Chapter 4Fourier TransformIn this chapter, we introduce a special kind of linear transformations called calledFourier transform to solve linear PDEs on unbounded domains. This methodis associated to the French physicist and mathematician Joseph Fourier whoworked extensively on heat equations. As we will see, this transform reduces a PDEin (t; x)- domain to an ODE in (t; !)-domain. This transformation is very similarto Laplace transform we studied in the �rst volume of this book.

4.1 Basic de�nitionsWe �rst introduce the set of functions we work with in this chapter.

De�nition 4.1. A function f(x);¡1<x<1 is called piecewise continuous if

i. f(x) is continuous everywhere in any interval (¡R; R) except possibly at�nitely many points,

ii. and if z is a discontinuity point of f, then both left and right limits exist atz, that is,

f(z+) := limx!z+

f(x)<1; f(z¡) := limx!z¡

f(x)<1:

For example, the function

f(x)=

�1 x2Q0 x2Qc ; (4.1)

is not piecewise continuous (it is discontinuous everywhere). The function

f(x)=

(1

xx=/ 0

0 x=0;

is not piecewise continuous because f(0+); f(0¡) do not exist. The function

f(x)=

�sin(�/x) x=/ 00 x=0

; (4.2)

is not piecewise continuous because f(0+); f(0¡) do not exist. The function

f(x)=

�x sin(�/x) x=/ 00 x=0

;

1

is continuous and thus piecewise continuous.

De�nition 4.2. A function f(x); ¡1 < x < 1 is called piecewise continuouslydi�erentiable if f 0(x) is piecewise continuous function.

For example, the function f(x) = jxj is piecewise continuously di�erentiablefunction, in fact f is continuously di�erentiable (that is called C1) everywhere exceptat x=0. Notice that the right and left derivatives of f at x=0 exist. The function

g(x)=

(x2 sin(�/x) x=/ 00 x=0

;

is di�erentiable everywhere, however, g 0(x)=2x sin(1/x)¡cos(1/x) and thus g 0(0+)and g 0(0¡) do not exist. Hence, g(x) is not piecewise continuously di�erentiable. Itis simply veri�ed that the function

h(x)=

(x3 sin(�/x) x=/ 00 x=0

is continuously di�erentiable everywhere. The veri�cation s left as a simple exerciseto the reader.

De�nition 4.3. A function f(x);¡1<x<1 is called integrable if the followingrelation holds Z

¡1

1jf(x)j dx<1: (4.3)

A function f(x); ¡1 < x < 1 is called admissible if it is piecewise continuouslydi�erentiable and integrable.

The Fourier transform is usually de�ned for admissible functions, and for thisreason we usually assume in this chapter that all functions are admissible. However,as we will see later, the de�nition can be extended to a wider class of functions thanadmissible function.

De�nition 4.4. Suppose f(x) is a function (not necessarily admissible) de�nedin (¡1;1). Its Fourier transform Fff g that we sometimes denote by f(!), isde�ned by the following integral

F(f) := f(!)=

Z¡1

1f(x) e¡i!x dx; (4.4)

as long as the above integral converges.

Note that integral (4.4) is improper, and thus is not integrable in general. Inorder to make the de�nition more precise, we make the following important conven-tion Z

¡1

1f(x) e¡i!xdx= lim

R!1

Z¡R

R

f(x) e¡i!xdx: (4.5)

2 Fourier Transform

The integral (4.4) may diverge even if f is bounded. The following proposition givesa su�cient condition for the convergence of the integral (and it is important to notethat the condition is su�cient and not necessary by no mean).

Proposition 4.1. If a function f(x) is admissible then f (!) exists, and continuousfor all ! 2R, and furthermore

max!jf(!)j<1:

Proof. Since jf(x) e¡i!nxj= jf(x)j, (4.4) implies

jf(!)j �Z¡1

1jf(x)j dx<1;

and thus f(!) converges for all !, and also

sup!

jf (!)�Z¡1

1jf(x)j dx<1:

Let !n be a sequence and !n!!, then

limn!1

f(!n)= limn!1

Z¡1

1f(x) e¡i!nxdx:

Again since jf(x) e¡i!nxj= jf(x)j, and f is integrable, we can pass the limit insidethe integral due to the dominant convergence theorem (see the appendix to thisbook), and write

limn!1

f (!n)=

Z¡1

1f(x) lim

n!1e¡i!nxdx=

Z¡1

1f(x) e¡i!xdx= f(!);

and this completes the proof. �

The following de�nition gives an inverse relation to the Fourier transform.

De�nition 4.5. Suppose f(!) is a function de�ned on ! 2 (¡1;1). The inverseFourier transform of f is de�ned by the following integral

F¡1(f)= 12�

Z¡1

1f(!) ei!xd!; (4.6)

as long as the integral exists. The integral is understood in the following senseZ¡1

1f(!) ei!xd!= lim

R!1

Z¡R

R

f (!) ei!xd!: (4.7)

What is the relation between F and F¡1? The following theorem answers thatquestion. For the proof, see the appendix of this chapter.

