Introduction to Orthogonal Transforms. with Applications in Data Processing and Analysis

Transcription

1 i Introduction to Orthogonal ransforms with Applications in Data Processing and Analysis

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3 Introduction to Orthogonal ransforms with Applications in Data Processing and Analysis June 8, 2009 i

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5 Contents iii cam b ridg e university press Cambridge, New York, Melbourne, Madrid, Cape own, Singapore, São Paulo Cambridge University Press he Edinburgh Building, Cambridge CB2 2RU, UK Published in the United States of America by Cambridge University Press, New York Information on this title: C Cambridge university Press 2007 his publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2007 Printed in the United Kingdom at the University Press, Cambridge A catalogue record for this publication is available from the British Library Library of Congress Cataloguing in Publication data ISBN-13 ISBN XXXXX-X hardback XXXXX-X hardback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

6 Contents Notation page v 1 Continuous ime Fourier ransform he Fourier Series Expansion Formulation of Fourier Expansion Physical Interpretation Properties of Fourier Series Expansion Fourier Expansion of ypical Functions Fourier ransform Formulation Physical Interpretation Relation to Fourier Expansion Properties of Fourier ransform Fourier ransform of ypical Functions Hilbert ransform and Analytic Signals he Uncertainty Principle Frequency Response Function of LI Systems 34 iv

7 Notation General notation iff j = 1 = e jπ/2 u + jv Re(u + jv) = u Im(u + jv) = v x n 1 x A m n A 1 A A if and only if the imaginary unit complex conjugate of u + jv, equal to u jv real part of u + jv imaginary part of u + jv an n by 1 column vector (bold face lower case letter), the transpose of x, a 1 by n row vector an m by n matrix of m rows and n columns the inverse of matrix A the transpose of matrix A the conjugate transpose of matrix A, i.e., A = A = A v

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9 1 Continuous ime Fourier ransform 1.1 he Fourier Series Expansion Formulation of Fourier Expansion As we have already seen in the previous chapter, the second-order differential operator D 2 over the interval [0, ] is a self-adjoint operator, and its eigenfunctions φ k (t) = e j2kπf0t / (k = 0, ±1, ±2, ) are orthonormal: < φ k (t), φ n (t) >= 1 e j2kπf0t e j2nπf0t dt (1.1) where ω 0 = 2πf 0 = 2π/. hese eigenfunctions form a complete orthogonal system that spans a function space over interval [0, ], and any periodic signal x (t) = x (t + ) in the space can be expressed as a linear combination of these basis functions: x (t) = k= X[k]φ k (t) = 1 k= X[k]e j2kπf 0t (1.2) Due to the orthogonality of these basis functions, the coefficients X[n] can be found by taking an inner product with φ n (t) = e j2nπf0t / on both sides of the equation above: < x (t), φ n (t) > = < x (t), e j2nπf0t / >= 1 X[k] < e j2kπf0t, e j2nπf0t > = k= k=0 X[k]δ[k n] = X[n] (1.3) i.e., the nth coefficient X[n] is the projection of function x (t) onto the nth basis function φ n (t): X[n] =< x (t), φ n (t) >= 1 x (t)e j2nπf0t dt (1.4) 1

10 2 Chapter 1. Continuous ime Fourier ransform Figure 1.1 Fourier series expansion of periodic signals Equations 1.2 and 1.4 form a pair of the Fourier series expansion: x (t) = 1 X[k] = 1 k= X[k]e j2kπf 0t x (t)e j2kπf 0t dt (1.5) As the signal and the basis functions are both periodic, the integral above can be over any interval of, such as from 0 to, or from /2 to /2. Note that when x (t) is real, we have X[ k] = 1 x (t)e j2kπf0t dt = X[k] (1.6) i.e., Re[X[ k]] = Re[X[k]], Im[X[ k]] = Im[X[k]] (1.7) In other words, when x (t) is real, the real part of X[k] is even and the imaginary part is odd. In practice, the constant scaling factor 1/ in the equations above is of little significance, and the Fourier series expansion pair could be expressed in some alternative forms such as: x (t) = X[k]e j2kπf0t = X[k]e jkω 0t X[k] = 1 k= k= x (t)e j2kπf 0t dt = 1 x (t)e jkω 0t dt (1.8) Now X[0] = x (t)dt/ has a clear interpretation, the average or the DC component of the signal. In some literatures, angular frequency ω 0 = 2πf 0 = 2π/ is preferred to use. But we will use either f 0 or ω 0 = 2πf 0 = 2π/ interchangeably, whichever is more convenient. he Fourier series expansion is a unitary transformation that converts a function x (t) in the vector space of all periodic time functions into a vector [, X[ 1], X[0], X[1], ] in another vector space. Moreover, the inner product of any two functions x (t) and y (t) remains the same before and after the

11 Continuous ime Fourier ransform 3 transformation: < x (t), y (t) >= = 1 = 1 = x (t)y (t)dt X[k]e j2kπf 0t k= l= k= l= k= l= X[k]Y [l] X[k]Y [l]δ[k l] = Y [l]e j2nπf 0t dt e j2(k l)πf 0t dt k= X[k]Y [k] =< X, Y > (1.9) In particular, if y (t) = x (t), the above becomes Parseval s identity x (t) 2 =< x (t), x (t) >=< X, X >= X 2 (1.10) indicating that the total energy or information contained in the signal is preserved by the Fourier series expansion, therefore the signal can be equivalently represented in either time or frequency domain Physical Interpretation he Fourier series expansion of a periodic signal x (t) can also be expressed in terms of sine and cosine functions: x (t) = X[k]e jkω0t = X[0] + [X[ k]e jkω0t + X[k]e jkω0t ] = X[0] + = X[0] + = X[0] + 2 k= k=1 [X[ k](cos kω 0 t j sin kω 0 t) + X[k](cos kω 0 t + j sin kω 0 t)] k=1 [(X[k] + X[ k]) cos kω 0 t + j(x[k] X[ k]) sin kω 0 t] k=1 (a k cos kω 0 t + b k sin kω 0 t) (1.11) k=1 where X[k] + X[ k] a k = = 1 x (t)[e jkω0t + e jkω0t ]dt = 1 x (t) cos kω 0 tdt 2 2 j(x[k] X[ k]) b k = = j x (t)[e jkω0t e jkω0t ]dt = 1 x (t) sin kω 0 tdt 2 2 (k = 1, 2, ) (1.12) his is an alternative form of the Fourier series expansion of x (t). Here we have used the Euler s formula: cos kω 0 = ejkω 0 + e jkω 0, sin kω 0 = ejkω 0 e jkω 0 2 2j (1.13)

