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Chapter 1

The Fourier Transform

1.1 Fourier transforms as integrals

There are several ways to dene the Fourier transform of a functionf:R! C. In this section, we dene it using an integral representation and state some basic uniqueness and inversion properties, without proof. Thereafter, we will consider the transform as being dened as a suitable limit of Fourier series, and will prove the results stated here. Denition 1Letf:R!R. The Fourier transform off2L1(R), denoted byF[f](:), is given by the integral:

F[f](x) :=1p2Z

1 1 f(t)exp(ixt)dt forx2Rfor which the integral exists. We have theDirichlet conditionfor inversion of Fourier integrals.

Theorem 1Letf:R!R. Suppose that (1)R1

1jfjdtconverges and (2)

in any nite interval,f,f0are piecewise continuous with at most nitely many maxima/minima/discontinuities. LetF=F[f]. Then iffis continuous at t2R, we have f(t) =1p2Z 1 1

F(x)exp(itx)dx:

This denition also makes sense for complex valuedfbut we stick here to real valued f 1 Moreover, iffis discontinuous att2Randf(t+0)andf(t0)denote the right and left limits offatt, then 12 [f(t+ 0) +f(t0)] =1p2Z 1 1

F(x)exp(itx)dx:

From the above, we deduce a uniqueness result:

Theorem 2Letf;g:R!Rbe continuous,f0;g0piecewise continuous. If

F[f](x) =F[g](x);8x

then f(t) =g(t);8t:

Proof: We have from inversion, easily that

f(t) =1p2Z 1 1

F[f](x)exp(itx)dx

1p2Z 1 1

F[g](x)exp(itx)dx

=g(t): 2 Example 1Find the Fourier transform off(t) = exp(jtj)and hence using inversion, deduce thatR1

0dx1+x2=2

andR1

0xsin(xt)1+x2dx=exp(t)2

; t >0.

SolutionWe write

F(x) =1p2Z

1 1 f(t)exp(ixt)dt 1p2 Z0 1 exp(t(1ix))dt+Z 1 0 exp(t(1 +ix)) r2

11 +x2:

Now by the inversion formula,

exp(jtj) =1p2Z 1 1

F(x)exp(ixt)dx

1 Z1

0exp(ixt) + exp(ixt)1 +x2dt

2 Z 1

0cos(xt)1 +x2dx:

2 Now this formula holds att= 0, so substitutingt= 0 into the above gives the rst required identity. Dierentiating with respect totas we may for t >0, gives the second required identity.2. Proceeding in a similar way as the above example, we can easily show that

F[exp(12

t2)](x) = exp(12 x2); x2R: We will discuss this example in more detail later in this chapter. We will also show that we can reinterpret Denition 1 to obtain the Fourier transform of any complex valuedf2L2(R), and that the Fourier transform is unitary on this space:

Theorem 3Iff;g2L2(R)thenF[f];F[g]2L2(R)and

Z 1 1 f(t)g(t)dt=Z 1 1

F[f](x)F[g](x)dx:

This is a result of fundamental importance for applications in signal process- ing.

1.2 The transform as a limit of Fourier series

We start by constructing the Fourier series (complex form) for functions on an interval [L;L]. The ON basis functions are e n(t) =1p2LeintL ; n= 0;1;; and a suciently smooth functionfof period 2Lcan be expanded as f(t) =1X n=1 12LZ L

Lf(x)einxL

dx e intL For purposes of motivation let us abandon periodicity and think of the func- tionsfas dierentiable everywhere, vanishing att=Land identically zero outside [L;L]. We rewrite this as f(t) =1X n=1eintL

12L^f(nL

which looks like a Riemann sum approximation to the integral f(t) =12Z 1

1^f()eitd(1.2.1)

3 to which it would converge asL! 1. (Indeed, we are partitioning the interval [L;L] into 2Lsubintervals, each with partition width 1=L.) Here, f() =Z 1 1 f(t)eitdt:(1.2.2)

Similarly the Parseval formula forfon [L;L],

Z L

Ljf(t)j2dt=1X

n=112Lj^f(nL )j2 goes in the limit asL! 1to thePlancherel identity 2Z 1 1 jf(t)j2dt=Z 1 1 j^f()j2d:(1.2.3) Expression (1.2.2) is called theFourier integralorFourier transformoff. Expression (1.2.1) is called theinverse Fourier integralforf. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert spaceL2[1;1] onto itself (or to another copy of it- self). We shall show that this is the case. Furthermore we shall show that the pointwise convergence properties of the inverse Fourier transform are somewhat similar to those of the Fourier series. Although we could make a rigorous justication of the the steps in the Riemann sum approximation above, we will follow a dierent course and treat the convergence in the mean and pointwise convergence issues separately.

