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sign convention you use. Any choice of the sign and the multiplication constant for the

Fourier transform can be used as long as:

(i) The product of the constants for the forward and backward transform is equal

to 1=2 and (ii) the sign of the exponent for the forward and the backward a

transform is opposite.

In this book, the Fourier transform pair (11.27) and (11.28) will mostly be used for the

Fourier transform from the space (x) domain to the wave number (k) domain.

Of course, the Fourier transform can also be used to transform a function in the time

(t) domain to the frequency (!) domain. Perhaps illogically the following convention will

used in this book for this Fourier transform pair:

Z1

F(!)e;i!t d!

f(t) = (11.42)

;1

1 Z 1 f(t)ei!t dt :

F(!) = 2 (11.43)

;1

The reason for this choice is that the combined Fourier transform from the (x t)-domain

to the (k !)-domain that is obtained by combining (11.27) and (11.42) is given by:

1

ZZ

F(k !)ei(kx;!t)dkd! :

f(x t) = (11.44)

;1

The function ei(kx;!t) in this integral describes a wave that moves for positive values of

k and ! in the direction of increasing values of x. To see this, let us assume we are at a

crest of this wave and that we follow the motion of the crest over a time t and that we

want to nd the distance x that the crest has moved in that time interval. If we follow

a wave crest, the phase of the wave is constant, and hence kx !t is constant.

;

CHAPTER 11. FOURIER ANALYSIS

138

Problem d: Show that this implies that x = c t, with c given by c = !=k. Why does

this imply that the wave moves with velocity c?

The exponential in the double Fourier transform (11.44) therefore describes for positive

values of ! and k a wave travelling in the positive direction with velocity c = !=k.

However, note that this is no proof that we should use the Fourier transform (11.44) and

not a transform with a di erent choice of the sign in the exponent. In fact, one should

realize that in the Fourier transform (11.44) one needs to integrate over all values of !

and k so that negative values of ! and k contribute to the integral as well.

Problem e: Use (11.28) and (11.43) to derive the inverse of the double Fourier transform

(11.44).

11.6 The convolution and correlation of two signals

There are di erent ways in which one can combine signals to create a new signal. In this

section the convolution and correlation of two signals is treated. For the sake of argument

the signals are taken to be functions of time, and the Fourier transform pair (11.42) and

(11.43) is used for the forward and inverse Fourier transform. Suppose a function f(t)

has a Fourier transform F(!) de ned by (11.42) and another function h(t) has a similar

Fourier transform H(!): Z1

H(!)e;i!t d! :

h(t) = (11.45)

;1

The two Fourier transforms F(!) and H(!) can be multiplied in the frequency domain,

and we want to nd out what the Fourier transform of the product F(!)H(!) is in the

time domain.

Problem a: Show that:

1

1 Z Z f(t )h(t )ei!(t1 +t2 ) dt dt :

F(!)H(!) = (11.46)

(2 )2 ;1 1 2 12

Problem b: Show that after a Fourier transform this function corresponds in the time

domain to:

1

Z1 1 ZZZ f(t )h(t )ei!(t1 +t2 ;t) dt dt d! :

F(!)H(!)e;i!t d! = (11.47)

12 12

(2 )2 ;1

;1

Problem c: Use the representation (11.31) of the delta function to carry out the integra-

tion over ! and show that this gives:

1

Z1 1 Z Z f(t )h(t ) (t + t t)dt dt :

F(!)H(!)e;i!t d! = 2 (11.48)

1 212 12

;1

;

;1

11.6. THE CONVOLUTION AND CORRELATION OF TWO SIGNALS 139

Problem d: The integration over t1 can now be carried out. Do this, and show that after

renaming the variable t2 to the result can be written as:

Z1 1 Z 1 f(t )h( )d = 1 (f h) (t) :

F(!)H(!)e;i!t d! = 2 (11.49)

;1 ;1 2

;

in the middle term is called the convolution of the functions f and h, this

The ;integral

operation is denoted by the symbol (f h). Equation (11.49) states that a multiplication

of the spectra of two functions in the frequency domain corresponds to the convolution

of these functions in the time domain. For this reason, equation (11.49) is called the

convolution theorem. This theorem is schematically indicated in the following diagram:

f(t) F(!)

!

h(t) H(!)

!

1 (f h) F(!)H(!)

2 !

Note that in the convolution theorem, a scale factor 1=2 is present in the left hand side.

This scale factor depends on the choice of the scale factors that one uses in the Fourier

transform, see section (11.5).

Problem e: Use a change of the integration variable to show that the convolution of f

and h can be written in the following two ways:

Z1 Z1

(f h) (t) = f(t f( )h(t )d :

)h( )d = (11.50)

;1 ;1

; ;

Problem f: In order to see what the convolution theorem looks like when a di erent scale

˜

factor is used in the Fourier transform de ne F(!) = C F(!) , and a similar scaling

for H(!). Show that with this choice of the scale factors, the Fourier transform of

˜˜

F(!)H(!) is in the time domain given by (1=2 C) (f h) (t). Hint: rst determine

the scale factor that one needs to use in the transformation from the frequency

domain to the time domain.

The convolution of two time series plays a very important role in exploration geophysics.

Suppose one carries out a seismic experiment where one uses a source such as dynamite to

generate waves that propagate through the earth. Let the source signal in the frequency

domain be given by S(!). The waves re ect at layers in the earth and are recorded by

geophones. In the ideal case, the source signal would have the shape of a simple spike, and

the waves re ected by all the re ectors would show up as a sequence of individual spikes.

In that case the recorded data would indicate the true re ectors in the earth. Let the signal

r(t) recorded in this ideal case have a Fourier transform R(!) in the frequency domain.

The problem that one faces is that a realistic seismic source is often not very impulsive.

If the recorded data d(t) have a Fourier transform D(!) in the frequency domain, then

this Fourier transform is given by

D(!) = R(!)S(!) : (11.51)

One is only interested in R(!) which is the earth response in the frequency domain, but

in practice one records the product R(!)S(!). In the time domain this is equivalent to

CHAPTER 11. FOURIER ANALYSIS

140

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