examples in LabVIEW 137
Measurement of reactive power
Block diagram
Front panel
138 7 Orthogonal Signals
Measurement of frequency
Block diagram
Front panel
7.5 Examples in LabVIEW 139
Chapter 8Modulations
Modulation realizes the transmission of low-frequency signal by means of a high-frequency carrier signal. The modulating signal contains information, while thecarrier signal carries it in a high-frequency range to the receiver. Modulation allowsthe selection of a modulated signal frequency such that:
• the signal is reliably handled by the receiver,• it will not cause interference with other low-frequency signals which are being
transmitted simultaneously and which have been assigned different carrierfrequencies.
The device performing the modulation is referred to as the modulator, whereasthe device performing demodulation is referred to as the demodulator. In the case ofbilateral communication, there is often a single device, which simultaneouslymodulates the transmitted signals and demodulates the received ones. Such a deviceis referred to as a modem, which is an abbreviation of modulator–demodulator.Currently, there are numerous different types of modulation in use. In the mostgeneral terms, they can be divided into three groups: analog, impulse, and digital.Analog modulations of amplitude (AM), phase modulation (PM), and frequencymodulation (FM) are the earliest to have been developed. Pulse code modulations,converting analog signals to digital prior to transmission, as well as digital mod-ulations used in the wireless transmission of global positioning system (GPS) data,have been developed much later, to meet the needs of digital transmissions. We willbe discussing the basic principles governing the modulation of signals, their mainproperties, and their applications.
On the transmitter side of the information system, the signal is modulated using acoder and a modulator. On the receiver side, the signal undergoes a demodulationprocess where the signal is reprocessed to its original form—Fig. 8.1.
The coder codes the information signal, e.g., coding its analog form into a binarysequence, while the modulator transforms that sequence to a form enabling itstransfer in the transmission channel. Figure 8.2 presents the modulation types andtheir main division.
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_8
141
8.1 Analog Modulations (AM)
In analog amplitude modulation (AM), the most commonly used type of carriersignal is the harmonic signal, whose amplitude changes in proportion to themodulating signal, containing the information. In this group of modulations, thereare several solutions. The main differences between these are the occurrence or non-occurrence of sidebands, and whether the carrier signal is suppressed or not sup-pressed. The modulating signal is crucial in determining the type of modulation.Below we will present the main properties for those types of modulation.
8.1.1 Double-Sideband Large Carrier Modulation (DSBLC)
For DSBLC, the modulating function mðtÞ is
mðtÞ ¼ 1þ xmðtÞ ð8:1Þ
Fig. 8.1 Process of signal modulation
Fig. 8.2 Classification of modulations
142 8 Modulations
while the low-frequency modulating signal carrying information is given by
xmðtÞ ¼ Am cosðxtÞ ð8:2Þ
The high-frequency carrier signal is
xcðtÞ ¼ Ac cosðXtÞ ð8:3Þ
Let xcaðtÞ denote the analytic form of carrier signal resulting from the Hilberttransform (6.12).
xcaðtÞ ¼ Ac cosðXtÞ þ j sinðXtÞ½ � ð8:4Þ
The analytic form of modulated signal is
xaðtÞ ¼ mðtÞxcaðtÞ ð8:5Þ
Substituting Eqs. (8.1) and (8.4) into Eq. (8.5), we have
xaðtÞ ¼ Ac½1þ xmðtÞ�½cosðXtÞ þ j sinðXtÞ� ð8:6Þ
The modulated signal xðtÞ represents the real part of xaðtÞ
xðtÞ ¼ Ac 1þ xmðtÞ½ � cosðXtÞ ð8:7Þ
Expanding Eq. (8.7), we have
xðtÞ ¼ Ac½cosðX tÞ þ Am cosðx tÞ cosðXtÞ� ð8:8Þ
Let us present Eq. (8.8) in the form of a sum
xðtÞ ¼ Ac cos Xt þ uð Þ þ Am cosðxtÞ cosðXtÞ¼ Ac cos Xt þ uð Þ þ Am
2cos ðX� xÞt½ � þ Am
2cos ðXþ xÞt½ � ð8:9Þ
From Eq. (8.9) results that the signal xðtÞ has three components: a carrier com-ponent with the amplitude of Ac rotating with the frequency of X, a positivecomponent, with an amplitude of Am=2 and frequency of ðXþ xÞt, and a negativecomponent, with an amplitude of Am=2 and a frequency of ðX� xÞt—Fig. 8.3.
The envelope EðtÞ of the signal xðtÞ is
EðtÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixðtÞ2 þ exðtÞ2q
¼ �Ac 1þ d cos xtð Þ½ � ð8:10Þ
where d ¼ Am=Ac is depth of modulation.In the case of DSBLC transmission, the transmitter must emit high energy, as it
transmits the carrier signal, as well as both sidebands, which occupy a wide
8.1 Analog Modulations (AM) 143
frequency range X� x=2;Xþ x=2ð Þ. This has the benefit of making signaldetection very simple and achievable using comparatively low-cost receivers(Fig. 8.4).
In DSBLC, the following three cases are possible:
• depth of modulation d\1—Fig. 8.5• depth of modulation for d[ 1—Fig. 8.6. In this case, the carrier signal is
overmodulated and the envelope EðtÞ of the signal reaches negative values• depth of modulation d ¼ 1, then
xðtÞ ¼ Ac 1þ cos xtð Þ½ � cos Xtð Þ ð8:11Þ
and envelopes (Fig. 8.7)
EðtÞ ¼ �Ac½1þ cosðxtÞ� ð8:12Þ
Fig. 8.3 Three component of the amplitude-modulated signal
Fig. 8.4 Spectra in DSBLC modulation
144 8 Modulations
Fig. 8.6 Amplitude-modulated signal for d[ 1
Fig. 8.5 Amplitude-modulated signal for d\1
8.1 Analog Modulations (AM) 145
8.1.2 Double Sideband with Suppressed CarrierModulation (DSBSC)
For DSBSC, the modulating function mðtÞ is
mðtÞ ¼ xmðtÞ ¼ Am cos xtð Þ ð8:13Þ
Hence, the analytic modulated signal Eq. (8.5) equals
xaðtÞ ¼ AcAm cos xtð Þ þ j sinðXtÞ½ � cosðXtÞ ð8:14Þ
The modulated real signal resulting from Eq. (8.5) is
xðtÞ ¼ AcAm cos Xtð Þ cos xtð Þ ð8:15Þ
Extending Eq. (8.15), we have
xðtÞ ¼ AcAm
2cos ðXþ xÞt½ � þ cos ðX� xÞt½ �f g ð8:16Þ
The envelopes EðtÞ of signal xðtÞ are
EðtÞ ¼ � AcAm cos xtð Þ½ � ð8:17Þ
From Eq. (8.16), it results that in DSBSC, the signal is composed of two sidebands:the upper and lower, while the carrier signal is nonexistent. The frequency band ofthe transmitted signal remains unchanged, while the power required for its trans-mission is definitely lower than in the case of DSBLC. Due to the absence of thecarrier signal in DSBSC, in order to reproduce the modulated signal, it is necessarythat each receiver generates its own modulated signal, with a high level of finetuning precision. For that reason, the cost of receivers of signals modulated inDSBSC is significantly greater than in DSBLC (Fig. 8.8).
Fig. 8.7 Amplitude-modulated signal for d ¼ 1
146 8 Modulations
8.1.3 Single-Sideband (SSB)
For the SSB, we have
mðtÞ ¼ 1þ xmðtÞ � j~xmðtÞ ð8:18Þ
where ~xmðtÞ is the Hilbert transform of xmðtÞ Eq. (8.2).The analytic modulated signal is (Fig. 8.9)
Fig. 8.8 Spectra in DSBSC modulation
Fig. 8.9 Spectra in SSB modulation
8.1 Analog Modulations (AM) 147
xaðtÞ ¼ Ac cosðXtÞ þ xmðtÞ cosðXtÞ � ~xmðtÞ sinðXtÞ½ �þ jAc sinðXtÞ þ xmðtÞ sinðXtÞ � ~xmðtÞ cosðXtÞ½ � ð8:19Þ
From Eq. (8.19), we obtain the modulated signal
xðtÞ ¼ Ac cosðXtÞ½ þ Am cos ðX� xÞt� ð8:20Þ
which, depending on the sign of x, contains the carrier signal and upper or lowersideband. Due to the minimum bandwidth necessary for the transmission of asignal, the SSB method provides an optimal form of modulation, requiring muchless power than DSBLC. However, it necessitates a composite and thereforeexpensive receiver. A substantial advantage of this type of modulation is the highlevel of energy savings in the transmitter and, even more importantly, the possibleincrease in the number of transmitters in the available frequency range.
8.1.4 Single Sideband with Suppressed Carrier (SSBSC)Modulation
In the case of SSBSC, we have
xðtÞ ¼ xmðtÞ � j~xmðtÞ ð8:21Þ
The analytic signal is given by
xaðtÞ ¼ Ac xmðtÞ cosðXtÞ � ~xmðtÞ sinðXtÞ½ �þ jAc xmðtÞ sinðXtÞ � ~xmðtÞ cosðXtÞ½ � ð8:22Þ
From Eq. (8.22), we obtain the modulated signal
xðtÞ ¼ AcAm cos ðX� xÞt½ � ð8:23Þ
which, depending on the sign of x, contains only the upper or lower sideband(Fig. 8.10).
