5 ms·
Signal strength does not decay exponentially, rather as distance^-3 for a dipole[1]. Most noise sources are probably well approximated by a dipole antenna[2]. S
by krastanov 10y ago
Signal strength does not decay exponentially, rather as distance^-3 for a dipole[1]. Most noise sources are probably well approximated by a dipole antenna[2]. Some fancy modern modulation scheme with extremely dense encoding of information might suffer stronger decay, but the article is about simple sources of noise, not about modern modulation schemes.
[1]: https://en.wikipedia.org/wiki/Dipole#Vector_form https://en.wikipedia.org/wiki/Dipole#Vector_form
[2]: https://en.wikipedia.org/wiki/Dipole_antenna#Hertzian_dipole https://en.wikipedia.org/wiki/Dipole_antenna#Hertzian_dipole
- Bromskloss 10y ago> rather as distance^-3 for a dipole Wait a minute. The radiated power in the far field goes like r^-2, right?
- kqr 10y agoThis is my intuition as well, because at a distance r around the source you have a sphere, and the area of the sphere will relate to the radius r as r^-2. That said, the equation linked in the comment does indeed contain an r^-3 factor. I don't know if it's cancelled out by something else – my physics game isn't strong enough.
- Bromskloss 10y agoThe expressions for the field include terms of different powers of r. At large distances (compared to the wavelength) from the transmitting antenna, only the r^-1 terms (which dies off with r more slowly than the others) remain with any appreciable strength. This is called the "far-field". The power per cross-sectional area is proportional to E·B (or, equivalently, to E^2 or to B^2), and therefore goes as r^-2.
- leni536 10y agoThe first link is for static dipole, no radiation. The second link shows that the for a dipole antenna there is a 1/r term in amplitude. It creates an 1/r^2 term in power so your intuition is right.