derivative using FFT properties

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Jonathan Ron
Jonathan Ron 2012-6-21
Hi, I am trying to make a calculation by taking the first deriviative without success, I've tried the next couple of methods:
%%%the data is a seismic shot gather
%%%size(data_TX,1)=1001 (number of time samples)
%%%siz (data_TX,2)=256 (number of increments)
data_FK=fft2(data_TX);
% setting the the time and spacial freq vectors
fq=linspace(0,1/dx,size(data_TX,1)-1/2*dt);
kx=linspace(0,1/dy,size(data_TX,2)-1/2*dx);
%1 method
[FQ,KX]=meshgrid(ifftshift(fq),ifftshift(kx));
Dt_data_FK=2i*pi*FQ.*data_FK;
Dx_data_FK=2i*pi*KX.*data_FK;
%in this case I need to transpose the matrices
FQ=FQ';
KX=KX';
Dt_data_FK=2i*pi*FQ.*data_FK;
Dx_data_FK=2i*pi*KX.*data_FK;
Dt_data_TX=ifft2(Dt_data_FK);
Dx_data_TX=ifft2(Dx_data_FK); %I get in this case a complex data set ?
%2nd method
data_FK=fftshift(fft2(data_TX));
Dt_data_FK=2*pi*1i*diag(fq)*data_FK;
Dx_data_FK=2*pi*1i*data_FK*diag(kx);
Dt_data_TX=ifft2(ifftshift(Dt_data_FK),'symmetric');
Dx_data_TX=ifft2(ifftshift(Dt_data_TX),'symmetric');
note: I need to make the calculation for a cube CUBE(k,j,i), k=1001 time samples, j=256 recievers,i=256 shots.
I tried at first to use fftn but all ways produced me with wrong results. the theory is right but i'm not sure exactly how to take the derivative.

