ACTIVITY 6 - Properties of the 2D Fourier Transform

Activity 6.A Familiarization with FT of Different 2D Patterns

This activity is similar to the previous one. We created different patterns, took their respective Fourier Transform and then display their FT modulus. This patterns were done with the aid of MS Paint and Scilab made it possible to compute their Fourier Transform.

The first pattern is a square aperture and beside it is its FT. The FT of a square aperture is similar to the FT of a circle (from Activity 5.A) but instead of an "airy disk" pattern, it has a central square maximum and maxima and minima intensities along its four sides. This pattern is know as the sinc function.

The next pattern is an annular aperture. Again, this is similar to the pattern obtained from a circular aperture except that at some fringes are not present due to the circular blocking at the center.
Then, a square annular aperture. This was analogous to the circular and annular aperture. The pattern obtained was almost the same with the square aperture and as expected, there are fringes missing. This is similar to a diffraction from a slit only that it is along x and y.

Next is a two-slit pattern. Along its side is its FT which yields from the cancellation of the upper and lower sections from the aperture. Thus, we see a horizontal pattern which is due to the interference coming from two slits.

The last pattern is a two-dot pattern or a double circular aperture. As has been said, the result looks like that of a circular aperture with maxima and minima intensities.

Activity 6.B Anamorphic Property of the Fourier Transform

Using Scilab, a 2D sinusoid in the x direction was created, with frequency that can be varied.

nx = 100; ny = 100;
x = linspace(-1,1,nx);
y = linspace(-1,1,ny);
[X,Y] = ndgrid(x,y);

f = 4 //frequency
z = sin(2*%pi*f*X);
imshow(z,[]);

And then, just like in previous part its Fourier Transform was taken and its FT modulus was displayed.

The figure above shows the source signal and its FT in 2D and 3D, respectively. The first figure shows a sinusoidal signal with frequency equals to 5 and below it is a signal with frequency 10. They corresponding FT shows that the peaks moves farther apart as the frequency increases. This is expected because the FT of a sinusoidal signal will have its peak on its frequency. Thus if you increase it, the peaks will farther along its axis.

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Then, a real image was simulated by adding a constant bias to the sinusoid created earlier. Its FT was also taken.

z = sin(2*%pi*20*X)+0.95
The figure above shows the sinusoidal signal with constant bias and its FT in 2D and 3D. On the contrary with the previous result, the FT of a sinusoidal signal with constant bias has 3 peaks. The two peaks with equal amplitude are from the frequency of the sinusoid and the peak at the center comes from the constant bias. We recall that the FT of a constant is a dirac delta. Also note that the sinusoid signal used has a frequency 20.
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If we suppose to have an image of an interferogram in a Young's Double Slit experiment, we can use this routine to extract its actual frequencies. We just have to find where the FT peaked. If a non-constant bias was added like those very low frequency sinusoids, we just neglect the peaks at low frequencies.

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The sinusoid was then rotated.

theta = 30;
z = sin(2*%pi*f*(Y*sin(theta) + X*cos(theta)));

As expected, the peaks of its FT are also rotated.

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Then a combination of sinusoids along x and y are investigated.

z = sin(2*%pi*4*X).*sin(2*%pi*4*Y);

Again, we expect this results, a four-peak plot. The first two peaks are due to the sinusoid along x and the last two are due to the sinusoid along y.

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Aside from a combination of sinusoids, we added rotated sinusoid to the signal.

z = sin(2*%pi*f*X).*cos(2*%pi*f*Y)+sin(2*%pi*f*(Y*sin(30) + X*cos(30)));

Before calculating and displaying its FT, I had a prediction that the rotated FT of the additional signal will add up to the four-peak FR of the two-sinusoid combination.

As I predicted, six peaks appear on the FT.

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I will give myself a grade of 8/10 for finishing the blog late. Although I was able to finish the activity during class hours, I'm having hard time doing blog because I have no access to the internet at home. Thanks again to Raffy and Gilbert for always answering my questions and for always having a good discussions.