8.1.5 Vestigial Sideband (VSB) Modulation
For VSB modulation, we have
xðtÞ ¼ xmðtÞ þ j ~xmðtÞkðtÞ½ � ð8:24Þ
148 8 Modulations
where kðtÞ is the impulse response of the filter attenuating the lower sideband.The analytic modulated signal is
xaðtÞ ¼ AcxmðtÞ cosðXtÞ � kðtÞ sinðXtÞ½ �þ jAcxmðtÞ½sinðXtÞ þ kðtÞ cosðXtÞ� ð8:25Þ
and thus,
xðtÞ ¼ AcAm cosðXtÞ cosðxtÞ � kðtÞ sinðXtÞ cosðxtÞ½ � ð8:26Þ
In VSB, the upper sideband is transmitted almost completely, whereas in the case ofthe lower sideband, only a trace amount is transmitted. In VSB systems, the DSBSCsignal is generated first and is then filtered through a filter attenuating the lowersideband. In VSB, due to the necessity of transmitting a partly attenuated sideband,a slightly wider transmission band is required than in DSBLC modulation.
Figure 8.11 presents the spectra of modulating the signal XðxÞ and modulatedsignal VSB where XV � X is the frequency of the partly attenuated sideband.
8.2 Angle Modulations
In the case of angle modulation, the amplitude of modulated signal is constant withtime. The angle of the modulated signal changes depending on the instantaneousvalue of the modulating signal. Relating to the angle changes, PM and FM areapplied in practice.
Fig. 8.10 Spectra of SSBSC with upper sideband
8.1 Analog Modulations (AM) 149
8.2.1 Phase Modulation (PM)
In PM, the modulated signal has the form
xðtÞ ¼ Ac cos Xt þ Am sinðxtÞ½ � ð8:27Þ
After extending Eq. (8.27), we get
xðtÞ ¼ Ac cosðXtÞ cos Am sinðxtÞ½ � � sinðXtÞ sin½Am sinðxtÞ�f g ð8:28Þ
The analytic form of the Eq. (8.28) is
xðtÞ ¼ Ac cosðXtÞ cos Am sinðxtÞ½ � � sinðXtÞ sin½Am sinðxtÞ�f gþ jAc cosðXtÞ sin Am sinðxtÞ½ � þ sinðXtÞ cos½Am sinðxtÞ�f g ð8:29Þ
which may be represented in the exponential form
xðtÞ ¼ AcejðX tþAm sinxtÞ ð8:30Þ
The instantaneous phase /ðtÞ of the modulated signal is
/ðtÞ ¼ Xt þ Am sinðxtÞ ð8:31Þ
This means that /ðtÞ with respect to carrier frequency X changes proportionately tothe modulated signal.
Fig. 8.11 Spectra of VSB modulation
150 8 Modulations
8.2.2 Frequency Modulation (FM)
A signal in FM has the form
xðtÞ ¼ Ac cos Xt þ Am
Z t
0
sin xtð Þdt24
35 ð8:32Þ
The analytic FM signal in exponential form
xðtÞ ¼ Acej½X tþAm
Rt0
sinðxtÞdt�ð8:33Þ
indicates that the instantaneous amplitude of the signal is constant, whereas theinstantaneous phase changes proportionally to the integral
R t0 sinðxtÞdt of the
modulating signal (Fig. 8.12).
8.3 Impulse Modulations
In impulse modulation, the carried signal is a sequence of impulses, while themodulating signal is a harmonic one (Fig. 8.13).
8.3.1 Pulse Width Modulation (PWM)
In PWM, the width of the impulse signal changes, while its frequency andamplitude remain constant (Fig. 8.14).
Fig. 8.12 Signals in PM modulation
8.2 Angle Modulations 151
PWM is most frequently used for changing the mean value of the signal, e.g., incontrolling brightness of lighting, in the control of DC systems, etc. In practicalapplications, the disadvantage of PWM is that signal switching generatesinterference.
8.3.2 Pulse Amplitude Modulation (PAM)
In PAM, depending on the value of the modulating signal, the amplitude of thecarrier impulse changes. The generation of the PAM signal is similar to sampling,where the modulated signal presents a sequence of samples of the modulatingsignal.
The modulated signal is generated by the product of signal xmðtÞ and xcðtÞ
xðtÞ ¼ xmðtÞ xcðtÞ ð8:36Þ
The sequence of samples for the modulated signal may be created by means of threetypes of sampling: ideal, real, and instantaneous.
Fig. 8.13 Signals in FM modulation
Fig. 8.14 PWM modulation
152 8 Modulations
8.3.3 PAM with Ideal Sampling
In PAM with ideal sampling, the sampled signal xcðtÞ has the form of impulses—Eq. (4.2), in which the sampling frequency results from Shannon’s theorem(Fig. 8.15).
From Eq. (4.3), we have the ideal sampling in the form
xiðtÞ ¼X1n¼0
xm nTp� �
d t � nTp� � ¼ xm tð Þ
X1n¼0
d t � nTp� � ð8:37Þ
8.3.4 PAM with Real Sampling
In PAM with real sampling, the carrier signal is a sequence of rectangular impulses—Fig. 8.16.
A single impulse and carrier signal are given by
Fig. 8.15 PAM with ideal sampling
8.3 Impulse Modulations 153
PsðtÞ ¼ 1 for 0\t\s0 elsewhere
�ð8:38Þ
and
xcrðtÞ ¼X1n¼0
PsðtÞnTp ð8:39Þ
Let us represent the signal (8.39) in the form of exponential Fourier series
xcrðtÞ ¼ sTp
X1n¼0
Sa npsTp
� �ejnxpt ð8:40Þ
Eq. (8.40) gives the modulated signal as
xrðtÞ ¼ sTp
xmðtÞX1n¼0
Sa npsTp
� �ejnxpt ð8:41Þ
Fig. 8.16 PAM with real sampling
154 8 Modulations
8.3.5 PAM with Instantaneous Sampling
In PAM, the carrier signal has the form of rectangular impulses, whose amplitudevalue at sampling moments depends on the instantaneous value of the modulatingsignal xmðtÞ—Fig. 8.17.
In PAM, the signal xcðtÞ is
xcðtÞ ¼X1n¼0
xm nTp� �
Ps t � nTp� � ¼X1
n¼0xm nTp
� �PsðtÞ � d t � nTp
� ��
¼ PsðtÞX1n¼0
xm nTp� �
d t � nTp� �" #
ð8:42Þ
Fig. 8.17 PAM with instantaneous sampling
8.3 Impulse Modulations 155
8.3.6 Pulse Duration Modulation (PDM)
In PDM, the width of the carrier signal impulses changes depending on theamplitude of the current sample of the modulating signal xmðtÞ:
The widths s nTsð Þ of successive impulses are
s nTsð Þ ¼ a0 þ a1xm nTsð Þ ð8:43Þ
where the constants a0 and a1 are selected to satisfy the inequality 0\s nTsð Þ\Ts.It is easy to see that the PDM is equivalent to analog the PM (Fig. 8.18).
8.3.7 Pulse Position Modulation (PPM)
In the PPM, depending on the current samples of the modulating signal xmðtÞ; theimpulse position changes, in relation to the nominal position n Tsð Þ—Fig. 8.19. ThePPM is achieved in a similar manner to analog FM.
8.3.8 Pulse Code Modulation (PCM)
PCM is the simplest way of converting an analog signal into a discrete one. Thesignal is sampled at regular time intervals and converted into digital form using anA/D converter (Fig. 8.20).
Fig. 8.18 Signals in PDM
156 8 Modulations
PCM is realized in two stages. In the first stage, the signal xmðtÞ is sampled bymeansof PAM, and in the second stage, it is quantized and coded in natural binary code.
Fig. 8.19 Signals in PPM
Fig. 8.20 Signals in PCM
8.3 Impulse Modulations 157
8.3.9 Differential Pulse Code Modulation (DPCM)
DPCM is based on the method applied in PCM and coding the difference betweenthe current and expected sample.
8.4 Digital Modulations
8.4.1 Modulation with Amplitude Shift Keying (ASK)
In ASK, the amplitude of a harmonic carrier signal is varied depending on thedigital value of a binary sequence. It is the equivalent of analog DSBLC.
ASKðtÞ ¼ Ac cosð2pftÞxmðtÞ ð8:44Þ
Figure 8.21 presents an ASK modulation, for which the digital modulatingsignal is the sequence of the bits 0010111010.
8.4.2 Modulation with Frequency Shift Keying (FSK)
In FSK, two subcarriers with frequencies of f0 or f1 are generated
FSKðtÞ ¼ Ac cosð2pf0tÞ for bit 0Ac cosð2pf1tÞ for bit 1
�ð8:45Þ
Figure 8.22 presents an FSK modulation for f1 [ f0 and the bit sequence0010111010.
Fig. 8.21 Digital modulating xmðtÞ signal and the modulated signal ASK(t)
158 8 Modulations
The frequency deviation in the FSK is
Df ¼ f1 � f0 ¼ 12tb ð8:46Þ
where tb is duration of the bit.If the phase of the signal is constrained to be continuous, we have a special case
of FSK named continuous-phase FSK modulation—CPFSK.For the deviation
Df ¼ 14tb ð8:47Þ
we have minimum shift keying modulation—MSK.In the case where the rectangular signal is approximated by signal of Gaussian
shape, we have the Gaussian minimum shift keying (GMSK) modulation. Incomparison with the rectangular impulse, it has a smaller sidebands and narrowersideband.
8.4.3 Phase Shift Keying (PSK) Modulation
In PSK, the phase of the harmonic modulated signal changes, depending on thedigital value of the modulating signal
PSKðtÞ ¼ Ac cosð2pft þ /1Þ for bit 0Ac cosð2pft þ /2Þ for bit 1
�ð8:48Þ
Figure 8.23 represents PSK modulation where the digital modulating signal is thesequence of the bits 0010110011.
Fig. 8.22 Digital modulating signal xmðtÞ and modulated signal FSK(t)
8.4 Digital Modulations 159
In the case of
BPSKðtÞ ¼ Ac cosð2pft � p=2Þ ¼ �Ac cosð2pftÞ for bit 0Ac cosð2pft þ p=2Þ ¼ Ac cosð2pftÞ for bit 1
�ð8:49Þ
we get the biphase shift keying modulation—BPSK (Fig. 8.24).In BPSK, the modulated signal consists of fragments of sine function, with a
period equal to the modulation impulse, and a frequency equal to the frequency ofthe carrier signal.
