Activity 6 - Properties of the 2D Fourier Transform

annulus

square

square annulus

double slit

two dots

Pattern (left) and corresponding 2D Fourier transform (right)

Some patterns were generated using Scilab and the corresponding (2D) Fourier transforms are shown above. (Patterns are symmetric).
Code for generating patterns:
Annulus:
x = [-64:1:64];
[X,Y] = meshgrid(x);
r = sqrt(X.^2 + Y.^2);
annulus = zeros(size(X,1), size(X,2));
annulus(find (r <=32)) = 1.0; annulus(find(r<=16)) = 0.0; Square: sq = zeros(129, 129); temp = ones(33, 33); sq(49:81, 49:81) = temp; Square Annulus: sqannulus = sq; temp2 = zeros(17, 17); sqannulus(57:73, 57:73) = temp2; Double Slit: slit = zeros(129, 129); slit(:, 55) = 1.0; slit(:, 75) = 1.0; Two Dots: dot = zeros(129, 129); dot( 65, 55) = 1.0; dot( 65, 75) = 1.0;
In this activity, the anamorphic property of the Fourier transform is analyzed. First, a 2D sinusoid (similar to a corrugated roof) is created. To examine the spatial frequencies of this image, the Fourier transform was obtained. Furthermore, the frequency of the sinusoid was varied (f = 4, 8 and 16). The results are shown below.

f = 4

f = 8

f=16
2D sinusoids (left) with the specified frequencies
and their corresponding Fourier transforms (right)

As noticed in the Fourier transform of the images, the peak values on the 2D-fft is moves farther along the axis perpendicular to the sinusoid lines as the frequency of the sinusoids are increased. This demonstrates that the Fourier transformation shows the spatial frequencies of an image. The sinusoids have frequencies along the vertical and it is presented in the same axis in the Fourier transforms.
Scilab code for generating 2D sinusoids and displaying the Fourier transforms (modulus):
nx = 100; ny = 100;
x = linspace(-1,1,nx);
y = linspace(-1,1,ny);
[X,Y] = ndgrid(x,y);
f = 4 ; //for varying frequency
sinusoid = sin(2*%pi*f*X); //the 2D sinusoid
fft_sinusoid = fft2(z); //obtain Fourier transform
imshow(abs(fftshift(fft_sinusoid)),[]); //display modulus of the transform

Notice, however, from the code above that the sinusoids generated have negative values (based on the sin() function of Scilab). Of course, this is not realistic because standard images taken are non-negative. One way to simulate this is by adding a constant bias to the sinusoids such that there are no negative values. This would shift the sinusoids and, as if, they are riding on a DC signal. Again, taking the Fourier transform, the obtained result is shown below.
(As a comparison, the unbiased (left) and its Fourier transform (bottom) is placed side by side with the biased (right) and its corresponding transform. Notice that the unbiased and biased 2D sinusoids are identical. This is the limitation of the imshow and imwrite functions because it requires normalized images, and therefore, also altering the information contained in the images because no negative values can be shown or written as an image.)

2D sinusoids without (left) and with (right) bias

Fourier transforms of the unbiased (left) and biased (right).

The resulting Fourier transform of the sinusoid with a constant bias shows a central maximum. The reason for this is because the aside from the sinusoid signal, there is also a DC signal which is the constant bias as an additive term. The frequency of a DC signal is actually zero which accounts for the central maximum.

As a consequence of non-negative values in the images, capturing interferograms (such as the result of a Young's Double Slit Experiment) would result also to a sinusoid but from positive minimum (can be zero) and maximum values. This is similar to a sinusoid with a constant bias. To determine the actual frequencies of the results, data can ,initially, be subtracted by the average of the minimum and maximum values. This would make the sinusoid oscillate symmetrically about zero and, in turn, a pure sinusoid signal is observed. Then, Fourier transform can be applied to obtain the actual frequencies of the interferogram.
If a non-constant bias is added it would be hard to determine the adjustment needed. However, for example, if low-frequency sinusoids are added to the interferogram, it can simply be subtracted from the Fourier transform. The center of the transform of the 2D image represents the low frequencies, and this can simply be removed. If the frequency of the unwanted signals added are known, it is much easier because it can easily just be subtracted. But for example, for noise, which is of unknown frequency, it is difficult. Usually, these are of low frequencies so the previous suggestion might work.

