Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?

The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.

Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).

The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.

So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.

The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.

A possible ansatz for calculating the radiation resistance \$R\$ of an antenna is:

Find an answer to the question: "How much power \$P\$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude \$V_0\$ (or \$I_0\$) is applied to the antenna?"

Then you get \$R = \frac{V_0^2}{2P}\$ (or \$=\frac{2P}{I_0^2}\$)

You get radiated power \$P\$ by integrating the Poynting vector \$\mathbf{S}\$ (=radiated power per area) over the sphere enclosing the antenna.

The Poynting vector is \$\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}\$ where \$\mathbf{E}\$ and \$\mathbf{B}\$ are electric/magnetic fields caused by the voltages and currents in your antenna.

You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.

All the answers name some valid points, but they fail to really answer the question which I want to repeat for clarity:

Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?

The Short & Simple Answer

These two impedances have no relation at all. They describe different physical phenomena: the antenna input impedance is not related to the 377 Ω free-space impedance. It is only by accident that the unit of both terms is the same (i,e., Ohms). Furthermore, 50 Ω is just a common value for characteristic impedances of transmission lines etc., see the other answers.

Basically, the input impedance of an antenna, any other resistance or reactance, and characteristic impedances are circuit-level descriptions for handling voltages and currents, while the free space wave impedance is for describing electric and magnetic fields. In particular, the (real-valued) 50 Ω input impedance means if you apply 50 V of voltage at the antenna feed, 1 A current will flow trough the antenna feed point. The free-space impedance has no relation to any antenna or material configuration. It describes the ratio of electric and magnetic fields in a propagating plane wave, which is approximated obtained in infinite distance to a radiating antenna.

The Longer Answer

The first impedance mentioned in the question is the input impedance of the antenna, which is a sum of radiation resistance, loss resistance and reactive components which are described as the imaginary part. It is related to currents \$I\$ and voltages \$V\$ at the feeding pont on a circuit-description level, i.e., $$R = \frac{V}{I}\,.$$ Changing the feeding point of the antenna, the value of this radiation resistance might change (this fact is employed e.g. for the matching of inset fed mircostrip patch antennas). The radiated fields, however, stay basically the same.

This impedance \$R\$ of the radiation resistance is the same kind as of a resistor or the transmission line characteristic impedance of coaxial lines or microstrip lines, since these are also defined via voltages and currents.

The radiation resistance is not a real resistance, it is just a model for the radiation case (i.e., operating the antenna to transmit power), where power gets lost from the circuit point of view since it is radiated away.

The second impedance is a wave impedance of the fields, which describes the ratios of electric (\$E\$) and magnetic (\$H\$) fields. The free-space impedance, for instance is given as $$ Z_{0,\mathrm{free\,space}} = \frac{E}{H} = \pi 119,9169832\,\Omega\approx377\,\Omega\,.$$ We can immediately see that fields and voltages have a relation that might change with geometry etc, or there might be no unique definition of voltages (e.g., in a hollow waveguide).

To make this lack of relation of these kinds of impedances more clear, an example might help. In the very simple case of the TEM wave inside of a coaxial cable, we know how to calculate the characteristic impedance the coaxial cable based on the geometry as $$Z_{0,\mathrm{coax}}=\frac{1}{2\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\ln\frac{r_{\mathrm{outer}}}{r_{\mathrm{inner}}}\,,$$ if we assume that the filling material is vacuum. This is a characteristic impedance (of the transmission line) for the currents and voltages of this line, and this is the kind of impedance which should be matched to the input impedance of an antenna.

However, having a look at the fields inside the cable, we find that the electric field has only the radial component (exact values are irrelevant in this context) $$E_r \propto \frac{1}{r \ln(r_{\mathrm{inner}}/r_{\mathrm{outer}})} \,.$$ More interestingly, the \$B\$ field has only a \$\phi\$-component which is a scaled version of the electric radial field $$B_\phi = \frac{k}{\omega}E_r=\frac{1}{c}E_r\,,$$ where \$c\$ is the speed of light, which is from free space (!) because the medium inside is free space. By using $$ B = \mu H\,,$$ we finally know the phi-component of the magnetic field as $$H_\phi =\frac{\sqrt{\epsilon}}{\sqrt{\mu}}E_r=Z_{0,\mathrm{free\,space}}E_r\,,$$ Therefore, the ratio of electric and magnetic fields is constant and only medium dependent; however, it does not depend on the geometry of the cable.

