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What you have derived here is the shift property of the Fourier transform, a translation

of a function over a time corresponds in the frequency domain to a multiplication with

exp i! :

i(t) I(!) (11.61)

!

i(t ) I(!) exp i!

; !

Problem h: Apply a Fourier transform to expression (11.59) for the output, use the shift

property (11.61) for each term and show that the output in the frequency domain is

related to the Fourier transform of the input by the following expression:

1

X

(;r)n ei!nT I(!) :

O(!) = (11.62)

n=0

Problem i: Use the theory of section (11.7) to show that the lter that describes the

generation of reverberations is in the frequency domain given by:

1

1 X (;r)n ei!nT :

G(!) = 2 (11.63)

n=0

Problem j: Since we know that the re ection coe cient r is less or equal to 1 (see prob-

lem a), this series is guaranteed to converge. Sum this series to show that

G(!) = 21 1 + rei!T :

1 (11.64)

CHAPTER 11. FOURIER ANALYSIS

146

This is a very useful result because it implies that the output and the input are in the

frequency domain related by

1

O(!) = 1 + rei!T I(!) : (11.65)

Note that the action of the reverberation leads in the frequency domain to a simple division

by (1 + r exp i!T). Note that this expression (11.65) has a similar form as equation

(2.32) of section (2.3) that accounts for the reverberation of waves between two stacks of

re ectors. This resemblance is no coincidence because the physics of waves bouncing back

and forth between two re ectors is similar.

Problem k: The goal of this section was to derive the dereverberation lter that produces

i(t) when o(t) is given. Use expression (11.65) to derive the dereverberation lter in

the frequency domain.

The dereverberation lter you have just derived is very simple in the frequency domain,

it only involves a multiplication of every frequency component O(!) with a scalar. Since

multiplication is a simple and e cient procedure it is attractive to carry out derever-

beration in the frequency domain. The dereverberation lter you have just derived was

developed originally by Backus 4].

The simplicity of the dereverberation lter hides a nasty complication. If the re ection

coe cient r and the two-way travel time T are exactly known and if the water bottom is

exactly horizontal there is no problem with the dereverberation lter. However, in practice

one only has estimates of these quantities, let these estimates be denoted by r0 and T 0 re-

spectively. The reverbarations lead in the frequency domain to a division by 1+r exp i!T

while the dereverberation lter based on the estimated parameters leads to a multiplica-

tion with 1 + r0 exp i!T 0 . The net e ect of the generation of the reverberations and the

subsequent dereverberation thus is in the frequency domain given by a multiplication with

1 + r0 exp i!T 0

1 + r exp i!T

Problem l: Show that when the re ection coe cients are close to unity and when the

estimate of the travel time is not accurate (T 0 = T) the term given above di ers

6

appreciably from unity. Explain that this implies that the dereverberation does not

work well.

In practice one does not only face the problem that the estimates of the re ection coef-

cients and the two-way travel time may be inaccurate. In addition the water bottom

may not be exactly at and there may be variations in the re ections coe cient along the

water bottom. In that case the action of the dereverberation lter can be signi cantly

degraded.

11.9 Design of frequency lters

In this section we consider the problem that a time series i(t) is recorded and that this

time series is contaminated with high-frequency noise. The aim of this section is to derive a

11.9. DESIGN OF FREQUENCY FILTERS 147

lter in the time domain that removes the frequency components with a frequency greater

than a cut-o frequency !0 from the time series. Such a lter is called a low-pass lter

because only frequencies components lower than the threshold !0 pass the lter.

Problem a: Show that this lter is in the frequency domain given by:

(

if !0

1

G(!) = (11.66)

j!j

if > !0

0 j!j

Problem b: Explain why the absolute value of the frequency should be used in this

expression.

Problem c: Show that this lter is in the time domain given by

Z !0

e;i!t d! :

g(t) = (11.67)

;!0

Problem d: Carry out the integration over frequency to derive that the lter is explicitly

given by

g(t) = 2!0 sinc (!0t) (11.68)

where the sinc-function is de ned by

sin (x) :

sinc (x) (11.69)

x

Problem e: Sketch the impulse response (11.68) of the low-pass lter as a function of

time. Determine the behaviour of the lter for t = 0 and show that the rst zero

crossing of the lter is at time t = =!0 .

The zero crossing of the lter is of fundamental importance. It implies that the width of

the impulse response in the time domain is given by 2 =!0 .

Problem f: Show that the width of the lter in the frequency domain is given by 2!0.

This means that when the the cut-o frequency !0 is increased, the width of the lter in

the frequency domain increases but the width of the lter in the time domain decreases.

A large width of the lter in the frequency domain corresponds to a small width of the

lter in the time domain and vice versa.

Problem g: Show that the product of the width of the lter in the time domain and the

width of the same lter in the frequency domain is given by 4 .

The signi cance of this result is that this product is independent of frequency. This implies

that the lter cannot be arbitrary peaked both in the time domain and the frequency

domain. This e ect has pronounced consequences since it is the essence of the uncertainty

relation of Heisenberg which states that the position and momentum of a particle can

never be known exactly, more details can be found in the book of Mertzbacher ?].

CHAPTER 11. FOURIER ANALYSIS

148

The lter (11.68) does actually not have very desirable properties, it has two basic

problems. The rst problem is that the lter decays only slowly with time. This means

that the lter is very long in the time domain, and hence the convolution of a time series

with the lter is numerically a rather ine cient process. This can be solved by making

the cuto of the lter in the frequency domain more gradual than the frequency cut-o

;n with n

de ned in expression (11.66), for example by using the lter G(!) = 1 + j!j !0

a positive integer.

Problem h: Does this lter have the steepest cuto for low values of n or for high values

of n? Hint: make a plot of G(!) as a function of !.

The second problem is that the lter is not causal. This means that when a function is

convolved with the lter (11.68), the output of the lter depends on the value of the input

at later times, i.e. the lter output depends on the input in the future.

Problem i: Show that this is the case, and that the output depends on the the input on

earlier times only when g(t) = 0 for t < 0.

A causal lter can be designed by using the theory of analytic functions shown in chapter

(12). The design of lters is quite an art, details can be found for example in the books

of Robinson and Treitel 51] or Claerbout 15].

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