Theorem 4.1. Assume that f(x) is an admissible function with the transform f(!).Then we have

F¡1(f )= f(x+)+ f(x¡)2

; (4.8)

4.1 Basic definitions 3

where f(x+) and f(x¡) are the right and left limits of f at x respectively.

4.2 Fourier transform of basic functionsAs it is common in physics, ! is usually interpreted as the frequency (or associated toit through the relation !=2�f) and therefore, f(!) provides us with the distributionof frequency components embedded in a function f(x). The best way to understandthis notion very well, is to study the transform of some basic functions. It is alsohelpful to note that f(x) is just a distribution of values of f in x-domain, and thatthis distribution is uniquely tied to the distribution of f over !-domain, that is thedomain of frequencies in the function f . Accordingly, we can see both f(x); f (!)representing same thing, one in x-domain and another in !-domain, with the relation

Fff g= f ;F¡1ff g= f:

4.2.1 A pulse functionLet pa(x) denote the pulse

pa(x)=

�1 ¡a<x<a0 otherwise

:

Clearly, pa is an admissible function, with the transform

pa(!)=

Z¡a

a

e¡i!xdx=2sin (a!)

!:

The function sin(x)x

is called sinc function and is denoted by snc(x). Thus, onecan write pa(!) = 2a snc(a!). The �gure (4.1) shows pa(!) in the interval (¡4�; 4�) for a few values of a. As it is observed from the �gure, ! extends in(¡1; 1) regardless of the fact that pa(x) is zero outside the interval [¡a; a].This is due to the fact that pa(x) changes suddenly at x = �a and thus bringshigh frequencies in play. Notice also that pa(!) looks narrower (containing lowerfrequencies and concentrating around !=0) for larger values of a. This fact conformsour expectation since if p(a) is smoother, then lower frequencies are more importantin pa(!). In other word, since for a> 1, the function pa(x) looks more smooth thanp1(x), the frequency distribution in pa(!) concentrates around ! = 0 compare top1(!).

−4π −2π 2π 4πω

1

2

p1(ω)

−4π−3π−2π −π π 2π 3π 4πω

1

2

3

4

p2(ω)

−4π−3π−2π −π π 2π 3π 4πω

1

2

3

4

5

6

p3(ω)

Figure 4.1.

4 Fourier Transform

4.2.2 Exponential functionThe function f(x)= e¡t jxj; t > 0 is admissible and its transform is

f(!)=

ZR

e¡t jxj e¡i!x dx=

Z¡1

0

etx e¡i!xdx+

Z0

1e¡tx e¡i!xdx=

2tt2+!2

:

Since f(x) is admissible and continuous, we obtain

e¡t jxj=F¡1�

2tt2+!2

�=t�

Z¡1

1 ei!x

t2+!2d!:

If we change x to ¡! and ! to x, we reach

e¡t j! j=t�

Z¡1

1 e¡i!x

t2+x2dx=

t�F�

1t2+ x2

�;

and thus

F�

1t2+ x2

�=�te¡t j! j:

4.2.3 The Gaussian functionLet gt(x)= e¡tx

2for t > 0. This function is called Gaussian. The transform is

gt(!)=

Z¡1

1e¡tx

2e¡i!xdx=

Z¡1

1e¡t

�x2+

i!

tx�dx= e

¡!2

4t

Z¡1

1e¡t

�x+

i!

2t

�2dx:

According to the following identityZ¡1

1e¡tx

2dx=

�t

r;

we derive

gt(!)=�t

re¡!2

4t : (4.9)

The left �gure in (4.2) shows gt(!) for t=1/4 and t=4. Observe again that gt(!)for t= 4 has wider range of frequencies than gt(!) for t= 1/4. Compare this factwith the shape of gt(x) shown at the right. Notice that g4(x) is wider in x-domainthat g1/4(x) while g4(!) is narrower than g1/4(!). For t=

1

2we have the symmetry

g1/2= 2�p

g1/2:

−10 −5 5 10ω

1

2

3

g(ω)

F{e−x

2/4}

F{e−4x

2}

−2 −1 1 2x

0. 5

1. 0

f(x)

e−x2/4

e−4x2

g(x)

Figure 4.2.

4.2 Fourier transform of basic functions 5

4.2.4 Dirac delta functionThe concept of Dirac delta function is explained in great detail in the �rst volumeof this book. Remember that the delta function �(x) has the following propertyZ

¡1

1f(x) �(x¡x0) dx= f(x0);

for any function f that is continuous at x0. Note that � is not a function in usualsense due to the following relation

lim"!0

Z¡"

"

�(x) dx=1!

