Chapter 1 The Fourier Transform - University Of Minnesota
Chapter 1The Fourier Transform1.1Fourier transforms as integralsThere are several ways to define the Fourier transform of a function f : R C. In this section, we define it using an integral representation and statesome basic uniqueness and inversion properties, without proof. Thereafter,we will consider the transform as being defined as a suitable limit of Fourierseries, and will prove the results stated here.Definition 1 Let f : R R. The Fourier transform of f L1 (R), denotedby F[f ](.), is given by the integral:Z 1f (t) exp( ixt)dtF[f ](x) : 2π for x R for which the integral exists. We have the Dirichlet condition for inversion of Fourier integrals.R Theorem 1 Let f : R R. Suppose that (1) f dt converges and (2)in any finite interval, f ,f 0 are piecewise continuous with at most finitely manymaxima/minima/discontinuities. Let F F[f ]. Then if f is continuous att R, we haveZ 1f (t) F (x) exp(itx)dx.2π This definition also makes sense for complex valued f but we stick here to real valuedf1
Moreover, if f is discontinuous at t R and f (t 0) and f (t 0) denote theright and left limits of f at t, thenZ 11[f (t 0) f (t 0)] F (x) exp(itx)dx.22π From the above, we deduce a uniqueness result:Theorem 2 Let f, g : R R be continuous, f 0 , g 0 piecewise continuous. IfF[f ](x) F[g](x), xthenf (t) g(t), t.Proof: We have from inversion, easily thatZ 1f (t) F[f ](x) exp(itx)dx2π Z 1F[g](x) exp(itx)dx 2π g(t).2Example 1 Find the Fourier transform of f (t) exp( t ) and hence usingR dxR x sin(xt)πinversion, deduce that 0 1 xdx π exp( t), t 0.2 2 and1 x220Solution We writeZ 1F (x) f (t) exp( ixt)dt2π Z 0Z 1exp( t(1 ix)) exp(t(1 ix))dt 2π 0r2 1 .π 1 x2Now by the inversion formula,Z 1exp( t ) F (x) exp(ixt)dx2π Z 1exp(ixt) exp( ixt) dtπ 01 x2Z2 cos(xt) dx.π 0 1 x22
Now this formula holds at t 0, so substituting t 0 into the above givesthe first required identity. Differentiating with respect to t as we may fort 0, gives the second required identity. 2.Proceeding in a similar way as the above example, we can easily showthat11F[exp( t2 )](x) exp( x2 ), x R.22We will discuss this example in more detail later in this chapter.We will also show that we can reinterpret Definition 1 to obtain theFourier transform of any complex valued f L2 (R), and that the Fouriertransform is unitary on this space:Theorem 3 If f, g L2 (R) then F[f ], F[g] L2 (R) andZ Z F[f ](x)F[g](x) dx.f (t)g(t) dt This is a result of fundamental importance for applications in signal processing.1.2The transform as a limit of Fourier seriesWe start by constructing the Fourier series (complex form) for functions onan interval [ πL, πL]. The ON basis functions areen (t) int1eL,2πLn 0, 1, · · · ,and a sufficiently smooth function f of period 2πL can be expanded as Z πL Xint1 inxf (x)e L dx e L .f (t) 2πL πLn For purposes of motivation let us abandon periodicity and think of the functions f as differentiable everywhere, vanishing at t πL and identicallyzero outside [ πL, πL]. We rewrite this asf (t) Xn eintL1 ˆ nf( )2πL Lwhich looks like a Riemann sum approximation to the integralZ 1f (t) fˆ(λ)eiλt dλ2π 3(1.2.1)