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Dr. Seis
Dr. Seis 2012-6-21
Based on the website below:
http://www.cl.cam.ac.uk/teaching/2000/ContMaths/JGD-notes/node11.html
I believe you would do something similar to the example below:
Nx = 64; % Number of samples collected along first dimension
Ny = 32; % Number of samples collected along second dimension
dx = 1; % x increment (i.e., Spacing between each column)
dy = .1; % y increment (i.e., Spacing between each row)
x = 0 : dx : (Nx-1)*dx; % x-distance
y = 0 : dy : (Ny-1)*dy; % y-distance
data_spacedomain = randn(Ny,Nx); % random 2D matrix
Nyq_kx = 1/(2*dx); % Nyquist of data in first dimension
Nyq_ky = 1/(2*dy); % Nyquist of data in second dimension
dkx = 1/(Nx*dx); % x-Wavenumber increment
dky = 1/(Ny*dy); % y-Wavenumber increment
kx = -Nyq_kx : dkx : Nyq_kx-dkx; % x-wavenumber
ky = -Nyq_ky : dky : Nyq_ky-dky; % y-wavenumber
data_wavenumberdomain = zeros(size(data_spacedomain)); % initialize data
% Compute 2D Discrete Fourier Transform
for i1 = 1 : Nx
for j1 = 1 : Ny
for i2 = 1 : Nx
for j2 = 1 : Ny
data_wavenumberdomain(j1,i1) = data_wavenumberdomain(j1,i1) + ...
(2i*pi*ky(j1))*(2i*pi*kx(i1))* ...
(data_spacedomain(j2,i2)*exp(-1i*(2*pi)*(kx(i1)*x(i2)+ky(j1)*y(j2))));
end
end
end
end
differentiated_data = ifft2(ifftshift(data_wavenumberdomain));
I have a program for performing this on a 1D dataset that you might leverage (in your own way) to produce the same effect on 2D datasets - see my answer in the post below:
http://www.mathworks.com/matlabcentral/answers/21700-finding-the-velocity-from-displacement
[EDIT 6/22]
Practically speaking, if you were going to compute the derivative of a 2D image, where your dimensions are either space-space or time-time, then you would need to use fft2, ifftshift, and meshgrid:
Nx = 64; % Number of samples collected along first dimension
Ny = 32; % Number of samples collected along second dimension
dx = 1; % x increment (i.e., Spacing between each column)
dy = .1; % y increment (i.e., Spacing between each row)
x = 0 : dx : (Nx-1)*dx; % x-distance
y = 0 : dy : (Ny-1)*dy; % y-distance
data_spacedomain = randn(Ny,Nx); % random 2D matrix
Nyq_kx = 1/(2*dx); % Nyquist of data in first dimension
Nyq_ky = 1/(2*dy); % Nyquist of data in second dimension
dkx = 1/(Nx*dx); % x-Wavenumber increment
dky = 1/(Ny*dy); % y-Wavenumber increment
kx = -Nyq_kx : dkx : Nyq_kx-dkx; % x-wavenumber
ky = -Nyq_ky : dky : Nyq_ky-dky; % y-wavenumber
% Compute 2D FFT
data_wavenumberdomain = fft2(data_spacedomain);
% Compute grid of wavenumbers
[KX, KY] = meshgrid(ifftshift(kx),ifftshift(ky));
% Compute 2D derivative
data_wavenumberdomain_differentiated = (2i*pi)^2*KX.*KY.*data_wavenumberdomain;
% Convert back to space domain
data_spacedomain_differentiated = ifft2(data_wavenumberdomain_differentiated );
If you only want to take the partial derivative of the the "image" with respect to one of your dimensions, then all you would do is:
d_MAT_x = 2i*pi*KX.*data_wavenumberdomain;
d_MAT_y = 2i*pi*KY.*data_wavenumberdomain;
Then just transform back into time domain using ifft2
[EDIT 6/27]
1. "Nx" and "Ny" are the number of columns and number of rows, respectively. Thus, you should have:
[KX,FQ]=meshgrid(ifftshift(kx),ifftshift(fq));
2. Your "fq" and "kx" may be defined wrong (I said in an earlier comment that if the number of rows or columns is an odd number then we need to do something different). Also, it seems strange that you would define "fq" in terms of "dx" and "dt"... and "kx" in terms of "dy" and "dx". Lets assume "dt" is the time increment and "dx" is the spatial increment... "Nt" is the number of time samples and "Nx" is the number of spatial samples, then:
Nyq_fq = 1/(2*dt);
Nyq_kx = 1/(2*dx);
dfq = 1/(Nt*dt);
dkx = 1/(Nx*dx);
if mod(Nt,2) == 0 % Nt is even
fq = -Nyq_fq : dfq : Nyq_fq-dfq;
else % Nt is odd
fq = [sort(-1*(dfq:dfq:Nyq_fq)) (0:dfq:Nyq_fq)];
end
if mod(Nx,2) == 0 % Nx is even
kx = -Nyq_kx : dkx : Nyq_kx-dkx;
else % Nx is odd
kx = [sort(-1*(dkx:dkx:Nyq_kx)) (0:dkx:Nyq_kx)];
end
Does this help solve things?
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Jonathan Ron
Jonathan Ron 2012-6-28
Hi Elige,
the fq and kx are fine, I don't have a problem with the odd number the result I'm getting is good now, using the meshgrid, except I have a couple of problems I can't really figure out
1. matlab defines the fft with -2*pi*i and not 2*pi*i. In order to get good results I need in both ways to multiply the derivative in (-1) after the ifft.
2. the partial derivative
Dx_data_TX=ifft2(Dx_data_FK);
gives me a complex data sets,usually something like (anynumber)+ 0.0000001i so when I take the real part it's ok, but still it is not suppose to be like that.
thanks for the help, if u have any ideas about these two problems let me know
Dr. Seis
Dr. Seis 2012-6-28
编辑：Dr. Seis 2012-6-28
1. Yes, the forward Fourier transform is "-2*pi*i". However, you are actually performing the derivative on the inverse Fourier transform:
d/dt( f(t) ) = d/dt( ifft(F(w) )
See the link below (about 4/5ths the way down):
http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html
So we should be using +2*pi*i. You should not need to multiply anything after the ifft... you really should consider using the way I define "fq" and "kx" because it looks like you are only defining the frequencies/wavenumbers associated with half the spectrum! You need to define the full spectrum which includes both negative and positive frequencies/wavenumbers.
2. This may be due to some precision issues to where the real part of the spectrum is not perfectly symmetrical (i.e., F(w) = F(-w) ), or the imaginary part of the spectrum is not perfectly anti-symmetrical (i.e., F(w) = -F(-w) ). But multiplying the frequency/wavenumber amplitudes by the wrong frequencies/wavenumbers will certainly not help.

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更多回答（1 个）

Francesco
Francesco 2013-1-18
Hello.
When I use the script shown after "EDIT 6/22" by Elige Grant, I seem to get complex numbers in the "data_spacedomain_differentiated" matrix. Shouldn't the signal just be real? Should I just extract the real component or am is that script somehow incomplete??
Cheers,
F

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