12 4 Chapter 1. Continuous ime Fourier ransform In particular, if we assume x (t) is real (all physical signals are real in reality), then X[ k] = X[k]. We have and X[k] + X[ k] X[k] + X[k] a k = = = Re[X[k]] 2 2 j(x[k] X[ k]) j(x[k] X[k]) b k = = = Im[X[k]] (1.14) 2 2 { { X[k] = a 2 k + b 2 k ak = X[k] cos X[k] X[k] = tan 1 b k /a k b k = X[k] sin X[k] (1.15) Now the Fourier series expansion of a real signal x (t) (Eq. 1.11) can be rewritten as: x (t) = X[0] + 2 (a k cos kω 0 t + b k sin kω 0 t) = X[0] + 2 k=1 X[k] (cos X[k] cos kω 0 t sin X[k] sin kω 0 t) k=1 = X[0] + 2 X[k] cos(kω 0 t + X[k]) k=1 (1.16) In other words, a real periodic signal x (t) can be constructed as a superposition of infinite sinusoids of (a) different frequencies kω 0, (b) different amplitudes X[k], and (c) different phases X[k]. In particular, when k = 0, the coefficient X[0] = x (t)dt/ is the average or DC component (offset) of the signal x (t); when k = 1, the sinusoid cos(ω0 t + φ 1 ) = cos(2πt/ + X[1]) has the same period as the signal x (t) and is therefore called the fundamental frequency of the signal; when k > 1, the frequency of the sinusoidal function cos(kω0 t + X[k]) is k times the frequency of the fundamental and is called the kth harmonic of the signal Properties of Fourier Series Expansion Here we consider only a few of the properties of the Fourier series expansion. Considered as a special case of the Fourier transform, Fourier series expansion share all of its properties, which will be discussed in a much greater extent later. Let x (t) be a periodic signal with period and X[k] be its Fourier series expansion coefficients. ime scaling: When x (t) is scaled in time by a factor of a > 0 to become x(at), its period becomes /a and its fundamental frequency becomes a/ =

13 Continuous ime Fourier ransform 5 af 0. If a > 1, the signal is compressed by a factor a and the frequencies of its fundamental and harmonics are a times higher; if a < 1, the signal is expanded and the frequencies of its fundamental and harmonics are a times lower. But in either case, the coefficients remain the same: x(at) = X[k]e jkaω 0t (1.17) k= ime shifting: A time signal x(t) shifted in time by t0 becomes y(t) = x(t t 0 ). Defining t = t t 0 we can get its Fourier coefficient as: Y [k] = 1 x(t t 0 )e jkω0t dt = 1 x(t )e jkω 0(t +t 0 ) dt = X[k]e jkω 0t 0 = X[k]e j2kπft 0 (1.18) Differentiation: he time derivative of x(t) is y(t) = d x(t)/dt = x (t) its Fourier coefficients can be found to be: Y [k] = 1 d dt x(t)e jkω 0t dt = 1 [e jkω 0t x(t) 0 + jkω 0 x(t)e jkω0t dt] = jkω 0 X[k] = jk 2π X[k] (1.19) Integration: he time integration of x(t) is y(t) = t x(τ)dτ (1.20) Note that y(t) is periodic only if the DC component or average of x(t) is zero, i.e., X[0] = 0 (otherwise it would accumulate over time by the integration to form a ramp). Since x(t) = y (t), according to the differentiation property, we have X[k] = jk 2π Y [k], i.e. Y [k] = X[k] (1.21) j2kπ Note that Y [0] can not be obtained from this formula as when k = 0, both the numerator and the denominator of Y [k] are zero. However, as the DC component of y(t), Y [0] can be found by the definition: Y [0] = 1 y(t)dt (1.22) Fourier Expansion of ypical Functions Constant: A constant x(t) = c can be expressed as a complex exponential x(t) = e j0t with arbitrary period, i.e., it is a zero-frequency or DC (direct current) component. he coefficient for this zero frequency is X[0] = c, while all other

14 6 Chapter 1. Continuous ime Fourier ransform coefficients for nonzero frequencies are zero. Alternatively, following the definition, we get X[k] = 1 { c k = 0 ce jkω0t dt = 0 k 0 Here we have used the fact that 1 e ±jkω0t dt = 1 e ±j2πkt/ dt = 1 [ cos( 2π /k t)dt ± j sin( 2π t)dt] = δ[k] /k he sinusoids have period /k and their integral over is zero, unless if k = 0 and cos 0 = 1 and dt/ = 1. Complex exponential: A complex exponential x(t) = e jω0t (with period = 2π/ω 0 ) with a coefficient X[1] = 1. We can also find X[k] by definition: c k = 1 e jω0t e jkω0t dt = 1 { 1 k = 0 e jω0(1 k)t dt = δ[k 1] = 0 k 0 Sinusoids: he cosine function x(t) = cos(2πf 0 t) = (e j2πf0t + e j2πf0t )/2 of frequency f 0 is periodic with = 1/f 0, and its Fourier coefficients are X[k] = 1 cos(2πf 0 t)e j2πkf0t dt = 1 2 [ 1 e j2π(k 1)f0t dt + 1 e j2π(k+1)f0t dt] = 1 (δ[k 1] + δ[k + 1]) (1.23) 2 In particular, when f 0 = 0, x(t) = 1 and X[k] = δ[k], an impulse at zero, representing the constant (zero frequency) value. Similarly, the Fourier transform of x(t) = sin(2πf 0 t) is: X[k] = 1 sin(2πf 0 t)e j2πkf0t dt = 1 2j [ 1 e j2π(k 1)f0t dt 1 e j2π(k+1)f0t dt] Square wave: Let x(t) be an odd square wave: = 1 (δ[k 1] δ[k + 1]) (1.24) 2j x(t) = { 1 0 < t < τ 0 τ < t < (1.25) he Fourier coefficients of this function are X[k] = 1 0 x(t)e jkω 0t dt = 1 τ 0 e jkω 0t dt = 1 j2kπ (1 e jkω 0τ ) (1.26)

15 Continuous ime Fourier ransform 7 In particular, as the DC component, X[0] = τ/. A sinc function is commonly defined as: sinc(x) = sin(πx) πx and the expression above for X[k] can be written as: X[k] = = e jkπτ / kπ e jkπτ / kπ, and lim sinc(x) = 1 (1.27) x 0 1 2j (ejkπτ / e jkπτ / ) sin(kπτ/ ) = τ sinc(kτ/ )e jkπτ / (1.28) In particular, if τ = /2, then X[0] = 1/2 and X[k] above becomes: X[k] = 1 j2kπ (1 e jkπ ) (1.29) Moreover, since e ±j2kπ = 1 and e ±j(2k 1)π = 1, all even terms X[±2k] = 0 become zero and the odd terms become: X[±(2k 1)] = ±1/jπ(2k 1), (k = 1, 2, ) (1.30) and the Fourier series expansion of the square wave becomes a linear combination of sinusoids: x(t) = X[k]e j2kπf 0t k= = X[0] + = π 1 [ jπ(2k 1) ej(2k 1)ω 0t 1 + jπ(2k 1) e j(2k 1)ω 0t ] k=1 k=1 = π [sin(ω 0t) 1 sin((2k 1)ω 0 t) 2k 1 + sin(3ω 0t) 3 + sin(5ω 0t) 5 + ] (1.31) If we remove the DC component of x(t) by letting X[0] = 0, the square wave become { 1/2 0 < t < /2 x(t) = (1.32) 1/2 /2 < t < and the square wave is an odd function composed of odd harmonics of sine functions (odd). Homework problem: If the square wave is shifted to the left by /4, it becomes an even function: { 1 t < /4 x (t) = (1.33) 0 /4 < t < /2

