《数字信号处理》教学参考资料（Numerical Recipes in C，The Art of Scientific Computing Second Edition）Chapter 13.9

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584 Chapter 13.Fourier and Spectral Applications ix=(int)x; if (x ==(float)ix)yy[ix]+=y; else 11o=LMIN(LMAX((1ong)(x-0.5*m+1.0),1),n-m+1); 1h1=11o+m-1: nden=nfac [m]; fac=x-ilo; for (j=ilo+1;j=ilo;j--){ nden=(nden/(j+1-ilo))*(j-ihi); yy[j]+y*fac/(nden*(x-j)); http://www.nr. readable files 83g 11-800 (including this one) granted for from NUMERICAL RECIPES IN 19881992 CITED REFERENCES AND FURTHER READING: Lomb,N.R.1976,Astrophysics and Space Science,vol.39,pp.447-462.[1] /Cambridge Barning,F.J.M.1963,Bulletin of the Astronomical Institutes of the Netherlands,vol.17,pp.22- 28.[2] server Vanicek,P.1971,Astrophysics and Space Science,vol.12,pp.10-33.[3] (Nort Scargle,J.D.1982,Astrophysical Journal,vol.263,pp.835-853.[4] computer, Ameri to make one paper UnN电.t THE Horne,J.H.,and Baliunas,S.L.1986,Astrophysical Journal,vol.302,pp.757-763.[5] Press,W.H.and Rybicki,G.B.1989,Astrophysica/Journal,vol.338,pp.277-280.[6] ART ictly proh Programs 13.9 Computing Fourier Integrals Using the FFT 6 Not uncommonly,one wants to calculate accurate numerical values for integrals of the form OF SCIENTIFIC COMPUTING (ISBN 198918920 I= eh(t)dt, (13.9.1) or the equivalent real and imaginary parts Numerica 10621 ^6 Ic= cos(wt)h(t)dt I= sin(wt)h(t)dt, (13.9.2) 43106 and one wants to evaluate this integral for many different values of w.In cases of interest,h(t) is often a smooth function,but it is not necessarily periodic in a,b,nor does it necessarily (outside Recipes go to zero at a or b.While it seems intuitively obvious that the force majeure of the FFT ought to be applicable to this problem,doing so turns out to be a surprisingly subtle matter, North as we will now see. Let us first approach the problem naively,to see where the difficulty lies.Divide the interval a,b into M subintervals,where M is a large integer,and define A=-a ，与三a+jA,h三h6),j=0,M (13.9.3) Notice that ho h(a)and hh(b),and that there are M+1 values hj.We can approximate the integral I by a sum, M-1 I≈△) hiexp(iwtj) (13.9.4) =0

584 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). ix=(int)x; if (x == (float)ix) yy[ix] += y; else { ilo=LMIN(LMAX((long)(x-0.5*m+1.0),1),n-m+1); ihi=ilo+m-1; nden=nfac[m]; fac=x-ilo; for (j=ilo+1;j=ilo;j--) { nden=(nden/(j+1-ilo))*(j-ihi); yy[j] += y*fac/(nden*(x-j)); } } } CITED REFERENCES AND FURTHER READING: Lomb, N.R. 1976, Astrophysics and Space Science, vol. 39, pp. 447–462. [1] Barning, F.J.M. 1963, Bulletin of the Astronomical Institutes of the Netherlands, vol. 17, pp. 22– 28. [2] Van´ıcek, P. 1971, ˇ Astrophysics and Space Science, vol. 12, pp. 10–33. [3] Scargle, J.D. 1982, Astrophysical Journal, vol. 263, pp. 835–853. [4] Horne, J.H., and Baliunas, S.L. 1986, Astrophysical Journal, vol. 302, pp. 757–763. [5] Press, W.H. and Rybicki, G.B. 1989, Astrophysical Journal, vol. 338, pp. 277–280. [6] 13.9 Computing Fourier Integrals Using the FFT Not uncommonly, one wants to calculate accurate numerical values for integrals of the form I = b a e iωth(t)dt , (13.9.1) or the equivalent real and imaginary parts Ic = b a cos(ωt)h(t)dt Is = b a sin(ωt)h(t)dt , (13.9.2) and one wants to evaluate this integral for many different values of ω. In cases of interest, h(t) is often a smooth function, but it is not necessarily periodic in [a, b], nor does it necessarily go to zero at a or b. While it seems intuitively obvious that the force majeure of the FFT ought to be applicable to this problem, doing so turns out to be a surprisingly subtle matter, as we will now see. Let us first approach the problem naively, to see where the difficulty lies. Divide the interval [a, b] into M subintervals, where M is a large integer, and define ∆ ≡ b − a M , tj ≡ a + j∆ , hj ≡ h(tj ), j = 0,...,M (13.9.3) Notice that h0 = h(a) and hM = h(b), and that there are M + 1 values hj . We can approximate the integral I by a sum, I ≈ ∆ M −1 j=0 hj exp(iωtj ) (13.9.4)

