I'm not an RF guy so excuse the naive questions. I took an antennas class in grad school and quickly learned that I should not be an RF engineer, but I'm trying to do some back-of-the-envelope calculations.

Assume we have two antennas in free space. One is the transmitter, the other the receiver. I'm curious how much power (in Watts) is transmitted to the receiver.

Let's say we have a high-power amplifier putting out ~100W (Ps) on our transmit antenna with a gain (G) of 40dB. The receiver is 10 meters (R) away, and is assumed to be in the far-field. We can then say that our transmitted power density [S=G*Ps/(4*pi*(R^2))] is 3.18W/m^2

If our effective aperture of our receiving antenna is 0.1W/m^2, and the efficiency is 100%, then we will get 0.318W of power to the receiver.

My questions are 1) have I done this math right and 2) is 0.1W/m^2 of effective aperture realistic for a directional receiving antenna?

 

The question is answered by Friis formula - www.ccs.neu.edu/home/rraj/Courses/6710/S10/Lectures/AntennasPropagation.pdf and https://en.wikipedia.org/wiki/Free-space_path_loss
The overall algorythm of calc is this: calculate how much E (V/m) is at the point of receiving antenna. Then taking in account G factor calculate how many microvolts antenna will harvest. Taking in account the antenna impedance, assess the current and - accordingly the received power.

 

I'm not an RF guy so excuse the naive questions. I took an antennas class in grad school and quickly learned that I should not be an RF engineer, but I'm trying to do some back-of-the-envelope calculations.

Assume we have two antennas in free space. One is the transmitter, the other the receiver. I'm curious how much power (in Watts) is transmitted to the receiver.

Let's say we have a high-power amplifier putting out ~100W (Ps) on our transmit antenna with a gain (G) of 40dB. The receiver is 10 meters (R) away, and is assumed to be in the far-field. We can then say that our transmitted power density [S=G*Ps/(4*pi*(R^2))] is 3.18W/m^2

If our effective aperture of our receiving antenna is 0.1W/m^2, and the efficiency is 100%, then we will get 0.318W of power to the receiver.

My questions are 1) have I done this math right and 2) is 0.1W/m^2 of effective aperture realistic for a directional receiving antenna?

You are fine up to working out the power flux at the receiver as the transmit power times the transmit antenna gain divided by \(4\pi r^2\)

The effective aperture of your receive antenna should be an area. An isotropic receiving antenna (gain = 0dBi) has an effective area of \(A_e=\frac{\lambda^2}{4\pi}\), so if your receiver antenna has no gain you will receive \(A_e\) times the power flux. If the receiver antenna has gain, simply multiply this by the gain.

If you write these steps out as one equation you get the Friis equation as Janis noted.

 

Why is this a function of the wavelength? This is surprising to me and not intuitively obvious, yet.

 

https://en.wikipedia.org/wiki/Antenna_aperture
https://www.dsprelated.com/showarticle/62.php

Another important issue is whether or not the antenna itself is included when discussing "propagation", and this winds up being a key point, if not the key point. Antenna response and antenna design generally is highly frequency dependent, but the propagation of an electromagnetic wave in free space is not. So whether or not one means to include antennas in a discussion regarding "propagation" makes all the difference in the world.

 

Why is this a function of the wavelength? This is surprising to me and not intuitively obvious, yet.

The antenna gain is also the directivity (for a low loss antenna anyway), which is the ability of the antenna to concentrate power in a particular direction (and by reciprocity, receive preferentially from a particular direction). As the antenna gets bigger, it gets more directive - think of an aperture, if it is short the pattern is broad, but as it gets longer the pattern gets narrower, simply as all the parts of the aperture only add in phase in a smaller range of directions. This directivity depends on the aperture size in wavelengths. An omni-directional antenna must be small relative to a wavelength. The usual derivation of the \(\frac{\lambda^2}{4\pi}\) uses the antenna gain derived from the fields of an infinitesimal dipole.