16 8 Chapter 1. Continuous ime Fourier ransform Show that its Fourier series expansion becomes x(t) = X[k]e jkω 0t k= = π [cos(ω 0t) 1 cos(3ω 0t) 3 + cos(5ω 0t) 5 composed of odd harmonics of cosine functions (even). riangle wave: A triangle wave can be defined as: + ] (1.34) x(t) = 2 t /, ( t /2) (1.35) his triangle wave can be obtained as an integral of the square wave defined in Eq with these modifications: (a) τ = /2, (b) DC offset X[0] set to zero, and (c) scaled by 4/. Now according to the integration property, the Fourier coefficients can be easily obtained as X[k] = 4 j2kπ e jkπ/2 kπ sin(kπ/2) = 2 j sin(kπ/2) (kπ) 2 e jkπ/2 (1.36) he DC offset is X[0] = 1/2. According to the time shift property, the complex exponential e jkπ/2 corresponds to a right-shifted signal x(t t 0 ) by t 0 = /4. If we shift the signal left by /4, then the complex exponential term in the expression of the coefficients disappear. he Fourier series expansion of such a triangle wave can be written as x(t) = X[k]e j2kπf0t = [X[k]e j2kπf0t + X[ k]e j2kπf0t ] k= = ( 2 j k=1 = π 2 k=1 k=1 sin(kπ/2) (kπ) 2 e j2kπf0t 2 sin(kπ/2) j (kπ) 2 e j2kπf0t ) sin(kπ/2) k 2 sin(2kπf 0 t) = π 2 [sin(2πf 0t) 1 9 sin(6πf 0t) sin(10πf 0t) ] (1.37) Sawtooth A sawtooth function is defined as x (t) = t/, (0 < t < ) (1.38) We first find X[0], the average or DC component: X[0] = 1 t e j0ω 0t dt = 1 2 (1.39) Next we find all remaining coefficients X[k] (k 0): X[k] = 1 t e jkω 0t dt (1.40)

17 Continuous ime Fourier ransform 9 In general, this type of integrals can be found using integration by parts: te at dt = 1 a 2 (at 1)eat + C (1.41) Here a = jkω 0 = j2kπ/ 0 and we get X[k] = 1 2 (jkω 0 ) 2 [( jkω 0t 1)e jkω 0t ] 0 = he Fourier series expansion of the function is x (t) = j [ 2kπ ejω 0t j 2kπ e jω 0t ] = π k=1 k=1 j 2kπ (1.42) 1 k sin(kω 0t) (1.43) Note that this sawtooth wave is an odd function and therefore it is composed of only odd sine functions. Homework problem: Consider a different version of the sawtooth wave: x (t) = t/, (0 < t < /2) (1.44) Impulse rain An impulse train, also called a comb function or sampling function, is a sequence of infinite unit impulse separated by time interval : comb(t) = δ(t n ) (1.45) n= As a periodic function with period, an impulse train can be Fourier expanded: comb(t) = Comb[k] e j2kπt/ (1.46) with coefficients: Comb[k] = 1 = 1 /2 /2 /2 /2 k= comb(t)e j2kπt/ dt = 1 /2 /2 n= δ(t n )e j2kπt/ dt δ(0)e j2kπt/ dt = 1, (k = 0, ±1, ±2, ) (1.47) he last equation is due to Eq.??. Substituting Comb[k] = 1/ back into the Fourier series expansion of comb(t), we can also express the impulse train as: comb(t) = δ(t n ) = 1 e j2kπt/ (1.48) n= k=

18 10 Chapter 1. Continuous ime Fourier ransform 1.2 Fourier ransform Formulation he Fourier series expansion is no longer applicable if the given signal x(t) is nonperiodic. In order to still be able to process and analyze the signal in frequency domain, the concept of the Fourier series expansion needs to be modified. First we make some minor modification of the Fourier series expansion pair in Eq. 1.5 by moving the factor 1/ from the second equation to the first one: 1 x (t) = X[k]ejkω 0t 1 = X[k]ej2kπf 0t k= k= X[k] = x (t)e jkω0t dt = x (t)e j2kπf0t dt (1.49) Now the coefficient X[k] is redefined so that its value is scaled by, and its dimensionality becomes that of the signal x (t) multiplied by time, or divided by frequency (the exponential term is dimensionless). A periodic signal x (t) can be converted into a non-periodic signal x(t) simply by increasing its period so that it approaches infinity as the limit. At the limit the following changes take place: ω0 = 2πf 0 = 2π/ 0, and the discrete frequencies kω 0 = 2kπf 0 for all k = 0, ±1, ±2, can be replaced by a continuous variable < ω = 2πf <. All basis functions φk (t) = e j2kπt/ for all k become non-periodic and uncountable φ f (t) = e j2πft for all f, and they now span a function space over (, ) containing all non-periodic functions x(t). he coefficients X[k] for the discrete frequency components φk (t) = e j2kπf 0t for all k can be replaced by a continuous weight function X(f) for the continuous and uncountable frequency component function φ f (t) = e j2πft for all f. Define f = 1/, then f df, and the summation in the first equation in Eq becomes an integral. Now the two equations in Eq become x(t) = lim [ 1 X[k]e j2kπf0t ] = k= X(f) = lim [ x(t)e j2kπf0t dt] = hese two equations form a Fourier transform pair: X(f) = F[x(t)] = x(t) = F 1 [X(f)] = x(t)e j2πft dt X(f) j2πft df x(t)e j2πft dt (1.50) X(f)e j2πft df (1.51)

19 Continuous ime Fourier ransform 11 For the integrals in Eq.1.51 to exist, the signal x(t) and its spectrum X(f) should both be square integrable (Eq.??): x(t) 2 dt < (1.52) i.e., x(t) L 2 and X(f) L 2 are energy signals. he second equation, the inverse Fourier transform, represents a non-periodic signal x(t) as a linear combination of an uncountable and infinite set of basis functions φ f (t) = e j2πft, weighted by a coefficient or weight function X(f), called the frequency spectrum of x(t), which can be obtained as the projection of the signal x(t) onto a basis function φ f (t) representing frequency f: X(f) = F[x(t)] =< x(t), φ f (t) >=< x(t), e j2πft >= x(t)e j2πft dt (1.53) his is the first equation, the forward Fourier transform. As the dimensionality of X(f) is that of the signal x(t) divided by frequency, it is actually a frequency density function, representing how the energy or information contained in the signal is distributed over the frequency axis. his integral is also called an integral transform, and φ f (t) = e j2πft, a function of two variables t and f, is called the kernel function of the transform. In some literatures, the angular frequency ω = 2πf is preferred and the above can be equivalently expressed as: X(ω) = F[x(t)] = x(t) = F 1 [X(ω)] = 1 2π x(t)e jωt dt X(ω)e jωt dω (1.54) We will use either ω 0 or 2πf 0 = 2π/ interchangeably, whichever is more convenient. Moreover, certain authors prefer to express the spectrum X(ω) as X(jω). But these are just some notational variations while the essential concepts underneath the notation remain the same. Using X(f) will make many expressions in time and frequency domains more symmetric, which is preferable as we will see in the future discussions. Similar to the Fourier series expansion, the Fourier transform is also a unitary transformation (heorem??): = = < x(t), y(t) >= X(f)Y (f )[ x(t)y(t)dt = [ e j2π(f f )t dt] df df = X(f)e j2πft df] [ Y (f )e j2πf t df ] dt X(f)Y (f )δ(f f )df df X(f)Y (f)df =< X(f), Y (f) > (1.55)