For free space inside the coaxial cable, the wave impedance is always approximately 377 Ω, while the characteristic impedance is geometry-dependent and can take any possible value from almost zero to extremely large values.

Conclusion & Final Remarks

If we look again at the example of the coaxial cable and leave it open at the end, achieving a characteristic impedance of ~377 Ω does not relate to anything about the fields. Any coaxial cable filled with air has a wave impedance of ~377 Ω, but this does not at all help to make the open piece of coaxial cable a good antenna. Therefore, a good definition of antenna does not relate at all to impedances, but reads

An antenna is a transducer from a guided wave to an unguided wave.

50 ohms is a convention. It's much more convenient if a room full of equipment all uses the same impedance.

Why is it the convention? Because coax is popular, and because 50 ohms is a good value for coax impedance, and it's a nice round number.

Why is it a good value for coax? The impedance of coax is a function of the ratio of the diameters of the shield and center conductor, and the dielectric material used:

$$ Z_0 = {138 \over \sqrt{\epsilon}} \log_{10}\left(D\over d\right) $$

Or rearranged algebraically:

$$ {D \over d} = 10^{\sqrt{\epsilon} Z_0 / 138} $$

where:

• \$Z_0\$ is the characteristic impedance of the coax
• \$\epsilon\$ is the dielectric constant (air is 1, PTFE is 2.1)
• \$D\$ is the diameter of the inside surface of the shield
• \$d\$ is the diameter of the outside surface of the center conductor

As the characteristic impedance increases, the center conductor must become smaller if the shield geometry and dielectric material remain constant. For \$Z_0 = 377\:\Omega \$, and PFTE dielectric:

$$ {D \over d} = 10^{\sqrt{2.1}\ 377 / 138} = 9097 $$

So for a coax cable with an outside diameter of 10 mm (RG-8, LMR-400, etc are approximately this size), the center conductor would have to be 10 mm / 9097 = 1.10 micrometers. That's impossibly fine: if it could even be manufactured with copper it would be extremely fragile. Additionally loss would be very high due to the high resistance.

On the other hand, the same calculation with \$Z_0 = 50\:\Omega \$ yields an inner conductor of approximately 3 mm, or 9 gauge wire. Easily manufactured, mechanically robust, and with sufficient surface area to result in acceptably low loss.

OK, so 50 ohms is a convention because it works for coax. But what about free space, which we can't change? Is that a problem?

Not really. Antennas are impedance transformers. A resonant wire dipole is a very easy to construct antenna, and it has a feedpoint impedance of 70 ohms, not 377.

It's not such a foreign concept. Air and other materials also have an acoustic impedance, which is the ratio of pressure to volume flow. It's analogous to electrical impedance which is the ratio of voltage to current. Somewhere in your house you probably have a speaker (perhaps a subwoofer) with a horn on it: that horn is there to take the very low acoustic impedance of air and transform it to something higher to better match the driver.

An antenna serves the same function, but for electric waves. The free space into which the antenna radiates has a fixed 377 ohm impedance, but the impedance at the other end depends on the geometry of the antenna. Previously mentioned, a resonant dipole has an impedance of 70 ohms. But bending that dipole so it forms a "V" instead of a straight line will decrease that impedance. A monopole antenna has half the impedance of the antenna: 35 ohms. A folded dipole has four times the impedance of the simple dipole: 280 ohms.

More complex antenna geometries can result in any feedpoint impedance you like, so while it would be technically possible to design an antenna with a feedpoint impedance of 377 ohms, but you wouldn't want to use it with coax for the reasons above. But perhaps twin-lead would work, though there wouldn't be any particular advantage to 377 ohm twin-lead.

At the end of the day, the antenna's job by definition is to convert a wave in one medium (free space) into a wave in another medium (a feedline). The two don't usually have the same characteristic impedance and so an antenna must be an impedance transformer to do the job efficiently. Most antennas transform to 50 ohms because most people want to use 50 ohm coax feedlines.