# fft2 results don't match the analytical Fourier transform results

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Seyedalireza Abootorabi 2023 年 3 月 10 日
コメント済み: Paul 2023 年 3 月 13 日
Hello everyone,
I am trying to implement fft2 and compare the results with the analytical Fourier trasnform. For that I have chosen a constant rectangular field with constant value (the constant c = 1) which its analytical Fourier trasnform already exist. However the results of my fft2 and analytival Fourier trasnform do not match each other. Can you help me to find out my mistake?
clear
clc
% Define the dimensions of the square field
Lx = 2; % length of the rectangle
Lz = 1; % width of the rectangle
x = 0:0.005:1; % x vector
z = 0:0.005:1; % z vector
xgrd = length(x);
zgrd = length(z);
kxgrd = length(x);
kzgrd = length(z);
dkx = 1/(x(2)-x(1)); % Sampling frequency in x
dkz = 1/(z(2)-z(1)); % smapling frequency in z
kxval = (-length(x)/2:length(x)/2-1)*(dkx/length(x)); % Frequency vector
kzval = (-length(z)/2:length(z)/2-1)*(dkz/length(z));
% Define the size of the rectangular field and the constant value
c = 1; % constant value
rect = zeros(xgrd,zgrd); % initialize the rectangular field to all zeros
rect(xgrd/4+1:3*xgrd/4,zgrd/4+1:3*zgrd/4) = c; % set the central region to the constant value
% Evaluate the analytical Fourier transform on a grid of u and v values
for indkx = 1:kxgrd
for indkz = 1:kzgrd
F_analytical(indkx,indkz) = Lx*Lz*sinc(pi*Lx*kxval(indkx))*sinc(pi*Lz*kzval(indkz)); % must match the fft2 results
end
end
func = rect; % cosine signal
% Compute the Fourier transform
func_wo_shift = fft2(func);
func_shift = fftshift(fft2(func));

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### 採用された回答

Paul 2023 年 3 月 12 日

Hi Seyedalireza,
The code below get closer, maybe all the way there, by addressing a few issues
% Define the dimensions of the square field
Lx = 2; % length of the rectangle
Lz = 1; % width of the rectangle
I multiplied by Lx and Lz to get the correct physical dimensions
x = Lx*(0:0.005:1); % x vector
z = Lz*(0:0.005:1); % z vector
Define dx and dz for use below
dx = (x(2)-x(1));
dz = (z(2)-z(1));
xgrd = length(x);
zgrd = length(z);
kxgrd = length(x);
kzgrd = length(z);
Defined in terms of dx and dz
dkx = 1/dx; % Sampling frequency in x
dkz = 1/dz; % smapling frequency in z
The next two lines are incorrect, I believe.
kxval = (-length(x)/2:length(x)/2-1)*(dkx/length(x)); % Frequency vector
kzval = (-length(z)/2:length(z)/2-1)*(dkz/length(z));
The middle of the frequency vector should be zero, which is important in this problem because that's where all the action occurs. But 0 doesn't show up.
kxval(99:103)
ans = 1×5
-1.2438 -0.7463 -0.2488 0.2488 0.7463
With numel(x) = numel(z) = 201, the frequency vectors should be
kxval = (-100:100)/201*dkx;
kzval = (-100:100)/201*dkz;
kxval(99:103)
ans = 1×5
-0.9950 -0.4975 0 0.4975 0.9950
% Define the size of the rectangular field and the constant value
c = 1; % constant value
rect = zeros(xgrd,zgrd); % initialize the rectangular field to all zeros
rect(xgrd/4+1:3*xgrd/4,zgrd/4+1:3*zgrd/4) = c; % set the central region to the constant value
Warning: Integer operands are required for colon operator when used as index.
Warning: Integer operands are required for colon operator when used as index.
That warning is troublesome, but it looks like rect comes out the way you want.
figure
xlabel('x'),ylabel('Z')
I think this is the correct expression for F_analytical, assuming we're only interested in its amplitude and not its phase. Note that sinc is defined as sinc(x) = sin(pi*x)/(pi*x);
% Evaluate the analytical Fourier transform on a grid of u and v values
for indkx = 1:kxgrd
for indkz = 1:kzgrd
% F_analytical(indkx,indkz) = Lx*Lz*sinc(pi*Lx*kxval(indkx))*sinc(pi*Lz*kzval(indkz)); % must match the fft2 results
F_analytical(indkx,indkz) = Lx/2*Lz/2*sinc(Lx/2*kxval(indkx))*sinc(Lz/2*kzval(indkz)); % must match the fft2 results
end
end
func = rect; % cosine signal
% Compute the Fourier transform
func_wo_shift = fft2(func);
func_shift = fftshift(fft2(func));
figure
pcolor(kxval,kzval,abs(F_analytical.')),colorbar
xlabel('Kx'); ylabel('Kz')
axis([-10 10 -10 10])
I'm going to assume that F_analytical is supposed to be the Continous Time Fourier Transform. If so, the result from fft2 has to be multiplied by dx*dz for comparison
figure
pcolor(kxval,kzval,abs(dx*dz*func_shift.')),colorbar
xlabel('Kx');ylabel('Kz')
axis([-10 10 -10 10])
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Seyedalireza Abootorabi 2023 年 3 月 13 日
thank you for the very clear answer.

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### その他の回答 (1 件)

Benjamin Thompson 2023 年 3 月 10 日
Your problem is 201 by 201 samples in size? Making the size a power of 2. So use 256.
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Paul 2023 年 3 月 13 日
Based on the doc page fft2, it doesn't pad the input unless specified via the optional second and third input arguments.

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