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?

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As far as I gathered from literature and discussions online, the antenna acts as an 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:

  • 4
    \$\begingroup\$ 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 \$\endgroup\$
    – PlasmaHH
    Oct 12 '18 at 7:35
  • \$\begingroup\$ 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. \$\endgroup\$
    – DonFusili
    Oct 12 '18 at 7:42
  • 6
    \$\begingroup\$ "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". \$\endgroup\$ Oct 12 '18 at 8:18
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Curd
    Oct 12 '18 at 8:25
  • 2
    \$\begingroup\$ @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. \$\endgroup\$
    – Curd
    Oct 12 '18 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.

  • 7
    \$\begingroup\$ 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. \$\endgroup\$
    – Puffafish
    Oct 12 '18 at 7:59
  • 8
    \$\begingroup\$ "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. \$\endgroup\$
    – Curd
    Oct 12 '18 at 8:17
  • 3
    \$\begingroup\$ 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. \$\endgroup\$ Oct 12 '18 at 8:31
  • 2
    \$\begingroup\$ 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? \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 14:05
  • 1
    \$\begingroup\$ @ahemmetter: yes, I think that properly sums it up. \$\endgroup\$
    – Curd
    Oct 12 '18 at 14:23

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 approximatly obtained in an 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 microstrip 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. (On a related note: using this resistance for the receive case is misleading, since there occurs no loss in the radation resistance. It is still important for matching, though.)

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 $$ Z_{0,\mathrm{free\,space}} = \frac{E}{H} = \pi 119,9169832\,\Omega\approx377\,\Omega\,.$$ (This exact value was used before 2019, see Wikipedia on the free-space impedance) 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.

  • \$\begingroup\$ "The first impedance mentioned in the question is the input impedance of the antenna, which is a sum of radiation resistance and losses." is not a correct statement. The input impedance of the antenna may also consist of a non-real component. Radiation resistance and efficiency losses are only real (purely resistive) terms. Many common antennas (including a strict definition of a 1/2 wavelength antenna) have a reactive impedance component. \$\endgroup\$
    – Glenn W9IQ
    Nov 30 '18 at 18:36
  • \$\begingroup\$ I should note that strictly speaking, the real part of antenna input impedance and the radiation resistance of the antenna can be quite different. A classic example is a non-center fed, 1/2 wavelength, dipole antenna. \$\endgroup\$
    – Glenn W9IQ
    Nov 30 '18 at 19:24
  • \$\begingroup\$ "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. " It also is not the "line impedance" nor the input impedance nor the characteristic impedance. \$\endgroup\$
    – Glenn W9IQ
    Nov 30 '18 at 20:22
  • \$\begingroup\$ @GlennW9IQ about the first comment: you are right, I forgot to mention the reactive input impedance parts. \$\endgroup\$
    – crateane
    Dec 1 '18 at 20:48
  • 1
    \$\begingroup\$ 2nd comment: this probably depends on how you define the radiation resistance. for me, the radiation resistance just changes in the non-center fed case and still is equal to the real part of the antenna input impedance, but now for a different kind of antenna \$\endgroup\$
    – crateane
    Dec 1 '18 at 20:49

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:

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


  • \$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.

  • \$\begingroup\$ 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). \$\endgroup\$ Oct 15 '18 at 15:17
  • \$\begingroup\$ True, I should have said it's an approximation. \$\endgroup\$
    – Phil Frost
    Oct 15 '18 at 15:20
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – crateane
    Oct 17 '18 at 11:53
  • \$\begingroup\$ @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. \$\endgroup\$
    – Phil Frost
    Oct 17 '18 at 12:52
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – crateane
    Oct 17 '18 at 14:45

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.