The BPSK may be represented in a so-called constellation diagram—Fig. 8.25.Similar to BPSK is differential phase shift keying differential PSK modulation
(DPSK), in which the phase changes by p if the binary value is 1, and remains thesame if the binary value is 0.
An extension of BPSK is the quadrature phase shift keying modulation (QPSK),consisting of two-bit coding on 4 orthogonal phase shifts, e.g., p=4; 3p=4; 5p=4,
Fig. 8.23 Digital modulating signal xmðtÞ and modulated signal PSK(t)
Fig. 8.24 Digital modulating signal xmðtÞ and modulated signal BPSK(t)
160 8 Modulations
and 7p=4: In one period of the carrier signal, two bits are coded, so that for a givencarrier frequency, QPSK allows data transmission at twice the speed of BPSK.
If the modulated signal is
QPSKðtÞ ¼ Ac cos½2pft þ ð2p� 1Þp=4� ð8:50Þ
we have for p ¼ 1; 2; 3; 4
QPSKðtÞ ¼ Ac cos½2pft þ p=4� for p ¼ 1
QPSKðtÞ ¼ Ac cos½2pft þ 3p=4� for p ¼ 2
QPSKðtÞ ¼ Ac cos½2pft þ 5p=4� for p ¼ 3
QPSKðtÞ ¼ Ac cos½2pft þ 7p=4� for p ¼ 4
ð8:51Þ
Let Eq. (8.50) be
QPSKðtÞ ¼ Ac cosð2pftÞ cosð2p� 1Þp=4� Ac sinð2pftÞ sinð2p� 1Þp=4 ð8:52Þ
Denoting
/1ðtÞ ¼ Ac cosð2pftÞ and /2ðtÞ ¼ Ac sinð2pftÞ ð8:53Þ
we finally get
QPSKðtÞ ¼ p=4½cos 2p� 1ð Þ/1ðtÞ � sin 2p� 1ð Þ/2ðtÞ� ð8:54Þ
The diagram of the constellation QPSK—Fig. 8.26 contains four points corre-sponding to the four possible two bits transmitted within one period of the carriersignal. The points are located symmetrically on the circle, the radius of whichequals the signal amplitude.
In BPSK, the distances between adjacent points on the constellation diagram areshorter, which gives a doubling of transmission speed, for a given carrier frequency.
Fig. 8.25 Constellation diagram for BPSK
8.4 Digital Modulations 161
8.4.4 Quadrature Amplitude Modulation (QAM)
QAM presents a group of modulations, in which changes to the modulating signalresult in changes to both the amplitude and phase of the modulated signal. Thus, itis a combination of ASK and PSK modulations. In QAM, the modulated signal isdivided into two parts, and the second part is shifted in relation to the first one bythe angle of p=2. Both parts of signal are individually modulated, before beingadded and transmitted.
The code data are formed according to the constellations diagram in a sequenceof binary data, which correspond to both amplitude and phase (Fig. 8.27).
Fig. 8.26 Diagram of constellation for QPSK signal
Fig. 8.27 Constellation diagram for 16-QAM modulation
162 8 Modulations
8.5 Examples in MathCad
1. DSBLC Modulation for m\1t :¼ 0; 0:01. . .10 x :¼ 2 X :¼ 10
Am :¼ 1 Ac :¼ 2
d :¼ Am
Ac
d ¼ 0:5
xcðtÞ :¼ Ac � cosðX � tÞxmðtÞ :¼ Am � cosðx � tÞxðtÞ :¼ Ac � ð1þ d � cosðx � tÞÞ � cosðX � tÞEðtÞ :¼ Ac þ Am � cosðx � tÞð Þ
0 2 4 6 8 102−
1−
0
1
2
xc )(t
t
8.5 Examples in MathCad 163
2. DSBLC Modulation, for m[ 1t :¼ 0; 0:01. . .10 x :¼ 2 X :¼ 10
Am :¼ 15 Ac :¼ 10
d :¼ Am
Ac
d ¼ 1:5
xðtÞ :¼ Ac � 1þ d � cosðx � tÞð Þ � cosðX � tÞEðtÞ :¼ Ac þ Am � cosðx � tÞð Þ
0 2 4 6 8 10− 1
− 0.5
0
0.5
1
xm )(t
t
0 2 4 6 8 10− 4
− 2
0
2
4
)(tx
)(tE
)(tE−
t
0 2 4 6 8 10−30
−20
−10
0
10
20
30
x(t)
Ε )(t
Ε )(t−
t
164 8 Modulations
3. DSBLC Modulation, for d ¼ 1t :¼ 0; 0:01. . .10 x :¼ 2 X :¼ 10
Am :¼ 1 Ac :¼ 1
xðtÞ :¼ Ac � 1þ cosðx � tÞð Þ � cosðX � tÞEðtÞ :¼ Ac � 1þ cosðx � tÞð Þ
4. Phase Modulationt :¼ 0; 0:01. . .10 x :¼ 1 X :¼ 4
Am :¼ 2 Ac :¼ 4
xmðtÞ :¼ Am � cosðx � tÞ xcðtÞ :¼ Ac � cosðX � tÞxðtÞ :¼ Ac � cos X � t þ Am � sinðx � tÞð Þ
0 2 4 6 8 10−2
−1
0
1
2
)( tx
)(tE
)(tE−
t
0 2 4 6 8 10−4
−2
0
2
4
)(tx
xm )(t
xc )(t
t
8.5 Examples in MathCad 165
5. Frequency Modulationt :¼ 0; 0:01. . .10 x :¼ 1 X :¼ 4
Am :¼ 2 Ac :¼ 4
xmðtÞ :¼ Am � cosðx � tÞ xc :¼ Ac � cosðX � tÞ
xðtÞ :¼ Ac � cos X � t þ Am �Z t
0
sinðx � tÞdt0@
1A
6. ASK Modulationt :¼ 0; 0:01. . .10 f :¼ 2
xmðtÞ :¼
0 if 0� t� 2
1 if 2\t� 3
0 if 3\t� 4
1 if 4\t� 7
0 if 7\t� 8
1 if 8\t� 9
0 if 9\t� 10
xcðtÞ :¼ cosð2 � p � f � tÞ
0 2 4 6 8 10−4
−2
0
2
4
)(tx
xm )(t
xc )(t
t
166 8 Modulations
7. FSK Modulationt :¼ 0; 0:01. . .10 f0 :¼ 1 f1 :¼ 2
xc1ðtÞ :¼ cosð2 � p � f0 � tÞxc2ðtÞ :¼ cosð2 � p � f1 � tÞ
xmðtÞ :¼
0 if 0� t� 2
1 if 2\t� 3
0 if 3\t� 4
1 if 4\t� 7
0 if 7\t� 8
1 if 8\t� 9
0 if 9\t� 10
FSK(tÞ :¼ FSK xc1ðtÞ if xmðtÞ ¼ 0
FSK xc2ðtÞ if xmðtÞ ¼ 1
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
xm )(t
t
0 2 4 6 8 10−1
−0.5
0
0.5
1
m(t)
)(tKSA
t
8.5 Examples in MathCad 167
8. PSK Modulation
t :¼ 0; 0:01. . .10 f :¼ 1 /1 :¼p2
/2 :¼ p
xmðtÞ :¼
0 if 0� t� 2
1 if 2\t� 3
0 if 3\t� 4
1 if 4\t� 6
0 if 6\t� 8
1 if 8\t� 10
xc1ðtÞ :¼ cosð2 � p � f � t þ /1Þxc2ðtÞ :¼ cosð2 � p � f � t þ /2Þ
PSK(tÞ :¼ PSK xc1ðtÞ if xmðtÞ ¼ 0
PSK xc2ðtÞ if xmðtÞ ¼ 1
0 2 4 6 8 10−1
−0.5
0
0.5
1
xm )(t
)(tKSP
t
0 2 4 6 8 10−1
−0.5
0
0.5
1
xm )(t
)(tKSF
t
168 8 Modulations
Chapter 9Convolution and Deconvolution
Convolution is one of the more important mathematical operations performed onboth analog and digital signals. The convolution joins together three signals: inputand output, as well as the signal characterizing the system which is the subject ofour studies.
A reverse transformation which allows us to determine an unknown input signalis referred to a deconvolution. In the domain of an automatic control, where theinput and output are usually known, deconvolution is used to identify the investi-gated system.
In this chapter, we will present the principle of convolution, its basic properties,as well as methods of convolving and deconvolving signals with data in digitalform. These transformations are performed by a digital signal processing(DSP)system, after the analog signals have been transformed to a digital form. Such asituation always takes place in computerized measurement systems provided withdata acquisition cards.
9.1 Analog and Digital Convolution
The bilateral convolution of the signals k(t) * x(t) is
yðtÞ ¼ kðtÞ � xðtÞ ¼Z1�1
kðt � sÞxðsÞds ð9:1Þ
for which it is assumed that k(t) and x(t) are absolutely integrable over the intervalð�1;1Þ:
A one-sided convolution of the signals kðtÞ � xðtÞ is
yðtÞ ¼ kðtÞ � xðtÞ ¼Z t
0
kðt � sÞxðsÞds ð9:2Þ
for which k(t) and x(t) are absolutely integrable in any interval of 0 � t1 \t2\1:
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_9
169
If k(t) or x(t) is a periodic signal kT(t) or xT(t), with the period of T, then itsconvolution with another signal is also periodic and is referred to as a circular orcyclic convolution
yðtÞ ¼ kðtÞ � xTðtÞ ¼Z1�1
kðsÞxTðt � sÞds ð9:3Þ
The convolution (9.3) in finite time interval is
yðtÞ ¼ kðtÞ � xTðtÞ ¼Zt0þtt0
kðt � sÞxTðsÞds ð9:4Þ
where t0 is the initial time.
