Log Periodic Dipole Antenna
SRFS Teleinfra manufacture and import LPDA Antenna to customers. Each dipole element of the LPDA is fed, but the phase is set at random. This prevents interference to any one beam, which may be undesirable for laser applications. At 0° this type of phase is desirable because it adds complexity when the laser beam passes through the angle. For example a beam passing through -30° is more difficult to use than a beam passing through +30°. In practice these angles are all very small. Since I was trying to use this technique to measure how well our sensors work, I was motivated by my interest in the behavior of a beam incident on a flat surface near an angle. Here is some notation I used:
As you can see, we have three vectors of length 2. At coordinates x and y where y = x/2, we have a vector x and two vectors at coordinates x and y and y. Also with coordinates z and w where z = x/2, we have a vector z and one vector at coordinates z and w. Our beam has been passed twice through each of these vectors. Now we calculate the phase of our beam. We first calculate the derivative of that phase with respect to the vector z (the derivative of z, and then take its inverse). If the value of the derivative is greater than or equal to 1, the beam is said to have “passed through” any line segment passing through z. As shown below, if z = -30°, the beam is passing at 0°.
Now let us add another beam to our equation: y + z = -30°/2. Just as before, z and w are chosen randomly to produce values between -30°/2 and 0°/2. Remember, z = x/2, so our beam is passing through x/2 or z/2. Let z’ = x/2 – w/2. It may not be possible to choose those values using just random numbers, but this is easy enough to show:
z’ = -1/2
This is also true if z is -70° or -90°, so we can continue down in angle from z’ until the beam only passes through z/2. Note that z/2 = -30°/2. When z/2 = -30°/2 our beam is passing through z/2! The next phase is set at z = 0, which the beam passes through 0/2! Again, we add two beams and calculate the resultant phase, this time as y + z =-30°/2. Again, we add the beam to the equation and calculate the phase. A second beam is added and we get:
y = -30°/2
Since the beam passed through z/2 or z/2 + w/2, the beam is now passing through either z/2 or w/2! This means that since z and w are independent variables in the above equations, z/2 and w/2 are constants. This means that z/2 = w/2. Now we can define a variable that describes how much beam of light passes through z/2. To do this, we simply multiply z/2 by w/2. Because z/2 = w/2, both z/2 and w/2 are constant to z/2. And the denominator then equals a constant c.
c = 100,000,000,000
Therefore z/2 = w/2 is constant to z/2. Therefore, we can define a constant term c that represents the amount of beam that passes through z/2. Thus, z/2 = c×w/2. Now we add two new beams. First let z’ = -30°/2, then z/2 = -c×w/2, then z/2 = -c×w/2. Let x’ = z/2 and y’ = y + z = -30°/2. Finally, we repeat the steps above. These steps allow us to calculate the phase of the beam and compute the intensity. Let me explain this. Note that I do not use y in the actual phase calculation since we are going to use the value of x/2, since y represents the phase of the beam. But of course that notation is useful for the readers who do. Now, let’s see what our amplitude is this time. This time, in accordance with the notation I mentioned, we use the same notation as before. Specifically, instead of z passing through w/2, we are passing through z/2 + w/2. Both z/2 and w/2 are constants to z/2 and w/2. So by this notation, z/2 = w/2. To compute our amplitude, we find z/2. Then, using the notation we used before, we can compute x/2, y/2, z/2, and y/2. After all these calculations, we finally compute the amplitude of our beam!
Our two amplitude plots are similar, but they have a different scale. My plot shows a slightly larger intensity than my other plot, due to how many beams were added, and I wanted to keep them distinct. From top-to-bottom, the graphs are nearly identical, except that the peaks are not quite exactly the same as they were in the earlier plots. Additionally, we see some changes to the peak heights in the latter plots. But I will look closer at these graphs in part two.