  • \$\begingroup\$ 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 \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 22:46
  • \$\begingroup\$ Note: MathJax is supported here. Using it might make your answer clearer. \$\endgroup\$ Oct 13 '18 at 9:49
  • \$\begingroup\$ What is your definition of a "resonant feedline"? "At resonance, the line impedance is purely resistive." cannot be the case since any real transmission line (i.e. with loss) must have a reactive component as part of the characteristic impedance. \$\endgroup\$
    – Glenn W9IQ
    Dec 2 '18 at 6:23

I'm doing my first steps in antenna and RF field. I was learning about Antenna Impedance when I found this question and I will try to answer it. Hopefully I have understood the question! Sorry if the answer looks stupid, I'm just a "BEGINNER" :)

You said "Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω?", I think the answer is already included in the question. Yes, it's the word "INPUT". The 50 Ohm is chosen as an input not as an output impedance, if we want to transmit or receive the maximum power between the coaxial line and the antenna we have to match their impedance.(in this case is 50 Ohm because of the standards) If you chose 377 Ohm as the input impedance of the antenna to match it to the air impedance you will lose the power transmission between the coaxial line and the antenna.
If we consider the antenna as an element of the circuit that has an input and an "output impedance" it will look as follows:


simulate this circuit – Schematic created using CircuitLab


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.

  • \$\begingroup\$ 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)? \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 14:08
  • \$\begingroup\$ @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. \$\endgroup\$ Oct 12 '18 at 14:16
  • \$\begingroup\$ What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is. \$\endgroup\$ Oct 12 '18 at 14:17
  • \$\begingroup\$ @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. \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 14:49
  • \$\begingroup\$ 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. \$\endgroup\$
    – Phil Frost
    Oct 13 '18 at 16:58

All this is good in theory but what works in practice is a different story. I have been a communications engineer for the better part of 50 years. What we have to keep in mind here is we are attempting to explain a device called an antenna and why it does or does not work, or how well it does or does not do its job. Yes a new student can usually make a functional device from all these calculations, however that is not always true. I have built some very exacting antennas from theory that simply performed very poorly if at all. A good example is the J pole the performance is often not at all what one would expect even if when hooked up to very fancy antenna test equipment i.e. VNA's, it looks like it should be a great radiator and receptor when in fact it was more of a dummy load. Practice and theory often don't intersect. 50 ohms has been mentioned, yes it is a great compromise between the worlds of 37.5 and 73 ohms and it works well for that, in fact 50 was chosen because it worked in practice and it was easy to build from existing materials. In particular 1/2 inch water pipe inserting insulators and a center conductor for use on US Navy ships for WWII. Isolation had to be had for the feedlines to go from the antennas on deck to the equipment located within the safety of the ship. Before WWII there were literally Shacks "Radio Shacks" and I don't mean the defunct electronics stores, built right out on the main deck so as to be able to conduct the antennas to the radios. Even in the newer (at the time) ships the radio room was built on the main deck on an outside wall. Now for obvious safety reasons in a war ship the radio room should never be on deck or easily exposed to enemy fire, equipment and personal safety was a must so coax was born. Yes there were theoretical applications before that but not in general practice, there was shielded wire in use but it was not coaxial nor did it need to be, but to conduct signals from above deck to below deck and vice versa a different feedline than twinlead or ladder line was needed, both to protect the signals coming and going but also to protect the personnel and other things like gunpowder from the RF. Antennas are much the same. I often see mention of 1/4 wave antennas mentioned, truth is there really is no such thing. Nearly all practical antennas are some sort of 1/2 wave dipole. In the case of the 1/4 wave the other half of the antennas is usually the car or some other ground plane. As for 377 ohms to 50 or any other impedance it is all about feed point and or literal angle of the antenna, such as the "V" antenna mentioned earlier. Take for example a 1/2 wave end fed antenna it needs somewhere between a 9:1 to a 12:1 Balun Transformer to make it match and work. As does the Off Center Fed Dipole. Now there is that magical and sometimes nasty word BalUn! It is very simply nothing bad or magical it is simply a matching transformer. Often used to go from a balanced feedline or antenna to a unbalanced feedline or antenna! Does the transformer know balanced from unbalanced, NO it does not. In fact it does not even know what the impedance is, it only knows ratios i.e. 1 to 1, 4 to 1 or 9 to 1. Again I point out practice is not THEORY, thousands upon thousands of 4:1 Baluns are in use all over the world matching 50 ohm devices (Radios) and feedlines usually coax to 300 400 and even 600 ohm antennas. Do they work, fantastically they do, are they text book correct, not on your life, but then again all this would be moot if it did not work in practice! So quit worrying about the numbers being correct they are at best guidelines, what works, WORKS! Besides 377 ohms is theoretical freespace and just like isotropic Virginia It Simply Does Not Exist!

