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

In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched. Free space can be regarded as a further element, since a transmitting antenna eventually should radiate all power from the transmission line into it.

Now, if the impedances in the transmission line and in the antenna are matched at 50 Ω, but the impedance of free space is 377 Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?

EDIT:

As far as I gathered from the answers, literature and discussions online, the antenna acts as a impedance transformer between the feed line and free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc). This means that there is no reflected power between antenna and free space, and the transition between antenna and free space is therefore matched.

The same should be true in the reverse direction for a receiving antenna (Reciprocity Principle): a wave in free space ($$\Z_0\$$) impinges onto an antenna, and the received power is fed into the transmission line (again through impedance transformation). At least in one paper (Devi et al., Design of a wideband 377 Ω E-shaped patch antenna for RF energy harvesting, Microwave and Optical Letters (2012) Vol. 54, No. 3, 10.1002/mop.26607) it was mentioned that a 377 Ω antenna with a separate circuit to match it to 50 Ω was used to "achieve a wide impedance bandwidth" with a high power level. If the antenna normally is already the impedance transformer, what is the matching circuit needed for then? Or alternatively, under what circumstances is the antenna not also the impedance transformer?

Some helpful sources and discussions I found:

• For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise – PlasmaHH Oct 12 at 7:35
• In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily. – DonFusili Oct 12 at 7:42
• "My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna". – Bruce Abbott Oct 12 at 8:18
• Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines. – Curd Oct 12 at 8:25
• @ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need. – Curd Oct 12 at 9:10

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.

• My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there. – Puffafish Oct 12 at 7:59
• "Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc. – Curd Oct 12 at 8:17
• At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals. – Chris Stratton Oct 12 at 8:31
• Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = \frac{E^2}{Z_0}$. And the antenna just transforms between both impedances. Is that more or less correct? – ahemmetter Oct 12 at 14:05
• @Faekynn: I wouldn't say they have relation because: suppose you submerge a 50Ω (air) antenna in water (or an other medium) its radiation resistance would change very well. – Curd Oct 15 at 18:10

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.

• Good answer. But the diameter on the inside surface of the shield of LMR-400 is 0.285" (7.2 mm). 10 mm is the diameter over the outer jacket. That makes your point even better, as now your conductor has to have a diameter of 8 µm (or about 80 AWG). – davidmneedham Oct 15 at 15:17
• True, I should have said it's an approximation. – Phil Frost Oct 15 at 15:20
• It is true as you state in your answer there wouldn't be any particular advantage to 377 ohm twin-lead. The reason is missing which I give in my answer: 377 Ohm line impedance or resistance is a ratio of voltage and current, whereas the 377 Ohm free space wave impedance is a ratio of electric and magnetic fields. So just same unit, but no relation. – Faekynn Oct 17 at 11:53
• @Faekynn It's the ratio of electric and magnetic fields in a transmission line also, if one considers the fields that exist between the conductors in the transmission line. – Phil Frost Oct 17 at 12:52
• yes that is correct but there the difference persists. The wave impedance of a coaxial cable filled with air is ~377 Ohm, but the line impedance is something with logarithm (diameters). So, also for the transmission line there are these two unrelated impedances. I tried to explain this in my answer. – Faekynn Oct 17 at 14:45

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 line impedances etc., see the other answers.

Basically, the input resistance of an antenna, any other resistance, and transmission lines impedances are circuit-level descriptions for handling voltages and currents, while the free space wave impedance is for electric and magnetic fields.

The first impedance mentioned in the question is the input impedance of the antenna, which is a sum of radiation resistance and losses. It is related to currents $$\I\$$ and voltages $$\V\$$ on a circuit-description level, i.e., $$R = \frac{V}{I}\,.$$ This impedance of the resistor is the same kind as the the transmission line impedance of coaxial lines or microstrip lines, since these are also defined via voltages and currents.

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 transmission line impedance 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 transmission line impedance for the currents and voltages of this line, and this is the line 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 line 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 line 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.

The radiation resistance, $$\\small R_r\$$, of a half-wave dipole is $$\\small 73\Omega\$$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.

$$\\small R_r\$$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.

• That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)? – ahemmetter Oct 12 at 14:08
• @ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to. – Chris Stratton Oct 12 at 14:16
• What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is. – Chris Stratton Oct 12 at 14:17
• @ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances. – ahemmetter Oct 12 at 14:49
• A half-wave dipole impedance is 73 + 43j. If the dipole is shortened slightly to make it resonant, the impedance goes down to about 70 ohms. – Phil Frost Oct 13 at 16:58

"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."