Remember that any usual function f , the following property holds

lim"!0

Z¡"

"

f(x) dx=0:

For this reason, � is called a generalized function. Working with generalized functionslike �(x) should be taken with extreme cautious since they show strange behaviorsometimes. The Fourier transform of � is de�ned as follows

Ff�g=Z¡1

1�(x) e¡i!xdx= e¡i!xjx=0=1:

For the inverse, we write

F¡1f1g= 12�

limR!1

Z¡R

R

ei!xd!= limR!1

sin(Rx)�x

:

It is left as an exercise to the reader to show that the sequence

�n(x)=sin(nx)�x

;

is a delta sequence function, that is,

limn!1

Z¡1

1f(x) �n(x) dx= f(0);

for arbitrary bounded and continuous function f(x). By the above justi�cation, wewrite

F¡1f1g= �(x):

4.2.5 Sine and cosine functionsAnother class of non-admissible functions with the Fourier transform is sine andcosine functions which is de�ned by the aid of the Diract delta function. First notethat we can write

F¡1f�(!¡!0)g=12�

Z¡1

1�(!¡!0) ei!xd!=

12�ei!0x:

6 Fourier Transform

According to the Euler formula ei�= cos(�)+ i sin(�), we can write

F¡1f�(!¡!0)+ �(!+!0)g=ei!0x+ e¡i!0x

2�=1�cos(!0x);

and therefore we write

Ffcos(!0x)g=� [�(!¡!0)+ �(!+!0)]: (4.10)

The geometrical interpretation of the above result is as follows. In fact, cos(!0x)is a periodic function with only one frequency component != !0, and this is whatformula (4.10) tells us. Even though, negative frequencies do not make sense fromphysical point of view, it is necessary in mathematics to keep formulation consistent.In fact, since, cos(!0x) is an even function, that is cos(!0x)=cos(¡!0x), the negativecomponent of frequency ¡!0 shows itself in the !-domain.

!0¡!0

��(!¡!0)��(!+!0)

!

Figure 4.3. Fourier transform of cos(!0x)

The transform of sine function sin(!0x) is de�ned similarly by the followingrelation

Ffsin(!0x)g=�i[�(!¡!0)¡ �(!+!0)]:

4.3 Energy and uncertainty principleWe discuss two notions related to the Fourier transform. Although, these notionshelps us to have a better understanding of the transform itself, the reader can skipthis section at the �rst reading.

4.3.1 Energy and Plancherel theoremIn physics and engineering, the energy of a function f(x) is de�ned by the followingintegral

E(f)=

ZR

jf(x)j2dx;

as long as the integral in bounded, that is, E(f)<1. Here we used for simplicitythe following notation Z

R

�Z¡1

1

4.3 Energy and uncertainty principle 7

The Plancherel theorem states a relationship between the energy of f(x) andthe energy of f(!).

Theorem 4.2. Assume that the function f is admissible and square integrable, thatis, E(f)<1, then the following relation holds

E(f)= 2�E(f): (4.11)

As an example, for f(x)= e¡jxj, we have E(f)= 1, and thus since f (!)= 2

1+!2,

we have

E

�2

1+!2

�=4

ZR

d!(1+!2)2

=2�:

It is simply seen that the 95 percent of the total energy of f(!) is distributed inthe frequency band ¡1.838� ! � 1.838; see the �gure (4.4). This cut o� value isusually used in electrical engineering to reconstruct an approximation of f(x).

−4 −3 −2 −1 1 2 3 4ω

1

2

f(ω)

Figure 4.4.

If we de�ne

f0.95(!)=

(f(!) ! 2 [¡1.838; 1.838]0 otherwise

;

then f0.95(x) is de�ned by the relation

f0.95(x)=12�

Z¡1.838

1.838

f(!) ei!xd!:

The result is shown in the �gure (4.5) along with the reconstruction based on the0.99 cut o� (for ¡3.373 � ! � 3.373). Observe that more frequency band, moreaccurate is the reconstruction of f(x). This method is extremely important in digitalcommunication and signal processing where the Fourier transfer of an electricalsignal is transmitted along communication lines instead of the signal itself. Forthis reason, one has to reconstructed the original signal at the destination from itsFourier transform.

8 Fourier Transform

−3 −2 −1 1 2 3

0. 2

0. 4

0. 6

0. 8

1. 0e−x

f0. 99

f0. 95

Figure 4.5. Reconstruction of f(x)= e¡x

4.3.2 Uncertainty principleHere, let us introduce one important result of the above property which is calleduncertainly principle. First, let us de�ned a measure for the dispersion of a functionf around x=0 (similarly for the dispersion of f around !=0). This is de�ned bythe following formula

D(f)=

�ZR

jxf(x)j2�1/2

/

�ZR

jf(x)j2�1/2

;

provided that both integrals exist.

Theorem 4.3. As long as D(f) is bounded, the following inequality that is calledthe uncertainty principle holds

D(f)D(f )� 12:

We can give an interpretation of the above theorem by the aid of informationtheory. According to this theory, the more certainty of f(x) (more concentrationaround x=0) leads to less certainty of f (wider around !=0) and vice versa. Thisis the reason why the inequality is called uncertainty principle. For a simple proofof the theorem see the problem set.

4.4 Properties of Fourier transform

The transform enjoys some pretty properties similar to of Laplace transform westudied in the �rst volume of the book. We need these properties when we studylinear PDEs on unbounded domains.

4.4.1 LinearityFor any a; b2R, the transform is linear in the following sense

Ffaf + bgg= a f(!)+ b g(!);

4.4 Properties of Fourier transform 9

as long as f and g exist. If f ; g are admissible functions, then it is simply veri�edthat af + bg is admissible and thus Ffaf + bgg exists.

