I am considering wireless transmission. Let us define a cosine signal with frequency 2.4 GHz, i.e. $$s(t) = A\cos(2\pi ft),$$ where $$f=2.4\times10^9.$$


Let us assume that when TX transmits the signal s(t) of power 50 W, RX receives the signal of power 0.5 W (that is, 49.5 W is attenuated in the air.) Saying more detail, transmitted signal and received signal are, respectively, $$s_{TX}(t)=10\cos(2\pi ft)$$ and $$s_{RX}(t) = \cos(2\pi ft).$$

Does the received signal represent voltage for load? I mean, if load is 1 Ohm resistor, is the power for resistor $$P_{Load}=\int_{T}\frac{s_{RX}(t)^2}{1}dt=\frac{1}{2}=0.5 W\quad?$$

If so, if I use 0.1 ohm, the power will be 5 W? (Received power is just 0.5...)

Please someone teach me where I have misunderstanding


2 Answers 2


If your transmitting antenna has low losses and is driven at the right frequency it will emit power efficiently. A straightforward simple antenna like a dipole will emit power evenly in all directions in one plane and, it will emit zero power in other directions. I mention this to set the scene.

The electromagnetic power emitted is made from two fields and those two fields are an electric field and a magnetic field measured in volts per metre and amps per metre respectively. The power of these two fields is simply volt.amps per square metre i.e. we talk about power in watts per square metre and the square metre part does represent the watts that flow through the air per square metre.

However, as you get further from the transmit antenna the power per square metre drops with distance squared. Think of a light bulb emitting light in all directions - your eyes have a certain area with which to capture that power and if you move away from the light the power per square metre reduces.

So a transmit antenna transmits a fixed amount of power and the further you are away, you receive less of that power because it is spreading out. It isn't attenuating, it is spreading out.

Now think about your receive antenna and importantly, don't think of it as a wire, think of it as a collector of power with a certain area - this is called the aperture of the antenna and is measured in square metres. Up close to the transmit antenna the energy density is greater and more watts are packed into a given area. Further away there are fewer watts per square metre so your receiver (with a fixed aperture) cannot receive the same power.

If your receive antenna's aperture "collects" 0.5 watts at position A it will collect one quarter of that power at position 2A.

There is no trade off with load resistance - you get the power that is collected by the receive antennas aperture. If of course your receive antenna is up really close to the transmit antenna you are in, what is called, the "near field" and this is much less predictable - you get induction and electric coupling effects and you can, under the right situation "load" the transmit antenna and produce much more difficult to understand relationships.

So, if you are just talking about EM radiation from an antenna you are in, what is called, the far-field. This nominally begins about 1 wavelength from the transmit antenna so, at 2.4 GHz this is round about 10 cm.

So, what I've largely said above, is about far-field power transfer and not near-field stuff. In the far-field the power you get is immutable by impedances - the receive antenna can only "collect" what is delivered and that is nearly the end of the story however, if you don't provide the correct impedance to the antenna you won't maximize the power potential. Any antenna has an impedance and this is vastly frequency dependent so, you are limited even more.

In short, in the far-field you get what power you are given and, you have to match impedances to maximize that power.

  • \$\begingroup\$ Very good explanation, you give me very good intuition. Thank you so much. \$\endgroup\$
    – God Danny
    Sep 19, 2017 at 4:02

The antenna has a design output impedance, usually 50 ohms, it is NOT a voltage source, but appears to be a voltage source in series with a resistor equal to its designed output impedance.... All RF work is better thought of as power transfer rather then voltage transfer as impedance is so easy to change.

For maximum power transfer you need to have a load that matches the output impedance of the antenna (Graph the power in a load for a voltage source in series with a resistor driving a variable load and you will find that maximum power transfer occurs when the source and load impedance are equal, seriously worth taking the time to do this). For sources having a complex source impedance the maximum power transfer occurs when the load is the complex conjugate of the source impedance.

Now one probably could design an antenna to match into 1 ohm, or 0.1 ohm, or 0.01 ohm (But things are getting VERY lossy because of secondary effects), but in all cases the power delivered will be equal (apart from the losses due to poor antenna efficiency), you will see much less voltage and more current as the system impedance drops.

Consider an antenna with a design output impedance of 1 ohm (Unlikely for all sorts of reasons, but go with it), model it as say a 1.414V voltage source in series with a 1 ohm resistor. Into a matched load (1 ohm) you get 0.707V (Voltage divider) and 0.707A, multiplying these gives you your 0.5W. If you load this up into 0.1 ohms you get ~0.13V @ 1.3A = 0.169W, hardly an improvement.

Now a free space path loss of only 20dB seems unlikely to me, with a 1m separation and isotropic antennas for example you have ~40dB of path loss at 2.4GHz, and you can add 20dB for every decade of path length, so you might want to look at your numbers.

  • \$\begingroup\$ Very kind explanation, nice, thank you. I actually want to grasp of how to change power signal to voltage signal, ignoring the impedance matching, i.e. ignoring reflecting effects and so on. \$\endgroup\$
    – God Danny
    Sep 15, 2017 at 1:53
  • \$\begingroup\$ Just use a very high impedance input (Good luck with that at 2.4GHz!), you get zero power transferred but can measure the voltage. \$\endgroup\$
    – Dan Mills
    Sep 19, 2017 at 10:08

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