20 12 Chapter 1. Continuous ime Fourier ransform Figure 1.2 Fourier transform of non-periodic and continuous signals Here we have used the fact (Eq.??) that: e j2π(f f )t dt = δ(f f ) (1.56) his equation has a major significance as it also illustrates that indeed the function family {φ f (t) = e j2πft, ( < f < )} forms an orthonormal basis that spans the function space, and any function in the space can be expressed as a linear combination of these basis functions. his is the very essence of the inverse Fourier transform given in Eq Replacing y(y) by x(t) in Eq.1.55 above, we get Parseval s equality: x(t) 2 =< x(t), x(t) >=< X(f), X(f) >= X(f) 2 (1.57) As a unitary transformation, the Fourier transform can be considered as a rotation of the basis functions of the function space. Before the Fourier transform, the function is represented as a linear combination of a uncountable set of standard basis functions δ(t τ), each for a particular time moment t = τ, weighted by the coefficient function x(τ) for the signal amplitude at the time moment: x(t) = x(τ)δ(t τ)dτ (1.58) After the transformation, the function is represented as a linear combination of a different set of orthonormal basis functions φ f (t) = e j2πft for all frequencies f, weighted by the coefficient function X(f) for the amplitude of each frequency component: x(t) = X(f)e j2πft df (1.59) he representations of the signal in time domain by x(t) and in frequency domain by X(f) are equivalent, in the sense that the total amount of energy or information is preserved due to the Parseval s equality: x(t) = X(f) Physical Interpretation In general the spectrum X(f) of a time signal x(t) is complex and can be expressed in either Cartesian or polar form: X(f) = X r (f) + jx j (f) = X(f) e j X(f) (1.60)

21 Continuous ime Fourier ransform 13 where { X(f) = Xr 2(f) + X2 j (f) X(f) = tan 1 X j (f)/x r (f), { Xr (f) = X(f) cos X(f) X j (f) = X(f) sin X(f) (1.61) If the signal x(t) is real, we have X(f) = x(t)e j2πft dt = x(t)(cos 2πft j sin 2πft)dt = X r (f) jx j (f) where the real part X r (f) is even and the imaginary part X j (f) is odd: therefore X(f) = as x(t) = = = 2 0 X r (f) = X j (f) = x(t) cos(2πft)dt = X r ( f) (1.62) x(t) sin(2πft)dt = X j ( f) (1.63) Xr 2(f) + X2 j (f) is even. Now the signal can be expressed X(f)e j2πft df = X(f) cos(2πft + X(f) e j2πft+ X(f) df X(f))df + j X(f) sin(2πft + X(f))df X(f) cos(2πft + X(f))df (1.64) he last equation is due to the fact that the integrand of the real term is even and that of the imaginary term is odd. We see that the Fourier transform expresses a real time signal as a superposition of infinitely many uncountable frequency components each with a different frequency f, magnitude X(f), and phase X(f) Relation to Fourier Expansion First consider the Fourier transform of the unit impulse function δ(t): F[δ(t)] = δ(t)e j2πft dt = e j2πf0 = 1 (1.65) where the second equal sign is due to Eq.??. Now let us see how the Fourier transform of a periodic function is related to its Fourier coefficients. Consider the Fourier series expansion of a periodic function x (t): x (t) = X[k]e j2kπt/ = X[k]e j2kπf 0t (1.66) k= k=

22 14 Chapter 1. Continuous ime Fourier ransform where f 0 = 1/ is the fundamental frequency and X[k] the expansion coefficient. he Fourier transform of this periodic function x (t) can be found to be: X(f) = x (t)e j2πft dt = [ X[k]e j2kπf0t ]e j2πft dt = k= k= X[k] e j2π(f kf0)t dt = k= X[k]δ(f kf 0 ) (1.67) where we have used the result of Eq.??. It is clear that the spectrum of a periodic function is discrete, in the sense that it is none-zero only at a set of discrete frequencies f = kf 0 where X(f) = X[k]δ(f kf 0 ). his result also illustrates an important point: while the dimensionality of the Fourier coefficient X[k] is the same as that of the function x (t), i.e., [X[k]] = [x (t)], the dimensionality of the spectrum is [X(f)] = [X[k]][t] = [X[k]] [f] (1.68) i.e., X(f) is a density function over frequency, only when integrated over frequency, will it become the coefficient: X[k]δ(f kf 0 )df = X[k] (1.69) When, x(t) becomes non-periodic and the gap f 0 = 1/ between two consecutive frequency components in its spectrum becomes zero, i.e., the discrete spectrum becomes continuous. Next, we consider how the Fourier spectrum X(t) of a signal x(t) can be related to the Fourier series coefficients of its periodic extension defined as: x (t) = x(t + n ) = x (t + ) (1.70) n= As x (t) is periodic, it can be Fourier expanded and the kth Fourier coefficient is: X [k] = 1 x (t)e j2πkt/ = 1 [ x(t + n ) ] e j2πkt/ dt = 1 0 n= 0 0 n= x(t + n )]e j2πkt/ dt (1.71) If we define τ = t + n, i.e., t = τ n, the above becomes: X [k] = 1 (n+1) x(τ)e j2πkτ / dτ e j2πnk = 1 n= n x(τ)e j2πkτ / dτ = 1 X( k ) (1.72)

23 Continuous ime Fourier ransform 15 (Note that e j2πnk = 1 as k and n are both integer.) his equation relates the Fourier transform X(f) of a signal x(t) to the Fourier series coefficient X [k] of the periodic extension x (t) of the signal. Now the Fourier expansion of x (t) can be written as: x (t) = k= X [k]e j2πkt/ = k= 1 X his equation is called Poisson summation formula. ( ) k e j2πkt/ (1.73) Properties of Fourier ransform In the following, we always assume x(t) and y(t) are two complex functions (real as a special case) and F[x(t)] = X(f) and F[y(t)] = Y (f). Linearity F[ax(t) + by(t)] = af[x(t)] + bf[y(t)] (1.74) he Fourier transform of a function x(t) is simply an inner product of the function with a kernel function φ f (t) = e j2πft (Eq.1.53). herefore due to the linearity of the inner product in the first variable, the Fourier transform is also linear. ime-frequency duality Proof: Defining t = t, we have if F[x(t)] = X(f), then F[X(t)] = x( f) (1.75) x(t) = F 1 [X(f)] = x( t ) = F 1 [X(f)] Interchanging variables t and f, we get x( f) = In particular, if x(t) = x( t) is even, we have X(f)e j2πft df (1.76) X(f)e j2πft df (1.77) X(t )e j2πft dt = F[X(t)] (1.78) if F[x(t)] = X(f), then F[X(t)] = x(f) (1.79) his duality is simply the result of the definition of the forward and inverse transforms in Eq. 1.51, which are highly symmetric between time and frequency. Consequently, many of the properties and transforms of typical functions have strong duality between the time and frequency domains.