  • \$\begingroup\$ Thanks for the answer! So you're saying impedance matching to free space is not necessary in practice? That seems to be the case, but the question was for what reason that is not an issue. I see from practice and Maxwell's equations that all power is radiated from an antenna if it is matched to the transmission line. But nevertheless, there is an impedance mismatch between two components, and that causes a reflection at a very basic physical level (not just some simplified model). So why do we not need to consider it here? Does the model break down for antennas? Are they transformers? \$\endgroup\$
    – ahemmetter
    Oct 13 '18 at 16:54
  • \$\begingroup\$ Antennas Can be considered transformers of a type. In fact some are in to the Magnetic relm such as the Single Turn small Magnetic Loop. RF is transformed into RF Fields i.e. E and H or into Magnetic Field in the case of the Magnetic Loop antenna. So yes I would say they can be called a transformer of a type. \$\endgroup\$ Nov 29 '18 at 18:54
  • 2
    \$\begingroup\$ Welcome to EE.SE, @Laurin. Paragraph breaks have been around for more than 50 years. Use 2 x <Enter> to break your wall of text into logical blocks. It will help legibility greatly. \$\endgroup\$
    – Transistor
    Nov 29 '18 at 19:00

"...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.

And to comment on the answer by Laurin Cavender WB4IVG: In theory, there is no difference between theory and practice.

  • \$\begingroup\$ 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. \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 16:13
  • \$\begingroup\$ 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. \$\endgroup\$ Oct 12 '18 at 16:22
  • \$\begingroup\$ 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? \$\endgroup\$
    – ahemmetter
    Oct 12 '18 at 16:27
  • \$\begingroup\$ "This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna)." is not correct. If the output impedance of the source (transmitter) matches the characteristic impedance of the transmission line (only) then there is no "re-reflection" back to the load. Otherwise there is a partial or total "re-reflection" towards the load. \$\endgroup\$
    – Glenn W9IQ
    Nov 30 '18 at 20:13

The half wave dipole antenna can be seen as an open circuit transmission line that has been folded out 1/4 wavelength from the end, which has some capacitance and inductance (which is considered parasitic, because the goal is the radiation resistance -- to radiate the power) along the transmission line, loss resistance in the material of the antenna along the transmission line. This parasitic capacitance is between the 2 poles, as well as a coupling to ground with conductive ground losses included in the loss resistance, and the self capacitance of both poles to virtual ground. For a transmitter, all radiated power is dissipated in the theoretical radiation resistance along the transmission line.

The characteristic impedance of the dipole gradually increases towards the tips, such that we assume there is no reflection. There is however a reflection at the end of the antenna at the open circuit load.

Like a transmission line, if a pole of the dipole is less than 1/4 wavelength then it acts as a capacitor if it is an open circuit. If it is 1/4 wavelength then it behaves like a short circuit equivalent to series resonance. If it is longer than 1/4 wavelength then it behaves like an inductor. This is illustrated below:


An inductor could be added in series with a less than 1/4 wavelength transmission line to make it resonant and behave like a short circuit

The diagram also shows how the resonant length (where the parasitic capacitance and inductance cancel out) is slightly less than 0.25 due to the smaller velocity factor in the antenna compared to free space.