13.9 Computing Fourier Integrals Using the FFT 585 which is at any rate first-order accurate.(If we centered the h,'s and the t,'s in the intervals, we could be accurate to second order.)Now for certain values of w and M,the sum in equation (13.9.4)can be made into a discrete Fourier transform,or DFT,and evaluated by the fast Fourier transform(FFT)algorithm.In particular,we can choose M to be an integer power of 2,and define a set of special w's by n (13.9.5) where m has the values m =0,1,...,M/2-1.Then equation (13.9.4)becomes M-1 I(wm)≈△euma h,e2rim时/M=△DFT(ho.hM-i月m (13.9.6) 三人0 8 Equation (13.9.6),while simple and clear,is emphatically not recommended for use:It is likely to give wrong answers! The problem lies in the oscillatory nature of the integral (13.9.1).If h(t)is at all smooth, and if w is large enough to imply several cycles in the interval a,b-in fact,m in equation (13.9.5)gives exactly m cycles-then the value of I is typically very small,so small that it is easily swamped by first-order,or even (with centered values)second-order,truncation error.Furthermore,the characteristic "small parameter"that occurs in the error term is not A/(b-a)=1/M,as it would be if the integrand were not oscillatory,but wA,which can be as large as for w's within the Nyquist interval of the DFT (cf.equation 13.9.5).The result is that equation (13.9.6)becomes systematically inaccurate as w increases. It is a sobering exercise to implement equation(13.9.6)for an integral that can be done 线 THE analytically,and to see just how bad it is.We recommend that you try it. Let us therefore turn to a more sophisticated treatment.Given the sampled points h,we ART can approximate the function h(t)everywhere in the interval a,b by interpolation on nearby hj's.The simplest case is linear interpolation,using the two nearest h's,one to the left and Programs one to the right.A higher-order interpolation,e.g.,would be cubic interpolation,using two points to the left and two to the right-except in the first and last subintervals,where we must interpolate with three hi's on one side,one on the other. The formulas for such interpolation schemes are (piecewise)polynomial in the inde- pendent variable t,but with coefficients that are of course linear in the function values hj.Although one does not usually think of it in this way,interpolation can be viewed as approximating a function by a sum of kernel functions(which depend only on the interpolation OF SCIENTIFIC COMPUTING (ISBN scheme)times sample values (which depend only on the function).Let us write 1988-19920 0（）+∑（之） (13.9.7) =0 j=endpoints Here (s)is the kernel function of an interior point:It is zero for s sufficiently negative or sufficiently positive,and becomes nonzero only when s is in the range where the Numerical Recipes 10-621 -43106 h;multiplying it is actually used in the interpolation.We always have (0)=1 and (m)=0,m=±l,±2，，since interpolation right on a sample point should give the sampled function value.For linear interpolation (s)is piecewise linear,rises from 0 to 1 (outside for s in (-1,0),and falls back to 0 for s in (0,1).For higher-order interpolation,(s)is Software. made up piecewise of segments of Lagrange interpolation polynomials.It has discontinuous 首 derivatives at integer values of s,where the pieces join,because the set of points used in the interpolation changes discretely As already remarked,the subintervals closest to a and b require different (noncentered) interpolation formulas.This is reflected in equation (13.9.7)by the second sum,with the special endpoint kernels (s).Actually,for reasons that will become clearer below,we have included all the points in the first sum (with kernel )so the ,'s are actually differences between true endpoint kernels and the interior kernel It is a tedious,but straightforward, exercise to write down all the(s)'s for any particular order of interpolation,each one consisting of differences of Lagrange interpolating polynomials spliced together piecewise. Now apply the integral operator dtexp(iwt)to both sides of equation (13.9.7). interchange the sums and integral,and make the changes of variable s=(t-)/A in the

13.9 Computing Fourier Integrals Using the FFT 585 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). which is at any rate first-order accurate. (If we centered the hj ’s and the tj ’s in the intervals, we could be accurate to second order.) Now for certain values of ω and M, the sum in equation (13.9.4) can be made into a discrete Fourier transform, or DFT, and evaluated by the fast Fourier transform (FFT) algorithm. In particular, we can choose M to be an integer power of 2, and define a set of special ω’s by ωm∆ ≡ 2πm M (13.9.5) where m has the values m = 0, 1,...,M/2 − 1. Then equation (13.9.4) becomes I(ωm) ≈ ∆eiωma M −1 j=0 hj e2πimj/M = ∆eiωma[DFT(h0 ...hM−1)]m (13.9.6) Equation (13.9.6), while simple and clear, is emphatically not recommended for use: It is likely to give wrong answers! The problem lies in the oscillatory nature of the integral (13.9.1). If h(t) is at all smooth, and if ω is large enough to imply several cycles in the interval [a, b] — in fact, ωm in equation (13.9.5) gives exactly m cycles — then the value of I is typically very small, so small that it is easily swamped by first-order, or even (with centered values) second-order, truncation error. Furthermore, the characteristic “small parameter” that occurs in the error term is not ∆/(b − a)=1/M, as it would be if the integrand were not oscillatory, but ω∆, which can be as large as π for ω’s within the Nyquist interval of the DFT (cf. equation 13.9.5). The result is that equation (13.9.6) becomes systematically inaccurate as ω increases. It is a sobering exercise to implement equation (13.9.6) for an integral that can be done analytically, and to see just how bad it is. We recommend that you try it. Let us therefore turn to a more sophisticated treatment. Given the sampled points hj , we can approximate the function h(t) everywhere in the interval [a, b] by interpolation on nearby hj ’s. The simplest case is linear interpolation, using the two nearest hj ’s, one to the left and one to the right. A higher-order interpolation, e.g., would be cubic interpolation, using two points to the left and two to the right — except in the first and last subintervals, where we must interpolate with three hj ’s on one side, one on the other. The formulas for such interpolation schemes are (piecewise) polynomial in the independent variable t, but with coefficients that are of course linear in the function values hj . Although one does not usually think of it in this way, interpolation can be viewed as approximating a function by a sum of kernel functions (which depend only on the interpolation scheme) times sample values (which depend only on the function). Let us write h(t) ≈ M j=0 hj ψ t − tj ∆ + j=endpoints hj ϕj t − tj ∆ (13.9.7) Here ψ(s) is the kernel function of an interior point: It is zero for s sufficiently negative or sufficiently positive, and becomes nonzero only when s is in the range where the hj multiplying it is actually used in the interpolation. We always have ψ(0) = 1 and ψ(m)=0, m = ±1, ±2,..., since interpolation right on a sample point should give the sampled function value. For linear interpolation ψ(s) is piecewise linear, rises from 0 to 1 for s in (−1, 0), and falls back to 0 for s in (0, 1). For higher-order interpolation, ψ(s) is made up piecewise of segments of Lagrange interpolation polynomials. It has discontinuous derivatives at integer values of s, where the pieces join, because the set of points used in the interpolation changes discretely. As already remarked, the subintervals closest to a and b require different (noncentered) interpolation formulas. This is reflected in equation (13.9.7) by the second sum, with the special endpoint kernels ϕj (s). Actually, for reasons that will become clearer below, we have included all the points in the first sum (with kernel ψ), so the ϕj ’s are actually differences between true endpoint kernels and the interior kernel ψ. It is a tedious, but straightforward, exercise to write down all the ϕj (s)’s for any particular order of interpolation, each one consisting of differences of Lagrange interpolating polynomials spliced together piecewise. Now apply the integral operator b a dt exp(iωt) to both sides of equation (13.9.7), interchange the sums and integral, and make the changes of variable s = (t − tj )/∆ in the