A second notation that we shall use is

F[f]() =1p2Z

1 1 f(t)eitdt=1p2^f() (1.2.4) F [g](t) =1p2Z 1 1 g()eitd(1.2.5) Note that, formally,F[^f](t) =p2f(t). The rst notation is used more often in the engineering literature. The second notation makes clear thatF andFare linear operators mappingL2[1;1] onto itself in one view, and Fmapping thesignal spaceonto thefrequency spacewithFmapping the frequency space onto the signal space in the other view. In this notation the

Plancherel theorem takes the more symmetric form

Z 1 1 jf(t)j2dt=Z 1 1 jF[f]()j2d:

Examples:

4

1. The box function (or rectangular wave)

(t) =8 :1 if < t < 12 ift=

0 otherwise:(1.2.6)

Then, since (t) is an even function andeit= cos(t)+isin(t), we have () =p2F[]() =Z 1 1 (t)eitdt=Z 1 1 (t)cos(t)dt Z cos(t)dt=2sin() = 2sinc: Thus sincis the Fourier transform of the box function. The inverse

Fourier transform is

Z 1 1 sinc()eitd= (t);(1.2.7) as follows from (??). Furthermore, we have Z 1 1 j(t)j2dt= 2 and Z1 1 jsinc ()j2d= 1 from (??), so the Plancherel equality is veried in this case. Note that the inverse Fourier transform converged to the midpoint of the discontinuity, just as for Fourier series.

2. A truncated cosine wave.

f(t) =8 :cos3tif < t < 12 ift=

0 otherwise:

Then, since the cosine is an even function, we have f() =p2F[f]() =Z 1 1 f(t)eitdt=Z cos(3t)cos(t)dt

2sin()92:

5

3. A truncated sine wave.

f(t) =sin3tift

0 otherwise:

Since the sine is an odd function, we have

f() =p2F[f]() =Z 1 1 f(t)eitdt=iZ sin(3t)sin(t)dt

6isin()92:

4. A triangular wave.

f(t) =8 :1 +tif1t0

1 if 0t1

0 otherwise:(1.2.8)

Then, sincefis an even function, we have

f() =p2F[f]() =Z 1 1 f(t)eitdt= 2Z 1 0 (1t)cos(t)dt 22cos
2: NOTE: The Fourier transforms of the discontinuous functions above decay as 1 forjj ! 1whereas the Fourier transforms of the continuous functions decay as 1

2. The coecients in the Fourier series of the analogous functions

decay as 1n ,1n

2, respectively, asjnj ! 1.

1.2.1 Properties of the Fourier transform

Recall that

F[f]() =1p2Z

1 1 f(t)eitdt=1p2^f() F [g](t) =1p2Z 1 1 g()eitd We list some properties of the Fourier transform that will enable us to build a repertoire of transforms from a few basic examples. Suppose thatf;gbelong toL1[1;1], i.e.,R1

1jf(t)jdt <1with a similar statement forg. We can

state the following (whose straightforward proofs are left to the reader): 6

1.FandFare linear operators. Fora;b2Cwe have

F[af+bg] =aF[f] +bF[g];F[af+bg] =aF[f] +bF[g]:

2. Supposetnf(t)2L1[1;1] for some positive integern. Then

F[tnf(t)]() =indnd

nfF[f]()g:

3. Supposenf()2L1[1;1] for some positive integern. Then

F [nf()](t) =indndt nfF[f](t)g:

4. Suppose thenth derivativef(n)(t)2L1[1;1] and piecewise contin-

uous for some positive integern, andfand the lower derivatives are all continuous in (1;1). Then

F[f(n)]() = (i)nF[f]()g:

5. Supposenth derivativef(n)()2L1[1;1] for some positive integer

nand piecewise continuous for some positive integern, andfand the lower derivatives are all continuous in (1;1). Then F [f(n)](t) = (it)nF[f](t):

6. The Fourier transform of a translation by real numberais given by

F[f(ta)]() =eiaF[f]():

7. The Fourier transform of a scaling by positive numberbis given by

F[f(bt)]() =1b

F[f](b

8. The Fourier transform of a translated and scaled function is given by

F[f(bta)]() =1b

eia=bF[f](b

Examples

7 We want to compute the Fourier transform of the rectangular box func- tion with support on [c;d]:

R(t) =8

:1 ifc < t < d 12 ift=c;d

0 otherwise:

Recall that the box function

(t) =8 :1 if < t < 12 ift=

0 otherwise:

has the Fourier transform ^() = 2sinc. but we can obtainRfrom by rst translatingt!s=t(c+d)2 and then rescalings!2dcs:

R(t) = (2dctc+ddc):

R() =42dcei(c+d)=(dc)sinc(2dc):(1.2.9)

Furthermore, from (??) we can check that the inverse Fourier transform of^RisR, i.e.,F(F)R(t) =R(t).

