## Tuesday, February 28, 2012

### Effective Cross-Section of a Dipole Antenna

Over the last couple of weeks I’ve been doing some pretty rough calculations to figure out the radiation from an oscillating charge. I’ve been in the ballpark, but today I think I’m like to get it right on the money. This is a calculation I figured out just over twenty years ago, and I’m pretty sure you won’t see it anywhere. I sketched it out last year in this article on the Crystal Radio, but I never did the numbers. That will be today’s project.

So far we’ve worked on the Larmor Formula, which gives you the radiation of an accelerating charge, and we’ve also tackled the radiation resistance of a short dipole. A charge in sinusoidal motion is just a special case of an oscillating charge, and I worked out a geometric argument which shows how to convert this case to an equivalent short dipole antenna. So these two formulas are really different versions of the same thing. Today I’m going to do a third version, which is also equivalent: the effective cross-sectional area of a small receiving antenna. If we know the power carried in the wave, this cross-section gives us the same information, in terms of absorbed power, as the radiation resistance. So if we know one, we know the other…in fact, we know all three.

The key to the calculation is this picture (see below), showing the interference of two wave patterns. The plane waves, moving from left to right, represent the incoming power from a distant transmitter. The circular waves, moving outwards, represent the response of the receiving antenna. What you have to understand is an antenna becomes a receiver by being a transmitter. The key is the shadow zone I’ve shown, where the two wave systems are close enough to be in phase with each other.

Everyone knows that a parabola is defined by the equation y = xˆ2, but not everyone knows this alternative definition: the locus of all points such that the difference of the distances to a given point (the focus) and a line not on that point (the directrix) is a constant. It’s called the “focus-directrix definition", and there are similar definitions for all the conic sections: circles, ellipses, and hyperbolas. For the antenna system shown in the diagram, the rough outline of the shadow zone is defined by the set of points where the phase difference of the two wave systems becomes 180 degrees, and according to the focus-directrix definition, that outline is a perfect parabola, Actually, it’s a paraboloid of rotation, if we look at it in three dimensions.

The calculation I’m about to do doesn’t rely on any details of electromagnetic theory, except two: I need to assume that the power in a wave is proportional to the square of its amplitude, and I will also assume that the radiated power depends only on that component of motion perpendicular to the angle of viewing. Putting these two assumptions together, for example, we would conclude that a short dipole puts out 50% of it’s peak output when viewed from a 45 degree angle. It’s a physically reasonable result, and it happens to be true.

We’re not going to worry about the details of tuning and matching: we’re going to focus on the resulting interference pattern. There’s a nice, quick-and-dirty way of ballparking the answer, and that is just to assume that inside the shadow zone, we have perfect interference; and outside, we don’t. It’s the cross-sectional area of the shadow zone that’s a bit dodgy: from the picture, you can see that it depends how far you are back from the receiving antenna, and there’s a square root relationship. In fact, if you are, say, seven wavelengths back of the antenna, the the cross-section of the paraboloid will be on the order of seven square wavelengths. And that will be enough to tell us the effective absorption cross-section of the antenna.

Here is how it goes. We back off reasonably far from the antenna, e.g. 20 wavelengths. The cross-section of the shadow is therefore on the order of 400  (oops) 20 square wavelengths. Now we ask: how much energy is being scattered by the receiving antenna? Suppose the incident field has unit amplitude, and let the power in the wave be one watt per square wavelength. What is the amplitude of the scattered field? Well, it can be almost whatever we want, because we control it by how we tune and match the antenna. Let’s say, for example, that we tune the antenna so the amplitude of the scattered field at a distance of 20 wavelengths is 1% of the incident field. That means the power density is1/10,000 watts/(squ. wavelength), since power goes as the square of amplitude. Multiplying this over the spherical surface at a radius of 20 wavelengths, we get a scattered power of… 4*pi*400/10000, or close to 0.51 watts.

Actually, the scattered power is a little less than that, because a short dipole does not radiate uniformly over a sphere. There’s something called the antenna gain, which we can actually calculate, but for now I’m going to assume you can look it up. The bottom line is you only radiate 2/3 as much as I calculated, so it’s actually 0.34 watts. But the real question is: where did that power come from? Answer: it was absorbed from the wave, and we should therefore see a weakening of the wave in the shadow zone.

And of course we do…it’s weakened by exactly 1%, because of the superposition of the two field patterns. Now, how much power do we take out of a wave at unit strength if we diminish its amplitude by 1%? Well, power goes as amplitude squared, so we actually reduce it by 2%; and this prevails over the cross-sectional area of the shadow, which we took to be 20 square wavelengths. You should be able to see that we have taken 0.4 watts out of the incoming wave.

What happened to the missing 60 milliwatts, I hear you ask? That’s the magic of an antenna. You have 60 milliwatts available to drive a load with, perhaps a small speaker.

Well, that’s convenient. Why don’t we gather more power, then? Instead of a paltry 1%, why not go for 2%? All you have to do is let twice as much current flow in your receiving antenna. If it’s just a question of changing resistors and tweaking a coil…

A nasty thing happens if you try to up your radiated amplitude to 2%. Radiated power goes as the square of the amplitude, so instead of pumping out 0.34 Watts, you’re now wasting 1.36 Watts in all directions. But the absorbed power, which happens in the shadow zone, is different. Instead of reducing the amplitude of the incident field in the shadow zone by one percent, giving you 98% power density, you now reduce it by two percent, giving you…96% power density. Your power absorption is only linear with the antenna current. You only absorb 0.8 watts, which isn’t enough to cover the wasted, re-radiated power, never mind driving a speaker.

Whenever you have two quantities, and one is linear and the other is quadratic, the linear quantity dominates first, and then the quadratic takes over. The receiving antenna works that way. At first, the more current you allow to flow, the more useful power you absorb. But then the wasted power starts to grow faster, and it dominates. There is a maximum useful power you can draw. I have drawn a graph of it here (I worked out the numbers so it agrees with the calcuations I've done so far:

You can see that the maximum useful power you can capture with this antenna is 0.22 Watts. SInce we are working with unit power density (1 watt/square wavelength) that also happens to be the amount of power flowing through 0.22 square wavelengths, and so we call that the absorption cross-section. Or we can also talk about the scattering cross-section, which is basically when you short-circuit the antenna and maximize the current. That’s of course four times as much power. (You can see why from basic circuit theory and load matching!) Either way, once you’ve got the cross-section, you’ve implicitly determined the radiation resistance. You just have to convert wave power to volts-per-meter, mutliply by the length of the antenna, and push your numbers around until everything agrees with everything else.

Actually, looking it up on Wikipedia I see they give an actual cross-section (they call it “effective aperture) of 0.11 square wavelengths, which is just half of what I got. It’s not surprising I’m out, since I used the coarsest imaginable approximation for the effective cross-section of the shadow zone. There’s an integration that gives me an exact value, and oddly enough, in this case I’m actually capable of doing the calculus. It’s just that the math tends to get in the way of the physics, and it’s so much easier to describe what you’re doing when you keep it simple.

Still, I promised I was going to try and get this one exact for a change, so maybe we’ll come back tomorrow and see if we can’t work in that missing factor of 2.