586 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). first sum, s = (t − a)/∆ in the second sum. The result is I ≈ ∆eiωa W(θ) M j=0 hj eijθ + j=endpoints hjαj (θ) (13.9.8) Here θ ≡ ω∆, and the functions W(θ) and αj (θ) are defined by W(θ) ≡ ∞ −∞ ds eiθsψ(s) (13.9.9) αj (θ) ≡ ∞ −∞ ds eiθsϕj (s − j) (13.9.10) The key point is that equations (13.9.9) and (13.9.10) can be evaluated, analytically, once and for all, for any given interpolation scheme. Then equation (13.9.8) is an algorithm for applying “endpoint corrections” to a sum which (as we will see) can be done using the FFT, giving a result with high-order accuracy. We will consider only interpolations that are left-right symmetric. Then symmetry implies ϕM−j (s) = ϕj (−s) αM−j (θ) = eiθMα* j (θ) = eiω(b−a) α* j (θ) (13.9.11) where * denotes complex conjugation. Also, ψ(s) = ψ(−s) implies that W(θ) is real. Turn now to the first sum in equation (13.9.8), which we want to do by FFT methods. To do so, choose some N that is an integer power of 2 with N ≥ M + 1. (Note that M need not be a power of two, so M = N − 1 is allowed.) If N>M + 1, define hj ≡ 0, M + 1 4M (see below). We can now rewrite equation (13.9.8) in its final form as I(ωn)=∆eiωna W(θ)[DFT(h0 ...hN−1)]n + α0(θ)h0 + α1(θ)h1 + α2(θ)h2 + α3(θ)h3 + ... + eiω(b−a) α* 0(θ)hM + α* 1(θ)hM−1 + α* 2(θ)hM−2 + α* 3(θ)hM−3 + ... (13.9.13) For cubic (or lower) polynomial interpolation, at most the terms explicitly shown above are nonzero; the ellipses (...) can therefore be ignored, and we need explicit forms only for the functions W, α0, α1, α2, α3, calculated with equations (13.9.9) and (13.9.10). We have worked these out for you, in the trapezoidal (second-order) and cubic (fourth-order) cases. Here are the results, along with the first few terms of their power series expansions for small θ: Trapezoidal order: W(θ) = 2(1 − cos θ) θ2 ≈ 1 − 1 12 θ2 + 1 360 θ4 − 1 20160 θ6 α0(θ) = − (1 − cos θ) θ2 + i (θ − sin θ) θ2 ≈ − 1 2 + 1 24 θ2 − 1 720 θ4 + 1 40320 θ6 + iθ 1 6 − 1 120 θ2 + 1 5040 θ4 − 1 362880 θ6 α1 = α2 = α3 = 0

13.9 Computing Fourier Integrals Using the FFT 587 Cubic order: w间)=()B-4ms0+cs291-品0+品的 /6+92 23 a00=-42+502)+6+62)8cos9-cos20+,-120+603)+6+92)sin29 604 604 86 3+ 22a6+0（+05-飞 -2835 04+ 467775 a1(0=143-g2)-76+0)cos0+:300-56+92)sim0 604 609 25920006+0(7、1 2+ 11 13 83 3456 （72-1681 72576 5987520 鱼 20=-48-02)+26+9)0s0+:-120+26+g2)sim0 1.00 304 304 7 1 284+立一8°+0-90十206 83 11 13 64800 90720 +7484400 a3(0)=28-69)-6+02)os0+00-6+02)sm0 604 604 1 7 .11g2+01020-2592009+8 1 24-180 360840 82+ 11 13 6 America Press. 362880 29937600 ART The program dftcor,below,implements the endpoint corrections for the cubic case. Given input valuesofw,A,a,b,and an array with the eight values ho,...,hs,hM-3,...,hM, it returns the real and imaginary parts of the endpoint corrections in equation (13.9.13),and the factor W().The code is turgid,but only because the formulas above are complicated.The formulas have cancellations to high powers of 0.It is therefore necessary to compute the right- hand sides in double precision,even when the corrections are desired only to single precision. to dir It is also necessary to use the series expansion for small values of 6.The optimal cross-over value of 6 depends on your machine's wordlength,but you can always find it experimentally SCIENTIFIC COMPUTING (ISBN as the largest value where the two methods give identical results to machine precision. 19841820 #include void dftcor(float w,float delta,float a,float b,float endpts[], float *corre,float *corim,float *corfac) For an integral approximated by a discrete Fourier transform,this routine computes the cor- idge.org uurggoglrion Numerical Recipes 10.621 43106 rection factor that multiplies the DFT and the endpoint correction to be added.Input is the angular frequency w,stepsize delta,lower and upper limits of the integral a and b,while the array endpts contains the first 4 and last 4 function values.The correction factor W()is (outside returned as corfac,while the real and imaginary parts of the endpoint correction are returned North Software. as corre and corim. void nrerror(char error_text []) float a0i,aor,ali,alr,a2i,a2r,a3i,a3r,arg,c,cl,cr,s,sl,sr,t; float t2,t4,t6; double cth,ctth,spth2,sth,sth4i,stth,th,th2,th4,tmth2,tth4i; th=y*delta: if (a >b lI th 0.0e0 Il th 3.1416e0)nrerror("bad arguments to dftcor"); if (fabs(th)<5.0e-2){ Use series. t=th; t2=t*t; t4=t2*t2; t6=t4*t2;

