Last month, in our ongoing discussion of meeting the ANSI specifications for electromagnetic radiation safety, we looked at shielding the electric field and shielding the magnetic field in a perfect conductor. While researching this, I noticed some similarities between electromagnetic radiation and transmission lines. In reviewing my college text on electromagnetics, I found that they discuss both of these (radiation in a medium or free space and propogation down a transmission line), but did not develop many analogies between them. Perhaps I'm seeing similarities that are not there! I found Smith charts to be an excellent way to visualize how a transmission line acts, so I've tried to apply them to radiation. Several people I've spoken with (including a physics instructor) had not heard of this approach, so it may not be valid, but I thought I'd put it out for your comment.
If we assume the outside world (outside the shielded box) can be represented by an infinitely long transmission line with a characteristic impedance of 377 ohms (intrinsic impedance of a vacuum), then our shield transmission line (the sheet of aluminum around the phaser) has a load of 377 ohms, since the input impedance of an infinitely long transmission line is its characteristic impedance, independent of the "load" at the end of the infinitely long line. This 377 ohms could then be "rotated" around the Smith chart an appropriate distance to determine the impedance at the input of the shield transmission line. Two factors affect how we rotate this about the Smith chart. The first is the velocity of the electromagnetic wave in the aluminum. The speed of an electromagnetic wave is 1/(ue)2 meters per second. This changes to C/(ue)2 if we use relative permeability and permittivity (C is velocity of light in a vacuum). I do not have the numbers for aluminum, but it appears that the velocity of an electromagnetic wave through copper is about 3.22 meters per second. If we assume aluminum is similar, the wave travels very slowly through the aluminum, causing a 6 mm "long" transmission line to be several wavelengths long (1863.354 wavelengths at 1 MHz, using the velocity factor for copper), causing us to "rotate" about the Smith chart several times. If this line were lossless, the "input impedance" could still be relatively high, since the input impedance of a lossless line "reappears" every half wavelength down the line. With a 377 ohm load on a lossless line that is some integer multiple of a half wavelength, the input impedance of the line would be 377 ohms, matching the impedance of the free space inside the phaser cabinet. This would allow the wave to propogate through the "shield" with no reflection. Not very desirable for a shield! However, the aluminum (or copper) is lossy. On a Smith chart, loss in the transmission line is handled by reducing the radius of the (constant SWR) circle an amount corresponding to the loss. This turns the circle into a spiral, closing in on the characteristic impedance of the line (the center of the spiral) as we move farther from the load. For aluminum, the signal is attenuated to 1/e for each .0814/f2 meters we go into the aluminum (the "skin depth"). At 1 MHz, the skin depth is 18.4 micrometers. The 1/e corresponds to a loss of about 8.7 dB. Our 6 mm thick aluminum has a loss of about 2,832 dB. By the time the "constant VSWR circle" has spiraled around 3,727 times (each rotation is 0.5 wavelengths) and in 2,832 dB, the input impedance of our line (the chunk of aluminum) is about the same as the characteristic impedance of the line (or the intrinsic impedance of the medium), which is around 2 milliohms at an angle of 45 degrees. The angle is introduced by the loss in the medium (a lossless medium or line has an angle of zero degrees). This phase angle represents the phase between the electric and magnetic fields, where a positive angle indicates the electric field leads the magnetic at by this amount at some point in space. The phase angle is not really important in this analogy, however.
At this point, the alumimum side of a phaser cabinet might be modeled as a transmission line that has a characteristic impedance of 2 milliohms, is 1863.354 wavelengths long and has a loss of 2,832 dB. The "load" on one end of the line is 377 ohms (the free space on the outside of the phaser cabinet). The impedance seen on the generator side of the line (the inside of the phaser cabinet) is about the same as the characteristic impedance of the line, since the line has so much loss. Inside the phaser cabinet, we have more free space (with its intrinsic impedance of 377 ohms). We can model the space inside the phaser cabinet as another transmission line with a characteristic impedance of 377 ohms. It is terminated by the impedance at the input of the transmission line that models the aluminum side of the cabinet (2 milliohms). The termination of the 377 ohm line by the 2 milliohm load results in almost all of the signal being reflected and very little being transmitted. About .001 % of the signal (voltage or current) is transmitted across the air to aluminum boundary while the remainder (about 100%) is reflected back into the cabinet. Finally, that portion of the signal that is transmitted is attenuated 2,832 dB before getting to the outside of the sheet of aluminum. At that point, the signal finds the free space outside the phaser cabinet, which again has an intrinsic impedance of 377 ohms. This 377 ohms is a severe mismatch for the 2 milliohm line, resulting in transmission of about .001% of the signal into the space surrounding the phaser, with the remainder reflected back into the aluminum, where it is eventually dissipated (it loses 2,832 dB each time it goes from one side to the other).
Next month we'll finish off this discussion with suggested actions you can take to insure your station meets the ANSI specifications.