I apologise for the wording of my question, it will no doubt come across as confusing. If there is anything you are unsure about please feel free to ask me.


As a radio wave moves further away from the source, attenuation takes place. Why is it when this attenuation takes place the amount of distance that the weak electrical field now covers increases, compared to when it was closer to the source (shown from the image)? Is there an equation that I can use to tell me this? (coulomb's law ). If you are still confused by what i am asking, i have attached an image. enter image description here As you can see the inverse law is taken into account as the radio wave moves away from the source. My question is how do i find out the the area covered on the last step (When it is 1/9). I have searched through the internet but no equation seems to be giving me what i need. (I want to know the total area that the final 3d covers in the picture, the part where it has the letters A, not the distance from the original source).

  • \$\begingroup\$ Dispersal is a much better word for what's happening than attenuation. The same amount of signal energy is spread over a wider area as you get further from the source, so the signal reaching a fixed-size receiving antenna must be less. \$\endgroup\$ – The Photon Apr 13 '16 at 0:57
  • \$\begingroup\$ So to find out the area covered by 3d(distance) would be SG/4*π = r^2, where SG is signal streght? Would the signal streght be inside the surface area be uniform? \$\endgroup\$ – Bad programmer Apr 13 '16 at 1:39
  • 1
    \$\begingroup\$ The signal isn't necessarily uniform, but whatever the pattern of the signal emission vs azimuth and elevation at 1 km from the source (assuming that's far enough to be in the far field), it will be the same but spread out over 100 times as much area at 10 km. \$\endgroup\$ – The Photon Apr 13 '16 at 1:53
  • \$\begingroup\$ The E field and H field of a proper RF signal fall with distance not distance squared. However, when E and H are multiplied to get watts per sq metre, power clearly falls with distance squared. \$\endgroup\$ – Andy aka Apr 13 '16 at 7:43

The dispersal is a natural consequence of conservation of energy and the fact that waves (and/or photons) travel in straight lines away from a central point (called the phase center of the antenna). The RF wave emanating from an antenna is a form of traveling energy. If you construct a spherical surface around the whole antenna, that surface will capture all the energy emitted by the antenna, no matter how large or small the surface is. Since this energy must be the same as what was transmitted from the antenna (due to conservation of energy) I think you can see that the energy per square meter of surface has to diminish as you move away from the antenna, because otherwise the total amount of energy inside the sphere would depend on the size of the sphere, and that would violate conservation of energy. In real life, the energy can be absorbed by matter inside the sphere rather than by the surface. But that does not invalidate the basic idea. In such an absorptive environment, the "dispersal" would be more than the r^2 law predicts. In vacuum the r^2 law works perfectly.

When people talk about the gain of an antenna, what they are talking about is how much stronger the signal is compared to an isotropic antenna (which radiates equally in all directions). Antennas with gain are directional, meaning that instead of radiating outward spherically, equal in all directions, the energy is concentrated, either horizontally or vertically or both. Even when concentrated, the signal still disperses according to R^2. If you move outward in a straight line from the phase center of the antenna, you will still observe this R^2 law. But if you move perpendicular to that line, you may move into a stronger or weaker portion of the antenna beam.

I hope this all makes sense. I tried to avoid any real math, since you are striving to understand based on intuition.


Imagine blowing up a balloon. If the balloon is a perfect sphere, the surface area is:


Note the squared term. The surface area of the balloon represents the signal strength. As a signal radiates out, its 'density' also decreases, which is why the signal is weaker.

The picture that you showed is only a small section of a sphere. There aren't any antennas that I know of that disperse as a sphere, but no antenna disperses as the graphic that you posted. Most antennas have a 'map' that tells how effective they are in each direction. This graphic is just for illustrative purposes.

  • \$\begingroup\$ Thank you for your input, i have to let you know that i am a very slow student, so forgive me if i am asking a stupid question. So you are saying that i would need to find the signal strength at point 3D, which i will call SG. Than using the information i can find the squared for the radius of the circle, So SG/4*π = r^2? \$\endgroup\$ – Bad programmer Apr 13 '16 at 0:47
  • \$\begingroup\$ An antenna that "disperses as a sphere" is called an isotropic antenna, and cannot exist (except in simplified models), because electromagnetic waves have polarization and you can't have a polarization vector of nonzero length on all of a sphere. \$\endgroup\$ – Kevin Reid Apr 13 '16 at 2:02
  • \$\begingroup\$ @Badprogrammer The graphic that you showed is a concept illustration. Find the strength at d and you should be able to predict the strength anywhere else in the radiation pattern, they use 2d and 3d b/c they are convenient for illustration. Check out the wikipedia article. In that article, there is a radiation pattern pic that is roughly illustrative of what you would see on a real antenna. \$\endgroup\$ – slightlynybbled Apr 13 '16 at 2:49
  • \$\begingroup\$ Taking the balloon example further: Let's say the balloon weighs 10 gram. if you blow up the balloon to where its surface area is 1 sq. meter, then the density of latex on the surface is 10g/m^2. If you continue blowing it up to 10 sq. m, the density of latex will be 1g/m^2. As the latex "travels out", its total weight remains 10g but the density gets less. \$\endgroup\$ – neonzeon Dec 14 '16 at 22:10

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