13.9 Computing Fourier Integrals Using the FFT 587 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). Cubic order: W(θ) = 6 + θ2 3θ4 (3 − 4 cos θ + cos 2θ) ≈ 1 − 11 720 θ4 + 23 15120 θ6 α0(θ) = (−42 + 5θ2) + (6 + θ2)(8 cos θ − cos 2θ) 6θ4 + i (−12θ + 6θ3) + (6 + θ2) sin 2θ 6θ4 ≈ − 2 3 + 1 45 θ2 + 103 15120 θ4 − 169 226800 θ6 + iθ 2 45 + 2 105 θ2 − 8 2835 θ4 + 86 467775 θ6 α1(θ) = 14(3 − θ2) − 7(6 + θ2) cos θ 6θ4 + i 30θ − 5(6 + θ2) sin θ 6θ4 ≈ 7 24 − 7 180 θ2 + 5 3456 θ4 − 7 259200 θ6 + iθ 7 72 − 1 168 θ2 + 11 72576 θ4 − 13 5987520 θ6 α2(θ) = −4(3 − θ2) + 2(6 + θ2) cos θ 3θ4 + i −12θ + 2(6 + θ2) sin θ 3θ4 ≈ − 1 6 + 1 45 θ2 − 5 6048 θ4 + 1 64800 θ6 + iθ − 7 90 + 1 210 θ2 − 11 90720 θ4 + 13 7484400 θ6 α3(θ) = 2(3 − θ2) − (6 + θ2) cos θ 6θ4 + i 6θ − (6 + θ2) sin θ 6θ4 ≈ 1 24 − 1 180 θ2 + 5 24192 θ4 − 1 259200 θ6 + iθ 7 360 − 1 840 θ2 + 11 362880 θ4 − 13 29937600 θ6 The program dftcor, below, implements the endpoint corrections for the cubic case. Given input values of ω, ∆, a, b, and an array with the eight values h0 ,...,h3, hM−3,...,hM , it returns the real and imaginary parts of the endpoint corrections in equation (13.9.13), and the factor W(θ). The code is turgid, but only because the formulas above are complicated. The formulas have cancellations to high powers of θ. It is therefore necessary to compute the righthand sides in double precision, even when the corrections are desired only to single precision. It is also necessary to use the series expansion for small values of θ. The optimal cross-over value of θ depends on your machine’s wordlength, but you can always find it experimentally as the largest value where the two methods give identical results to machine precision. #include void dftcor(float w, float delta, float a, float b, float endpts[], float *corre, float *corim, float *corfac) For an integral approximated by a discrete Fourier transform, this routine computes the correction factor that multiplies the DFT and the endpoint correction to be added. Input is the angular frequency w, stepsize delta, lower and upper limits of the integral a and b, while the array endpts contains the first 4 and last 4 function values. The correction factor W(θ) is returned as corfac, while the real and imaginary parts of the endpoint correction are returned as corre and corim. { void nrerror(char error_text[]); float a0i,a0r,a1i,a1r,a2i,a2r,a3i,a3r,arg,c,cl,cr,s,sl,sr,t; float t2,t4,t6; double cth,ctth,spth2,sth,sth4i,stth,th,th2,th4,tmth2,tth4i; th=w*delta; if (a >= b || th 3.1416e0) nrerror("bad arguments to dftcor"); if (fabs(th) < 5.0e-2) { Use series. t=th; t2=t*t; t4=t2*t2; t6=t4*t2;

588 Chapter 13. Fourier and Spectral Applications *c0rfac=1.0-(11.0/720.0)*t4+(23.0/15120.0)*t6: a0r=(-2.0/3.0)+t2/45.0+(103.0/15120.0)*t4-(169.0/226800.0)*t6; a1r=(7.0/24.0)-(7.0/180.0)*t2+(5.0/3456.0)*t4-(7.0/259200.0)*t6; a2r=(-1.0/6.0)+t2/45.0-(5.0/6048.0)*t4+t6/64800.0; a3r=(1.0/24.0)-t2/180.0+(5.0/24192.0)*t4-t6/259200.0: a01=t*(2.0/45.0+(2.0/105.0)*t2-(8.0/2835.0)*t4+(86.0/467775.0)*t6); a11=t*(7.0/72.0-t2/168.0+(11.0/72576.0)*t4-(13.0/5987520.0)*t6); a21=t*(-7.0/90.0+t2/210.0-(11.0/90720.0)*t4+(13.0/7484400.0)*t6) 31=t*(7.0/360.0-t2/840.0+(11.0/362880.0)*t4-(13.0/29937600.0)*t6); else Use trigonometric formulas in double precision. cth=cos(th); sth=sin(th); http://www.nr read able files Permission is ctth=cth*cth-sth*sth; stth=2.0e0*sth*cth; th2-th*th; th4=th2*th2: tmth2=3.0e0-th2; spth2=6.0e0+th2: (including this one) 19881992 sth41=1.0/(6.0e0*th4) tth4i=2.0e0*sth4i; 11-800-872 *corfac=tth4i*spth2*(3.0e0-4.0e0*cth+ctth); aOr=sth4i*(-42.0e0+5.0e0*th2+spth2*(8.0e0*cth-ctth)); a0i=sth4i*(th*(-12.0e0+6.0e0*th2)+spth2*stth); alr=sth4i*(14.0e0*tmth2-7.0e0*spth2*cth); ali=sth4i*(30.0e0*th-5.0e0*spth2*sth); a2r=tth4i*(-4.0e0*tmth2+2.0e0*spth2*cth): a21=tth4i*(（-12.0e0*th+2.0e0*spth2*sth); (North America to any server computer,is users to make one paper /Cambridge University Press.Programs from NUMERICAL RECIPES IN THE a3r=sth4i*(2.0e0*tmth2-spth2*cth); ART a3i=sth4i*(6.0e0*th-spth2*sth); cl=aOr*endpts[1]+alr*endpts [2]+a2r*endpts [3]+a3r*endpts [4]; sl=aoi*endpts[1]+ali*endpts [2]+a2i*endpts [3]+a3i*endpts [4]; copy for their cr=aOr*endpts [8]+alr*endpts [7]+a2r*endpts [6]+a3r*endpts [5]; sr =-aoi*endpts[8]-ali*endpts[7]-a2i*endpts[6]-a3i*endpts[5]; st st arg=w*(b-a); c=cos(arg); s=sin(arg); *corre=cl+c*cr-s*sr; *corim=sl+s*cr+c*sr; rectcustser OF SCIENTIFIC COMPUTING(ISBN 0-521 Since the use of dftcor can be confusing,we also give an illustrative program dftint which uses dftcor to compute equation (13.9.1)for general a,b,w,and h(t).Several points within this program bear mentioning:The parameters M and NDFT correspond to M and N in the above discussion.On successive calls,we recompute the Fourier transform only if 1988-1992 by Numerical Recipes -431085 a or b or h(t)has changed. Since dftint is designed to work for any value of w satisfying wA #include "nrutil.h" #define M 64 #define NDFT 1024 #define MPOL 6 #def1neTW0PI(2.0*3.14159265)

