With the use of my good friend Scilab, I'll show you what happen to different patterns when operated by Fourier Transform (FT). Recall that we can directly get the Fourier Transform in Scilab by using the function fft2() and shifting its quadrants using fftshift(). (Remember to take the absolute value first using abs() before shifting.)
The patterns I used here are 128x128 images of square, annulus, square annulus, two symmetric slits along the x-axis and two symmetric dots along the x-axis. Results are shown below
Figure 1. (A) square (B) annulus (C) square annulus (D) two symmetric slits
(E) two symmetric dots (F-J) Corresponding FT of A to E
Just in case you wonder what the FT really looks like because the pattern produced is not really obvious, we can play a trick by creating smaller versions of A to E (not image size wise) and take their FT's.
Figure 2. (A) square (B) annulus (C) square annulus (D) two symmetric slits
(E) two symmetric dots (F-J) Corresponding FT of A to E
Now that looks clearer now. Figure 1 and 2 provides us a small bank of the Fourier Transform of some common basic patterns observable in real life optical systems. These correspondences are often called Fourier Transform (FT) pairs. Oftenly, it is helpful and saves time to know the FT pair of common patterns beforehand instead of deriving each everytime you need it.
We now proceed to the anamorphic property of Fourier Transform. Anarmophic according to [2] is the production of an optical effect along mutually perpendicular radii. Basically, we'll talk about the FT of sinusoids and the corresponding modifications to it when we try distorting the sinusoids.
As a start, let's observe the FT of a simple sinusoid with varying frequency of the form z = sin(2*pi*f*x).
Figure 3. (A-F) Sinusoids of frequency 2, 4, 6, 8, 10 and 20 Hz.
(G-L) Complementary FT of A to F.
We can see that the FT of a sinusoid produces two identical peaks symmetric about the center of the image and along the axis of the sinusoid. This is indeed true since FT decomposes a signal to its consequent frequencies. Thus, the two peaks represent two frequencies for a sinusoid which are equal and just negation of each other. On the other hand, we can observe that as the frequency of the sinusoid increases, the peaks move upward/downward. This is because the derived component frequencies go higher and moves further away from the center.
If we try adding a constant bias to a sinusoid or technically shifting its value to the positive region, the outcome is
Figure 4. (A) Sinusoid of frequency 4Hz with
added constant bias of 1. (B) FT of A.
In figure 4B, we notice another peak present in the center of the image (try clicking the image to enlarge). The reason for the center peak is that another component was separated by FT with 0 Hz frequency corresponding to the added constant bias. This is factual because a sinusoid (sine) with 0 frequency is a constant. This observation can be useful in finding actual frequencies in interferograms with DC biases. Just remove the 0 frequency peak and the observed frequencies will then be correct.
Suppose however a non-constant bias(another sinusoid with low frequency) is added, what happens to the FT?
Figure 5. (A) Sinusoid of frequency 4Hz with
added 0.5Hz sinusoid bias. (B) FT of A.
Looking at figure 5, we can still apply the same reasoning of removing the peaks very close to 0 to get the actual frequencies of an interferogram.
We further move one step higher by rotating our sinusoid with respect to different angles. Just change the sinusoid to the form
theta_deg = 30;
theta = theta_deg*(%pi/180);
z = sin(2*%pi*f*(y*sin(theta) + x*cos(theta)));
theta = theta_deg*(%pi/180);
z = sin(2*%pi*f*(y*sin(theta) + x*cos(theta)));
Note: The conversion of theta_deg to theta is necessary because the argument supported by Scilab trigonometric functions is radians.
Figure 6. (A-F) Sinusoids of frequency 4Hz rotated by 30, 60,
90, 120, 150 and 180 degrees. (G-L) FT's of A to F.
We can observe that as the rotation angle increases, the peaks move counterclockwise about an angle equal to the rotation angle. The important artifact to remember is that the rotation did not change the presence of two frequencies as observed in figure 3H. The FT simply rotated as a result of the rotation of the sinusoid.
If we are then to create a sinusoid as a result of multiplication of two perpendicular sinusoids,
z = sin(2*%pi*4*x).*sin(2*%pi*4*y);
the following result happens
Figure 7. (A) Sinusoid as a result to multiplication of 4Hz
sinusoids along x and y. (B) FT of A.
The result shown in figure 7B suggests that the peak frequencies doubled instead of the usual two and they went off the center axes. The doubling phenomena happened because the two frequencies corresponding for each sinusoid along a certain position paired up. In particular, the observed peaks correspond to combinations of 4 and -4Hz along the x-direction with the 4 and -4Hz along the y-direction. The locations would then be at (4,4), (4,-4), (-4,4) and (-4,-4).
As curiosity-driven people, we can try combining the sinusoid described in figure 7A with one rotated sinusoid as shown
Figure 8. (A-G) Figure 7A with added single rotated
sinusoid with angles 0, 30, 60, 90,120, 150, 180
degrees respectively.
A good prediction would be that the FT will four frequencies as shown in figure 7B can still be seen with additional frequencies rotated. Basically, just integrate figure 7B with figures 6G-6L individually.
Figure 9. (A-G) FT of figures 8A-G, respectively.
Aha! We had a correct prediction! What if we try to incorporate 6 sinusoids rotated an angle 30, 60, 90, 120, 150, and 180 at the same time to figure 7A. I must say that the resulting FT should look like a superposition of figures 9B-G. And since the chosen angles somewhat form half of a full rotation, I think the FT would have a circle in the middle.
Figure 10. (A) Combination of 1 multiplication of two perpendicular
sinusoids and 6 rotated sinusoids. (B) FT of A.
Indeed, the formed pattern has a circle inside!!
In conclusion, this activity was pretty much a good one for understanding Fourier Transform deeper. The key idea is that whenever we combine sinusoids and produce complicated patterns, the Fourier Transform decomposes the frequencies we used to provide a clearer picture on the operations we used to create the pattern.
For this activity, I would give myself a grade of 10.0 for producing all the outputs necessary and for giving clear figures that explains the objectives.
References:
[1] 'Properties of the 2D Fourier Transform', 2010 Applied Physics 186 manual by Dr. Maricor Soriano
[2] http://www.thefreedictionary.com/anamorphic
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