4.4.2 Frequency shift

If f(!) is the transform of f(x), we have

Fff(x) cos(!0x)g=12

ZR

f(x)ei!0x e¡i!x+12

ZR

f(x)e¡i!0x e¡i!x:

The �rst integral is just f (!¡!0) and the second one is f (!+!0), and �nally

Fff(x) cos(!0x)g=12f(!¡!0)+

12f (!+!0):

Let us see how this property helps us to calculate some integrals. Consider thefollowing integral

I =

Z0

1cos(!0x)1+x2

dx:

We can write

2I = lim!!0

F�cos(!0x)1+ x2

�;

and since

F�

11+x2

�= �e¡j! j;

we obtain

F�cos(!0x)1+x2

�=�2fe¡j!¡!0j+ e¡j!+!0jg:

Letting !! 0, we obtain I =�

2e¡!0.

Perhaps the most important application of the frequency shift in engineering isthe frequency modulation of digital communication. For a simple presentation of theapplication, suppose that n users are simultaneously using a common transmittingline as shown in �gure (4.6)

U1

Vn

V2

V1

U2

Un

U3 V3

center 2center 1

Figure 4.6.

10 Fourier Transform

Assume that the maximum band width dedicated to all users is !�. All signalsare collected at a local center (the center 1) and are transmitted along a high bandwidth line (the bold line shown in the �gure). Since all signals are mixed in timedomain, we need to separate them at the destination (the local center 2) and sendeach signal to its pair. To do this, each signal is multiplied by cos(k!0t) for some!0 � !� at center-1, and is transmitted along the line. Note that their Fouriertransform are separated in !-domain as shown in �gure (4.7). At the destination,these signals are �ltered by appropriate �lters and send to their pairs.

!

4!03!02!0!0

Figure 4.7.

4.4.3 Shift in x

There is a beautiful duality between shift in x-domain and !-domain. In face, wehave the following relation

Fff(x¡x0)g=ZR

f(x¡x0) e¡i!x dx= e¡i!x0ZR

f(y) e¡i!y dy= e¡i!x0 f(!):

This property is widely used in the analysis of linear time invariant (LTI) systems.

4.4.4 Di�erentiation and integrationIf f(x) is admissible and di�erentiable, then we have

Fff 0g= i! f(!); (4.12)

as long as f(!) exists. The property is simply proved by the formal de�nition (4.4).Similarly, if F (x), the anti-derivative of f , that is,

F (x)=

Z0

x

f(t) dt; (4.13)

is admissible, then by the relation FfF 0g= f (!) and (4.12) we have

FfF g= 1i!

f(!): (4.14)

The formula (4.14) should be used with cautious for non-admissible functions. Forexample, we have Z

¡1

x

�(x)dx=

�1 x> 00 x< 0

:=u(x):

Thus, one may be tempted to write

Ffu(x)g= 1i!Ff�g= 1

i!;

4.4 Properties of Fourier transform 11

that is not correct. In fact, according to the inverse transform, we we should have

F¡1fu(!)g(0)= 12[u(0+)+u(0¡)] =

12;

however, we have

F¡1�1i!

�(0)=

12�i

limR!1

Z¡R

R d!!=0:

For this reason, we take the transform of u(x) as follows

u(!)=1i!+ ��(!):

Note that by (4.8), we have

F¡1�1i!+ ��(!)

�(0)=

12�

limR!1

Z¡R

R

��(!) d!=12:

It is left as an exercise to reader to check the following property

F¡1�1i!+��(!)

�(x)=

�1 x> 00 x< 0

:

4.4.5 Multiplication by x

Assume that f(x) is admissible, then we have

Ffxf(x)g= idd!

f (!);

as long as f is di�erentiable. The justi�cation is as follows

dd!

f(!)=dd!

ZR

f(x) e¡i!xdx=¡iZR

xf(x) e¡i!xdx=¡iFfxf(x)g:

Example 4.1. Let us calculate the transform of g(x)= x

1+x2. As we saw before, we

have

F�

11+x2

�=�e¡j! j:

Therefore, we have

Fn

x1+ x2

o= i�e¡j! j sgn(!):

Notice that the transform of the function f(x)= 1

1+x2is not di�erentiable at !=0.

The following theorem gives a su�cient condition that the Fourier transformof a function is continuously di�erentiable. The proof is given in the appendix tothis chapter.

Theorem 4.4. Assume that f is admissible function and decays exponentially at�1, that is,

jf(x)j<�e¡� jxj; �; � > 0

12 Fourier Transform

for jxj>M for some su�ciently large M. Then f(!) is continuously di�erentiable.

4.4.6 Expansion and contractionConsider the function f(ax) and assume that a> 0 is a constant. As we know, thefactor a causes a contraction or expansion on f(x) based on the value of a (for a>1the contraction and for 0<a< 1, expansion). The FT of f(ax) is

Fff(ax)g=Z¡1

1f(ax) e¡i!x dx=

1a

Z¡1

1f(y) e¡i!y/a dy=

1af�!a

�:

If a< 0, we have

Fff(ax)g=¡1af�!a

�;

and thus for a=/ 0, we obtain

Fff(ax)g= 1jaj f

�!a

�:

Therefore, if f(x) is contracted by the factor jaj>1, its Fourier transform expandsby the factor 1

aand vice versa. We have already seen a few important examples of

this property, in particular the Gaussian functions gt(x)= e¡tx2.

