Part I Preliminaries

Transcription

1 Part I Preliminaries

2 Chapter An Elementary Introduction to the Discrete Fourier ransform his chapter is intended to provide a brief introduction to the discrete Fourier transform (DF) It is not intended to be comprehensive; instead, through a simple example, it provides an illustration of how the computation that is the subject of this bookarises, and how its results can be used he DF arises in a multitude of other contexts as well, and a dozen more DF-related applications, together with information on a number of excellent references, are presented in Chapter 6 in Part II of this book Readers familiar with the DF may safely skip this chapter A major application of Fourier transforms is the analysis of a series of observations: x l,l=0,,n ypically, N will be quite large: 0000 would not be unusual he sources of such observations are many: ocean tidal records over many years, communication signals over many microseconds, stockprices over a few months, sonar signals over a few minutes, and so on he assumption is that there are repeating patterns in the data that form part of the x l However, usually there will be other phenomena which may not repeat, or repeat in a way that is not discernably cyclic his is called noise he DF helps to identify and quantify the cyclic phenomena If a pattern repeats itself m times in the N observations, it is said to have Fourier frequency m o make this more specific, suppose one measures a signal from time t =0to t = 680 in steps of 5 seconds, giving 73 observations he measurements might appear as shown in Figure How does one make any sense out of it? As shown later, the DF can help Complex Numbers Effective computation of the DF relies heavily on the use of complex numbers, so it is useful to review their basic properties his material is elementary and probably wellknown to most readers of this book, but it is included for completeness Historically, complex numbers were introduced to deal with polynomial equations, such as x +=

3 Figure Example of a noisy signal , which have no real solutions Informally, they can be defined as the set C of all numbers of the form a + jb where a and b are real numbers and j = Addition, subtraction, and multiplication are performed among complex numbers by treating them as binomials in the unknown j and using j = to simplify the result hus (a + jb)+(c + jd)=(a + c)+j(b + d) and (a + jb) (c + jd)=(ac bd)+j(ad + bc) For the complex number z = a+jb, a is the real part of z and b is the imaginary part of z he zero element of C is 0+0i, and the additive inverse of z = a + jb is a + i( b) he multiplicative inverse z is z = a jb a + b he complex conjugate of z = a + jb is denoted by z and is equal to a jb he modulus of z, denoted by z, is z z = a + b Some additional facts that will be used later are e z = e (a+jb) = e a e jb and e jb = cos b + j sin b hus, Re(e z )=e a cos b and Im(z) =e a sin b

4 Just as a real number can be pictured as a point lying on a line, a complex number can be pictured as a point lying in a plane With each complex number a + jb one can associate a vector beginning at the origin and terminating at the point (a, b) hese notions are depicted in Figure Figure Visualizing complex numbers Instead of the pair (a, b), one can use the length (modulus) together with the angle the number makes with the real axis hus, a+jb can be represented as r cos θ + jr sin θ = re jθ, where r = z = a + b and θ = arctan(b/a) his representation of a complex number is depicted in Figure 3 Figure 3 Polar representation of a complex number Multiplication of complex numbers in polar form is straightforward: if z = a+jb = r e jθ and z = c + jd = r e jθ, then z z = r r e j(θ+θ) he moduli are multiplied together, and the angles are added Note that if z = e jθ, then z = for all values of θ rigonometric Interpolation Suppose a function f(θ) is defined on the interval (0, π), with f assumed to be periodic on the interval; thus, f(θ) =f(θ ± π)