9.2 Properties of Convolution
1. Commutativity
kðtÞ � xðtÞ ¼ xðtÞ � kðtÞ ð9:5Þ
2. Associativity
½kðtÞ � xðtÞ� � zðtÞ ¼ kðtÞ � ½xðtÞ � zðtÞ� ð9:6Þ
3. Distributivity over addition
kðtÞ � ½xðtÞ þ zðtÞ� ¼ kðtÞ � xðtÞ þ kðtÞ � zðtÞ ð9:7Þ
4. Associativity for multiplication
c½kðtÞ � xðtÞ� ¼ ½ckðtÞ� � xðtÞ ¼ kðtÞ � ½cxðtÞ� ð9:8Þ
where c is a constant.
If k(t) and x(t) have Laplace transforms, then the relation between them and theirconvolution is expressed by the Borel theorem
L½kðtÞ � xðtÞ� ¼L½kðtÞ�L½xðtÞ� ð9:9Þ
170 9 Convolution and Deconvolution
A similar relation occurs for Fourier transforms
F½kðtÞ � xðtÞ� ¼ F½kðtÞ� F½xðtÞ� ð9:10Þ
Because of the commutativity, the convolution integrals (9.1) and (9.2) may beexpressed in an equivalent form
yðtÞ ¼ kðtÞ � xðtÞ ¼Z1�1
kðsÞxðt � sÞds ð9:11Þ
and
yðtÞ ¼ kðtÞ � xðtÞ ¼Z t
0
kðsÞxðt � sÞds ð9:12Þ
If in the convolution integrals k(t) is the kernel, that is the response of system toDirac delta dðtÞ, and x(t) is the input, then y(t) represents the impulse response.Such a response is commonly used in the analysis of the properties of variousdynamic systems, and we will therefore review the basic relations related to thatimpulse.
The example of the Dirac delta is shown in Fig. 9.1, while the basic relations aregiven by the formulae [9.13–9.23].
xðtÞ dðtÞ ¼ xð0Þ dðtÞ ð9:13Þ
xðtÞ dðt � t0Þ ¼ xð0Þ dðt � t0Þ ð9:14Þ
A � et dðtÞ ¼ A dðtÞ ð9:15Þ
et cos t dðtÞ ¼ dðtÞ ð9:16Þ
A � sin t dðtÞ ¼ 0 ð9:17Þ
Fig. 9.1 Example of dðtÞ
9.2 Properties of Convolution 171
dð�tÞ ¼ dðtÞ ð9:18ÞZ1�1
adðtÞdt ¼ aZ1�1
dðtÞdt ¼ a a 2 < ð9:19Þ
Zþ1�1
f ðtÞ dðtÞdt ¼ f ð0Þ ð9:20Þ
Zþ1�1
f ðtÞ dðt � t0Þdt ¼ f ðt0Þ ð9:21Þ
Z t
0
f ðsÞdðt � sÞds ¼Z t
0
f ðt � sÞdðsÞds ¼ f ðtÞ ð9:22Þ
ddt1ðtÞ ¼ dðtÞ ð9:23Þ
where 1ðtÞ is a unit step signal.For signals k(t) and x(t) given in analytic form, there is no special problem in
calculating the convolution integral (9.2). Below, we will discuss the method forthe calculation in digital of the convolution for signals k(t) and x(t) presented inFig. 9.2.
In order to calculate the integral (9.2), we will shift signal kð�sÞ to the right,starting from zero, by a step equal to D, that is by D; 2D; 3D . . . to T. Then, for eachshift, we will multiply the signal spectral lines kðD� sÞ, kð2D� sÞ; kð3D� sÞ . . .by the corresponding values of the spectral lines xðsÞ: Multiplying the total sum of
Fig. 9.2 Signals xðtÞ; kðsÞ; kð�sÞ
172 9 Convolution and Deconvolution
the products by D, we arrive at the convolution integral in digital form. The zerospectral line y0 of the convolution equals (Fig. 9.3)
y0 ¼ k0 x0D ð9:24Þ
The first spectral line of the convolution for kð�sÞ shifted to the right by D fromt = 0 equals (Fig. 9.4)
y1 ¼ ðk1 x0 þ k0 x1ÞD ð9:25Þ
The product (9.25) can be easily illustrated by a figure, representing the multi-plication of respective spectral lines (Fig. 9.5).
The second spectral line of the convolution for kð�sÞ shifted to the right by 2Dfrom t = 0 equals (Figs. 9.6 and 9.7)
y2 ¼ ðk2 x0 þ k1 x1 þ k0 x2ÞD ð9:26Þ
Fig. 9.3 Zero spectral line of the convolution
Fig. 9.4 First spectral line of the convolution
Fig. 9.5 Multiplication of spectral lines k0, k1 by spectral lines x0, x1 for the first spectral line ofconvolution y1
9.2 Properties of Convolution 173
Third spectral line of convolution for kð�sÞ shifted to the right by 3D from t = 0equals (Figs. 9.8 and 9.9)
y3 ¼ ðk3 x0 þ k2 x1 þ k1 x2 þ k0 x3ÞD ð9:27Þ
Fig. 9.7 Multiplication of spectral lines k0 − k2 by spectral lines x0 − x2 for the second spectralline of convolution y2
Fig. 9.8 Third spectral line of convolution. Signals kðt � 3DÞ and x(t)
Fig. 9.9 Multiplication of spectral lines k0 − k3 by spectral lines x0 − x3 for the third spectral lineof convolution y3
Fig. 9.6 Second spectral line of convolution. Signals kðt � 2DÞ and x(t)
174 9 Convolution and Deconvolution
For the nth spectral line of convolution for kð�sÞ shifted to the right byðn� 1ÞD, we get the digital form of the convolution in the form
yn ¼ DXni¼0
ki xn�i n ¼ 0; 1; 2; . . .; N � 1 ð9:28Þ
where n ¼ T=D.The last equation may be represented in matrix form. We then have
y0y1y2...
yN�1
2666664
3777775 ¼
k0 0 0 . . . 0k1 k0 0 . . . 0k2 k1 k0 . . . 0... ..
. ...
. . . ...
kN�1 kN�2 kN�3 . . . k0
2666664
3777775
x0x1x2...
xN�1
2666664
3777775D ð9:29Þ
The convolution (9.2) can also be easily realized using DFT
yn ¼ 1nþ 1
Xnm¼0
Xni¼0
xie�j2pnþ1mi
Xni¼0
kie�j2pnþ1mi
" #ej
2pnþ1mn D ð9:30Þ
The two-sided convolution for digital data is
yn ¼XN�1
i¼ðn�Nþ1ÞxikN�iþmD for n ¼ N; N þ 1; . . .; 2N � 2; m ¼ n� N ð9:31Þ
which, in the matrix form for vectors of length N, takes the form
y0y1y2...
yN�1...
y2N�3y2N�2
26666666666664
37777777777775¼
k0 0 0 . . . 0 0k1 k0 0 . . . 0 0k2 k1 k0 . . . 0 0... ..
. ...
. . . ... ..
.
kN�1 kN�2 kN�3 . . . k1 k0... ..
. ...
. . . ... ..
.
0 0 0 . . . kN�1 kN�20 0 0 . . . 0 kN�1
26666666666664
37777777777775
x0x1x2...
xN�2xN�1
266666664
377777775D ð9:32Þ
The convolution Eq. (9.31) with the use of DFT is realized by the formula
YðejxÞ ¼ XðejxÞKðejxÞ ð9:33Þ
9.2 Properties of Convolution 175
for which
YðejxÞ ¼ FðynÞ; XðejxÞ ¼ FðxnÞ; KðejxÞ ¼ FðknÞ ð9:34Þ
and
ReYðejxÞ ¼ ReXðejxÞReKðejxÞ � ImXðejxÞImKðejxÞImYðejxÞ ¼ ImXðejxÞReKðejxÞ þ ReXðejxÞImKðejxÞ ð9:35Þ
where F is the Continuous Fourier transform.Resulting from Eq. (9.35), the output signal yn is
yn ¼ IF ½ReYðejxÞ þ jImYðejxÞ�D ð9:36Þ
where IF is the inverse Fourier transform.It is also possible to determined yn as
yn ¼ 12N � 1
X2N�2i¼0
xiX2N�2v¼0
kvX2N�2m¼0
ej2p
2N�1mðn�v�iÞ" #
D ð9:37Þ
9.3 Continuous and Digital Deconvolution
Deconvolution is used in order to determine the input signal ~xðtÞ if k(t) and y(t) areknown or to determine the signal ~kðtÞ and if we know the input x(t) and output y(t).The first case deals with measurements, in which ~xðtÞ is the unknown measuredsignal, k(t) is the impulse response of the measurement system, and y(t) is the signalbeing measured. In the second case, we are determining the unknown ~kðtÞ on thebasis of the known input x(t) and the known output y(t) of the system beinginvestigated. We will now present the method for determining the signal ~xðtÞ.
For the digital values of ~kn and ~yn, successive spectral lines of deconvolutionresult directly from the Eqs. (9.24)–(9.28). Thus, we have the zero spectral line ofdeconvolution
~x0 ¼ y0k0D
; k0 6¼ 0 ð9:38Þ
• first spectral line
~x1 ¼ y1k0D� k1k0D
~x0; k0 6¼ 0 ð9:39Þ
176 9 Convolution and Deconvolution
• second spectral line
~x2 ¼ y2k0D� k2k0D
~x0 � k1k0D
~x1 ¼ y2k0D� k2~x0 þ k1~x1
k0D; k0 6¼ 0 ð9:40Þ
• third spectral line of deconvolution
~x3 ¼ y3k0D� k3k0D
~x0 � k2k0D
~x1 � k1k0D
~x2 ¼ y3k0D� k3~x0 þ k2~x1 þ k1~x2
k0D; k0 6¼ 0
ð9:41Þ
and nth spectral line
~xn ¼ y0k0D
for n ¼ 0; k0 6¼ 0
~xn ¼ ynk0D�Pn�1
i¼1 ki~xn�ik0D
; n ¼ 1; 2; 3; . . .;N � 1; k0 6¼ 0ð9:42Þ
It is easy to see that the deconvolution in matrix form, for n = N − 1 is
~x0~x1~x2...