24 16 Chapter 1. Continuous ime Fourier ransform Multiplication (Plancherel) theorem x(t)y(t)dt = X(f)Y (f)df (1.80) his is Eq. 1.55, indicating the Fourier transform is a unitary transformation that conserves inner product. In particular, letting y(t) = x(t), we get Parseval s identity: x(t) 2 dt = x(t)x(t)dt = X(f)X(f)df = X(f) 2 df (1.81) where x(t) 2 represents how the signal energy is distributed over time, while X(f) 2 represents how the signal energy is distributed over frequency, and X(f) 2 = S x (f) is defined as the power density spectrum (PDS). Complex conjugate F[x(t)] = X( f) (1.82) Proof: aking the complex conjugate of the inverse Fourier transform, we get: x(t) = = X(f)e j2πft df = X(f)e j2πft df X( f )e j2πf t df = F 1 [X( f)] (1.83) (1.84) where we have defined f = f. Symmetry Let us consider some symmetry properties of the Fourier transform in both time and frequency domains. First note that: X(f) = x(t)e j2πft dt = x(t) cos(2πft)dt j x(t) sin(2πf t)dt = X e (f) + jx o (f) (1.85) where X e (f) and X o (f) are the even and odd components of X(f): X e (f) = X o (f) = x(t) cos(2πft)dt = X e ( f) (1.86) x(t) sin(2πft)dt = X o ( f) (1.87) We can also represent the spectrum in terms of its real and imaginary parts: X(f) = Re[X(f)] + jim[x(f)] = X r (f) + jx j (f) (1.88) Note, however, here X r (f) and X j (f) do not necessarily correspond to X e (f) and X o (f), respectively, as x(t) is in general assumed to be complex. Real signal:

25 Continuous ime Fourier ransform 17 If x(t) is real, then both X e (f) and X o (f) are real, and they become respectively the real and imaginary parts of X(f): { Xe (f) = X r (f) = X r ( f) (1.89) X o (f) = X j (f) = X j ( f) As the real part of X(f) is even, and the imaginary part is odd, i.e., the spectrum X(f) of a real signal is a Hermitian function satisfying: X( f) = X r ( f) + jx j ( f) = X r (f) jx j (f) = X(f) (1.90) he symmetry property of spectrum X(f) indicates that in frequency domain, only half of the data is independent (fifty percent redundancy), which is of course the natural consequence of the fact that only half of the data in time domain, the real part, is independent, as the imaginary part is all zero. As the spectrum of a real signal symmetric (real part even and imaginary odd), it can be reconstructed by inverse transform by only half of the spectrum for f > 0: x(t) = = = 0 0 X(f)e j2πft df = 0 X( f)e j2πft df + 0 X(f)e j2πft df + X(f)e j2πft df 0 X(f)e j2πft df [X(f)e j2πft + X(f)e j2πft ]df (1.91) Moreover, depending on whether x(t) is even or odd, we have the following results: If x(t) = x( t) is even (X j (f) = X o (f) = 0), X(f) = X r (f) = X e (f) is real and even; If x(t) = x( t) is odd (X j (f) = X e (f) = 0), X(f) = X j (f) = X o (f) is imaginary and odd. Imaginary signal: If x(t) is imaginary, then both X e (f) and X o (f) are imaginary, and they become respectively the imaginary and real parts of X(f): { Xe (f) = X j (f) = X j ( f) (1.92) X o (f) = X r (f) = X r ( f) and we have X( f) = X r ( f) + jx j ( f) = X r (f) + jx j (f) = X(f) (1.93) i.e., the spectrum of an imaginary signal is anti-hermitian. Moreover, depending on whether x(t) is even or odd, we have the following results: If x(t) = x( t) is even (X r (f) = X o (f) = 0), X(f) = X j (f) = X e (f) is imaginary and even;

26 18 Chapter 1. Continuous ime Fourier ransform able 1.1. Symmetry Properties of Fourier ransform x(t) = x r (t) + jx i (t) X(f) = X r (f) + jx j (f) x(t) = x r (t) real X r (f) = X r ( f) even, X j (f) = X j ( f) odd x r (t) = x r ( t) real, even X r (f) = X r ( f) real, even x r (t) = x r ( t) real, odd X j (f) = X j (f) imaginary, odd x(t) = x j (t) imaginary X r (f) = X r ( f) odd, X j (f) = X j ( f) even x j (t) = x j ( t) imaginary, even X j (f) = X j ( f) imaginary, even x j (t) = x j ( t) imaginary, odd X r (f) = X r ( f) real, odd If x(t) = x( t) is odd (Im[X(f)] = X e (f) = 0), X(f) = X r (f) = X o (f) is real and odd. ime reversal F[x( t)] = X( f) (1.94) i.e., if the signal x(t) is flipped in time with respect to the origin t = 0, its spectrum X(f) is flipped in frequency with respect to the origin f = 0, Proof: F[x( t)] = x( t)e j2πft dt = x(t )e j2πft dt = X( f) (1.95) where we have assumed t = t. In particular, when x(t) = x(t) is real, F[x( t)] = X( f) = ime and frequency scaling Proof: Let u = at, i.e., t = u/a F[x(at)] = x(t)e j2πft = x(t)e j2πft = X(f) (1.96) F[x(at)] = 1 a X(f ), (a > 0) (1.97) a x(at)e j2πft dt = x(u)e j2πfu/a d( u a ) = 1 a X(f a )(1.98) If a < 1, the signal is stretched and its spectrum is compressed and scaled up. When a 0, x(at) is so stretched that it approaches a constant, and its spectrum is compressed and scaled up to the extent that it approaches an impulse. On the other hand, if a > 1, then the signal is compressed and its spectrum is stretched and scaled down. When a, we redefine the signal as a x(at) with spectrum X(f/a), the signal becomes an impulse and its spectrum X(f/a) becomes a constant. ime and frequency shift F[x(t ± t 0 )] = e ±j2πft 0 X(f) (1.99) Proof: F[x(t ± t 0 )] = x(t ± t 0 )e j2πft dt (1.100)

27 Continuous ime Fourier ransform 19 Let t = t ± t 0, then t = t t 0, dt = dt, the above becomes F[x(t ± t 0 )] = x(t )e j2πf(t t 0 ) dt = e ±j2πft 0 X(f) (1.101) A time shift t 0 of the signal corresponds to a phase shift 2πft 0 for every frequency component e j2πft. Note that as the phase shift is proportional to the frequency, a higher frequency component will shift more so that the relative positions of the harmonics remain the same, and the shape of the signal as a superposition of these harmonics remains the same when shifted. applying the time-frequency duality to the time shift property, we get the frequency shift property: F 1 [X(f ± f 0 )] = e j2πf 0t x(t) (1.102) Correlation: he cross-correlation between two functions x(t) and y(t) is defined as his property states: r xy (t) = x(t) y(t) = x(τ)y(τ t)dτ (1.103) F[r xy (t)] = F[x(t) y(t)] = X(f)Y (f) (1.104) Proof: As F[x(τ)] = X(f) and F[y(τ t)] = Y (f)e j2πft, we apply the multiplication theorem to get: where r xy (t) = = x(τ)y(τ t)dτ = X(f)Y (f)e j2πft df S xy e j2πfτ df = F 1 [S xy (f)] (1.105) S xy (f) = X(f)Y (f) (1.106) is the cross power density spectrum, which is the Fourier transform of the cross-correlation r xy (t). In particular, when x(t) = y(t), we have: r x (τ) = x(t)x(t τ)dt = S x e j2πfτ df = F 1 [S x (f)] (1.107) where r x (t) is the auto-correlation and S x (f) = X(f) 2 is the power density spectrum of x(t) defined previously. Convolution theorem: As discussed in the previous chapter, the convolution of two functions x(t) and y(t) is defined as: x(t) y(t) = x(τ)y(t τ)dτ (1.108)