to which it would converge as L . (Indeed, we are partitioning the λinterval [ L, L] into 2L subintervals, each with partition width 1/L.) Here,Z ˆf (t)e iλt dt.(1.2.2)f (λ) Similarly the Parseval formula for f on [ πL, πL],Z πL X1 ˆ n 22 f (t) dt f ( ) 2πL L πLn goes in the limit as L to the Plancherel identityZ Z 22π f (t) dt fˆ(λ) 2 dλ. (1.2.3) Expression (1.2.2) is called the Fourier integral or Fourier transform of f .Expression (1.2.1) is called the inverse Fourier integral for f . The Plancherelidentity suggests that the Fourier transform is a one-to-one norm preservingmap of the Hilbert space L2 [ , ] onto itself (or to another copy of itself). We shall show that this is the case. Furthermore we shall show thatthe pointwise convergence properties of the inverse Fourier transform aresomewhat similar to those of the Fourier series. Although we could makea rigorous justification of the the steps in the Riemann sum approximationabove, we will follow a different course and treat the convergence in the meanand pointwise convergence issues separately.A second notation that we shall use isZ 11f (t)e iλt dt fˆ(λ)(1.2.4)F[f ](λ) 2π 2πZ 1 F [g](t) g(λ)eiλt dλ(1.2.5)2π Note that, formally, F [fˆ](t) 2πf (t). The first notation is used moreoften in the engineering literature. The second notation makes clear that Fand F are linear operators mapping L2 [ , ] onto itself in one view, andF mapping the signal space onto the frequency space with F mapping thefrequency space onto the signal space in the other view. In this notation thePlancherel theorem takes the more symmetric formZ Z 2 f (t) dt F[f ](λ) 2 dλ. Examples:4
1. The box function (or rectangular wave) 1 if π t π1if t πΠ(t) 20otherwise.(1.2.6)Then, since Π(t) is an even function and e iλt cos(λt) i sin(λt), wehaveZ Z iλtΠ(t) cos(λt)dtΠ(t)edt Π̂(λ) 2πF[Π](λ) Zπcos(λt)dt π2 sin(πλ) 2π sinc λ.λThus sinc λ is the Fourier transform of the box function. The inverseFourier transform isZ sinc(λ)eiλt dλ Π(t),(1.2.7) as follows from (?). Furthermore, we haveZ Π(t) 2 dt 2π andZ sinc (λ) 2 dλ 1 from (?), so the Plancherel equality is verified in this case. Notethat the inverse Fourier transform converged to the midpoint of thediscontinuity, just as for Fourier series.2. A truncated cosine wave. cos 3t if π t π 21if t πf (t) 0otherwise.Then, since the cosine is an even function, we haveZ Z π iλtˆf (λ) 2πF[f ](λ) f (t)edt cos(3t) cos(λt)dt π2λ sin(λπ).9 λ25
3. A truncated sine wave. sin 3t if π t π0otherwise.f (t) Since the sine is an odd function, we haveZZ iλtfˆ(λ) 2πF[f ](λ) f (t)edt i πsin(3t) sin(λt)dt π 6i sin(λπ).9 λ24. A triangular wave. 1 t if 1 t 0 1if 0 t 1f (t) 0otherwise.(1.2.8)Then, since f is an even function, we haveZ Z 1 iλtˆf (t)edt 2(1 t) cos(λt)dtf (λ) 2πF[f ](λ) 02 2 cos λ.λ2NOTE: The Fourier transforms of the discontinuous functions above decayas λ1 for λ whereas the Fourier transforms of the continuous functionsdecay as λ12 . The coefficients in the Fourier series of the analogous functionsdecay as n1 , n12 , respectively, as n .1.2.1Properties of the Fourier transformRecall that1F[f ](λ) 2πZ 1f (t)e iλt dt fˆ(λ)2π Z 1F [g](t) g(λ)eiλt dλ2π We list some properties of the Fourier transform that will enable us to build arepertoire of transformsR from a few basic examples. Suppose that f, g belong1to L [ , ], i.e., f (t) dt with a similar statement for g. We canstate the following (whose straightforward proofs are left to the reader):6