588 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). *corfac=1.0-(11.0/720.0)*t4+(23.0/15120.0)*t6; a0r=(-2.0/3.0)+t2/45.0+(103.0/15120.0)*t4-(169.0/226800.0)*t6; a1r=(7.0/24.0)-(7.0/180.0)*t2+(5.0/3456.0)*t4-(7.0/259200.0)*t6; a2r=(-1.0/6.0)+t2/45.0-(5.0/6048.0)*t4+t6/64800.0; a3r=(1.0/24.0)-t2/180.0+(5.0/24192.0)*t4-t6/259200.0; a0i=t*(2.0/45.0+(2.0/105.0)*t2-(8.0/2835.0)*t4+(86.0/467775.0)*t6); a1i=t*(7.0/72.0-t2/168.0+(11.0/72576.0)*t4-(13.0/5987520.0)*t6); a2i=t*(-7.0/90.0+t2/210.0-(11.0/90720.0)*t4+(13.0/7484400.0)*t6); a3i=t*(7.0/360.0-t2/840.0+(11.0/362880.0)*t4-(13.0/29937600.0)*t6); } else { Use trigonometric formulas in double precision. cth=cos(th); sth=sin(th); ctth=cth*cth-sth*sth; stth=2.0e0*sth*cth; th2=th*th; th4=th2*th2; tmth2=3.0e0-th2; spth2=6.0e0+th2; sth4i=1.0/(6.0e0*th4); tth4i=2.0e0*sth4i; *corfac=tth4i*spth2*(3.0e0-4.0e0*cth+ctth); a0r=sth4i*(-42.0e0+5.0e0*th2+spth2*(8.0e0*cth-ctth)); a0i=sth4i*(th*(-12.0e0+6.0e0*th2)+spth2*stth); a1r=sth4i*(14.0e0*tmth2-7.0e0*spth2*cth); a1i=sth4i*(30.0e0*th-5.0e0*spth2*sth); a2r=tth4i*(-4.0e0*tmth2+2.0e0*spth2*cth); a2i=tth4i*(-12.0e0*th+2.0e0*spth2*sth); a3r=sth4i*(2.0e0*tmth2-spth2*cth); a3i=sth4i*(6.0e0*th-spth2*sth); } cl=a0r*endpts[1]+a1r*endpts[2]+a2r*endpts[3]+a3r*endpts[4]; sl=a0i*endpts[1]+a1i*endpts[2]+a2i*endpts[3]+a3i*endpts[4]; cr=a0r*endpts[8]+a1r*endpts[7]+a2r*endpts[6]+a3r*endpts[5]; sr = -a0i*endpts[8]-a1i*endpts[7]-a2i*endpts[6]-a3i*endpts[5]; arg=w*(b-a); c=cos(arg); s=sin(arg); *corre=cl+c*cr-s*sr; *corim=sl+s*cr+c*sr; } Since the use of dftcor can be confusing, we also give an illustrative program dftint which uses dftcor to compute equation (13.9.1) for general a, b, ω, and h(t). Several points within this program bear mentioning: The parameters M and NDFT correspond to M and N in the above discussion. On successive calls, we recompute the Fourier transform only if a or b or h(t) has changed. Since dftint is designed to work for any value of ω satisfying ω∆ #include "nrutil.h" #define M 64 #define NDFT 1024 #define MPOL 6 #define TWOPI (2.0*3.14159265)