28 20 Chapter 1. Continuous ime Fourier ransform Note that if y(t) = y( t) is even, then x(t) y(t) = x(t) y(t) is the same as the correlation. he convolution theorem states: Proof: = = F[x(t) y(t)] = X(f) Y (f) (1.109) F[x(t)y(t)] = X(f) Y (f) (1.110) F[x(t) y(t)] = [ x(τ)y(t τ)dτ]e j2πft dt x(τ)e j2πfτ y(t τ)e j2πf(t τ ) dt dτ Similarly, we can also prove: x(τ)e j2πfτ Y (f)dτ = X(f)Y (f) (1.111) F[x(t)y(t)] = X(f) Y (f) (1.112) In particular, as shown in Eq.??, the output y(t) of an LI system can be found as the convolution of its impulse response h(t0 and the input x(t) y(t) = h(t) x(t). Now according to the convolution theorem, the output of the system can be more conveniently obtained in frequency domain by a multiplication: Y (f) = H(f)X(f) (1.113) where X(f) and Y (f) are respectively the spectra of the input x(t) and the output y(t), and H(f) = F[h(t)], the Fourier transform of the impulse response function h(t), is the frequency response function of the system. ime derivative: Proof: d dt x(t) = d dt = F [ d x(t)] = j2πfx(f) = jωx(ω) (1.114) dt X(f)e j2πft df = X(f) d dt ej2πft df j2πfx(f)e j2πft df = F 1 [j2πfx(f)] (1.115) i.e., F[x (t)] = j2πfx(f). Repeating this process we get F [ dn dt n x(t)] = (j2πf)n X(f) (1.116) Frequency derivative: he proof is very similar to the above. F [t x(t)] = j d df X(f) F[t n x(t)] = j n 1 d n (2π) n X(f) (1.117) df n

29 Continuous ime Fourier ransform 21 ime integration: According to the time derivative property, F [x (t)] = j2πfx(f), therefore it seems only natural that the Fourier transform of the integral of a signal should be: t F [ x(τ)dτ] = 1 X(f) (1.118) j2πf However, this is only a partial result. o find the complete Fourier transform of the integration, let us first consider the sign function sgn(t) and the unit step function u(t) defined as: sgn(t) = 1 t < 0 0 t = 0 1 t > 0 u(t) = 1 [sgn(t) + 1] = 2 0 t < 0 1/2 t = 0 1 t > 0 (1.119) i.e., sgn(t) = 2u(t) 1. According to the time derivative property of the Fourier transform: F[x(t)] = 1 j2πf F[ d x(t)] (1.120) dt we have and F[u(t)] = 1 j2πf F[ d dt u(t)] = 1 j2πf F[δ(t)] = 1 j2πf F[ 1 2 sgn(t)] = 1 j2πf F[ d dt = 1 j2πf F[δ(t)] = 1 j2πf sgn(t) ] = 1 2 j2πf F[ d dt [u(t) 1 2 ]] (1.121) (1.122) Why do these two different signals have the same spectrum? his is obviously due to the fact that the constant difference 1/2 is lost in the derivative operation. o reflect this constant difference in frequency domain as well as in time domain, the spectrum of u(t) should include an additional term to represent its non-zero DC component. Specifically, as the sign function has a zero DC component, its spectrum is still the same as above: F[ 1 2 sgn(t)] = 1 1, or F[sgn(t)] = j2πf jπf (1.123) But for the unit step function u(t) = [sgn(t) + 1]/2, a delta function δ(f)/2 will need to be included in its spectrum to represent its DC component in time domain: F[u(t)] = F[ 1 1 (sign(t) + 1)] = 2 j2πf + 1 δ(f) (1.124) 2 Now we ready to show that the Fourier transform of a time integration is: F[ t x(τ)dτ] = 1 j2πf X(f) + 1 X(0)δ(f) (1.125) 2

30 22 Chapter 1. Continuous ime Fourier ransform Proof: he integral of a signal x(t) can be considered as its convolution with u(t): x(t) u(t) = x(τ)u(t τ)dτ = Due to the convolution theorem, we have F[ t t 1 x(τ)dτ] = F[x(t) u(t)] = X(f)[ j2πf δ(ω)] = 1 j2πf x(τ)dτ (1.126) X(f) + X(0) 2 δ(f) (1.127) Fourier ransform of ypical Functions Unit impulse he Fourier transform of the unit impulse function is given in Eq.1.65 according to the definition of the Fourier transform: F[δ(t)] = δ(t)e j2πft dt = 1 (1.128) Unit step he Fourier transform of a unit step is also given before (Eq.1.124), U(f) = F[u(t)] = 1 j2πf + 1 δ(f) (1.129) 2 As the unit step is the time integral of the unit impulse: u(t) = t δ(t)dt (1.130) their Fourier spectra are related according the time integration property. Sign function Also, as discussed before (Eq.1.123), the Fourier transform of the sign function is: F[sgn(t)] = 1 (1.131) jπf Constant he Fourier transform of constant 1 can be obtained according to the property of time-frequency duality, based on the Fourier transform of the unit impulse: F[1] = e j2πft dt = δ(f) (1.132) Due to the property of time-frequency scaling, if the time function x(t) is scaled by a factor of 1/2π to become x(t/2π), its spectrum X(f) will become 2πX(2πf) = 2πX(ω). Specifically in this case, if we scale the constant 1 as a time function by 1/2π (still the same), its spectrum X(f) = δ(f) can be expressed as a function of angular frequency X(ω) = 2πδ(ω).

31 Continuous ime Fourier ransform 23 Sinusoids he Fourier transform of a cosine function x(t) = cos(2πf 0 t) of frequency f 0 is: F[cos(2πf 0 t)] = F[e j2πf 0t + e j2πf 0t ] = 1 2 e j2π(f f 0)t dt e j2π(f f 0)t dt = 1 2 [δ(f f 0) + δ(f + f 0 )] (1.133) Similarly, the Fourier transform of x(t) = sin(2πf 0 t) is: F[sin(2πf 0 t)] = 1 2j [δ(f f 0) δ(f + f 0 )] (1.134) Exponential decay A right-sided exponential decay e at u(t) (a > 0) is defined as: F[e at u(t)] = = 0 e at e j2πft 1 dt = a + j2πft e (a+j2πf)t 0 1 a + j2πf = a j2πf a 2 + (2πf) 2 (1.135) Next consider a left-sided exponential decay function, which is the timereversal version of the right-sided. According time reversal property F[x( t)] = X( f), we get: F[e at u( t)] = 1 a j2πf = a + j2πf a 2 + (2πf) 2 (1.136) Finally consider two-sided exponential decay e a t which is the sum of the previous two signals: F[e a t 1 ] = a + j2πf + 1 a j2πf = 2a a 2 + (2πf) 2 (1.137) Rectangular function and sinc function A rectangular function is defined as { 1 0 < t < τ/2 rect τ (t) = (1.138) 0 otherwise which can be considered as the difference between two unit step functions: rect(t) = u(t + τ/2) u(t τ/2) (1.139) Due to the properties of linearity and time shift, the spectrum of rect τ (t) can be found to be F[rect(t)] = F[u(t + τ/2)] F[u(t τ/2)] = ejπfτ j2πf e jπfτ j2πf = τ sin(πfτ) = τ sinc(fτ) (1.140) πfτ his spectrum is zero at f = k/τ for any integer k. If we let τ, the time function is a constant 1 and its spectrum an impulse function. If we