~xN�1
2666664
3777775 ¼
1k0D
y0y1y2...
yN�1
2666664
3777775�
0 0 0 . . . 0k1 0 0 . . . 0k2 k1 0 . . . 0... ..
. ...
. . . ...
kN�1 kN�2 kN�3 . . . 0
2666664
3777775
~x0~x1~x2...
0
2666664
3777775
0BBBBB@
1CCCCCA ð9:43Þ
Applying Continuous Fourier transform, we can determine the deconvolutionbased on Eq. (9.44)
Re~XðejxÞ ¼ ReYðejxÞReKðejxÞ þ ImYðejxÞImKðejxÞ½ReKðejxÞ�2 þ ½ImKðejxÞ�2 ð9:44Þ
and
Im~XðejxÞ ¼ ImYðejxÞReKðejxÞ � ReYðejxÞImKðejxÞ½ReKðejxÞ�2 þ ½ImKðejxÞ�2 ð9:45Þ
hence, the input signal ~xn is
~xn ¼ IF½Re~XðejxÞ þ jIm~XðejxÞ�D
ð9:46Þ
9.3 Continuous and Digital Deconvolution 177
Signal ~xn may also be obtained using the state equation (2.102). For systems oforder higher than one, we have
~xn ¼ 1w1
ynþ1 � u1;1yn � u1;2y2n � � � � � u1;mymn� � ð9:47Þ
where
y2nþ1 ¼ w2~xn þ u2;1yn þ u2;2y2n þ � � � þ u2;mymn
..
.
ymnþ1 ¼ wm~xn þ um;1yn þ um;2y2n þ � � � þ um;mymn
ð9:48Þ
In Eq. (9.47), ~xn is determined on the basis of yn at instants n and n + 1, whileauxiliary variables y2n; . . .; ymn at instant n − 1.
Calculation of Eqs. (9.47) and (9.48) requires a knowledge of the initial valuesof the variables y2n : This value is assumed to be equal to zero.
For systems of the first order, it is not possible to calculate the deconvolution in arecurrent way. In this case, the deconvolution algorithm has the form
~xn ¼ A1ynþ1 þ A2yn; n ¼ 0; . . .;TD� 2 ð9:49Þ
and
A1 ¼ 1
1� e�DT0
; A2 ¼ �e�DT01� e
�DT0
ð9:50Þ
where T0 is time constant of the system.
9.4 Deconvolution for Low-Pass System
Determination of the signal ~xn on the basis of the kernel kn and output yn of thesystem Eq. (9.47) may be applied without any restrictions for low-pass and high-pass systems. Eqs. (9.38)–(9.46) may be used only for high-pass systems for whichk0 ≠ 0. In order to use this equation for low-pass system, for which k0 = 0, we canshift the elements of kn vector by a constant value ν multiplying them by 1ðt þ mÞ:To simplify the calculations, it is convenient to apply the step response charac-teristics, as these have a lower overshoot than the impulse response and do notassume negative values. For these characteristics, the following formulae are used
178 9 Convolution and Deconvolution
~xn ¼ y0h0
for n ¼ 0; h0 6¼ 0
~xn ¼ ~xn�1 þ yn � ~x0hnh0
�Xni¼2
~xnþ1�i � ~xn�i½ �hih0
for n ¼ 1; 2; . . .;N � 1; h0 6¼ 0
ð9:51Þ
9.5 Conjugate Operator and Maximum Integral SquareCriterion
Let us present the integral square criterion by means of a scalar product
I2ðxÞ ¼ Kx;Kxh i ¼ y; yh i ð9:52Þ
or
I2ðxÞ ¼ K�Kx; xh i ð9:53Þ
where Kx represents the convolution integral and K*is the conjugate for K.Criterion I2(x) can be given in the equivalent form
I2ðxÞ ¼ y;Kxh i ¼ x;K�yh i ð9:54Þ
Let us present Eq. (9.54) as follows
ZT0
yðtÞZ t
0
k t � sð Þ xðsÞ ds dt ¼ZT0
xðtÞ½K�y�dt ð9:55Þ
Changing the limits of integration of the internal integral on the left-hand side ofthe Eq. (9.55) to [0, T] and, at the same time, multiplying it by 1ðt � sÞ, we get
ZT0
yðtÞZT0
kðt � sÞ 1ðt � sÞ x ðsÞ ds dt ¼ZT0
xðtÞ½K�y�dt ð9:56Þ
which, after changing the order of integration and replacing t by s, gives
ZT0
xðtÞZT0
kðs� tÞ 1ðs� tÞ yðsÞ ds dt ¼ZT0
xðtÞ½K�y�dt ð9:57Þ
9.4 Deconvolution for Low-Pass System 179
Taking into account that the integral in (9.57) has the value of zero for s\t, wecan present it in the form
ZT0
xðtÞZTt
kðs� tÞ yðsÞ ds dt ¼ZT0
xðtÞ½K�y�dt ð9:58Þ
From Eq. (9.58), it follows that the conjugate operator K*y is
K�y ¼ZTt
kðs� tÞ yðsÞ ds ð9:59Þ
Thus,
K�Kx ¼ZTt
kðs� tÞZs
0
kðs� vÞxðvÞdv24
35 ds ð9:60Þ
Equation (9.60) allows us to determine the input signal xðtÞ ¼ x0ðtÞ, maximizingcriterion I2(x)
I2ðx0Þ ¼ supfI2ðxÞ : x 2 Xg ð9:61Þ
where X is a set of signals constrained in amplitude
xðtÞ� 1 ð9:62Þ
From the condition of optimality, we have
@I2ðxÞ@x x0 ; x� x0j
� �� 0 ð9:63Þ
After simple transformation Eq. (9.63) yields
K�Kx0; xh i � K�Kx0; x0h i ð9:64Þ
in which the right-hand side represents the maximum. The left-hand side ofEq. (9.64) reaches a maximum making both sides equal if a signal has the form
xðtÞ ¼ x0ðtÞ ¼ sgn ½K�Kx0ðtÞ� ð9:65Þ
and has the maximum permissible amplitude
xðtÞj j ¼ 1 ð9:66Þ
180 9 Convolution and Deconvolution
Substituting Eq. (9.60) into Eq. (9.65), we have
x0ðtÞ ¼ sgnZTt
kðs� tÞZs
0
kðs� mÞ x0ðmÞ dm0@
1A ds
24
35 ð9:67Þ
Equation (9.67) enables the determination of switching moments for the signalx0(t) after solving the system of integral equations resulting from its extension inconsecutive time intervals. Let us assume that the consecutive intervals t1, t2, …, tnin [0, T] correspond to the zeroing of the function under a sgn in Eq. (9.67) and thatthe first switching occurs between +1 and −1. It can be easily checked that theswitching moments resulting from (9.67) represent the system of equations
Xnl¼i
Ztlþ1tl
kðs� tiÞXl
m¼0ð�1mÞ
Ztmþ1tm
kðs� mÞdm0@
1A ds ¼ 0; i ¼ 1; 2; . . .; n ð9:68Þ
where t0 = 0, tn+1 = T, tmþ1 ¼ s for m = l, and n—number of switches.The upper value of the index n is not given at advance, but it is being con-
secutively increased until the criterion I2(x0) reaches a maximum.Examples of the equations for three switching instants in t1, t2, and t3, resulting
from Eq. (9.68), are as follows:
Rt2t1
kðs� t1ÞRt10kðs� vÞdv� Rs
t1
kðs� vÞdv" #
ds
þ Rt3t2
kðs� t1ÞRt10kðs� vÞdv� Rt2
t1
kðs� vÞdvþ Rst2
kðs� vÞdv" #
ds
þ RTt3
kðs� t1ÞRt10kðs� vÞdv� Rt2
t1
kðs� vÞdvþ Rt3t2
kðs� vÞdv� Rst3
kðs� vÞdv" #
ds ¼ 0
ð9:69Þ
Rt3t2
kðs� t2ÞRt10kðs� vÞdv� Rt2
t1
kðs� vÞdvþ Rst2
kðs� vÞdv" #
ds
þ RTt3
kðs� t2ÞRt10kðs� vÞdv� Rt2
t1
kðs� vÞdvþ Rt3t2
kðs� vÞdv� Rst3
kðs� vÞdv" #
ds ¼ 0
ð9:70Þ
9.5 Conjugate Operator and Maximum … 181
ZTt3
kðs� t3ÞZt10
kðs� vÞdv�Zt2t1
kðs� vÞdvþZt3t2
kðs� vÞdv24
�Zs
t3
kðs� vÞdv35ds ¼ 0
ð9:71Þ
9.6 Examples in MathCad
Analog and digital convolution for low-pass second-order system
T :¼ 5 D :¼ 0:01
t :¼ 0;D; . . .T
a :¼ 2 b :¼ 0:1 x0 :¼ 4 f :¼ 0:3
KðsÞ :¼ �a � s2s2 þ 2 � b � x0 � sþ x2
0
kðtÞ :¼ �a � e�b�x0�t
b2 � 1�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2
0 � b2 � x20
q� sin
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�x2
0 � b2 � 1� � � tq
0 1 2 3 4 5−10
−5
0
5
10
k(t)
t
xðtÞ :¼ sinð3 � p � f � tÞ þ sinð5 � p � f � tÞ þ cosð7 � p � f � tÞ
182 9 Convolution and Deconvolution
0 1 2 3 4 5−2
−1
0
1
2
3
x(t)
tAnalog convolution
yðtÞ :¼Z t
0kðt� v) � x(v)dv
0 1 2 3 4 5−10
−5
0
5
10
y(t)
t
Discretization of signals k(t), x(t), and time T
KD :¼for i 2 0. . . TD� 1
KDi kði � DÞKD
��������XD :¼
for i 2 0. . . TD� 1