13.9 Computing Fourier Integrals Using the FFT 589 The values of M,NDFT,and MPOL are merely illustrative and should be optimized for your particular application.M is the number of subintervals,NDFT is the length of the FFT (a power of 2),and MPOL is the degree of polynomial interpolation used to obtain the desired frequency from the FFT. void dftint(float (*func)(float),float a,float b,float w,float *cosint, float *sinint) Example program illustrating how to use the routine dftcor.The user supplies an external function func that returns the quantity h(t).The routine then returnsfcos(wt)h(t)dt as cosint andsin(wt)h(t)dt as sinint. void dftcor(float w,float delta,float a,float b,float endpts], 常 float *corre,float *corim,float *corfac); void polint(float xa[],float ya[],int n,float x,float *y,float *dy); void realft(float data],unsigned long n,int isign); static int init=0; nted for 18881992 int j,nn; static float aold =-1.e30,bold =-1.e30,delta,(*funcold)(float); 1800 static float data[NDFT+1],endpts[9]; float c,cdft,cerr,corfac,corim,corre,en,s; 872 Cambridge float sdft,serr,*cpol,*spol,*xpol; 7423 from NUMERICAL RECIPES I cpol=vector(1,MPOL); spol-vector(1,MPOL); xpol=vector(1,MPOL); if (init !=1 I a !aold b !bold func !funcold){ (North America to any server computer, to make one paper e University Press. 2 THE Do we need to initialize,or is only w changed? init=1; ART aold=a: bold=b; Programs funcold=func; delta=(b-a)/M; Load the function values into the data array. st st for(j=1;j<=M+1;j++) data[j]=(*func)(a+(j-1)*delta); for (j=M+2;j<=NDFT;j++) Zero pad the rest of the data array. to dir Copyright (C) data[i]=0.0; for(J=1;j=4;j+)( Load the endpoints endpts[i]-data[i]; ectcustser OF SCIENTIFIC COMPUTING(ISBN endpts[j+4]=data [M-3+j]; 18881920 realft(data,NDFT,1); v@cam realft returns the unused value corresponding to wN/2 in data[2].We actually want Numerical Recipes 10-621 this element to contain the imaginary part corresponding to wo,which is zero. data[2]=0.0; -43108 Now interpolate on the DFT result for the desired frequency.If the frequency is an wn, i.e.,the quantity en is an integer,then cdft=data[2*en-1],sdft=data[2*en],and you (outside could omit the interpolation. en=w*delta*NDFT/TWOPI+1.0; nn=IMIN(IMAX((int)(en-0.5*MPOL+1.0),1),NDFT/2-MPOL+1);Leftmost point for the North Software. for (j=1;j<=MPOL;j++,nn++){ interpolation. cpol[j]=data[2*nn-1]; Ame spol[j]=data[2*nn]; xpol[j]=nn; visit website machine polint(xpol,cpol,MPOL,en,&cdft,&cerr); polint(xpol,spol,MPOL,en,&sdft,&serr); dftcor(w,delta,a,b,endpts,&corre,&corim,&corfac); Now get the endpoint cor- cdft *corfac; rection and the mul- sdft corfac; tiplicative factor W(0) cdft +corre; sdft te corim

13.9 Computing Fourier Integrals Using the FFT 589 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). The values of M, NDFT, and MPOL are merely illustrative and should be optimized for your particular application. M is the number of subintervals, NDFT is the length of the FFT (a power of 2), and MPOL is the degree of polynomial interpolation used to obtain the desired frequency from the FFT. void dftint(float (*func)(float), float a, float b, float w, float *cosint, float *sinint) Example program illustrating how to use the routine dftcor. The user supplies an external function func that returns the quantity h(t). The routine then returns b a cos(ωt)h(t) dt as cosint and b a sin(ωt)h(t) dt as sinint. { void dftcor(float w, float delta, float a, float b, float endpts[], float *corre, float *corim, float *corfac); void polint(float xa[], float ya[], int n, float x, float *y, float *dy); void realft(float data[], unsigned long n, int isign); static int init=0; int j,nn; static float aold = -1.e30,bold = -1.e30,delta,(*funcold)(float); static float data[NDFT+1],endpts[9]; float c,cdft,cerr,corfac,corim,corre,en,s; float sdft,serr,*cpol,*spol,*xpol; cpol=vector(1,MPOL); spol=vector(1,MPOL); xpol=vector(1,MPOL); if (init != 1 || a != aold || b != bold || func != funcold) { Do we need to initialize, or is only ω changed? init=1; aold=a; bold=b; funcold=func; delta=(b-a)/M; Load the function values into the data array. for (j=1;j<=M+1;j++) data[j]=(*func)(a+(j-1)*delta); for (j=M+2;j<=NDFT;j++) Zero pad the rest of the data array. data[j]=0.0; for (j=1;j<=4;j++) { Load the endpoints. endpts[j]=data[j]; endpts[j+4]=data[M-3+j]; } realft(data,NDFT,1); realft returns the unused value corresponding to ωN/2 in data[2]. We actually want this element to contain the imaginary part corresponding to ω0, which is zero. data[2]=0.0; } Now interpolate on the DFT result for the desired frequency. If the frequency is an ωn, i.e., the quantity en is an integer, then cdft=data[2*en-1], sdft=data[2*en], and you could omit the interpolation. en=w*delta*NDFT/TWOPI+1.0; nn=IMIN(IMAX((int)(en-0.5*MPOL+1.0),1),NDFT/2-MPOL+1); Leftmost point for the for (j=1;j<=MPOL;j++,nn++) { interpolation. cpol[j]=data[2*nn-1]; spol[j]=data[2*nn]; xpol[j]=nn; } polint(xpol,cpol,MPOL,en,&cdft,&cerr); polint(xpol,spol,MPOL,en,&sdft,&serr); dftcor(w,delta,a,b,endpts,&corre,&corim,&corfac); Now get the endpoint correction and the multiplicative factor W(θ). cdft *= corfac; sdft *= corfac; cdft += corre; sdft += corim;