1. F and F are linear operators. For a, b C we haveF[af bg] aF[f ] bF[g],F [af bg] aF [f ] bF [g].2. Suppose tn f (t) L1 [ , ] for some positive integer n. ThenF[tn f (t)](λ) indn{F[f ](λ)}.dλn3. Suppose λn f (λ) L1 [ , ] for some positive integer n. ThenF [λn f (λ)](t) indn{F [f ](t)}.dtn4. Suppose the nth derivative f (n) (t) L1 [ , ] and piecewise continuous for some positive integer n, and f and the lower derivatives areall continuous in ( , ). ThenF[f (n) ](λ) (iλ)n F[f ](λ)}.5. Suppose nth derivative f (n) (λ) L1 [ , ] for some positive integern and piecewise continuous for some positive integer n, and f and thelower derivatives are all continuous in ( , ). ThenF [f (n) ](t) ( it)n F [f ](t).6. The Fourier transform of a translation by real number a is given byF[f (t a)](λ) e iλa F[f ](λ).7. The Fourier transform of a scaling by positive number b is given byλ1F[f (bt)](λ) F[f ]( ).bb8. The Fourier transform of a translated and scaled function is given by1λF[f (bt a)](λ) e iλa/b F[f ]( ).bbExamples7
We want to compute the Fourier transform of the rectangular box function with support on [c, d]: 1 if c t d1if t c, dR(t) 20 otherwise.Recall that the box function 1 if π t π1if t πΠ(t) 20otherwise.has the Fourier transform Π̂(λ) 2π sinc λ. but we can obtain R from2πand then rescaling s d cs:Π by first translating t s t (c d)2R(t) Π(2πc dt π).d cd c2πλ4π 2 iπλ(c d)/(d c)esinc().(1.2.9)d cd cFurthermore, from (?) we can check that the inverse Fourier transformof R̂ is R, i.e., F (F)R(t) R(t).R̂(λ) Consider the truncated sine wave sin 3t if π t πf (t) 0otherwisewith 6i sin(λπ)fˆ(λ) .9 λ2Note that the derivative f 0 of f (t) is just 3g(t) (except at 2 points)where g(t) is the truncated cosine wave cos 3t if π t πif t π 12g(t) 0otherwise.We have computedĝ(λ) 2λ sin(λπ).9 λ2so 3ĝ(λ) (iλ)fˆ(λ), as predicted. Reversing the example above we can differentiate the truncated cosinewave to get the truncated sine wave. The prediction for the Fouriertransform doesn’t work! Why not?8
1.2.2Fourier transform of a convolutionThe following property of the Fourier transform is of particular importancein signal processing. Suppose f, g belong to L1 [ , ].Definition 2 The convolution of f and g is the function f g defined byZ f (t x)g(x)dx.(f g)(t) Note also that (f g)(t) of variable.R f (x)g(t x)dx, as can be shown by a changeLemma 1 f g L1 [ , ] andZ Z f g(t) dt Z f (x) dx g(t) dt. Z f (x)g(t x) dx dt Z Sketch of proof:ZZ f g(t) dt Z Z g(t x) dt f (x) dx Z f (x) dx. g(t) dt 2Theorem 4 Let h f g. Thenĥ(λ) fˆ(λ)ĝ(λ).Sketch of proof:Z Z iλtĥ(λ) f g(t)edt Z iλx Z Z iλ(t x)g(t x)ef (x)e f (x)g(t x)dx e iλt dt Zdt dx fˆ(λ)ĝ(λ).29f (x)e iλx dx ĝ(λ)