When it is 1/4 wavelength i.e. in a half wavelength dipole then the short circuit appears as 0 resistance at the feedpoint. The quarter wave impedance transformer equation shows that the feedpoint impedance of 1/4 wave transmission line is inversely proportional to the load impedance which is a large real number i.e. the open circuit resistance, and is therefore effectively 0. Therefore, only the loss resistance and radiation resistance can be seen.

When the incident wave reflects off the open circuit, the transmitted wave has double the voltage but 0 current. This is produced from the incident voltage and current that are in phase and the reflected voltage with current 180 degrees out of phase with voltage. However, this also creates a standing wave is created that pulsates 90 degrees out of phase between peaks at double the incident current and peaks at double the incident voltage, whose magnitudes are therefore related by the characteristic impedance of the line, except they are now 90 degrees out of phase, so there is an imaginary relationship. As the wave enters the far field, the fields transition gradually into phase, and the characteristic impedance of the antenna gradually transitions to the characteristic impedance of free space.

When the reflected wave reaches the feedpoint it looks like below:

enter image description here

The reflected transmitted wave arrives at the feedpoint just as the other 1/4 wavelength is arriving. Here, voltage cancels out, but the negative current in one direction and positive current in the opposite direction sum to produce a negative current of double the magnitude (the original magnitude being the current through a circuit with impedance equal to the characteristic impedance of the antenna + source) -- this will reflect off the mismatch between the antenna feedpoint characteristic impedance and the source impedance (which could be matched to the input impedance, but in a transmitter is ideally as small as possible for maximum efficiency, but there will be a reflection in both cases) back up to the antenna again -- the final pulse of current will settle at the value of current according to ohms law where the antenna is seen as a short circuit, therefore current is Vsource / source impedance, and the voltage after the source impedance is 0V.

The input impedance of the antenna is the return or reflection impedance when the reflection reaches the feedpoint before it hits the feedpoint, and the impedance it presents when the reflections reach equilibrium. On a DC current, or half wave transformer, this is always the value of the load impedance, i.e. open circuit, but it is not when you have an AC current and the antenna is not a multiple of half wavelength.

Because of this, only the radiation resistance and loss resistance can be seen from the feedpoint, because they attenuate the incident and reflected waves, to give a final input impedance of 0 + loss resistance + radiation resistance.

The radiation resistance increases as the dipole gets longer, but is also a complicated nonlinear function of the length of the dipole, shown below on the Dipole Antenna Wikipedia page as Rdipole.

enter image description here

The input impedance, radiation resistance and directivity are shown to have a relationship with dipole length below:

enter image description here

The loss resistance and the radiation resistance increases as the dipole gets longer for obvious reasons.

The radiation efficiency of the antenna is based on how much power is radiated vs. how much power is lost in the source resistance and the antenna loss resistance. I.e. in the transmitter, the radiation resistance is the desired source of power dissipation, with the RF circuitry and antenna loss resistance being undesired, but in the receiver, the RF circuitry is the desired source of power dissipation, with the other 2 being undesired.

The instantaneous power available at the terminals of a receiving antenna is the power density (irradiance) multiplied by the effective aperture of the antenna (100% of this power is transferred to the load when impedance matched):

$$\frac{{E_{(V/m)}}^2}{377_{(\Omega)}} A_{e_{(m^2)}} ~~(W) = I_{(A)}^2R_{L_{(\Omega)}} ~~ (W) $$

The maximum value is:

$$\frac{{V_{tot}}^2}{{4R_{A_{(\Omega)}} }}~~(W)$$

i.e. 1/4 of the instantaneous real power that would be received and absorbed by real part of the antenna impedance (loss + radiation resistance) in a resonant antenna if the antenna impedance were the only impedance in the circuit i.e. the antenna terminals were shorted. The above maximum value will be received if the antenna is resonant (as there will be no reactive power, increasing vdrop over RA in the above equation), and impedance matched to a load of the same resistance. Adding a load of the same impedance halves the current, which halves the real power, and they'd share half of that power each.