32 24 Chapter 1. Continuous ime Fourier ransform Figure 1.3 Rectangular function and its spectrum divide both sides of the equation above by τ and let τ 0, the time function becomes an impulse and its spectrum a constant. On the other hand, in frequency domain, an ideal low-pass filter is defined as: { 1 f < fc H lp (f) = (1.141) 0 f > f c then according to time-frequency duality, its time impulse response is h lp (t) = sin(2πf ct) πt = 2f c sinc(2f c t) (1.142) Note that the impulse response h lp (t) is nonzero for t < 0, indicating that the ideal low-pass filter is not causal (response before the input δ(0) at t = 0). In other words, an ideal low-pass filter is impossible to implement in real-time, but it can be trivially realized off-line in frequency domain. riangle function triangle(t) = { 1 t /τ t < τ 0 t τ (1.143) his triangle function (with width 2τ) is the convolution of two square functions (with width τ) scaled by 1/τ: triangle(t) = 1 τ rect(t) rect(t) (1.144) its Fourier transform can be conveniently obtained based on the convolution theorem: F[triangle(t)] = 1 τ F[rect(t) rect(t)] = 1 τ τ sinc(f) τ sinc(f) = τ sin2 (fτ) (1.145)

33 Continuous ime Fourier ransform 25 Figure 1.4 riangular function and its spectrum Alternatively, the spectrum of the triangle function can be obtained by the definition. As this is an even function, its Fourier transform is τ F[triangle(t)] = 2 (1 t/τ) cos(2πft)dt 0 τ = 2[ cos(2πft)dt 1 τ t cos(2πft)dt] = 1 0 τ 0 πf [sin(2πfτ) 1 τ t d sin(2πft)] τ 0 = 1 πf [sin(2πfτ) t τ sin(2πft) τ τ sin(2πft)dt] = 1 τ 2τ(πf) 2 cos(2πft) τ 0 = 0 1 2τ(πf) 2 (1 cos(2πfτ)) = τ sin2 (πfτ) (πfτ) 2 = τ sinc 2 (fτ) (1.146) his spectrum is zero at f = k/τ for any integer k. Gaussian function Consider the Gaussian function x(t) = e π(t/a)2 /a. Note that in particular when a = 2πσ 2, x(t) becomes the normal distribution with variance σ 2 and mean µ = 0. he spectrum of x(t) is: X(f) = F[ 1 a e π(t/a)2 ] = 1 e π(t/a)2 e j2πft dt = 1 e π((t/a)2 +j2ft) dt a a = 1 a eπ(jaf)2 e π[(t/a)2 +j2ft+(jaf) 2] dt = e π(af)2 e π(t/a+jaf)2 d(t/a + jaf) = e π(af)2 (1.147) he last equation is due to the identity e πx2 dx = 1. We see that the Fourier transform of a Gaussian function is another Gaussian function, and the area underneath either x(t) or X(f) is unity. Moreover, If we let a 0, x(t) will approach δ(t), while its spectrum e π(af)2 approaches 1. On the other hand, if we rewrite the above as X(f) = F[x(t)] = F[e π(t/a)2 ] = ae π(af)2 (1.148) and let a, x(t) approaches 1 and X(f) approaches δ(f).

34 26 Chapter 1. Continuous ime Fourier ransform Figure 1.5 Gaussian function and its spectrum Impulse train As discussed before the impulse trains a sequence of infinite unit impulse separated by time interval : comb(t) = δ(t n ) (1.149) n= First, we find the Fourier transform of this function: Comb(f) = comb(t)e j2kπft dt = [ δ(t n )]e j2kπft dt = n= n= δ(t n )]e j2kπft dt = n= e j2nπf (1.150) On the other hand, we also realize that the impulse train is a periodic function and its Fourier series expansion is (Eq. 1.48): comb(t) = 1 e j2kπt/ (1.151) k= Applying the Fourier transform to this expression of comb(t), we get 1 Comb(f) = e j2kπt/ e j2πft dt = 1 k= k= e j2π(f k/ )t dt = 1 k= δ(f k/ )(1.152) We see that the spectrum of an impulse train in time domain with interval is another impulse train in frequency domain with interval f 0 = 1/. Equating the two expressions of Comb(f) in Eqs and 1.152, we get Comb(f) = e j2nπf = 1 δ(f k/ ) (1.153) n= k=

35 Continuous ime Fourier ransform 27 which can also be obtained from Eq by the time-frequency duality. hese two mathematically equivalent formulae, sometimes called Poisson formula, are useful while discussing impulse trains. Example 1.1: In radio and V broadcasting, a carrier wave c(t) = cos(2πf c t) with radio frequency (RF) f c is first modulated by the audio or video signal s(t) before it is transmitted. In particular, in amplitude modulation (AM), the modulation is carried out as a multiplication by a modulator (mixer): x(t) = s(t)c(t) = s(t) cos(2πf c t) = s(t) 1 2 [ej2πf ct + e j2πfct ] (1.154) his multiplication in time domain corresponds to a convolution in frequency domain: X(f) = S(f) C(f) = S(f) 1 2 [δ(f f c) + δ(f + f c )] = 1 2 [S(f f c) + S(f + f c )] (1.155) We note that the bandwidth occupied by the signal is f = 2f m, twice the highest frequency contained in the signal. his modulated signal with RF frequency is transmitted and then received by a radio or V receiver, where the audio or video signal needs to be separated from the carrier wave by a demodulation process, which can be easily implemented by another multiplication: y(t) = x(t) cos(2πf c t) = s(t) cos 2 (2πf c t) = s(t) 2 + s(t) cos(4πf ct) 2 (1.156) he signal s(t) can then be obtained by a low-pass filter to remove the higher frequency component at 2f c. Note that this demodulation method requires the sinusoid cos(2πf c t) used in the demodulator of the receiver is synchronous with that used in the modulator of the transmitter. If there is a phase difference between the two, the trigonometry relation used in the equation above is no longer valid. For this reason, alternative methods exist for the purpose of demodulation. his process of both modulation and demodulation in frequency domain is illustrated in Fig Hilbert ransform and Analytic Signals he Hilbert transform of a time function x(t) is another time function, denoted by ˆx(t), defined as the following convolution with 1/πt: H[x(t)] = ˆx(t) = x(t) 1 πt = 1 π x(τ) t τ dτ = 1 π x(t τ) dτ (1.157) τ As the integrand is not integrable due to its pole at τ = 0, the integral of the Hilbert transform is defined in the sense of the Cauchy principal value of the

36 28 Chapter 1. Continuous ime Fourier ransform Figure 1.6 AM modulation and demodulation integral defined as: H[x(t)] = 1 ɛ π lim[ ɛ 0 x(t τ) dτ + τ ɛ x(t τ) dτ] (1.158) τ In particular, if x(t) = c is a constant, the sum of the two integrals above is zero, indicating the Hilbert transform, as a linear operator, will remove the DC component of the signal. he Hilbert transform can be much more conveniently discussed in frequency domain as a multiplication corresponding to the time convolution in Eq o do so, we assume X(f) = F[x(t)] and find the spectrum of 1/πt by applying the property of time-frequency duality to the Fourier transform of the sign function sgn(t) (Eq.1.123): F 1 [ 1 ] = j Sgn(f) = j πt 1 (f < 0) 0 (f = 0) 1 (f > 0) j (f < 0) = 0 (f = 0) j (f > 0) (1.159) Now the Hilbert transform can be expressed in frequency domain as a multiplication: jx(f) (f < 0) ˆX(f) = F[ˆx(t)] = j Sgn(f) X(f) = 0 (f = 0) (1.160) jx(f) (f > 0) he effect of the Hilbert transform applied of a signal x(t) becomes very clear: it multiplies the negative part of the signal spectrum X(f) by j = e jπ/2, (a rotation of π/2 in complex plane) and the positive part by j = e jπ/2 (a rotation of π/2). herefore the Hilbert transform is also called a quadrature filter.

