I'm starting with the assumption that a higher frequency EM wave is more energetic than a lower frequency one and thus requires more energy (and thus more power) to transmit.

In my naive model of a transmitter, there's an oscillator circuit, an amplifier and an antenna. It doesn't seem to be the case that there's a proportional relationship between frequency and power in an oscillator circuit so am assuming that the extra power must be consumed by either the amplifier or antenna circuitry.

Is the power consumed by an oscillator circuit not proportional to the frequency of oscillation, ceteris paribus?

In what stage of the generation/transmission of higher frequency EM waves does the extra power get consumed ?

Assuming wide enough bandwidth in the amplifier + antenna for the frequencies under consideration (ie 3 dB point is far enough out or ideal no parasitic system), where does the extra power get consumed to justify conservation of energy and Einstein-Planck (E=hv)?

  • \$\begingroup\$ Oscillators themselves consist of amplifiers that are required to maintain a positive feedback condition. This is likely where the energy will be consumed. \$\endgroup\$ – user57794 Nov 8 '14 at 5:05
  • \$\begingroup\$ The amplifier can be considered discrete - separate from the oscillator. Yes, the amplifier consumes power but that's the case for all frequencies. This doesn't answer the question. \$\endgroup\$ – Sridhar Nov 8 '14 at 5:59
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    \$\begingroup\$ You are mixing up different ideas. Photons of higher frequency radiation have more energy per photon. But the power of an EM signal is strictly a matter of its voltage and current. All sine waves of the same amplitude have the same ability to do work (as long as we are in the theoretical world where we ignore parasitic losses and such). Imagine a transmitter in free space. When it transmits at a power of 1W, then no matter the frequency, the Voltage amplitude of the EM wave will be the same. At higher frequencies, the photons may be more energetic, but there will be fewer of them. \$\endgroup\$ – mkeith Nov 8 '14 at 6:49
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    \$\begingroup\$ In RF spectrum, the wave INTENSITY could be proportional to the square of the amplitude (either E or M field), because particles in this frequencies have less energy than the kinetic energy of atoms and electrons due to their thermal motion. "Most energetic" you mean the energy of some eV obtaned as reaction energy. In higher frequencies (i.e light) the intensity of the wave is the energy per photon times the number of photons per unit area. The entropy paradox explained by the intensity and phase of photons. \$\endgroup\$ – GR Tech Nov 8 '14 at 6:55
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    \$\begingroup\$ I am disengaging because you seem intent on mystifying something that is not mystical. I have not offered an answer because your basic premise is incorrect. The appearance of a paradox is based on some misconstruction on your part of the relationship between the energy of a photon and the photon flow rate. \$\endgroup\$ – mkeith Nov 8 '14 at 19:57

Here is an engineers and not a physics forum, so I will try to give a simple answer.

A radio (electrical) oscillator working like a fast switch opening and closing a circuit those producing a series of pulses (very simplified). It does not charge atoms to change their energy level and then to release photons that have energy related with their frequency. So in radio oscillator it is not necessary to add power as the frequency increase. Few meters from an ordinary RF antenna you will have EM radiation or photons with very low energy (not intensity!) hundred thousand lower than a UV photon.

But in –let’s say- X-ray tube yes you have to increase the voltage (even DC) to accelerate electrons and to obtain high-energy photons from specific metals (i.e. 100kV for 100eV photon).

It is clearly different mechanism.

In any case don’t mix-up classical electrodynamics and quantum mechanics.

  • \$\begingroup\$ Your earlier comment about intensity is the only plausible explanation. For a given level of power, higher (lower) frequency waves will be generated at a lower (higher) intensity. \$\endgroup\$ – Sridhar Nov 10 '14 at 3:29
  • \$\begingroup\$ I don't see a difference between generating higher frequency radio waves versus generating x-rays. Higher frequency is higher frequency and requires more power to generate a certain given intensity. \$\endgroup\$ – Sridhar Nov 10 '14 at 3:30
  • \$\begingroup\$ According to your reasoning mind, a hanfull of an enriched radioactive material, it is not possible to have this enormous power (nuclear), because there is no another higher power to stimulate this. In pyrotechnics this is we are called "ignition". The formula E=hv confuse you. Use the precice one v=(E2-E1)/h which E1 and E2 is the energy levels of electrons, and h the Plank's constant off cource. Regarding RF ordinary frequencies, please keep in mind that the thermal noise covers the photon's small energy on this frequencies. \$\endgroup\$ – GR Tech Nov 10 '14 at 11:32
  • \$\begingroup\$ Forget about E=h\$\nu\$. Even for an electric dipole, power radiated goes like \$ \omega^{4} \$ so definitely frequency dependent. That extra power must come from the driving process - not the oscillator circuit so has to be the amplifier/antenna. \$\endgroup\$ – Sridhar Nov 10 '14 at 12:28
  • \$\begingroup\$ Forget about this...forget about that....I hope you will come back after some decades with a better explanation, that may be you allraedy have \$\endgroup\$ – GR Tech Nov 10 '14 at 13:04

The answer I found most convincing was on the physics stackexchange.

The argument is that for given input power there are fewer resistive losses when driving an emitter/antenna at higher frequency resulting in higher intensity (Poynting vector) of a higher frequency emitter.


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