XDi xði � DÞXD
��������TD :¼
for i 2 0. . . TD� 1
TDi i � DTD
��������
Digital convolution
YD :¼
for i 2 0. . . TD� 1Yi;0 0for j 2 0. . .iYi;0 Yi;0 þ XDj;0 � KDi�j;0
������Y Y � DY
������������
9.6 Examples in MathCad 183
0 1 2 3 4 5−10
−5
0
5
10
YΔ
TΔDigital response of second-order system
T :¼ 5 D :¼ 0:01
t :¼ 0;D; . . .T
a :¼ 2 b :¼ 0:1 x0 :¼ 4 f :¼ 0:3
KðsÞ :¼ a � x20
s2 þ 2 � b � x0 � s + x20
xðtÞ :¼ sinð3 � p � f � tÞ þ sinð5 � p � f � tÞ þ cosð7 � p � f � tÞ
yðtÞ :¼Z t
0
kðt � vÞ � uðvÞdv
Discretization of signals k(t), x(t), T, and y(t)
KD :¼for i 2 0. . . TD� 1
KDi kði � DÞKD
��������XD :¼
for i 2 0. . . TD� 1
XDi xði � DÞXD
��������TD :¼
for i 2 0. . . TD� 1
TDi i � DTD
��������
184 9 Convolution and Deconvolution
YD :¼
for i 2 0. . . TDYi;0 0
for j 2 0. . .i
Yi;0 Yi;0 þ XDj;0 � KDi�j;0
�������Y Y � DY
��������������
k1 :¼
for i 2 0. . . TD� 1
K10;0 YD0;0XD0;0
if i ¼ 0
if i[ 0K1i;0 0
for j 2 0. . .i� 1
K1i;0 K1i;0 þK1j;0 �XDi�j;0
XD0;0
K1i;0 YDi;0XD0;0� K1i;0
�����������
�����������������K1 K1
D
K
�������������������������
0 1 2 3 4 5−10
−5
0
5
10
K1
TΔState equation in deconvolution
T :¼ 5 D :¼ 0:01
t :¼ 0;D; . . .T
a :¼ 2 b :¼ 0:1 x0 :¼ 4 f :¼ 0:3
kðtÞ :¼ �a � e�b�x0�t
b2 � 1�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2
0 � b2 � x20
q� sin
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�x2
0 � b2 � 1� � � tq
xðtÞ :¼ sinð3 � p � f � tÞ þ sinð5 � p � f � tÞ þ cosð7 � p � f � tÞ
9.6 Examples in MathCad 185
yðtÞ :¼Z t
0
kðt � vÞ � xðvÞdv
A :¼ 0 1�x2
0 �2 � b � x0
� �B :¼ 0
a � x20
� �
Dicretization of signals k(t), y(t) and T
KD: =
for i 2 0. . . TD� 1
KDi kði � DÞ
KD
���������YD: =
for i 2 0. . . TD� 1
YDi yði � DÞ
YD
���������
TD: =
for i 2 0. . . TD� 1
TDi i � D
TD
���������
e
0 1�x2
0 �2 � b � x0
� ��D! 0:99920223539382994143 0:0099574506405935111096
�0:1593192102494617775 0:99123627488135513254
� �
U ¼ 0:99920223539382994143 0:0099574506405935111096�0:1593192102494617775 0:99123627488135513254
� �
ZD0
e
0 1�x2
0 �2 � b � x0
� ��kdk � 0
a � x20
� �! 0:001595529212340117141
0:31863842049899235551
� �
w :¼ 0:0015955292123401171410:31863842049899235551
� �
186 9 Convolution and Deconvolution
Determination of input signal
X1 :¼
Y20;0 0
for k 2 0. . . TD� 2
X1k;0 1W0;0
YDðkþ1Þ;0 � U0;0 � YDk;0 � U0;1Y2k;0h i
Y2kþ1;0 U1;0 � YDk;0 þ U1;1Y2k;0 þW1;0 � X1k;0
������X1
���������������
0 1 2 3 4 5−2
−1
0
1
2
3
X1
x(t)
TΔ t,
Conjugate operator
KðsÞ :¼ 12 � � � sþ 1
kðtÞ :¼ 3 � e� t2 D :¼ 0:01 T :¼ 20
t :¼ 0;D; . . .T
0 5 10 15 200
1
2
3
k(t)
t
9.6 Examples in MathCad 187
XðsÞ :¼ 1sþ 2
XðtÞ :¼ e�2�t
KðsÞ � XðsÞ invlaplace ! �2 � e� t2 � e�
3�t2 � 1
� ZT0
ZTt
kðs� tÞ �Z s
0kðs� tÞ � xðvÞdv
� �ds � xðtÞdt ¼ 1:8
ZT0
�2 � e� t2 � e�
3�t2 � 1
� h i2dt ¼ 1:8
188 9 Convolution and Deconvolution
Chapter 10Reduction of Signal Disturbance
In the previous chapters, where we covered signal analysis, we often used trans-forms made up of specific mathematical operations, containing variables and theirderivatives and integrals. We assumed then that signals we were analyzing had anideal, non-disturbed form. Such situations, however, do not exist in practice, assignals always are more or less disturbed. The error produced by disturbance can beso significant, in comparison with the measurement signal, that in practice,achieving a meaningful measurement is impossible. When a signal is differentiated,the disturbance is also differentiated. All disturbances are then amplified, and as aresult, the signal that is generated is even more disturbed. In such cases, situationsmay occur in which the disturbance will be greater than the signal itself. In order toreduce the disturbance, various methods are used, among which two are worthnoting: filtration by means of time windows and adoption of the Kalman filtermethod. In both cases, additivity of the disturbance is assumed. The method uti-lizing time windows refers to an analog procedures, in which reduction of distur-bance is executed thanks to application of special windows, to which the derivativeof disturbed signal is transmitted, whereas in the Kalman filter method, a recurrentalgorithm, based on a minimum variance estimator, is used. In the latter case, themeasurement system is represented by means of discrete equations, and the dis-turbed signal is assumed to have the properties of white noise. We will deal firstwith the method of disturbance reduction by means of time windows and then withthe Kalman filter method.
10.1 Time Windows in Reduction of Disturbance
The method deals with the reduction of disturbance for the m-th order system,described by linear differential equation with constant coefficients
Xmk¼0
akyðkÞðtÞ ¼ xðtÞ ð10:1Þ
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9_10
189
where x(t) is the input, y(k)(t) is the k-th derivative of the output signal, and ak is thek-th constant coefficient. Let us consider the integral
~yðtÞ ¼Ztþd
t�d
yzðsÞgðs� tÞds ð10:2Þ
where disturbed signal is the sum
yzðtÞ ¼ yðtÞ þ zðtÞ ð10:3Þ
and g(t) is the time window (Fig 10.1).For successful reduction of the disturbed signal, the window g(t) must fulfill the
following conditions:
• At the ends of intervals (t − δ) and (t + δ), the window and its derivatives mustreach zero.
gðkÞðt � sÞ ¼ gðkÞðt þ sÞ ¼ 0; k ¼ 0; 1; 2; . . . ð10:4Þ
• In the middle of the range, the window should have a maximum value.• The window must meet the condition
Zþd
�d
gðtÞdt ¼ 1 ð10:5Þ
These requirements are met, for example, by Nuttall windows
gðs� tÞ ¼ npd
cospp2d
s� tð Þh i
; p ¼ 1; 2; 3; . . . ð10:6Þ
Fig. 10.1 Reduction of disturbance by means of a time window
190 10 Reduction of Signal Disturbance
where
np ¼ p4
Zp2
0
cospðuÞdu
8><>:
9>=>;
�1
ð10:7Þ
or triangular windows
gðs� tÞ ¼ 1� s� td
��� ���h ip; p ¼ 1; 2; 3; . . . ð10:8Þ
It is easy to check that the k-th derivative of ~yðtÞ in Eq. (10.2) gives
~yðkÞz ðtÞ ¼ ð�1ÞkZtþd
t�d
yðsÞgðkÞðs� tÞdsþ ð�1ÞkZtþd
t�d
zðsÞgðkÞðs� tÞds ð10:9Þ
from which it appears that differentiation of the disturbance carries over to windowg(t).
Let us estimate the right part of sum in Eq. (10.9). Then, we have
Ztþd
t�d
zðsÞgðkÞðs� tÞds � supt�d� s� tþd
gðkÞðs� tÞh i Ztþd
t�d
zðsÞds ð10:10Þ
Assuming that z(t) is the randomsignal, changing quickly its value and the signwith respect to gðkÞðtÞ; we get
Ztþd
t�d
zðsÞds � 0 ð10:11Þ
so we have
~yðkÞn � ð�1ÞkZtþd
t�d
yðsÞgðkÞðs� tÞds ð10:12Þ
The last equation describes the effect of the reduction of the k-th time differ-entiated signal.
