Fourier Theory in Seismic Processing

(From Liner and Ikelle and Amundsen)

• Temporal aliasing• Spatial aliasing

Also with notes from http://

lcni.uoregon.edu/fft.ppt

http://www.falstad.com/fourier/j2/

Fourier series

• Periodic functions and signals may be expanded into a series of sine and cosine functions

The Nyquist Frequency

• The Nyquist frequency is equal to one-half of the sampling frequency.

• The Nyquist frequency is the highest frequency that can be measured in a signal.

The Fourier Transform

• A transform takes one function (or signal) and turns it into another function (or signal)

The Fourier Transform

• A transform takes one function (or signal) and turns it into another function (or signal)

• Continuous Fourier Transform:

The Fourier Transform

• The input signal gives the proper weight to all the cosines and sines

• Continuous Fourier Transform:

2

2

ifth f h t e dt

ifth t h f e df

Famous Fourier Transforms

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.5

0

0.5

1

1.5

-100 -50 0 50 1000

1

2

3

4

5

6

Sinc function

Square wave

The Fourier Transform

2

2

iftu f u t e dt

iftu t u f e df

12( )

02

Tt

u tT

t

2( )

21

2

iftu f u t e dt

T

i te dtT

axde axaedx

axe axdx e ka

,

2

2

Ti te

Ti

1 2 2

1cos sin

2 2

1 cos sin2 2

T Ti i

e ei

T Ti

i

T Ti

i

cos sinie i

cos sin

cos sini

ie i

,and

1cos

2T

i

sin2

1 cos2

Ti

T

i

sin2T

i

1 1sin sin

2 2

1 1sin sin

2 22

sin2

2

T Ti i

i i

T Ti i

i iT

ii

i

i

sin2

sin22

sin2

2

sin2

2

sinc2

T

T

T

T

TT

TT

• Mathematica Plot

• A transform takes one function (or signal) and turns it into another function (or signal)

• The Discrete Fourier Transform:

Discrete Fourier Transform

1 20

11 20

N ikn NH h eknkN ikn Nh H enNk n

Fast Fourier Transform

• The Fast Fourier Transform (FFT) is a very efficient algorithm for performing a discrete Fourier transform

• FFT algorithm published by Cooley & Tukey in 1965

• In 1969, the 2048 point analysis of a seismic trace took 13 ½ hours. Using the FFT, the same task on the same machine took 2.4 seconds!

Sampling Real Data

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-8

-6

-4

-2

0

2

4

6

8

5*sin (24t)

Amplitude = 5

Frequency = 4 Hz

seconds

A sine wave

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-8

-6

-4

-2

0

2

4

6

8

5*sin(24t)

Amplitude = 5

Frequency = 4 Hz

Sampling rate = 256 samples/second

seconds

Sampling duration =1 second

A sine wave signal

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2sin(28t), SR = 8.5 Hz

An undersampled signal

Sample Rates

What is the fewest number of times I need to sample this waveform per second?

?

?

?

Sample Rates

Sample Rates

Sample Rates

Sample Rates

What is the fewest number of times I need to sample this waveform per second?

At least twice per wavelength or period!

OTHERWISE ….

Undersampled waveforms

True frequency (f -true)

Am

plit

ud

e

Reconstructed frequency

(f -aliased)

ff

Oversampled waveforms

= True frequency (f -true)

Am

plit

ud

e

Reconstructed frequency

frequency is unaliased

Nyquist frequency

Nyquist frequency = 1 / twice the sampling rate

Minimum sampling rate must be at least twice the desired frequency

E.g., 1000 samples per second for 500Hz,

2000 samples per second for 1000 Hz

Oversampled waveformsA

mp

litu

de Nyquist frequency

In practice we are best oversampling by double the required minimum

i.e. 1000 samples per second for a maximum of 500 Hz

i.e., 2000 samples per second for a maximum of 1000 Hz

Oversampling is relatively cheap.

Spatial frequency, or wavenumber (k) is the number of cycles per unit distance.

One spatial cycle or wavenumber = frequency/velocity.

Each wavenumber must be sampled at least twice per wavelength

(two CMP’s per wavelength)

Spatial aliasing

1

2( )kN CMPspacing

IN PRACTICE each wavenumber must be sampled at least four times per minimum

wavelength (two CMP’s per wavelength)

Spatial aliasing

However, dip (theta) as well as frequency and velocity event changes the number of cycles per distance, so

4sin

lambdaCMPinterval

Liner, 9.7,p.192

Spatial aliasing

4sin

lambdaCMPinterval

x

V t

limitsinV t

x

For aliasing NOT to occur, delta(t) must be less than T/2

Spatial aliasing

limitsin2

VT

x

lim 2sinit

VTx

Geophone Spacing and Spatial Aliasing

K=0

1/4 wavelength shift per trace

total shift across array=3/4 wavelength

K=+ or -ve?

1/4 wavelength shift per trace

total shift across array=3/4 wavelength

K=?

1/2 wavelength shift per trace

total shift across array=3/2 wavelength

K=0

3/4 wavelength shift per trace

total shift across array=2 1/4 wavelength