590 Chapter 13. Fourier and Spectral Applications c=delta*cos(w*a); Finally multiply by A and exp(iwa). s-delta*sin(w*a); *cosint=c*cdft-s*sdft; *sinint=s*cdft+c*sdft; free_vector(cpol,1,MPOL); free_vector(spol,1,MPOL); free_vector(xpol,1,MPOL); Sometimes one is interested only in the discrete frequencieswm of equation (13.9.5), the ones that have integral numbers of periods in the interval a,b.For smooth h(t),the value of tends to be much smaller in magnitude at these w's than at values in between, since the integral half-periods tend to cancel precisely.(That is why one must oversample for interpolation to be accurate:I(w)is oscillatory with small magnitude near the wm's.)If you 8 want these wm's without messy (and possibly inaccurate)interpolation,you have to set N to a multiple of M(compare equations 13.9.5 and 13.9.12).In the method implemented above. however,N must be at least M+1,so the smallest such multiple is 2M,resulting in a factor ~2 unnecessary computing.Alternatively,one can derive a formula like equation (13.9.13) but with the last sample function h=h(b)omitted from the DFT,but included entirely in the endpoint correction for h.Then one can set M N (an integer power of 2)and get the special frequencies of equation (13.9.5)with no additional overhead.The modified formula is Iunm）=△eimaw(O)DFT(ho..haM-n +ao(0)ho+a1(0)h1+a2(0)h2+a3(0)h3 (13.9.14) 是ga 令 Press. +e(-)[A(0)hat +ai(0)hM-1+az(0)hM-2+as(0)hM-3 Programs where 0≡wm△andA(o)is given by SCIENTIFIC A(0)=-ao(0) (13.9.15) 6 for the trapezoidal case,or A0)==6+1102)+6+02)c0s2 604 ._iImlao(0)] (13.9.16) 1 8-0+14175 1106-iImlao(0)] for the cubic case. 10521 Factors like W()arise naturally whenever one calculates Fourier coefficients of smooth Numerical functions,and they are sometimes called attenuation factors[1].However,the endpoint 431 corrections are equally important in obtaining accurate values of integrals.Narasimhan Recipes and Karthikeyan [2]have given a formula that is algebraically equivalent to our trapezoidal formula.However,their formula requires the evaluation of no FFTs,which is unnecessary. (outside The basic idea used here goes back at least to Filon [3]in 1928(before the FFT!).He used Software. Simpson's rule (quadratic interpolation).Since this interpolation is not left-right symmetric. North two Fourier transforms are required.An alternative algorithm for equation(13.9.14)has been given by Lyness in[4];for related references,see [5].To our knowledge,the cubic-order formulas derived here have not previously appeared in the literature. Calculating Fourier transforms when the range of integration is (-oo,oo)can be tricky. If the function falls off reasonably quickly at infinity,you can split the integral at a large enough value of t.For example,the integration to +oo can be written eith(t)dt=eith(t)dt+ eith(t)dt ()(b)e(e (13.9.17) iw (iw)2

590 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). c=delta*cos(w*a); Finally multiply by ∆ and exp(iωa). s=delta*sin(w*a); *cosint=c*cdft-s*sdft; *sinint=s*cdft+c*sdft; free_vector(cpol,1,MPOL); free_vector(spol,1,MPOL); free_vector(xpol,1,MPOL); } Sometimes one is interested only in the discrete frequencies ωm of equation (13.9.5), the ones that have integral numbers of periods in the interval [a, b]. For smooth h(t), the value of I tends to be much smaller in magnitude at these ω’s than at values in between, since the integral half-periods tend to cancel precisely. (That is why one must oversample for interpolation to be accurate: I(ω) is oscillatory with small magnitude near the ωm’s.) If you want these ωm’s without messy (and possibly inaccurate) interpolation, you have to set N to a multiple of M (compare equations 13.9.5 and 13.9.12). In the method implemented above, however, N must be at least M + 1, so the smallest such multiple is 2M, resulting in a factor ∼2 unnecessary computing. Alternatively, one can derive a formula like equation (13.9.13), but with the last sample function hM = h(b) omitted from the DFT, but included entirely in the endpoint correction for hM . Then one can set M = N (an integer power of 2) and get the special frequencies of equation (13.9.5) with no additional overhead. The modified formula is I(ωm)=∆eiωma W(θ)[DFT(h0 ...hM−1)]m + α0(θ)h0 + α1(θ)h1 + α2(θ)h2 + α3(θ)h3 + eiω(b−a) A(θ)hM + α* 1(θ)hM−1 + α* 2(θ)hM−2 + α* 3(θ)hM−3 (13.9.14) where θ ≡ ωm∆ and A(θ) is given by A(θ) = −α0(θ) (13.9.15) for the trapezoidal case, or A(θ) = (−6 + 11θ2) + (6 + θ2) cos 2θ 6θ4 − iIm[α0(θ)] ≈ 1 3 + 1 45θ2 − 8 945 θ4 + 11 14175 θ6 − iIm[α0(θ)] (13.9.16) for the cubic case. Factors like W(θ) arise naturally whenever one calculates Fourier coefficients of smooth functions, and they are sometimes called attenuation factors[1]. However, the endpoint corrections are equally important in obtaining accurate values of integrals. Narasimhan and Karthikeyan [2] have given a formula that is algebraically equivalent to our trapezoidal formula. However, their formula requires the evaluation of two FFTs, which is unnecessary. The basic idea used here goes back at least to Filon [3] in 1928 (before the FFT!). He used Simpson’s rule (quadratic interpolation). Since this interpolation is not left-right symmetric, two Fourier transforms are required. An alternative algorithm for equation (13.9.14) has been given by Lyness in [4]; for related references, see [5]. To our knowledge, the cubic-order formulas derived here have not previously appeared in the literature. Calculating Fourier transforms when the range of integration is (−∞,∞) can be tricky. If the function falls off reasonably quickly at infinity, you can split the integral at a large enough value of t. For example, the integration to + ∞ can be written ∞ a eiωth(t) dt = b a eiωth(t) dt + ∞ b eiωth(t) dt = b a eiωth(t) dt − h(b)eiωb iω + h (b)eiωb (iω)2 −··· (13.9.17)