L2 convergence of the Fourier transform1.3In this book our primary interest is in Fourier transforms of functions inthe Hilbert space L2 [ , ]. However, the formal definition of the Fourierintegral transform,Z 1f (t)e iλt dt(1.3.10)F[f ](λ) 2π doesn’t make sense for a general f L2 [ , ]. If f L1 [ , ] then f isabsolutely integrable and the integral (1.3.10) converges. However, there are1square integrable functions that are not integrable. (Example: f (t) 1 t .)How do we define the transform for such functions?We will proceed by defining F on a dense subspace of f L2 [ , ]where the integral makes sense and then take Cauchy sequences of functionsin the subspace to define F on the closure. Since F preserves inner product,as we shall show, this simple procedure will be effective.First some comments Ron integrals of L2 functions. If f, g L2 [ , ] then the integral (f, g) f (t)g(t)dt necessarily exists, whereas the integral (1.3.10) may not, because the exponential e iλt is not an element of L2 .However, the integral of f L2 over any finite interval, say [ N, N ] doesexist. Indeed for N a positive integer, let χ[ N,N ] be the indicator functionfor that interval: 1 if N t Nχ[ N,N ] (t) (1.3.11)0otherwise.Then χ[ N,N ] L2 [ , ] soZNRN Nf (t)dt exists because f (t) dt ( f , χ[ N,N ] ) f L2 χ[ N,N ] L2 f L2 2N NNow the space of step functions is dense in L2 [ , ], so we can find aconvergent sequence of step functions {sn } such that limn f sn L2 0.Note that the sequence of functions {fN f χ[ N,N ] } converges to f pointwiseas N and each fN L2 L1 .Lemma 2 {fN } is a Cauchy sequence in the norm of L2 [ , ] and limn f fn L2 0.Proof: Given 0 there is step function sM such that f sM 2 .Choose N so large that the support of sM is contained in [ N, N ], i.e.,10
sM (t)χ[ N,N ] (t) sM (t) for all t. Then sM fN 2 R s (t) f (t) 2 dt sM f 2 , so MRN N sM (t) f (t) 2 dt f fN (f sM ) (sM fN ) f sM sM fN 2 f sM .2Here we will study the linear mapping F : L2 [ , ] L̂2 [ , ] fromthe signal space to the frequency space. We will show that the mapping isunitary, i.e., it preserves the inner product and is 1-1 and onto. Moreover,the map F : L̂2 [ , ] L2 [ , ] is also a unitary mapping and is theinverse of F:F F IL2 ,FF IL̂2where IL2 , IL̂2 are the identity operators on L2 and L̂2 , respectively. Weknow that the space of step functions is dense in L2 . Hence to show thatF preserves inner product, it is enough to verify this fact for step functionsand then go to the limit. Once we have done this, we can define Ff for anyf L2 [ , ]. Indeed, if {sn } is a Cauchy sequence of step functions suchthat limn f sn L2 0, then {Fsn } is also a Cauchy sequence (indeed, sn sm Fsn Fsm ) so we can define Ff by Ff limn Fsn . Thestandard methods of Section 1.3 show that Ff is uniquely defined by thisconstruction. Now the truncated functions fN have Fourier transforms givenby the convergent integralsZ N1f (t)e iλt dtF[fN ](λ) 2π Nand limN f fN L2 0. Since F preserves inner product we have Ff FfN L2 F(f fN ) L2 f fN L2 , so limN Ff FfN L2 0. We writeZ N1F[f ](λ) l.i.m.N F[fN ](λ) f (t)e iλt dt2π Nwhere ‘l.i.m.’ indicates that the convergence is in the mean (Hilbert space)sense, rather than pointwise.We have already shown that the Fourier transform of the rectangular boxfunction with support on [c, d]: 1 if c t d1if t c, dRc,d (t) 20 otherwise.isF[Rc,d ](λ) 4π 22πλeiπλ(c d)/(d c) sinc().d c2π(d c)11