This is your assumption. And it is correct, but not in the case of antennas.

Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.

So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.

• That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length. – ahemmetter Oct 12 at 16:13
• ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern. – Baruch Atta Oct 12 at 16:22
• Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different? – ahemmetter Oct 12 at 16:27

This question is a good example of over interpreting electrical engineering rules that were devised to make the physics more manageable in practical contexts. Impedance simply isn't that important.

The energy of a radio wave is embodied in the electric and magnetic fields distributed in a spatial volume. Maxwell's equations establish requirements for the relationships among those fields, and the homogenous equations imply that a disturbance from equilibrium will propagate. The latter is evident from the fact that the wave equation is easily derived from the fundamental equations.

In the wave equation there is an implied velocity of propagation that is the reciprocal of the square root of the product of the magnetic permeability and electric permittivity of the medium of propagation.

The square root of the quotient of those two quantities has units of impedance, and when the medium in question is a vacuum or air, it is called the 'radiation impedance of free space'.

This phrase refers to the ease (or difficulty) of establishing a non-equilibrium electro-magnetic disturbance. Loosely, it is a measure of the capacity of a volume of the medium to store energy in electro-magnetic form. More energy requires more volume or you risk non-linear breakdown. Very loosely, we are quantifying how hard it is to push energy into the system.

In a transmission line, say an old fashioned twin lead, we have a similar situation with different boundary conditions. The energy in the line is stored (transiently) in the oscillating electric field between conductors and the oscillating magnetic field about the conductors. This energy can propagate in two directions. If you have equal amounts of energy propagating in both directions, you have resonance or a standing wave. If you have matched terminations, energy leaves the line when it gets to the end and does not reflect or propagate back. It is important to understand that the power is transmitted in the insulator, not the conductors. The conductors are present only to provide boundary conditions, and the charge carriers in the conductors oscillate essentially in place, providing terminals for electric fields, and coupling the electric and magnetic fields. These ideas apply equally well to coaxial lines, but it is easier to visualize in a twin lead.

Like free space, a transmission line has a characteristic impedance that is a measure of its capacity to temporarily store energy distributed along its length. This impedance is dependent upon the geometry of the conductors (boundary conditions) and the relative permeability and permittivity of the materials from which the line is fabricated. Likewise, there is a characteristic propagation velocity that is typically a substantial fraction of the velocity of light in a vacuum.

The requirement for 'matching' impedances arises from the physics of wave reflection. Obviously any reflected energy is not propagated out of the system. A match eliminates reflected energy. It is important to realize that broadband matches are difficult. Matches are typically tuned to the specific design frequency of the system, and out of band signals may exhibit significant reflections.

In a resonant feed line, this fact is exploited by driving the line at its resonant frequency. At resonance, the line impedance is purely resistive. The difficulty is, you need to control the feed line length precisely, and it is only useful at its resonant frequency.

A more practical compromise is to match impedance. Then the feed line may be any reasonable length, and the signal may be a composition of many frequencies, or many independent signals, within the limitations of the bandwidth of the match.

A simple antenna like a dipole is operated at resonance. It is a resonant feedline. It therefore presents a purely resistive characteristic impedance (dependent on geometry and physics) at its design frequency. A line matched to that impedance will deliver all of its energy to the antenna. The antenna, being a resonant feedline, in turn delivers all of its energy to the next system, which is typically free space. It does this because at its design frequency, there is no reactive impedance. If you need to push more energy, you need to drive the antenna harder, which raises the peak voltages and currents in the antenna, which increases the amount of energy pushed out into free space during a given cycle. Obviously there are limitations imposed by non-linear breakdown.

A broadband antenna is really just a lossy feedline. Within its design bandwidth, all energy is radiated by the time an oscillation reaches the end of the feedline. Such antennas typically embody conical geometry in some form, with the low frequency limit set by the base of the cone and the high frequency limit set by practal limits on the pointiness of the cone.

• Thanks for the answer! If we take the optical analog to the feed line/antenna/free space system, we can consider different slabs of transparent media with different refractive indices. Lets assume the first interface is matched and provides no reflection: the energy is in the second ("antenna") medium and forms a standing wave (for example a Fabry-Perot resonance). Eventually of course the energy in the cavity is radiated into the third medium (free space). What would change if the antenna medium and free space medium have the same $n$? There is no cavity and all radiation is transmitted – ahemmetter Oct 12 at 22:46
• Note: MathJax is supported here. Using it might make your answer clearer. – Peter Mortensen Oct 13 at 9:49