The effective aperture of the antenna:

$$A_e ~~ (m^2) = \frac{{{\lambda_{(m)}}^2}G_{\text{pract}}}{4\pi} = \frac{{{\lambda_{(m)}}^2}\epsilon (1-|Г|^2)D}{4\pi} ~~ (m^2) $$

The effective aperture of the antenna is the effective aperture of an isotropic antenna, basically the maximum cross section of the fresnel zone of an isotropic antenna, multiplied by the directivity and efficiency of the antenna. The directivity is a function of the length of the dipole as well as the height off of the ground where a dipole exists in antiphase due to reflections off the ground being 180 degrees out of phase + the greater distance and hence acts as a physical dipole array. Directivity is the unit solid angle with the highest power's power compared to the power in a unit solid angle in an isotropic antenna if the total power radiated were radiated from an isotropic antenna.

enter image description here

The directivity can be approximated as Ptot/beamwidth / Ptot/4pi = 4pi/beamwidth i.e. the solid angle of a sphere over the solid angle of the half power beamwidth, meaning effective aperture physically gets bigger when half power beamwidth gets smaller. This is the perfect case where all the isotropic power is restricted to a spherical sector where every unit solid angle in the sector has the same power and outside of it is 0, I.e. where 100% of the power is in the beam, therefore the power U in any unit solid angle within the sector can approximated as Ptot/beamwidth, so the equation becomes Ptot/beamwidth / Ptot/4pi because 100% of power is in the beam and 100% of power is in the isotropic sphere. When 100% of power isn't in the beam then you don't know the total power and you have to use the equation above on a solid angle by solid angle basis. The gain of the antenna is always equal to or less than the directivity of the antenna. Gain dbi is decibels with respect to a lossless isotropic antenna (100% efficiency so gain is based on directivity) and dbd is with respect to a half wave dipole. dbd + 2.15 = dbi.

In order for 100% of the instantaneous power available at the terminals of a receiving antenna to be transferred to the load, the antenna needs to be made resonant either by shortening the electrical length of the antenna to half wavelength, which is about 0.48 free space wavelength due to the velocity factor of the antenna (0.5 free space wavelength i.e. physical antenna length dipole is 73.1+42.5j ohms; ~0.48 wavelength is 73 ohms), and then matching the resonant antenna to the load using a real quarter wave transformer; or matching a non resonant antenna's input impedance to the load, which can be done using a 2 stage impedance transformer, which presents the input impedance of the RF circuitry as the conjugate of the antenna, by first quarter wave transforming the load impedance and then adding an imaginary component.

enter image description here

VSWR and reflection result when the antenna is not resonant or reflectionless matched (has reactive power) and when it is not correctly matched to the load. The reflection coefficient multiplied by the incident wave gives the amount reflected, and the amount transmitted is the sum of the incident and reflected waves. VSWR gives the ratio between the minimum and maximum voltage peak in the standing wave. The total reflection coefficient of all reflection coefficients multiplied together is included in the definition of the effective aperture of the antenna, and modifies the classic definition of ϵ (efficiency).

Seeing as the signals are quantised by an ADC with a high input impedance, you'd think a conjugate match wouldn't be required at the receiver, and that received voltage should be maximised and that impedance bridging would be desired, but actually this would result in less power being transferred to the load, and it is equivalent noise power that is added by attenuators in the circuit, so proportionally more noise voltage would be added for the same current. The conjugate match is still important for the RF front end.

enter image description here

A reflectionless match means Zin is ZS and a conjugate match means means Zin is ZS* (it doesn't matter what way the arrows are facing, it only matters where the tip of the arrow is -- input impedance = output impedance once reflections reach equilibrium). In this case the source is the antenna radiation and loss resistance.

When transmitting, it is desirable to have a low source impedance such that the transmission is as efficient as possible. We don't want maximum power transfer in this case because then half of the transmitted power would be wasted. To increase the transmitted power, the voltage is simply increased.


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