and that F (F)Rc,d (t) Rc,d (t). (Since here we are concerned only withconvergence in the mean the value of a step function at a particular pointis immaterial. Hence for this discussion we can ignore such niceties as thevalues of step functions at the points of their jump discontinuities.)Lemma 3(Ra,b , Rc,d )L2 (FRa,b , FRc,d )L̂2for all real numbers a b and c d.Proof:Z(FRa,b , FRc,d )L̂2 N F[Ra,b ](λ)F[Rc,d ](λ)dλ d eiλt dt dλ limN N2πc Z d Z Neiλt limF[Ra,b ](λ) dλ dt.N c2π NNow the inside integral is converging to Ra,b as N in both the pointwiseand L2 sense, as we have shown. ThusZ d(FRa,b , FRc,d )L̂2 Ra,b dt (Ra,b , Rc,d )L2 .Z ZF[Ra,b ](λ)c2Since any step functions u, v are finite linearof indicatorP combination Pfunctions Raj ,bj with complex coefficients, u j αj Raj ,bj , v k βk Rck ,dkwe haveXαj β k (FRaj ,bj , FRck ,dk )L̂2(Fu, Fv)L̂2 j,k Xαj β k (Raj ,bj , Rck ,dk )L2 (u, v)L2 .j,kThus F preserves inner product on step functions, and by taking Cauchysequences of step functions, we have theTheorem 5 (Plancherel Formula) Let f, g L2 [ , ]. Then f 2L2 Ff 2L̂2(f, g)L2 (Ff, Fg)L̂2 ,In the engineering notation this readsZZ 2πf (t)g(t)dt fˆ(λ)ĝ(λ)dλ. 12
Theorem 6 The map F : L̂2 [ , ] L2 [ , ] has the followingproperties:1. It preserves inner product, i.e.,(F fˆ, F ĝ)L2 (fˆ, ĝ)L̂2for all fˆ, ĝ L̂2 [ , ].2. F is the adjoint operator to F : L2 [ , ] L̂2 [ , ], i.e.,(Ff, ĝ)L̂2 (f, F ĝ)L2 ,for all f L2 [ , ], ĝ L̂2 [ , ].Proof:1. This follows immediately from the facts that F preserves inner productand F[f ](λ) F [f ](λ).2.(FRa,b , Rc,d )L̂2 (Ra,b , F Rc,d )L2as can be seen by an interchange in the order of integration. Thenusing the linearity of F and F we see that(Fu, v)L̂2 (u, F v)L2 ,for all step functions u, v. Since the space of step functions is dense inL̂2 [ , ] and in L2 [ , ]2Theorem 71. The Fourier transform F : L2 [ , ] L̂2 [ , ] isa unitary transformation, i.e., it preserves the inner product and is 1-1and onto.2. The adjoint map F : L̂2 [ , ] L2 [ , ] is also a unitarymapping.3. F is the inverse operator to F:F F IL2 ,FF IL̂2where IL2 , IL̂2 are the identity operators on L2 and L̂2 , respectively.13
Proof:1. The only thing left to prove is that for every ĝ L̂2 [ , ] there is af L2 [ , ] such that Ff ĝ, i.e., R {Ff : f L2 [ , ]} L̂2 [ , ]. Suppose this isn’t true. Then there exists a nonzero ĥ L̂2 [ , ] such that ĥ R, i.e., (Ff, ĥ)L̂2 0 for all f L2 [ , ].But this means that (f, F ĥ)L2 0 for all f L2 [ , ], so F ĥ Θ.But then F ĥ L2 ĥ L̂2 0 so ĥ Θ, a contradiction.2. Same proof as for 1.3. We have shown that FF Ra,b F FRa,b Ra,b for all indicator functions Ra,b . By linearity we have FF s F Fs s for all step functions s. This implies that(F Ff, g)L2 (f, g)L2for all f, g L2 [ , ]. Thus([F F IL2 ]f, g)L2 0for all f, g L2 [ , ], so F F IL2 . An analogous argument givesFF IL̂2 .21.4The Riemann-Lebesgue Lemma and pointwise convergenceLemma 4 (Riemann-Lebesgue) Suppose f is absolutely Riemann integrablein ( , ) (so that f L1 [ , ]), and is bounded in any finite subinterval [a, b], and let α, β be real. ThenZ limf (t) sin(αt β)dt 0.α Proof: Without loss of generality, we can assume that f is real, because wecan break up the complex integral into its real and imaginary parts.1. The statement is true if f Ra,b is an indicator function, forZ Z b 1cos(αt β) ba 0Ra,b (t) sin(αt β)dt sin(αt β)dt α aas α .14