5 Now consider constructing a trigonometric polynomial p(θ) to interpolate f(θ) of the form () p(θ) =a 0 + a k cos kθ + b k sin kθ k= his function has n + coefficients, so it should be possible to interpolate f at n + points In the applications considered in this book, the points at which to interpolate are always equally spaced on the interval: () θ l = lπ n +, l =0,,,n Let x l = f(θ l ), and consider an example with n = hen the interpolation conditions are x l = p(θ l ), or x l = a 0 + a cos θ l + b sin θ l + a cos θ l + b sin θ l, l =0,,,4 his leads to the system of equations cos θ 0 sin θ 0 cos θ 0 sin θ 0 cos θ sin θ cos θ sin θ cos θ sin θ cos θ sin θ cos θ 3 sin θ 3 cos θ 3 sin θ 3 cos θ 4 sin θ 4 cos θ 4 sin θ 4 Recall that e jθ = cos θ + j sin θ, which implies that a 0 a b a b = x 0 x x x 3 x 4 cos θ = ejθ + e jθ and sin θ = ejθ e jθ j Using these in () with n = yields ( a ) ( p(θ) = a 0 + e jθ a ) ( ) + e jθ b + e jθ j ( a ) ( + e jθ a ) ( ) + e jθ b + e jθ j ( ) ( ) a + jb = e jθ a + jb + e jθ ( ) ( a jb +a 0 + e jθ a jb + ) e jθ ( b j ( b j Giving the coefficients names corresponding to the powers of e jθ yields ) e jθ ) e jθ (3) p(θ) =X e jθ + X e jθ + X 0 + X e jθ + X e jθ Note that the coefficients appear in complex conjugate pairs When the x l are real, it is straightforward to show that this is true in general (See the next section) Recall (see ()) that the points at which interpolation occurs are evenly spaced; that is, θ l = lθ Let ω = e jθ = e jπ n+ hen all e jθ l can be expressed in terms of ω: e jθ l = e jlθ = ω l,l=0,,,n

6 Also, note that ω l = ω l±(n+) and ω l = ω l±(n+) For the example with n =, ω = e jπ 5, and the interpolation condition at θ l in (3) is f(θ l )=x l = p(θ l )=X ω l + X ω l + X 0 ω 0 + X ω l + X ω l Using the fact that ω l = ω (n+ l), and renaming the coefficients similarly (X l X n+ l ), the interpolation condition at x l becomes x l = X 0 + X ω l + X ω l + X 3 ω 3l + X 4 ω 4l, which has to be satisfied for l =0,,,4: ω ω ω 3 ω 4 ω ω 4 ω 6 ω 8 (4) ω 3 ω 6 ω 9 ω ω 4 ω 8 ω ω 6 his can be written as a matrix equation MX = x X 0 X X X 3 X 4 = x 0 x x x 3 x 4 It will be useful to have some additional properties of ω First note that +ω + ω + + ω n =0 his can be established by observing that the expression on the left side is a geometric sum equal to ω n+ ω, and this quantity is zero because ω n+ = For integers r and s one can show in a similar way that { 0 if r s (5) ω (kr ks) = n + ifr = s k=0 hese simple results make solving MX = x easy o begin, let ω ω ω 3 ω 4 M = ω ω 4 ω 6 ω 8 ω 3 ω 6 ω 9 ω ω 4 ω 8 ω ω 6 hen using (5) above, together with the fact that ω l = ω l, shows that MM is

7 hus, M = 5M, and X 0 X X = (6) 5 X 3 X 4 ω ω ω 3 ω 4 ω ω 4 ω 6 ω 8 ω 3 ω 6 ω 9 ω ω 4 ω 8 ω ω 6 It is a simple exercise to carry out this development for general n, yielding the following formula for the DF: x 0 x x x 3 x 4 (7) X r = n + l=0 x l ω rl,r =0,,,n Similarly, the inverse DF (IDF) has the form (8) x l = r=0 X r ω rl,l=0,,,n 3 Analyzing the Series What information can the X r provide? As noted earlier for the example with n =, when the given data x are real, the X r appear in complex conjugate pairs o establish this, note that (9) X n+ r = Writing (8) as n + l=0 x l ω (n+ r)l = x l = X 0 + n + l=0 x l ω rl = n + (X r ω rl + X n+ r ω l(n+ r)) r= l=0 x l ω rl = X r and using the fact that ω l(n+ r) = ω rl = ω rl and X n+ r = X r, x l can be expressed as (0) ( x l = X 0 + Xr ω rl + X r ω rl) r= Recall that if a and b are complex numbers, then a +ā =Re(a), ab = a b e j(θa+θ b) and Re(ab) = a b cos(θ a + θ b ) Using these, (0) can be written as x l = X 0 + Re(X r ω rl ) = X 0 + r= r= ( ) rlπ X r cos n + + φ r,