4.5 Convolution and trigonometric transform

We discuss two important notions related to the Fourier transform. We extensivelyuse them in solving linear PDEs in next chapter.

4.5.1 ConvolutionIn the �rst volume of this book, we discussed in great detail the concept of convo-lution and its importance to the study of linear time invariant (LTI) systems. Thereader is referred to the chapter of Laplace transform of the �rst volume. Recallthat the convolution of two functions f ; g is de�ned by the following formula

(f � g)(x)=ZR

f(y) g(x¡ y)dy;

as long as the integral exist. It is simply seen that if f ; g are admissible then theconvolution exists. In fact, since f ; g are bounded (why?), we can write��������Z

R

f(y) g(x¡ y)dy

���������max jg jZR

jf(y)j dy <1:

On the other hand, if f ; g are admissible, their transform exists. We have thefollowing important property

Fff � gg= f(!) g(!): (4.15)

4.5 Convolution and trigonometric transform 13

The convolution formula is simply proved by the following justi�cation. Accordingto the Fubini theorem, we can change the order of the integrals and write

Fff � gg=ZR

f(y)

�ZR

g(x¡ y) e¡i!x dx

�dy:

Taking z= x¡ y, we obtainZR

g(x¡ y) e¡i!xdx=e¡i!yZR

g(z) e¡i!z dz;

and thus

Fff � gg=ZR

f(y)e¡i!y dy

ZR

g(z) e¡i!z dz= f(!)g(!):

We accept the formula (4.15) for non-admissible functions, like Dirac delta func-tions as well.

4.5.2 Sine and cosine transformConsider a function f(x) for x> 0. To de�ne the transform of f(x), we can extendit on (¡1;1) in two di�erent ways: a) as an even extension fev(x), b) as an oddextension fodd(x). The de�nition of the even extension is as follows

fev(x)=

�f(x) x> 0f(¡x) x< 0

;

and the odd extension is de�ned as

fodd(x)=

�f(x) x> 0¡f(¡x) x< 0

:

For the fev(x), we have

Fffevg=Z¡1

1fev(x) fcos(!x)+ i sin(!x)g dx;

where we used the Euler formula for ei!x. It is simply seen thatZ¡R

R

fev(x) sin(!x) dx=0; 8R> 0;

and therefore

Fffevg=Z¡1

1fev(x) cos(!x) dx=2

Z0

1f(x) cos(!x) dx:

The integral in the right hand side of the above formula is called the cosine Fourieintegral

fc(!)=

Z0

1f(x) cos(!x) dx:

Since fc(!) is even with respect to !, we can write

f(x)=12�

Z¡1

1Fffevg ei!xd!=

1�

Z¡1

1fc(!) cos(!x) d!:

14 Fourier Transform

Similarly for the odd extension fodd(x), we have

Fffddg=¡iZ¡1

1fodd(x) sin(!x) dx=¡2i

Z0

1f(x) sin(!x) dx;

and we de�ne the sine Fourier integral of f(x) as

fs(!)=

Z0

1f(x) sin(!x) dx:

It is simply seen that fs(¡!)=¡fs(!), and thus

f(x)=12�

Z¡1

1Fffoddg ei!xd!=

1�

Z¡1

1fs(!) sin(!x) d!:

In summary, we have the following relation for functions f(x) de�ned on [0;1)

fc(!)=

Z0

1f(x) cos(!x) dx; f(x)=

1�

Z¡1

1fc(!) cos(!x) d!; forx> 0: (4.16)

fs(!)=

Z0

1f(x) sin(!x) dx; f(x)=

1�

Z¡1

1fc(!) sin(!x) d!; forx> 0: (4.17)

Overall, for a function f(x); x � 0 (or x � 0), we can de�ne two di�erent transfor-mations: sine and cosine transformations. Each of them are appropriate for di�erentapplications.

Example 4.2. Consider f(x)= e¡x for x� 0. The sine transform of f is

fs(!)=

Z0

1e¡x sin(!x) dx=

!1+!2

:

It is simply seen that

fsn

x1+x2

o=�2e¡!; ! > 0:

4.6 Higher dimensional transform

We can de�ne the transform for multi-variable functions as well. Let f :R2!R bea well de�ned function. The 2D transform of f is de�ned by the following expression

f(!x; !y)=

ZR2

f(x; y) e¡i!xx e¡i!yy dxdy;

as long as the integral exists. Here !x; !y are just tow parameters (you can interpretthem as the frequency components of f with respect to x; y respectively). Similarly,the inverse transform is de�ned by the formula

f(x; y)=F¡1ff g= 14�2

ZR2

f(!x; !y) ei!xx ei!yy d!xd!y:

4.6 Higher dimensional transform 15

Example 4.3. Consider the Gaussian function g(x; y)= e¡(x2+y2)/2. The transform

of g is

g(!x; !y)=

ZR2

e¡(x2+y2)/2 e¡i!xx e¡i!yy dxdy=�Z

R

e¡x2/2 e¡i!xxdx

��ZR

e¡y2/2 e¡i!yy dy

�=2�e¡(!x

2+!y2)/2:

Remark 4.1. If f(x; y) is odd with respect to x then

f (!x; !y)=

ZR2

f(x; y) e¡i!xx e¡i!yy=¡iZR2

f(x; y) sin(!xx) e¡i!yy;

and thus

f (¡!x; !y)=¡f(!x; !y):

Similarly if f is odd with respect to y then f(!x;¡!y) = ¡f(!x; !y). Also if f iseven with respect to x then f(¡!x; !y)= f (!x; !y).