2D rotated sinusoid (left) and the corresponding Fourier transform (right)

Scilab code to generate rotated sinusoid:
nx = 100; ny = 100;
x = linspace(-1,1,nx);
y = linspace(-1,1,ny);
[X,Y] = ndgrid(x,y);
f = 4; //frequency of sinusoid
theta = 30; //angle of rotation in radians (+ for counterclockwise)
z = sin(2*%pi*f*(Y*sin(theta) + X*cos(theta))); //generate rotated sinusoid

Further analysis was done in the Fourier transform of a sinusoid. This time, first rotating the sinusoid and then, taking the transform. As can be seen from the figure above, the axis of the sinusoid was rotated. As a consequence, the axis of the Fourier transform (in frequency space) is also rotated. The result demonstrates the property of the Fourier transform for obtaining the spatial frequencies in an image. Because the sinusoid is along a rotated axis, the spatial frequencies (which is actually the frequency of a component sinusoid of an image) would be along the same rotated axis.

It is also interesting to look at a combination (multiplication) of different sinusoids in X and Y. An easier way to visualize this is by providing a combination of different sinusoids in succession as presented below.

Sinusoid in X (left) with f = 4, and the corresponding Fourier transform (right)

Sinusoid in Y (left) with f = 4, and the corresponding Fourier transform (right)

Combination of the two previous sinusoid (left), and the corresponding Fourier transform (right)

Combination of the previous sinusoid with a sinusoid in X with f = 8 (left)
and the corresponding Fourier transform (right)

Combination of the previous sinusoid with a sinusoid in Y with f = 8 (left)
and the corresponding Fourier transform (right)

Notice that upon combining the sinusoids, the Fourier transform of the result looks like a spread of one of the Fourier transform to the Fourier transform of the other sinusoid in the combination. This is what basically happens in convolution. Now, it is known (also from previous activities) that the Fourier transform of the convolution of two functions can be obtained by multiplying the Fourier transforms of the two functions. The reverse is seen here. Initially, there is a multiplication of two images or sinusoid functions (combination of two sinusoids). The Fourier transform of the combination can be obtained by convolving the Fourier transforms of the two images or functions.

Scilab code for generating the combination of sinusoids in X and Y:
nx = 100; ny = 100;
x = linspace(-1,1,nx);
y = linspace(-1,1,ny);
[X,Y] = ndgrid(x,y);

z = sin(2*%pi*4*X); //sinusoid in X with f = 4
z1 = sin(2*%pi*4*Y); //sinusoid in Y with f = 4
z2 =sin(2*%pi*4*X).* sin(2*%pi*4*Y); //combination of sinusoid in X and Y with f = 4.
//z and z1 are simply multiply element by element

z3 = z2.*sin(2*%pi*8*X); //combination of sinusoid in X and Y with f =4 and sinusoid in X with f =8.

z4 = z3.*sin(2*%pi*8*Y); //combination of sinusoid in X and Y with f =4 and sinusoid in X and Y with f =8.

One important property of Fourier transform is linearity. This can be demonstrated by having an image that is a sum of different sinusoid signals. Intuitively, a linear transformation acting on a sum can be distributed into a sum of the transforms of each added terms. So for example, if the combination of X and Y sinusoids is added with a rotated sinusoid (X sinusoid rotated by 30 radians), the transform of the sum is basically the sum of the transform of the X and Y sinusoids and the transform of the rotated sinusoid. Indeed, this is verified by the results below.

Combination of sinusoids in X and Y with f = 4 (left)
and the corresponding Fourier transform (right)

Rotated sinusoid in X (angle of rotation = 30 radians) with f = 4 (left)
and the corresponding Fourier transform (right)

Sum of the two previous sinusoid function images (left)
and the corresponding Fourier transform (right).
The transform, indeed, looks like the addition of the Fourier transform of the two previous sinusoid function images.

For this activity, I would like to give myself a grade of 10 for a job well done (and ahead of time).
I would also like to acknowledge Winsome Chloe Rara for her help in doing this activity and the guidance of our professor Dr. Maricor Soriano.