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[tensor20]

I am reviewing some EM theory for my final exams: Fresnel formulas for reflection and transmission off interface.
If the interface is between air and a pure, perfect metal, perfect reflection of all the incident energy will take place.

In simulations, when you use your NEC program, do you assume the antennas are made of perfect conductors? If so, only a surface current is generated on the metal surface....
But it takes energy to set that surface charge in motion. How can all the energy then be reflected? Some of the incident energy should be used up for the surface current generation...

In real life applications, antennas can surely accept, receive plenty of energy from an incident EM field. Of course, real antennas have a finite conductivity (not infinite)....but, again, the higher the conductivity, the lower the Ohmic losses, but all the lower the energy captured by the antenna if the metal reflects so well...

thanks,
tensor20

[Schubert]

Your thinking is a little off.

"But it takes energy to set that surface charge in motion. How can all the energy then be reflected? Some of the incident energy should be used up for the surface current generation... "

The energy is simply transferred. The energy exists in the electromagnetic wave that is incident on the surface. This energy induces a surface current on the metal, which then induces the reflected wave. How much energy does it take to create the current on a perfect conductor? You may notice power loss is I^2*R...R=0 so nothing is lost there. It just transfers the energy to the reflected wave.

When you have an antenna, it is different. The antenna has a radio connected to it. This radio will have an impedance (typically 50 Ohms). In this situation, when the wave hits the antenna, it induces a current on the antenna. However in this case we have current flow which will travel to the radio (which is why antenna's impedance is matched to the radio). Then we don't have reflection of the current to a reflected wave, but the incident wave is transferred to a traveling current to the antenna terminals.

If you disconnect the radio then, so the antenna is open-circuited, the current has nowhere to flow and all the power will be reflected back.

[tensor20]

I see, thanks Schubert!

I read a book chapter on Fresnel formulas, with reflectivity and transmissivity, and the interface air-perfect metal is discussed.
An incident field gets completely reflected. That is where I got my information...

But as you explain, the antenna is a different story...The impedance matching allows for the current to flow down to the load.
However, that does not mean that the incident field is able to penetrate the perfect metal antenna.
Energy simply flows along the antenna to the load, guided by the surface current....

thanks
tensor20