Properties of the 1D transform hold for higher transform as well. For example,if f is smooth enough, we have

Ff@xf g= i!xf(!x; !y); Ff@xyf g=¡!x!yf(!x; !y):

Generally speaking, if f :Rn!R is smooth enough and moreoverZRn

jf(x1; :::; xn)j dV <1;

then f is de�ned by the following integral

f(!1; :::; !n)=

ZRn

f(x1; :::; xn) e¡i!1x1 ���e¡i!nxndV ;and also

f(x1; :::; xn)=1

(2�)n

ZRn

f(!1; :::; !n) ei!1x1 ���ei!nxn d!:

We also need the following proposition for our subsequent study of linear partialdi�erential equations.

Proposition 4.2. Assume that u(t; x) is continuously di�erentiable with respect tot and has the transform with respect to x

u(t; !)=

ZR

u(t; x) e¡i!xdx (4.18)

If @tu(t;x) is admissible in R, then u(t;!) is continuously di�erentiable with respectto t and we have

@tu(t; !)=

ZR

@tu(t; x) e¡i!xdx:

16 Fourier Transform

Proof. For arbitrary h2R, we have

u(t+h; !)¡ u(t; !)h

=

ZR

u(t+h; x)¡u(t; x)h

e¡i!xdx=

ZR

@tu(� ; x) e¡i!xdx;

where � is in the segment <t; t+h> by the intermediate value theorem. Therefore,

limh!0

u(t+ h; !)¡ u(t; !)h

= limh!0

ZR

@tu(� ; x) e¡i!xdx=

ZR

limh!0

@tu(� ; x) e¡i!x dx:

We passed the limit inside integral due to the dominant convergence theorem (seethe appendix of the book). Since @tu is continuous, we obtain

@tu(t; !) := limh!0

u(t+h; !)¡ u(t; !)h

=

ZR

@tu(t; x) ei!xdx:

The continuity of @tu(t; !) with respect to t follows from the continuity and theintegrability of @tu(t; x) and the dominant convergence theorem. �

4.7 Reading: Nyquist-Shanon rate

Let f(t) be a an electrical signal to be transmitted through a long transmission line.In real applications, the sampled signal (the digital signal) is transmitted instead off(t). At the destination, the whole signal f(t) is recovered by the aid of sampledsignal. The question is this: What should the sampling rate be that the whole signalis re-constructable without any loss? The answer is given by Nyquist.

Assume that the signal f(t) has a limited frequency band, say, ¡!0�!�!0. AsNyquist and later, Shanon have shown, the original function can be reconstructedby the sampled signal if the rate of the sampling is equal or greater than !0

�. This is

an important result in the digital signal transmission. From the information pointof view, sampling a signal with the rate of twice of highest frequency mode of thesignal, preserves all information in that signal.

By Fourier series, we know that f has the representation

f(!)=X

n=¡1

1�!0f(n�/!0) e

¡in�!/!0; (4.19)

where f(n�/!0) are the data sampled at tn=n�/!0. Remember that !0=2�f0, forthe frequency f0 and thus tn=n/2f0. Thus, in order to reconstruct f(!), one needsonly the set of values ff(tn)gn=¡11 . The proof of (4.19) is based on the notion ofFourier series that we sill study in detail in the chapter.

Let us see how to reconstruct f(t) from its Fourier series. As we know, theFourier series of f(t) is periodic; see the �gure (4.8). In order to �lter the principalpart f(!) (shown with the blue) from the Fourier series and to reconstruct f(t),one has to �lter the series by multiplying it by the pulse function

P!0(!)=

�1 ! 2 (¡!0; !0)0 otherwise

:

4.7 Reading: Nyquist-Shanon rate 17

!0

!

f(!)P!0

¡!0

Figure 4.8.

Therefore, the original function f(t) can be reconstructed by the inverse trans-form of g(!). In fact, we have

f(t)=F¡1ff(!)P!0(!)g=X

n=¡1

112!0

f

�n�!0

�Z¡!0

!0

e¡in�!/!0 ei!t d!=

Xn=¡1

1

f

�n�!0

�sin(!0t¡n�)!0t¡n�

:

Example 4.4. Consider the function

f(t)=2(sin(t)¡ t cos(t))

�t3:

The Fourier transform of f is

f(!)=

(1¡!2 ¡1�!� 10 otherwise

:

For !0=1, the sequences

F = ff(n�)gn=¡11 ;

captures all information in f(t). In fact, the reconstruction is done by the series

f (t)=X

n=¡1

1

f(n�)sin(t¡n�)t¡n� ;

which converges fast to the original function. In the �gure (4.9), the original functionis shown with only 3 terms of the series.