10.1 Time Windows in Reduction of Disturbance 191
10.2 Signal Reconstruction
In many practical applications, it is necessary to reconstruct the input signal in asituation, in which the output signal is disturbed. Using the time window, it is easyto realize this. In order to prove the above, let us put x(t) in the place y(t) inEq. (10.2). Thus, we have
~xðtÞ ¼Ztþd
t�d
xðsÞgðs� tÞds ð10:13Þ
Substituting the left hand side of Eq. (10.1) in the place of x(t) in Eq. (10.13) andthen changing y(t) to yz(t), we get the reconstructed input signal
~xðtÞ ¼Ztþd
t�d
yzðsÞXmk¼0
�1ð ÞkakgðkÞðs� tÞ" #
ds ð10:14Þ
10.3 Kalman Filter
This filter uses an algorithm that performs the recurrent determination of theminimum variance estimate of the state vector of a linear discretediscrete dynamicsystem, on the basis of measurements of its output. The Kalman filter algorithm is
x½k þ 1� ¼ AðkÞx½k� þ BðkÞu½k� þ w½k�y½k� ¼ CðkÞx½k� þ DðkÞu½k� þ v½k�; k ¼ 0; 1; 2; . . . ð10:15Þ
where u½k� is the vector of input signals with m coordinates; x½k� and x½k þ 1� arestate vectors with n coordinates for moments k and k + 1; yðkÞ is the vector ofoutput signals with p coordinates; w½k� is the vector of system noise with n coor-dinates; v½k� is the vector of measurement noise with p coordinates; AðkÞ—state-transition matrix with dimensions n � n; BðkÞ is the control matrix with dimen-sions n � m; CðkÞ is the output matrix with dimensions p � n; and DðkÞ is thetransition matrix with dimensions p � m.
Figure 10.2 shows the block diagram of the system represented by Eq. (10.15).For the Kalman filter, it is assumed that both the measurement and processing
within the system is disturbed by noise, with a Gaussian distribution.In the Kalman filter synthesis, the following assumptions are made:
1. The deterministic component of input u(k) is equal to zero.2. Due to lack of control, the state variable is close to zero.
192 10 Reduction of Signal Disturbance
E x½k�f g ¼ 0 ð10:16Þ
3. Disturbances w½k� and v½k� have the properties of discrete white noise, i.e., theyare not correlated and have a zero expected value and constant covariance.
E w½k�wT ½k�� � ¼ R½k�; i ¼ k0; i 6¼ k
�ð10:17Þ
Efv½k�vT ½k�g ¼ Q½k�; i ¼ k0; i 6¼ k
�ð10:18Þ
where R½k� and Q½k� are matrices of disturbance covariance.4. State and measurement errors are not correlated.
E v½k�wT ½k�� � ¼ 0 ð10:19Þ
5. Estimation errors do not depend on measurements
E x½k� � x̂½k�ð ÞvT ½k�� � ¼ 0 ð10:20Þ
which requires that the vector x̂½k� depends, in a random way on observation,until the step k − 1.
6. Matrix DðkÞ ¼ 0:
The above assumptions allow us to modify the state equation (10.15) to thefollowing form:
x½k þ 1� ¼ AðkÞx½k� þ BðkÞu½k�y½k� ¼ CðkÞx½k� þ v½k� ð10:21Þ
Figure 10.3 presents the block diagram corresponding to Eq. (10.21).
Fig. 10.2 Block diagram of a discrete dynamic system
10.3 Kalman Filter 193
Kalman filtration is based on the assumption that at k − 1 discrete moment, thestate estimator x̂ k � 1; k � 1½ � and covariance P k � 1; k � 1ð Þ are obtained, whilefor k moment, the value of the estimator x̂ k; k � 1½ � is predicted, as well as the valueof the covariance P k; k � 1ð Þ; relating to it. If the obtained results differ from thosepredicted in the previous step, then a correction is introduced to the prediction forthe moment k + 1 executed in step k.
The Kalman filter equations, resulting from the above assumptions, are dividedinto two categories:
1. Time update equations, which predict the system state at the discrete moment k,on the basis of an estimate at the moment k − 1. They follow the algorithmpresented below:
(a) Project the state ahead
x̂½k; k � 1� ¼ AðkÞx̂½k � 1; k � 1� þ BðkÞu½k � 1� ð10:22Þ
where x̂ k � 1; k � 1½ � and x̂ k; k � 1½ � are, respectively, the a priori estimate(before measurement) and a posteriori estimate (after measurement) of thestate vector.
(b) Project the error covariance ahead
P k; k � 1ð Þ ¼ AðkÞP k � 1; k � 1ð ÞATðkÞ þ R½k� ð10:23Þ
where
P k � 1; k � 1ð Þ ¼ E e½k � 1; k � 1�eT ½k � 1; k � 1�� � ð10:24Þis the matrix of a priori covariance of the error vector
e k � 1; k � 1½ � ¼ x k � 1½ � � x̂ k � 1; k � 1½ � ð10:25Þ
whereas
P k; k � 1ð Þ ¼ E e½k; k � 1�eT ½k; k � 1�� � ð10:26Þ
Fig. 10.3 Schematic diagram of Kalman filtration
194 10 Reduction of Signal Disturbance
and
e k; k � 1½ � ¼ x½k� � x̂ k; k � 1½ � ð10:27Þ
are the matrix of a posteriori covariance of the error vector (10.25).The vectors in Eqs. (10.25) and (10.27) show the difference between theactual value of the state vector and its estimate and constitute the measureof assessment error for the state vector.
2. Equations of measurement updates which, on the basis of the current observa-tion data, introduce a correction to the prediction
(a) Compute the Kalman gain
KðkÞ ¼ P k; k � 1ð ÞCTðkÞ QðkÞ þ CðkÞPðk; k � 1ÞCTðkÞ� ��1 ð10:28Þ
(b) Update the estimate with measurement y[k]
x̂½k� ¼ x̂ k; k � 1½ � þK k; kð Þ y½k� � CðkÞx̂½k; k � 1�f g ð10:29Þ
(c) Update the error covariance
PðkÞ ¼ I�Kðk; kÞCðkÞ½ �P k; k � 1ð Þ ð10:30Þ
Figure 10.4 presents the algorithm according to which the Kalman filter resultingfrom Eqs. (10.22)–(10.30) is executed.
In the Kalman filter, the equations updating time and measurements are realizedin a cycle, for subsequent moments k, which allows us to estimate the process statex̂½k� due to minimum error of Eq. (10.27).
Fig. 10.4 Algorithm of function for Kalman filter
10.3 Kalman Filter 195
In numerical calculations to determine initial parameters, where a priori infor-mation about the process is missing, it is necessary to assume zero initial values ofthe state vector which estimates the covariance matrix P k � 1; k � 1ð Þ:
10.4 Examples in MathCad
Application of Nuttall window to the filtration of a signal with disturbance
T :¼ 5 D :¼ 0:01
t :¼ 0; D; . . .;T
xðtÞ :¼ e�t � sinð2 � tÞ þ e�2t � sinð3 � tÞ
0 1 2 3 4 5−0.5
0
0.5
1
)(tx
tDisturbance
zðtÞ :¼ 0:3 � sinð50 � tÞ þ 0:5 � sinð90 � tÞ � e�0:5t
0 1 2 3 4 5−1
−0.5
0
0.5
1
)(tz
t
196 10 Reduction of Signal Disturbance
xzðtÞ :¼ xðtÞ þ zðtÞ
0 1 2 3 4 5−1
0
1
2
xz )(t
tNuttall window
p :¼ 5 d :¼ 0:01
np :¼ p4�
Zp2
0
cosðuÞpdu
264
375�1
gðtÞ :¼ npd� p � t2 � dh ip
0 1 2 3 4 50
5
10
15
)(tg
t
10.4 Examples in MathCad 197
Zd
�d
gðsÞds ¼ 1
Filtration of signal with disturbance
x1ðtÞ :¼Ztþd
t�d
xzðsÞ � gðs�tÞds
0 1 2 3 4 5−0.5
0
0.5
1
x1 )(t
)(tx
tReproduction of input signal x(t)
T :¼ 5 D :¼ 0:01
t :¼ 0; D; . . .; T
xðtÞ :¼ sinð5 � tÞ þ sinð7 � tÞ
0 1 2 3 4 5−2
−1
0
1
21.933
−1.933
x(t)
50 t
198 10 Reduction of Signal Disturbance
Impulse response of second-order system
a :¼ 1 x0 :¼ 5 b :¼ 0:2
kðtÞ :¼ a � x0ffiffiffiffiffiffiffiffiffiffiffiffiffi1� b2
p � e�b�x0�t � sin x0 �ffiffiffiffiffiffiffiffiffiffiffiffiffi1� b2
q� t
0 1 2 3 4 5−2
0
2
4
k(t)
t
yðtÞ :¼Z t
0
xðsÞ � kðt � sÞds
0 1 2 3 4 5−4
−2
0
2
4
y(t)
tDisturbance
zðtÞ :¼ 0:3 � sinð50 � tÞ þ 0:5 � sinð90 � tÞ � e�0:5t
10.4 Examples in MathCad 199
0 1 2 3 4 5−1
−0.5
0
0.5
1
)(tz
t
yðtÞ :¼ yðtÞ þ zðtÞ
0 1 2 3 4 5−4
−2
0
2
4
yz t( )
tNuttall window
r :¼ 4 d :¼ 0:294
np :¼ p4�
Zp2
0
cosðuÞpdu
0B@
1CA
�1
gðtÞ :¼ npd� p � t2 � dh ip
200 10 Reduction of Signal Disturbance
0 1 2 3 4 50
1
2
3
4
5
)(tg
t
g1ðtÞ :¼ddtgðtÞ
g2ðtÞ :¼d2
dt2gðtÞ
x1ðtÞ :¼Ztþd
t�d
yzðsÞ � gðs� tÞ � 2 � bx0
� g1ðs� tÞ þ 1x2
0� g2ðs� tÞ
� �ds
0 1 2 3 4 5−2
−1
0
1
2
x1 )(t
)(tx
t
10.4 Examples in MathCad 201
10.5 Kalman Filter in LabVIEW
202 10 Reduction of Signal Disturbance
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Bibliography 205