37 Continuous ime Fourier ransform 29 As the Hilbert transform of a time function is still a time function, it can be applied to a signal x(t) multiple times, and the result is most conveniently obtained in frequency domain: In particular, as Sgn 2 (f) = 1, we have F[H n [x(t)]] = [ j Sgn(f)] n X(f) (1.161) [ j Sgn(f)] 2 = 1, [ j Sgn(f)] 3 = j Sgn(f), [ j Sgn(f)] 4 = 1 (1.162) Correspondingly in time domain, we have: H[x(t)] = ˆx(t), H 2 [x(t)] = x(t), H 3 [x(t)] = ˆx(t), H 4 [x(t)] = x(t) (1.163) In other words, applying the Hilbert transform to x(t) once we get H[x(t)] = ˆx(t), and applying the transform three more times we get the original signal back, i.e., this is the inverse Hilbert transform: { H[x(t)] = x(t) 1/πt = ˆx(t) H 1 [ˆx(t)] = H 3 (1.164) [ˆx(t)] = H[ˆx(t)] = x(t) Example 1.2: Consider a simple sinusoid: cos(2πf 0 t) = ej2πf 0t + e j2πf 0t 2 = 1 2 ej2πf 0t e j2πf 0t (1.165) Here the coefficients for f = f 0 > 0 and f = f 0 < 0 are both 1/2. When the Hilbert transform is applied to the signal, the coefficient 1/2 for f < 0 is rotated by 90 to become e jπ/2 /2 while the other 1/2 for f > 0 is rotated by 90 to become e jπ/2 /2, i.e., the transformed signal becomes: H[cos(2πft)] = e j2π e j2πf0t + ejπ/2 2 2 e j2πf 0t = sin(2πft) (1.166) Similarly we have H[sin(2πft)] = cos(2πft), H[ cos(2πft)] = sin(2πft), H[ sin(2πft)] = cos(2πft) (1.167) Next let us consider the concept of analytic signals. A real-valued signal x a (f) is said to be analytic if its Fourier spectrum X a (f) = F[x a (t)] is zero when f < 0. Given a usual signal x(t), we can always construct an analytic signal by multiplying its spectrum X(f) = F[x(t)] with a step function 2u(f) in frequency domain: 0 (f < 0) X a (f) = X(f)2u(f) = X(0) (f = 0) (1.168) 2X(f) (f > 0)

38 30 Chapter 1. Continuous ime Fourier ransform and the corresponding analytic signal can be obtained as x a (t) = F 1 [X a (f)] = F 1 [X(f)] F 1 [2u(f)] = x(t) [δ(t) + j πt ] = x(t) + j x(t) 1 = x(t) + j ˆx(t) (1.169) πt where the inverse Fourier transform of the unit step spectrum u(f) is F 1 [u(f)] = 1 j2πt δ( t) = j 2πt + 1 δ(t) (1.170) 2 which can be obtained by the time-frequency duality applied to the spectrum of the unit step u(t) in time given in Eq Alternatively, an analytic signal can also be initially defined in time domain by Eq.1.169, and if we take the Fourier transform on both sides, we have X a (f) = X(f) + j ˆX(f) jx(f) (f < 0) = X(f) + j 0 (f = 0) jx(f) (f > 0) = 0 (f < 0) X(0) (f = 0) 2X(f) (f > 0) (1.171) where ˆX a (f) = F[ˆx(t)]. When the signal x(t) is real, the real and imaginary parts of its spectrum are even and odd, respectively (X r (f) = X r ( f) and X j (f) = X j ( f)), i.e., its spectrum X(f) is Hermitian: X(f) = X( f) (1.172) his means that the corresponding analytic signal x a (t) = x(t) + j ˆx(t) will still contain the complete information in x(t), even though the negative half of its spectrum is suppressed to zero. In fact the original spectrum X(f) can also be reconstructed from X a (f). When f > 0, obviously we get X(f) = X a (f)/2 from Eq When f < 0, we have X(f) = X( f) = X( f ) = 1 2 X a( f ) (1.173) Combining these two cases, we have: X(f) = 1 { Xa (f) (f > 0) 2 X a ( f ) (f < 0) = X a(f) + X a ( f) 2 (1.174) the last equality is due to the fact that X a ( f) = 0 when f > 0 and X a (f) = 0 when f < 0. Example 1.3: Given x(t) = cos(ω 0 t), we can construct an analytic signal x a (t) = x(t) + j ˆx(t) = cos(ω 0 t) + j sin(ω 0 t) = e jω 0t (1.175)

39 Continuous ime Fourier ransform 31 Figure 1.7 Single sideband modulation using Hilbert transform with spectrum X a (f) = δ(ω ω 0 ). Similarly, if y(t) = sin(ω 0 t), the corresponding analytic signal is y a (t) = y(t) + j ŷ(t) = sin(ω 0 t) j cos(ω 0 t) = je jω 0t (1.176) with spectrum Y a (f) = jδ(ω ω 0 ). In both cases, the negative half of the spectrum is zero. Example 1.4: From the previous discussion regarding the modulation and demodulation in AM broadcasting, we see that the bandwidth ω = 2ω m taken by a transmission is twice of the highest frequency contained in the signal, one sideband of ω m on each side of the carrier frequency ω c (double sideband). However, in order to efficiently use the broadcasting spectrum as a limited resource, it is desirable to minimize the bandwidth needed for the radio or V transmission. And single-sideband modulation (SSB) is such a method by which the bandwidth is reduced by half (from 2ω m to ω m ). One way to implement SSB is to use the idea of the Hilbert transform and analytic signals, taking advantage of the fact that the negative half of the spectrum of an analytic signal is completely zero and therefore does not need to be transmitted. Specifically, an analytic signal is first constructed based on the real signal s(t) to be transmitted: s a (t) = s(t) + j ŝ(t) (1.177) where ŝ(t) = H[x(t)] is the Hilbert transform of s(t). Using this analytic signal to modulate a carrier frequency 2πf c, represented as an complex exponential e j2πf c, we get s a (t)e j2πf ct and then transmit its real part: x(t) = Re[s a (t)e j2πf ct ] = Re[(s(t) + j ŝ(t)) (cos(2πf c t) + j sin(2πf c t))] = s(t) cos(2πf c t) ŝ(t) sin(2πf c t) = x 0 (t) x 1 (t) (1.178) where x 0 (t) = s(t) cos(2πf c t) and x 1 (t) = ŝ(t) sin(2πf c t) are two modulated RF signals with 90 phase difference. he block diagram of the single sideband modulation is illustration in Fig.1.7. o show that the bandwidth of the modulated