−20 −10 10 20t

0. 1

0. 2

f(t)

Figure 4.9.

18 Fourier Transform

ProblemsProblem 4.1. Use Fourier theorem and conclude the following relationsZ

¡1

1sinc(x) dx=�;Ffsinc(x)g= � p(!);

where p(x) is the pulse function in [¡1; 1]. This exercise shows that a low band �lter in !-domain corresponds to a sinc function in x-domain.

Problem 4.2. Find the Fourier transform of following functions

a) f(x) =1

a2+ x2

b) f(x) = ei!0x for jxj< 1 and f(x)= 0 in jxj> 1

c) f(x) =(e¡x x> 00 x< 0

d) f(x) =(xe¡x x> 00 x< 0

e) f(x) =�1 a<x< b0 otherwise

Problem 4.3. Find the inverse transform of the following functions

a) f(!)=1

!2¡ 2!+2

b) f(!)= e¡j! j¡i!

Problem 4.4. Verify the following relation

Ffcos(x2)g= �p

cos�!2¡�4

�:

Find a similar relation for Ffsin(x2)g.

Problem 4.5. Find cosine transform of the following functions

a) f(x) = e¡x

b) f(x) =xe¡x

Problem 4.6. Assume that f (k) are admissible for k=0; :::; n. Find a formula for Fff (n)g.

Problem 4.7. Show the following relation

F¡1�1i!+ ��(!)

�(x)=

�1 x> 00 x< 0

:

Problem 4.8.

a) Show the following relation for t> 0

F

(e¡tjxj

jxjp )

= 2�p

t+ t2+!2pt2+!2

!1/2:

b) Conclude the following relation

F(

1

jxjp )

=2�j! j

r:

c) Find the following integral Z0

1cosxx

p dx:

Problem 4.9. Use the transform of the function f(x)= e¡jxj to show

�e=

ZR

cosx1+x2

dx:

4.7 Reading: Nyquist-Shanon rate 19

Also prove

lims!1

ZR

cos (sx)1+x2

dx=0:

Problem 4.10. Consider the function g(x) = 2p

e¡2jxj.

a) Find the energy of g(x) and verify the Plancherel identity.

b) Find the frequency band of the 95 percent cut o� energy of g(x).

c) Draw g(x) and the reconstructed function based on the band you found in the part (b).

Problem 4.11. We give a proof for the uncertainty principle.RRx2jf(x)j2RRjf(x)j2

RR!2jf(!)j2RRjf(!)j2

� 14:

a) Without loss of generality, assume kf k = 1 (why is this plausible?) Then due toPlancherel identity (4.11), kf k= 2�

p. On the other hand, we have Fff 0g= i! f(!)

and thus ZR

!2f(!)=

ZR

jf 0(x)j2:We show �Z

R

x2jf(x)j2�1/2�Z

R

jf 0(x)j2�1/2

� 2�p

2:

b) Use Cauchy-Schwarz inequality to show�ZR

x2jf(x)j2�1/2�Z

R

jf 0(x)j2�1/2

�ZR

xf(x) f 0(x):

and the equality hold only if xf(x)=�f 0(x) for some �2R.

c) Show that ZR

xf(x) f 0(x)� 12

and conclude the uncertainty principle. Show that the equality holds only if f(x)=ceax2

where c; a are some constants.

Problem 4.12. Suppose f is a Fourier transformable function and f is its transform.

a) If f is even function then f (¡!)= f(!).

b) If f is an odd function then f(¡!)=¡f(!).Problem 4.13.

a) Verify that the tent function

f(x)=

8<: x+1 ¡1�x� 01¡x 0�x� 10 otherwise

is the convolution of the pulse function

p(x)=

�1 ¡1/2�x� 1/20 otherwise

:

b) Use convolution to �nd the transform of f(x).

Problem 4.14. Suppose f ; g are real admissible functions.

a) Show the following relation

Fff(x)g(x)g= 12�f (!) � g(!):

20 Fourier Transform

b) Use the above relation to show the following version of Parseval relationZR

f(x)g(x)dx=12�

ZR

f(!) g(!) d!:

In particular, we have ZR

jf(x)j2dx= 12�

ZR

jf(!)j d!:

Problem 4.15.

a) Show the following relation for a> 0

F�sin(ax)

x

�=� pa(!);

where pa(!) is a pulse in (¡a; a) in !-domain.

b) Use Parseval relation to showZ0

1sin(ax) sin(bx)x2

=

8<:a�

2a< b

b�

2b <a

:

Problem 4.16. Assume that f is smooth, Fourier transformable andXn=¡1

1

f(n)<1:

Show the following relation which is called the Poisson summationXn=¡1

1

f(n) =X

n=¡1

1

f (2n�):

Hint: note that

f(n)=12�

Z¡1

1f(!) ein! d!;

and also that for m=¡1 to 1 integers, we have

f(2m�)=X

n=¡1

1

fn;2m;

where

fn;2m=12�

Z(2m¡1)�

(2m+1)�

f(!) e¡in! d!:

Problem 4.17. Use the result of above problem to show the following identityXn=¡1

11

1+n2= � coth�:

Problem 4.18. Assume that function f and its derivative f 0 are admissible. Prove theproperty (4.12). If the anti-derivative of f is admissible, show the property (4.14).

