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I do not understand a basic concept about Friis transmision equation between two antennas:

enter image description here

where Dt and Dr are the antenna directivities (with respect to an isotropic radiator) of the transmitting and receiving antennas respectively, λ is the wavelength and d is the distance between the antennas.

enter image description here

Let's consider for instance the case in which both are dipole antennas, whose pattern is drawn below:

enter image description here

How can I apply the previous formula to this case? Precisely, I have the following doubt:

1) Since dipoles are not isotropic radiators/receivers, their directivities will depend on the direction considered. So, I'd say that Pr will depend on direction. How do I find the total received power?

2) Part of the radiation from TX goes at also other directions (for instance at left). Is this fact considered in the Friis equation (and, if yes, which factor describes it?)? How can I distinguish, numerically, a situation like that in the previous picture, from something like this (a RX antenna, drawn in green, that surrounds the TX and so receives more power):

enter image description here

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  • \$\begingroup\$ With all these questions, I think you need to spend some time with Feynman's Lecture series. It's free on the web, now. Start at the first page of volume 1 and don't stop until you reach the last page of volume 2. You can hold short of volume 3, for now, I think. \$\endgroup\$ – jonk Jun 9 at 21:47
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The Friis equation considers transmission from a point source in free space emitting power spherically hence, the power emitted at a particular distance is in units of watts per square metre where the “square metre” part is that fraction of a sphere’s surface having a radius equivalent to the distance between transmitting antenna and receiving antenna.

Hence, all you need to know about the receiving antenna is its effective aperture (also described in square metres). The transmitting antenna is a simpler concept in the Friis equation.

Of course, the above is for an isotopic antenna which transmits power equally in all directions so, for a dipole or monopole, we know how much more power it can transmit or receive in its optimal direction and we add those dB to the Friis path loss equation to improve the path loss.

If there could be a receive antenna that totally surrounded a transmit antenna then the receive antenna would collect all the transmit power.

You need to google “effective aperture” and this tells you how much a receive antenna can collect when sat at the distance from the point source transmission. I think this might be the missing link in your learning.

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  • \$\begingroup\$ And where does the effettive aperture appear in Friis equation? Since Friis equation considers a point source emitting power spherically, how can this factor be absent? \$\endgroup\$ – Kinka-Byo Jun 9 at 21:53
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    \$\begingroup\$ It isn’t absent, it’s absolutely embodied into the equation. Look up the effective aperture Formula for an isotopic antenna. Try this : en.wikipedia.org/wiki/Antenna_aperture#Aperture_and_gain and you should see its remnants in the formula in your equation. \$\endgroup\$ – Andy aka Jun 9 at 22:21
  • \$\begingroup\$ Clear, thank you very much. Only a last question: since antennas' gains depend on the direction (let's say, Dr = Dr(θ,φ) and Dt = Dt (θ,φ), if we use for instance a spherical reference system), how do we evaluate the whole received power? Is it the sum of all the received powers, for each θ,φ? \$\endgroup\$ – Kinka-Byo Jun 9 at 22:48
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    \$\begingroup\$ I’m not sure what you mean. I think of the receiver like a small fishing net that captures only a fraction of the watts per square metre hitting it from the centre of a distant sphere. Not sure if that helps. \$\endgroup\$ – Andy aka Jun 9 at 23:48

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