8 where rlπ n+ is the phase angle of ωrl and φ r is the phase angle of X r hus, after computing the coefficients X r,r =0,,,n, the interpolating function can be evaluated at any point in the interval [0, π] using the formula p(θ) =X 0 + X r cos(rθ + φ r ) r= In many applications, it is the amplitudes (the size of X r ) that are of interest hey indicate the strength of each frequency in the signal omake this discussion concrete, consider the signal shown in Figure, where the 73 measurements are plotted Using Matlab, one can compute and plot X, as shown on the left in Figure 4 Note that apart from the left endpoint (corresponding Figure 4 Plot of X for the example in Figure to X 0 ), the plot of the entire X is symmetric, as expected; as shown above, the X r occur in complex conjugate pairs, with X r = X n+ r he plot on the right in Figure 4 contains the first 30 components of X so that more detail can be seen It suggests that the signal has two dominant Fourier frequencies: 0 and 30 4 Fourier Frequency Versus ime Frequency As θ goes from 0 to π, cos(rθ + φ r ) makes r cycles Suppose the x l are collected over an interval of seconds As θ goes from 0 to π, t goes from 0 to hus, cos(rθ+φ r ) oscillates at r/ cycles per second Making a change of variable yields () p(t) =X 0 + r= ( ( r ) ) X r cos π t + φ r he usual practice is to express the phase shift φ r in terms of a negative shift in time t hus, () is often written in the form () p(t) =X 0 + r= ( ( r ) ) X r cos π (t t r ) Matlab is a proprietary commercial product of he MathWorks, Inc, Natick, MA Web URL:

9 where t r = φ r π ( r ) Returning to the signal shown in Figure, recall that the 73 data elements were collected at intervals of 5 seconds over a period of = 680 seconds hus, since the dominant Fourier frequencies in the signal appear to be 0 and 30, the dominant frequencies in cycles per second would be and cycles per second Figure 5 contains aplot of the first 40 amplitudes ( X r ) against cycles per second Figure 5 Plot of amplitudes against cycles per second for the example in Figure Filtering a Signal Suppose X d is much larger than the other coefficients If one assumes the other frequencies are the result of noise, one can clean up the signal by setting all but X d to zero hus, () might be replaced by the filtered signal ( p clean (t) =X 0 + X d cos π ( ) ) d t + φ d Of course there may be several apparently dominant frequencies, in which case more than one of the elements of X would be retained As an illustration, again consider the example of Figure he dominant signals appear to be Fourier frequency 0 and 30 Discarding all elements of X for which X r < 06 yields a cleaned up signal Evaluating () from t =0tot = 50 yields the signal shown on the right in Figure 6 he plot on the left is the corresponding part of the original signal shown in Figure here is vast literature on digital filtering, and the strategy described here is intended only to illustrate the basic idea For a comprehensive introduction to the topic, see errell [03]

10 Figure 6 Plot of part of the original and clean signals for the example in Figure 6 How Often Does One Sample? In performing the analysis of a time series, one has the values of a certain (unknown) function f(t) at equally spaced intervals of time Let δ be the time interval between successive observations in seconds hen /δ is called the sampling rate his is the number of observations taken each second If the sampling rate is /δ, what frequencies can the Fourier transform reliably detect? An intuitive argument is as follows Consider a pure cosine signal with frequency, sampled over =3seconds as shown in Figure 7 In the representation Figure 7 A pure cosine signal p(t) =X 0 + r= ( ( r ) ) X r cos π t + φ r, it is evident that in order to be able to represent this signal, the cos(π ( 3 ) t + φ3 ) term must be present his implies that one needs n 3orn + 7 hat is, more than samples per second, or at least 7 sample points Another way to lookat it is as follows One needs to sample often enough to detect all the oscillations if the true signal is to be detected In order to detect oscillations up to s cycles per second, one must sample at a rate at least s times per second In most practical situations, there is a range of frequencies of interest, and the sampling rate is chosen accordingly

11 7 Notes and References he trigonometric polynomial p(θ) =a 0 + a k cos kθ + b k sin kθ k= was used in Section to interpolate a periodic function f(θ) his polynomial has n + coefficients, and by interpolating f(θ) at the n + primitive roots of unity, a DF of length N =n + results DFs of length N = s are the most convenient and most efficient to compute Such a DF can be obtained by using a a trigonometric polynomial having a slightly different form than the one above, namely (3) p(θ) = a 0 + a n+ cos (n +)θ + a k cos kθ + b k sin kθ k= his polynomial has n + coefficients; thus, n can be chosen so that N =n + is a power of two he derivation of the DF using (3) is similar to the derivation done for the case where N is odd, and is left as an exercise