Problem 4.19. Find the transform of the following functions

a) f(x) = e¡jxj¡jy j

b) f(x; y) = e¡tjxj¡sjy j

c) f(x; y) = e¡x2¡y2

d) f(x; y) =1

(a2+ x2)(b2+ y2)

Problem 4.20. Assume that f(x; y) is smooth function of order 2 and integrable. Find theformula of the transform Ff�f g.

4.7 Reading: Nyquist-Shanon rate 21

Appendix A

A.1 A proof of the Fourier theorem

We need the following Fubini theorem

Theorem A.1. Assume that f(x; y) is an integrable function, i.e.,

I =

ZR

ZR

jf(x; y)j dA<1:

If integrals

I1=

ZR

�ZR

f(x; y) dx

�dy; I2=

ZR

�ZR

f(x; y) dy

�dx;

exist, then I = I1= I2.

Now we can prove the Fourier theorem.

Theorem A.2. Assume that f(x) is admissible with the Fourier transform f(!),then we have

12�

limn!1

Z¡n

n

f (!) ei!xd!=12[f(x+)+ f(x¡)]:

Proof. Since f(!) is continuous, then for any n> 0, the integral

In(x)=

Z¡n

n

f(!) ei!xd!;

exists. On the other hand, we have

In(x)=

Z¡n

n�Z

R

f(�) e¡i�! d�

�ei!xd!=

Z¡n

nZR

f(�) ei!(x¡�) d�d!:

By Fubini theorem, we can switch the integration as

In(x)=

ZR

f(�)

�Z¡n

n

ei!(x¡�) d!

�d�:

But we saw before Z¡n

n

ei!(x¡�) d!=2sinn(x¡ �)

x¡ �;

and then

In(x)= 2

ZR

f(�)sinn(x¡ �)

x¡ �d�:

23

The integral in the right hand can be written as

In(x)= 2

Z¡1

x

f(�)sinn(x¡ �)

x¡ �d�+2

Zx

1f(�)

sinn(x¡ �)x¡ �

d�:

The �rst integral in the right hand side is simpli�ed asZ¡1

x

f(�)sinn(x¡ �)

x¡ �d�=

Z¡1

x f(�)¡ f(x¡)x¡ �

sinn(x¡ �)d�+

f(x¡)

Z¡1

x sinn(x¡ �)x¡ �

d�:

But we saw Z¡1

x sinn(x¡ �)x¡ �

d�=�2;

and thusZ¡1

x

f(�)sinn(x¡ �)

x¡ �d�=

Z¡1

x f(�)¡ f(x¡)x¡ �

sinn(x¡ �)d�+ f(x¡)�2:

Since f(x) is admissible, then ��������f(�)¡ f(x¡)x¡ �

��������<1;

and the Riemann-Lebesgue lemma implies

limn!1

Z¡1

x f(�)¡ f(x¡)x¡ �

sinn(x¡ �) d�=0;

and thus

limn!1

Z¡1

x

f(�)sinn(x¡ �)

x¡ �d�= f(x¡)

�2

Similarly the second integral gives

limn!1

Zx

1f(�)

sinn(x¡ �)x¡ �

d�=�2f(x+);

and therefore we have

12�

limn!1

In(x)=1�

limn!1

ZR

f(�)sinn(x¡ �)

x¡ �d�=

12(f(x+)+ f(x¡));

and this completes the proof. �

A.2 A proof of the theorem (4.4)

We �rst prove f(!) is continuous. Fix !, then

lim"!0

f (!+ ")= lim"!0

ZR

f(x)e¡i!xe¡i"xdx=

ZR

f(x) e¡i!x lim"!0

e¡i!" dx= f(!):

24

That the limit can be passed inside the integration follows from the dominant con-vergence theorem (see the appendix). Now we show that f is di�erentiable. Fix !and write

f(!+ ")¡ f(!)"

=

ZR

f(x) e¡i!xe¡i"x¡ 1

"dx=¡i

ZR

xf(x) e¡i!x e¡ix"0

for some "0 in the segment (0; ") (if "> 0). The condition of the exponentially decayof f at in�nity implies that there is R> 0 such that

jf(x)j ��e¡� jxj; jxj>R;

for some constants �; � > 0. Note that this implies the function g(x)=xf(x) to beadmissible. We have

lim"!0

ZR

xf(x) e¡i!x e¡ix"0=

ZR

xf(x) e¡i!x lim"!0

e¡ix"0=

ZR

xf(x) e¡i!x:

The possibility to pass the limit inside the integral follows from the dominant con-vergence theorem again. Therefore

lim"!0

f (!+ ")¡ f (!)"

=¡iZR

xf(x) e¡i!xdx=¡ig(!):

Since g is continuous, it follows that g is admissible. This argument applied forhigher order derivatives. That f(x) decay exponentially at in�nity, implies thatfunctions gn(x)=xnf(x) are admissible.

A.2 A proof of the theorem (4.4) 25