# How many watts does a 40 MHz radio frequency signal create?

I am new to the study of electromagnetic frequencies and electrical engineering as a whole.

I am trying to understand how to convert a radio frequency into energy, or calculate the power density of a radio signal. i.e. ->

40 MHz = (x) Watts / meters^2 | Solve for x

Anyone have any idea?

• Frequency is irrelevant here. Commented Jan 19, 2020 at 22:17
• Does your question make sense? Think about it. Your WiFi router and your microwave both operate on 2.4GHz. Commented Jan 19, 2020 at 22:26
• A light bulb would conduct very little power as the filament inductance would reflect some of the energy to the 50 Ohm transmission line and transmitter. and the resistance would be too high and reflect even more energy and the impedance is not what the Tx expects. Commented Jan 19, 2020 at 23:15
• Answer is easily found on Web looking for FRIIS LOSS antenna-theory.com/basics/friis.php Commented Jan 19, 2020 at 23:33
• What measurement is used in a frequency then to measure it's power? i.e. how can I tell the difference between a microwave and wifi signal? Commented Jan 20, 2020 at 16:15

This is like asking "how bright is a red light?". It's as bright as it is. You can make a bright red light or a dim red light.

As another answer points out, the energy per photon of an electromagnetic signal depends on the frequency of the signal. But you can make a brighter or dimmer (higher or lower power) source at any frequency by emitting more or fewer photons.

The energy of a photon is E = $$\h c\over\lambda\$$ where h is Planck's constant and c is the speed of light. So E = $$\h f\$$ where f is the frequency.

h is exactly 6.62607015$$\\times 10^{−34}\$$ J-s, by definition.

By the way, frequency can be measured in hertz (Hz). You wrote hz.

You can have few or many photons, so the energy per photon is not useful at all in knowing the total amount of power in an electromagnetic signal, but it is useful if you compare it with energy required to ionize atoms or excite phosphor etc.

• While nothing in this answer is incorrect, I don't see how it helps OP understand what they're trying to understand. Commented Jan 19, 2020 at 22:31
• The energy of the photon increases the Friis loss thru the radiated medium, while moisture increases that loss. and water significant more. There are very low frequency wavelengths VLF that get bonus reflections from the ionosphere. around the earth upwards to HF and moon bouncing with VHF. Commented Jan 19, 2020 at 23:23
• I remember reading somewhere (forgot where, though), that shorter-wavelength light was more energetic than lower-wavelength light and for that reason, Audi started using orange-ish reds in their cockpit illumination (i.e. to keep the driver's pupils from contracting too much when looking at the instruments during night drives). If that's true, then frequency would not be irrelevant to the question "how much energy is transported by this light beam"? What am I misunderstanding? Commented Jan 20, 2020 at 12:31
• @sixtyfive, That's more a physiology question. Astronomers also use (dim) red light, for example to read charts during an observation, to avoid spoiling their night vision. It's probably because the rod cells responsible for night vision aren't sensitive to red light, so won't be "washed out" by it. (Rod cell response curve here. Commented Jan 20, 2020 at 16:59
• @ThePhoton There is likely an underlying physics (or maybe it would be classed as photochemical) reason why the retinal structures that have evolved to sense very low light levels are blind to lower energy photons. As you say, at (amateur astronomy) star parties they issue red filters to tape over headlights to minimize the disruption caused if someone has to leave after dark. Commented Jan 20, 2020 at 17:16