Electronic – Radiation resistance of an antenna

antennaelectromagnetism

Radiation resistance of an antenna is defined as $$ R_R=\dfrac{P_R}{0.5I^2} $$ where $$P_R$$ is total radiated power, and $$I$$ is the antenna input current.
Antenna's far field is given as $$ E = Iπ(a_θ+ja_Φ)f(θ,Φ)\dfrac{e^-jkr}{r} $$
where $$ f(θ,Φ) $$ is $$ f(θ,Φ)=[cos^2(θ)+1]$$
In this context, how one can find radiation resistance of an antenna whose far field is given above?

Best Answer

I guess the term 0.5 in your first equation comes from the common habit to use the peak values of sinusoidal quantities. If the current were RMS value 0.5 would be omitted.

You must calculate the total radiated power through a spherical surface (or any closed surface around the antenna) in the far field region. In the far field the H is E/Zo where Zo=377Ohm, both E and H are perpendicular with each other and with the propagation direction. Thus the intensity i.e. power/area is simply (abs(E)^2)/Zo on a distant sphere. That's the simplest case of using the Poynting radiation intensity vector.

abs(E) is the maginitude of the complex E. The complex numbers are used to have E as a phasor. The phase lags as the distance increases.

You select some radius or calculate with symbols and integrate as surface integral the intensity (abs(E)^2)/Zo over the sphere. That's the power for your first formula.

The integration can be tricky with symbols, but it can be done as purely numerically even in Excel. You calculate with 1 Ampere. Unfortunately surface integrals must be understood well. The needed math belongs to first year math studies for academic engineering students. Calculus books present it.

Not asked: Radiation resistance cannot be measured, it's purely a thinking tool. Antenna's complex impedance is divided to 3 parts:

  • The reactance which generates into the space the non-propagating near-field
  • the radiation resistance, antenna's current generates the radiated power in that imagined resistor
  • the loss resistance, antenna's current generates there heat; it covers as well losses in the resistance of the metal and losses in insulation materials.
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