2. The statement is true if f is a step function, since a step function is afinite linear combination of indicator functions.3. The statement is true if f is bounded and Riemann integrable on thefinite interval [a, b] and vanishes outside the interval. Indeed givenany 0 there exist two step functions s (Darboux upper sum) and s(Darboux lower sum) with support in [a, b] such that s(t) f (t) s(t)Rbfor all t [a, b] and a s s 2 . ThenbZf (t) sin(αt β)dt aZbbZ[f (t) s(t)] sin(αt β)dt s(t) sin(αt β)dt.aaNowZ bZ bZ b [f (t) s(t)] sin(αt β)dt f (t) s(t) dt s s 2aaaand (since s is a step function, by choosing α sufficiently large we canensureZ b s(t) sin(αt β)dt .2aHenceZ bf (t) sin(αt β)dt afor α sufficiently large.4. The statement of the lemma is true in general. IndeedZ Z af (t) sin(αt β)dt f (t) sin(αt β)dt Z bZ f (t) sin(αt β)dt af (t) sin(αt β)dt .bGiven 0 we can choose a and b such the first and third integralsare each 3 , and we can choose α so large the the second integral is 3 . Hence the limit exists and is 0.2The sinc function has a delta-function property:Lemma 5 Let c 0, and F (x) a function on [0, c]. Suppose15
F (x) is piecewise continuous on [0, c] F 0 (x) is piecewise continuous on [0, c] F 0 ( 0) exists.ThenZlimL 0δsin LxπF (x)dx F ( 0).x2Proof: We writeZ cZ cZ csin LxF (x) F ( 0)sin kxF (x)dx F ( 0)dx sin Lx dx.xxx000( 0)Set G(x) F (x) Ffor x [0, δ] and G(x) 0 elsewhere. Since F 0 ( 0)xexists it follows that G L2 . hence, by the Riemann-Lebesgue Lemma, thesecond integral goes to 0 in the limit as L . HenceZ cZ csin Lxsin LxlimF (x)dx F ( 0) limdxL 0L 0xxZ Lcπsin udu F ( 0). F ( 0) limL 0u2For the last equality we have used our evaluation (?) of the integral of thesinc function. 2Theorem 8 Let f be a complex valued function such that1. f (t) is absolutely Riemann integrable on ( , ).2. f (t) is piecewise continuous on ( , ), with only a finite number ofdiscontinuities in any bounded interval.3. f 0 (t) is piecewise continuous on ( , ), with only a finite number ofdiscontinuities in any bounded interval.4. f (t) f (t 0) f (t 0)2at each point t.Letfˆ(λ) Z f (t)e iλt dt be the Fourier transform of f . ThenZ 1f (t) fˆ(λ)eiλt dλ2π for every t ( , ).16(1.4.12)
Proof: For real L 0 set Z LZ L Z 1iλt iλxfˆ(λ)e dλ fL (t) f (x)edx eiλt dλ2π L L Z L Z Z 1iλ(t x) f (x)edλ dx f (x) L (t x)dx,2π L where LZ L1if x 0iλxπ L (x) e dλ sin Lxotherwise.2π LπxHere we have interchanged the order of integration, which we can since theintegral is absolutely convergent. IndeedZ LZ Z L
Expression (1.2.2) is called the Fourier integral or Fourier transform of f. Expression (1.2.1) is called the inverse Fourier integral for f. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to anoth
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(d) Fourier transform in the complex domain (for those who took "Complex Variables") is discussed in Appendix 5.2.5. (e) Fourier Series interpreted as Discrete Fourier transform are discussed in Appendix 5.2.5. 5.1.3 cos- and sin-Fourier transform and integral Applying the same arguments as in Section 4.5 we can rewrite formulae (5.1.8 .
FT Fourier Transform DFT Discrete Fourier Transform FFT Fast Fourier Transform WT Wavelet Transform . CDDWT Complex Double Density Wavelet Transform PCWT Projection based Complex Wavelet Transform viii. . Appendix B 150 Appendix C 152 References 153 xiii.
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