Consider the truncated sine wave

f(t) =sin3tift

0 otherwise

with ^f() =6isin()92: Note that the derivativef0off(t) is just 3g(t) (except at 2 points) whereg(t) is the truncated cosine wave g(t) =8 :cos3tif < t < 12 ift=

0 otherwise:

We have computed

^g() =2sin()92: so 3^g() = (i)^f(), as predicted. Reversing the example above we can dierentiate the truncated cosine wave to get the truncated sine wave. The prediction for the Fourier transform doesn't work! Why not? 8

1.2.2 Fourier transform of a convolution

The following property of the Fourier transform is of particular importance in signal processing. Supposef;gbelong toL1[1;1]. Denition 2The convolution offandgis the functionfgdened by (fg)(t) =Z 1 1 f(tx)g(x)dx:

Note also that (fg)(t) =R1

1f(x)g(tx)dx, as can be shown by a change

of variable.

Lemma 1fg2L1[1;1]and

Z 1 1 jfg(t)jdt=Z 1 1 jf(x)jdxZ 1 1 jg(t)jdt:

Sketch of proof:

Z 1 1 jfg(t)jdt=Z 1 1 Z1 1 jf(x)g(tx)jdx dt Z 1 1 Z1 1 jg(tx)jdt jf(x)jdx=Z 1 1 jg(t)jdtZ 1 1 jf(x)jdx: 2

Theorem 4Leth=fg. Then

h() =^f()^g():

Sketch of proof:

h() =Z 1 1 fg(t)eitdt=Z 1 1 Z1 1 f(x)g(tx)dx e itdt Z 1 1 f(x)eixZ1 1 g(tx)ei(tx)dt dx=Z 1 1 f(x)eixdx^g() ^f()^g(): 2 9

1.3L2convergence of the Fourier transform

In this book our primary interest is in Fourier transforms of functions in the Hilbert spaceL2[1;1]. However, the formal denition of the Fourier integral transform,

F[f]() =1p2Z

1 1 f(t)eitdt(1.3.10) doesn't make sense for a generalf2L2[1;1]. Iff2L1[1;1] thenfis absolutely integrable and the integral (1.3.10) converges. However, there are square integrable functions that are not integrable. (Example:f(t) =11+jtj.)

How do we dene the transform for such functions?

We will proceed by deningFon a dense subspace off2L2[1;1] where the integral makes sense and then take Cauchy sequences of functions in the subspace to deneFon the closure. SinceFpreserves inner product, as we shall show, this simple procedure will be eective. First some comments on integrals ofL2functions. Iff;g2L2[1;1] then the integral (f;g) =R1

1f(t)g(t)dtnecessarily exists, whereas the inte-

gral (1.3.10) may not, because the exponentialeitis not an element ofL2. However, the integral off2L2over any nite interval, say [N;N] does exist. Indeed forNa positive integer, let[N;N]be the indicator function for that interval: [N;N](t) =1 ifNtN

0 otherwise.(1.3.11)

Then[N;N]2L2[1;1] soRN

Nf(t)dtexists because

Z N Njf(t)jdt=j(jfj;[N;N])j jjfjjL2jj[N;N]jjL2=jjfjjL2p2N <1 Now the space of step functions is dense inL2[1;1], so we can nd a convergent sequence of step functionsfsngsuch that limn!1jjfsnjjL2= 0. Note that the sequence of functionsffN=f[N;N]gconverges tofpointwise asN! 1and eachfN2L2\L1. Lemma 2ffNgis a Cauchy sequence in the norm ofL2[1;1]andlimn!1jjf f njjL2= 0. Proof: Given >0 there is step functionsMsuch thatjjfsMjj<2 ChooseNso large that the support ofsMis contained in [N;N], i.e., 10 s

M(t)[N;N](t) =sM(t) for allt. ThenjjsMfNjj2=RN

NjsM(t)f(t)j2dtR1

1jsM(t)f(t)j2dt=jjsMfjj2, so

jjffNjjjj(fsM)+(sMfN)jj jjfsMjj+jjsMfNjj 2jjfsMjj< : 2 Here we will study the linear mappingF:L2[1;1]!^L2[1;1] from the signal space to the frequency space. We will show that the mapping is unitary, i.e., it preserves the inner product and is 1-1 and onto. Moreover, the mapF:^L2[1;1]!L2[1;1] is also a unitary mapping and is the inverse ofF: F

F=IL2;FF=I^L2

whereIL2;I^L2are the identity operators onL2and^L2, respectively. We know that the space of step functions is dense inL2. Hence to show that Fpreserves inner product, it is enough to verify this fact for step functions and then go to the limit. Once we have done this, we can deneFffor any f2L2[1;1]. Indeed, iffsngis a Cauchy sequence of step functions such that lim n!1jjfsnjjL2= 0, thenfFsngis also a Cauchy sequence (indeed, jjsnsmjj=jjFsnFsmjj) so we can deneFfbyFf= limn!1Fsn. Thequotesdbs_dbs10.pdfusesText_16