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If we have two dipole antennas, it is well known that in order to transfer maximal power, two dipole antennas should be parallel and on the same height, which means that line that connects their middle points is perpendicular to both antennas. I wonder now, if two dipoles are parallel but not on the same height, how is received power calculated?

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You'd need to consider the angle between the two antennas and the distance between them. Dipole antennas do not radiate uniformly into $4\pi$. If you're looking up or down at the poles of the antenna, you will see no radiation (ideally). Looking at a direction transverse to this, the radiation is at a maximum. The angular dependence of the far (electric) field is given by:

$$E_\theta \propto \frac{sin~\theta}{r \lambda}$$

the power radiated per unit surface area is just:

$$P/A \propto E_\theta^2 \propto \frac{sin^2~\theta}{r^2 \lambda^2}$$

These equations just come from applying Maxwell's equations to an alternating current in the dipole antenna geometry. So you can see that the power per unit area at the receiver depends on the relative distance ($r$) and angle ($\theta$) between the transmitting and receiving antennas. Here, $\theta = 0$ and $\theta = \pi$ correspond to the poles of the antenna. This should be enough to get you a relative power loss given an offset in height if you already know the power radiated in the parallel case.

The diagram below illustrates the angles, height offset and the distance r:

enter image description here

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  • $\begingroup$ How is angle between two antennas calculated? In the case of two dipole antennas for example? $\endgroup$ – Nexy_sm Oct 19 '14 at 21:04
  • $\begingroup$ @Nexy_sm I've updated the post with an image to clarify $\endgroup$ – user3814483 Oct 20 '14 at 2:01
  • $\begingroup$ If the antennas are not parallel (as shown on the figure), should polarization mismatch term be added into equation? $\endgroup$ – Nexy_sm Oct 21 '14 at 6:58
  • $\begingroup$ @Nexy_sm If the polarizations were mismatched you'd have to throw in another cos(x) term where x is the angle of mismatch (e.g., in the drawing above x = 0, both antenna poles are in the same plane). This is because you're essentially picking off the component of the transmitted electric field that matches the orientation of your receiving antenna. The factor cos(x) is for the electric field E so the power dependence picks up cos^2(x) $\endgroup$ – user3814483 Oct 22 '14 at 19:43

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