Chapter 1 The Fourier Transform

1 Chapter 1 The Fourier Fourier transforms as integralsThere are several ways to define the Fourier Transform of a functionf: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 1 Letf:R R. The Fourier Transform off L1(R), denotedbyF[f](.), is given by the integral:F[f](x) :=1 2 f(t) exp( ixt)dtforx Rfor which the integral exists. We have theDirichlet conditionfor inversion of Fourier 1 Letf:R R.

2 Suppose that (1) |f|dtconverges and (2)in any finite interval,f,f are piecewise continuous with at most finitely manymaxima/minima/discontinuities. LetF=F[f]. Then iffis continuous att R, we havef(t) =1 2 F(x) exp(itx)dx. This definition also makes sense for complex valuedfbut we stick here to real valuedf1 Moreover, iffis discontinuous att Randf(t+ 0)andf(t 0)denote theright and left limits offatt, then12[f(t+ 0) +f(t 0)] =1 2 F(x) exp(itx) the above, we deduce a uniqueness result:Theorem 2 Letf,g:R Rbe continuous,f ,g piecewise continuous. IfF[f](x) =F[g](x), xthenf(t) =g(t), : We have from inversion, easily thatf(t) =1 2 F[f](x) exp(itx)dx=1 2 F[g](x) exp(itx)dx=g(t).

3 2 Example 1 Find the Fourier Transform off(t) = exp( |t|)and hence usinginversion, deduce that 0dx1+x2= 2and 0xsin(xt)1+x2dx= exp( t)2, t > writeF(x) =1 2 f(t) exp( ixt)dt=1 2 [ 0 exp(t(1 ix))dt+ 0exp( t(1 +ix))]= 2 11 + by the inversion formula,exp( |t|) =1 2 F(x) exp(ixt)dx=1 [ 0exp(ixt) + exp( ixt)1 +x2dt]=2 0cos(xt)1 + this formula holds att= 0, so substitutingt= 0 into the above givesthe first required identity. Differentiating with respect totas we may fort >0, gives the second required in a similar way as the above example, we can easily showthatF[exp( 12t2)](x) = exp( 12x2), x will discuss this example in more detail later in this will also show that we can reinterpret Definition 1 to obtain theFourier Transform of any complex valuedf L2(R), and that the Fouriertransform is unitary on this space.

4 Theorem 3 Iff,g L2(R)thenF[f],F[g] L2(R)and f(t)g(t)dt= F[f](x)F[g](x) is a result of fundamental importance for applications in signal The Transform as a limit of Fourier seriesWe start by constructing the Fourier series (complex form) for functions onan interval [ L, L]. The ON basis functions areen(t) =1 2 LeintL, n= 0, 1, ,and a sufficiently smooth functionfof period 2 Lcan be expanded asf(t) = n= (12 L L Lf(x)e inxLdx) purposes of motivation let us abandon periodicity and think of the func-tionsfas differentiable everywhere, vanishing att= Land identicallyzero outside [ L, L].

5 We rewrite this asf(t) = n= eintL12 L f(nL)which looks like a Riemann sum approximation to the integralf(t) =12 f( )ei td ( )3to which it would converge asL . (Indeed, we are partitioning the interval [ L,L] into 2 Lsubintervals, each with partition width 1/L.) Here, f( ) = f(t)e i tdt.( )Similarly the Parseval formula forfon [ L, L], L L|f(t)|2dt= n= 12 L| f(nL)|2goes in the limit asL to thePlancherel identity2 |f(t)|2dt= | f( )|2d .( )Expression ( ) is called theFourier integralorFourier ( ) is called theinverse Fourier integralforf. The Plancherelidentity suggests that the Fourier Transform is a one-to-one norm preservingmap of the Hilbert spaceL2[ , ] onto itself (or to another copy of it-self).

6 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 second notation that we shall use isF[f]( ) =1 2 f(t)e i tdt=1 2 f( )( )F [g](t) =1 2 g( )ei td ( )Note that, formally,F [ f](t) = 2 f(t). The first notation is used moreoften in the engineering literature.

7 The second notation makes clear thatFandF are linear operators mappingL2[ , ] onto itself in one view, andFmapping thesignal spaceonto thefrequency spacewithF mapping thefrequency space onto the signal space in the other view. In this notation thePlancherel theorem takes the more symmetric form |f(t)|2dt= |F[f]( )|2d .Examples:41. The box function (or rectangular wave) (t) = 1 if < t < 12ift= 0otherwise.( )Then, since (t) is an even function ande i t= cos( t) +isin( t), wehave ( ) = 2 F[ ]( ) = (t)e i tdt= (t) cos( t)dt= cos( t)dt=2 sin( ) = 2 sinc .Thus sinc is the Fourier Transform of the box function.

8 The inverseFourier Transform is sinc( )ei td = (t),( )as follows from (??). Furthermore, we have | (t)|2dt= 2 and |sinc ( )|2d = 1from (??), so the Plancherel equality is verified in this case. Notethat the inverse Fourier Transform converged to the midpoint of thediscontinuity, just as for Fourier A truncated cosine (t) = cos 3tif < t < 12ift= , since the cosine is an even function, we have f( ) = 2 F[f]( ) = f(t)e i tdt= cos(3t) cos( t)dt=2 sin( )9 A truncated sine (t) ={sin 3tif t the sine is an odd function, we have f( ) = 2 F[f]( ) = f(t)e i tdt= i sin(3t) sin( t)dt= 6isin( )9 A triangular (t) = 1 +tif 1 t 0 1if 0 t 10otherwise.}

9 ( )Then, sincefis an even function, we have f( ) = 2 F[f]( ) = f(t)e i tdt= 2 10(1 t) cos( t)dt=2 2 cos : The Fourier transforms of the discontinuous functions above decayas1 for| | whereas the Fourier transforms of the continuous functionsdecay as1 2. The coefficients in the Fourier series of the analogous functionsdecay as1n,1n2, respectively, as|n| . of the Fourier transformRecall thatF[f]( ) =1 2 f(t)e i tdt=1 2 f( )F [g](t) =1 2 g( )ei td We list some properties of the Fourier Transform that will enable us to build arepertoire of transforms from a few basic examples. Suppose thatf,gbelongtoL1[ , ], , |f(t)|dt < with a similar statement forg.

10 We canstate the following (whose straightforward proofs are left to the reader) are linear operators. Fora,b Cwe haveF[af+bg] =aF[f] +bF[g],F [af+bg] =aF [f] +bF [g].2. Supposetnf(t) L1[ , ] for some positive integern. ThenF[tnf(t)]( ) =indnd n{F[f]( )}.3. Suppose nf( ) L1[ , ] for some positive integern. ThenF [ nf( )](t) =indndtn{F [f](t)}.4. Suppose thenth derivativef(n)(t) L1[ , ] and piecewise contin-uous for some positive integern, andfand the lower derivatives areall continuous in ( , ). ThenF[f(n)]( ) = (i )nF[f]( )}.5. Supposenth derivativef(n)( ) L1[ , ] for some positive integernand piecewise continuous for some positive integern, andfand thelower derivatives are all continuous in ( , ).