13.10 Wavelet Transforms 591 The splitting point b must be chosen large enough that the remaining integral over(b,)is small.Successive terms in its asymptotic expansion are found by integrating by parts.The integral over(a,b)can be done using dftint.You keep as many terms in the asymptotic expansion as you can easily compute.See[6]for some examples of this idea.More powerful methods,which work well for long-tailed functions but which do not use the FFT, are described in [7-9]. CITED REFERENCES AND FURTHER READING: Stoer,J.,and Bulirsch,R.1980,Introduction to Numerical Analysis(New York:Springer-Verlag), p.88.[1] Narasimhan,M.S.and Karthikeyan,M.1984,IEEE Transactions on Antennas Propagation, vol.32,pp.404-408.[2] Filon,L.N.G.1928.Proceedings of the Royal Society of Edinburgh,vol.49,pp.38-47.[3] Giunta,G.and Murli,A.1987,ACM Transactions on Mathematical Software,vol.13,pp.97-107. 4] ICAL Lyness,J.N.1987,in Numerical Integration,P.Keast and G.Fairweather,eds.(Dordrecht: Reidel).[5] Pantis,G.1975,Journal of Computational Physics,vol.17,pp.229-233.[6] RECIPES Blakemore,M.,Evans,G.A.,and Hyslop,J.1976,Journal of Computational Physics,vol.22. pp.352-376.7] 9 Lyness,J.N.,and Kaper,T.J.1987,SIAM Journal on Scientific and Statistical Computing,vol.8. pp.1005-1011.[8] Thakkar,A.J.,and Smith,V.H.1975,Computer Physics Communications,vol.10,pp.73-79.[9] 邑白 9 13.10 Wavelet Transforms Like the fast Fourier transform(FFT),the discrete wavelet transform(DWT)is a fast,linear operation that operates on a data vector whose length is an integer power of two,transforming it into a numerically different vector of the same length.Also like the FFT,the wavelet transform is invertible and in fact orthogonal-the inverse transform,when viewed as a big matrix,is simply the transpose of the transform. Both FFT and DWT,therefore,can be viewed as a rotation in function space,from Numerical Recipes 10621 43106 the input space(or time)domain,where the basis functions are the unit vectors ei, or Dirac delta functions in the continuum limit,to a different domain.For the FFT. this new domain has basis functions that are the familiar sines and cosines.In the (outside wavelet domain,the basis functions are somewhat more complicated and have the North fanciful names“mother functions'”and“wavelets. Of course there are an infinity of possible bases for function space,almost all of them uninteresting!What makes the wavelet basis interesting is that,unlike sines and cosines,individual wavelet functions are quite localized in space;simultaneously, like sines and cosines,individual wavelet functions are quite localized in frequency or (more precisely)characteristic scale.As we will see below,the particular kind of dual localization achieved by wavelets renders large classes of functions and operators sparse,or sparse to some high accuracy,when transformed into the wavelet domain.Analogously with the Fourier domain,where a class of computations,like convolutions,become computationally fast,there is a large class of computations

13.10 Wavelet Transforms 591 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). The splitting point b must be chosen large enough that the remaining integral over (b,∞) is small. Successive terms in its asymptotic expansion are found by integrating by parts. The integral over (a, b) can be done using dftint. You keep as many terms in the asymptotic expansion as you can easily compute. See [6] for some examples of this idea. More powerful methods, which work well for long-tailed functions but which do not use the FFT, are described in [7-9]. CITED REFERENCES AND FURTHER READING: Stoer, J., and Bulirsch, R. 1980, Introduction to Numerical Analysis (New York: Springer-Verlag), p. 88. [1] Narasimhan, M.S. and Karthikeyan, M. 1984, IEEE Transactions on Antennas & Propagation, vol. 32, pp. 404–408. [2] Filon, L.N.G. 1928, Proceedings of the Royal Society of Edinburgh, vol. 49, pp. 38–47. [3] Giunta, G. and Murli, A. 1987, ACM Transactions on Mathematical Software, vol. 13, pp. 97–107. [4] Lyness, J.N. 1987, in Numerical Integration, P. Keast and G. Fairweather, eds. (Dordrecht: Reidel). [5] Pantis, G. 1975, Journal of Computational Physics, vol. 17, pp. 229–233. [6] Blakemore, M., Evans, G.A., and Hyslop, J. 1976, Journal of Computational Physics, vol. 22, pp. 352–376. [7] Lyness, J.N., and Kaper, T.J. 1987, SIAM Journal on Scientific and Statistical Computing, vol. 8, pp. 1005–1011. [8] Thakkar, A.J., and Smith, V.H. 1975, Computer Physics Communications, vol. 10, pp. 73–79. [9] 13.10 Wavelet Transforms Like the fast Fourier transform (FFT), the discrete wavelet transform (DWT) is a fast, linear operation that operates on a data vector whose length is an integer power of two, transforming it into a numerically different vector of the same length. Also like the FFT, the wavelet transform is invertible and in fact orthogonal — the inverse transform, when viewed as a big matrix, is simply the transpose of the transform. Both FFT and DWT, therefore, can be viewed as a rotation in function space, from the input space (or time) domain, where the basis functions are the unit vectors e i, or Dirac delta functions in the continuum limit, to a different domain. For the FFT, this new domain has basis functions that are the familiar sines and cosines. In the wavelet domain, the basis functions are somewhat more complicated and have the fanciful names “mother functions” and “wavelets.” Of course there are an infinity of possible bases for function space, almost all of them uninteresting! What makes the wavelet basis interesting is that, unlike sines and cosines, individual wavelet functions are quite localized in space; simultaneously, like sines and cosines, individual wavelet functions are quite localized in frequency or (more precisely) characteristic scale. As we will see below, the particular kind of dual localization achieved by wavelets renders large classes of functions and operators sparse, or sparse to some high accuracy, when transformed into the wavelet domain. Analogously with the Fourier domain, where a class of computations, like convolutions, become computationally fast, there is a large class of computations

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