Index
AAmplitude, 1, 118, 143, 144, 149, 151, 155,
156, 161, 162, 180, 182Analysis
asynchronous, 72discrete, 70, 72spectral, 71–73synchronous, 70
Approximation, 101Associativity, 170Associativity for multiplication, 170Attenuation, 49Autocorrelation, 5
BBandwidth, 148Binary data, 162Bit duration, 159Borel theorem, 170
CCapacitor, 31Carrier
amplitude, 152frequency, 150, 161
Cauchy, 107Characteristic
frequency, 94Code
natural binary, 157Coder, 141Commutativity, 170Complex
conjugate, 83coordinate, 112number, 82
Componentfundamental, 4harmonic, 4negative, 143positive, 143
Conditionacceptability, 98initial, 33optimality, 180
Conjugate, 128, 179operator, 179, 180, 187
Constellationdiagram, 160, 161QPSK, 161
Covariance, 193–195Convergence
abscissa, 22area, 22
ConverterA/D, 81, 156
Convolution, 46, 86, 169, 170, 173–176analog, 169, 182, 183bilateral, 169circular, 170cyclic, 170digital, 100, 169, 182, 183integral, 171–173, 179one-sided, 169
Current, 30, 31, 125–127, 129digital, 127
DData acquisition card, 169Decimation, 101Decomposition, 91, 92, 102Deconvolution, 169, 176–178, 185
© Springer International Publishing Switzerland 2015E. Layer and K. Tomczyk, Signal Transforms in Dynamic Measurements,Studies in Systems, Decision and Control 16,DOI 10.1007/978-3-319-13209-9
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Deconvolution (cont.)algorithm, 178continuous, 176digital, 176
Demodulator, 141Denominator, 29, 34, 35, 92Derivative, 26, 86, 110, 121Detail, 102Digital
measurement, 176Dirac delta, 171Dirac impulse series, 47Dirichlet condition, 43, 44Distributivity over addition, 170Disturbance, 189–191, 193, 196, 199Domain
frequency, 97, 98, 108time, 25, 43, 85, 97
DSP, 169
EElectrical circuit, 31Energy
limited, 6, 13, 18Envelope, 110, 143, 144, 146Equation
differential, 26, 81linear, 81state, 30, 185
Errorcovariance, 195estimation, 193measurement, 193state, 193vector, 194, 195
Expander, 102
FFactor
distortion, 4filling, 4nonlinear distortion, 4peak, 4scaling, 97shape, 4shift, 97
Filteranalog, 94bank, 100, 102Butterworth, 94coefficient, 103digital, 81, 94, 100high-pass, 100, 104
ideal, 49Kalman, 189, 192, 194, 195, 202low-pass, 94, 100, 101
Fluctuation, 101Fourier series, 60, 63, 64, 154Frequency, 132, 143, 144
band, 146deviation, 159
Functiondensity, 5even, 77harmonic horizontal, 113harmonic verticall, 113odd, 78weight, 117, 121
HHelix
circular, 113conic, 115
IInitial condition, 30Input, 29, 30, 171Integral, 22, 23, 50, 98, 189
convolution, 107internal, 179maximum, 179square error, 121, 124, 179
Integration, 46, 62, 179limit, 179
KKalman gain, 195Kernel, 171, 178Kirchhoff’s law, 31, 32Kronecker delta, 100
LLabVIEW, 105, 135, 202Linearity, 25, 45, 85
MMagnitude, 48Mallat algorithm, 101MathCad, 18, 38, 66, 74, 112, 129, 163, 182,
196Matrix
control, 192feedthrough, 29form, 175, 177state-transmission, 192
208 Index
Measurementactive power, 137current, 125, 135frequency, 125, 128, 139noise, 192output, 192power, 125reactive power, 138system, 176transition, 192voltage, 125, 136
Minimum variance, 192Modulation
AM, 141, 142analog, 141, 142angle, 142, 159ASK, 142, 158, 166BPSK, 142, 160, 161CPFSK, 142, 159depth, 143, 144digital, 141, 158DPCM, 142, 158DPSK, 142, 160DSBLC, 142–144, 146, 148, 158, 163–165DSBSC, 142, 146, 147FM, 141, 142, 151, 156, 166FSK, 142, 158, 159, 167GMSK, 142, 159impulse, 142, 151MSK, 142, 159PAM, 142, 152–155, 157PCM, 142, 156, 157PDM, 142, 156PM, 141, 142, 150, 151, 156, 165PPM, 142, 156PSK, 142, 159, 168Pulse code, 141PWM, 142, 151QAM, 142, 162QPSK, 142, 160SSB, 147SSBSC, 142, 148, 149VSB, 148, 150
Modulator, 141Mother wavelet, 97
unit, 30Multiplication, 25, 46, 62
NNumerator, 29, 34, 92Numerical
method, 125
OOrder, 21, 23, 27, 29, 34, 35, 91Ordinate, 84Orthogonal
component, 125Orthogonality, 129Orthonormal
mutually, 118polynomials, 121set, 123
Orthonormality, 119Output, 29, 30, 38, 101Overshoot, 178
PParseval equality, 46Period, 1Phase, 1, 110, 162
frequency, 110instantaneous, 150shift, 119
Pole, 27, 86, 91, 92Polynomial, 91, 121
approximating, 124Tchebyshev, 122
Poweractive, 126–129digital, 127limited, 10, 15reactive, 127–129
Probabilitydistribution, 3, 5
QQuantities
electrical, 124
RRadius, 84, 161Receiver, 141, 146Reconstruction, 102Residuum, 87, 88Resistor, 30Response
digital, 184impulse, 36, 100, 102, 149, 171, 176, 178,
199Ruth
method, 35table, 34, 35
Index 209
SSample
initial, 3Sampling
frequency, 152ideal, 152, 153impulses, 81instantaneous, 152, 155interval, 87, 89, 94moment, 36, 155principle, 82process, 81, 82rate, 94real, 152, 153system, 81
Scale, 45Shannon’s theorem, 153Sideband, 142, 143, 146, 159
attenuated, 149lower, 146, 159, 149upper, 146, 148, 149
Signalalmost periodic, 2amplitude, 161analog, 6, 10, 16, 156analytic, 108, 112, 113, 146–149bipolar rectangular, 11carrier, 141–144, 146, 153, 155, 160, 161comb, 12continuous, 129, 132cosinusoidal, 7, 47deterministic, 1, 3digital, 169digital modulating, 158, 160Dirac delta, 12, 18, 24, 46, 85discrete, 3, 13, 15, 18, 131, 133, 156discretization, 183, 184distributive, 12disturbance, 189disturbed, 189, 190energy, 117envelope, 110ergodic, 3exponential, 14, 48exponentially decreasing, 7, 16exponentially increasing, 10Gaussian, 9, 17, 48harmonic, 10, 15, 47high frequency, 141input, 49, 100, 169, 176, 177, 180, 187,
190, 192, 198low-frequency, 141measured, 176
modulated, 143–150, 152, 154, 159, 160modulating, 142, 149, 151, 152, 155, 156,
158, 159monoharmonic, 1, 2non-ergodic, 2non-periodic, 2non-stationary, 3, 97orthogonal, 117, 121orthonormal, 117, 118, 122output, 159, 176, 190, 192over modulated, 144periodic, 1, 170polyharmonic, 1power, 4quasi-orthogonal, 120random, 1, 3, 5, 191real, 146reconstructed, 192reconstruction, 192rectangular, 6, 13, 48, 49, 70Sa, 8, 14, 17, 47sampled, 153shifted, 124, 133sign, 47sinusoidal, 8, 16, 47, 125stationary, 3, 97transient, 2transmitted, 146triangular, 7, 13unipolar rectangular, 11unit, 2, 7, 15, 19unit step, 9, 47, 49, 85, 172
Spectra, 108, 147Spectral
first, 173, 176line, 172–174, 176, 177n-th, 177second, 173, 174, 177third, 174, 177zero, 173, 176
Spectrum, 44, 48, 55, 108Standard deviation, 6State, 29
equation, 36, 193estimator, 194vector, 192
Symmetry, 45System
discrete, 192dynamic, 171global positioning, 141high-pass, 178linear, 81, 192
210 Index
low-pass, 178, 182response, 171second order, 184, 199
TTime
constant, 178initial, 3, 170interval, 170, 181reversal, 86
Transfer function, 30, 35Transform
continuous Fourier, 43, 176, 177continuous Wavelet, 97discrete Fourier, 51discrete wavelet, 100, 102fast Fourier, 52Fourier, 43, 44, 46, 51, 97, 98, 171Hilbert, 107, 108, 110, 143inverse, 83, 90, 91inverse Fourier, 176inverse Laplace, 33inverse wavelet, 98Laplace, 21, 22, 24–26, 28–30, 43, 81, 82,
87, 170multistage wavelet, 101, 102short-time Fourier, 55, 74–76three stage wavelet, 105wavelet, 97Z, 81–85, 88–91, 105
Transformationbilinear, 94reverse, 169
Translation, 98Transmission
channel, 141Transmitter, 141, 143, 148
VValue
constant, 178expected, 3final, 86initial, 86mean, 3, 5, 98peak, 4RMS, 3
Variablecomplex, 21random, 5state, 30
Variance, 3, 6Voltage, 30, 125, 127, 129
digital, 127
WWavelet
Coiflet, 104Daubechies, 103Harr, 103Marr, 98Meyer, 99Morlet, 98norm, 98symplet, 104
Window, 51Barlett, 58Blackman, 59, 69discrete, 66exponential, 59flat top, 59, 69Gaussian, 59, 68Hamming, 59, 68Hanning, 57, 58, 67, 70–72Kaiser, 60Nuttall, 190, 196rectangular, 56, 58, 66, 70, 72, 74time, 55, 57, 189, 190, 192triangular, 58, 67, 191
Index 211
