antenna-theory.com

[AntennaMan]

What is the best way to match the impedance of an antenna with that of a signal generator and that of air. I'm using a 50 ohm coax, but I'm unsure due to the high impedance of air (~280 ohms). Should I just match the antenna with the signal generator, or should I factor in the impedance of air?

Thanks!

[bigSteve]

A lot of people ask this question. The problem is that they know about impedance matching is important for power transfer, and then there is an "intrinsic or characteristic impedance" of vacuum or air. So often the conjecture is that some power is lost due to impedance mismatch between the antenna (at 50 Ohms) and air (at 377 Ohms).

But, to put it simply, there's no mismatch and no need to match the antenna to 377 Ohms. Impedance mismatch implies that energy is reflected somehow. But if you hook your 50 Ohm source up to a 50 Ohm antenna (that is sitting in air) you will see a VSWR=1, or no reflection. The power radiates away, and there is no spot for reflection (or impedance mismatch) to occur.

If the antenna efficiency is the only thing that matters - how much power is radiated relative to the power you supply to the antennna. And it wouldn't be any better if you used a 300 Ohm antenna.

[dovo]

I'm not an RF guy but after searching the internet a lot about this issue I think I've found the answer (but please correct me if I'm wrong).

This webpages seems to explain why the impedance mismatch between the antenna material and air doesn't affect the total effective radiated power (even though of course whenever there's an impedance mismatch there will be reflections).

http://www.sm0vpo.com/antennas/anten.htm

It does get reflected back mostly due to the impedance mismatch however if the length of the antenna is correct the reflected wave will be in phase with the incident wave and so the radiation of the reflected wave will be the same and eventually after bouncing hundreds of times the power will all get transmitted to the 377 ohm air from the whatever ohm of the antenna.

[bigSteve]

No, that webpage is just hand waving nonsense.

Essentially, it doesn't matter that the ratio of your voltage to current on your transmission line is 50 Ohms and the ratio of the electric field to the magnetic field in the far field is 377 Ohms.

The 50 Ohms is only important because it ensures power is transferred to the antenna, which induces the currents on the dipole. The dipole then radiates via maxwell's equations.

Everyone hear's impedance of free space and thinks there needs to be some sort of impedance matching there. however, it is incorrect. If you match your trnasmission line to your antenna, you know there is not reflections back down the transmission line. So where are these reflections occuring you speak of? It doesn't make sense.

[Nojodes]

Actually just assuming the "equations" work out and don't worry about it does not explain what is really happening at all.

From the standpoint of transmission line theory, it would be impossible to fully power a load of 377 ohms when the transmission line characteristic impedance is 50 ohms!

You definitely want the source impedance to match the antenna impedance. This will assure power is delivered to the antenna, but why assume a 50 ohm antenna will radiate all of the input power to a 377 ohm load? Could it be the power that is not radiated is actually be consumed in wire losses, especially since you just matched your source impedance to the antenna impedance? Do the transmitters get hot? 50 ohm to 377 ohm has a reflection coefficient modulus of 0.76, which means you will only radiate half the power you input, the other half will be wasted in wire losses. That is if transmission line theory is not completely bogus.

This is a question I have wondered for a long time and no one seems to give it much thought. What is really happening? Would not a speaker push more air (radiate more power) if the speaker diaphragm is impedance matched with that of air?

I'm still waiting for a proper answer to this other than "it doesn't matter because the reflections go away".

[E Kafeman]

[Nojodes]

Thank you so much E Kafeman. This is exactly what I was thinking about the antenna being an impedance transformer but they never explain them in that way (that I have found).

My other question would be can you really get 98% power radiated?

[E Kafeman]

Hi Nojodes,
First, I am a bit lying in picture above. As it is a monopole, do it have half of antenna structure mirrored in ground, so half of the EM impedance belongs to mirrored, not visible antenna part. Makes no big difference, but it was easier to explain this way.


I do regularly make monopole antennas which measure almost 99% efficiency. They are used as reference antennas. A IFA antenna on a circular solid groundplane, lambda/2 in diameter can relative easy reach 98% efficiency.
The antenna at photo is typical somewhere around 80% efficiency, in free space.
Add some components, battery, lcd display, FR4 PCB, a not so perfect ground and a plastic enclosure, and efficiency drops to 60%-65 for a reasonable big groundplane, type cellphone.
When groundplane is much shorter then lambda/4, antenna efficiency drops further.

[Nojodes]

Is it possible to use the equation for a quarter wave transmission line (which they usually use for impedance matching for the feedline) but applied to antennas, since we have established that they are nothing more than an impedance transformer?

Zin = (Zo^2)/Zout,

Where Zo is the characteristic impedance of the antenna, and Zout is the output of the antenna, Zin is the input to the antenna. For efficient radiation we want Zo to be 377. A typical monopole has an input impedance of 36 ohm roughly. So we can determine Zout which will be how much top loaded capacitance the antenna needs to give you a characteristic (or wave impedance) match with free space. Zout here would be 3948 Ohms which is the output impedance the antenna sees at the top. Effectively this would be the required stray capacitance with the ground plane. If we were operating at 2MHz (not sure anyone operates at that anymore haha - this is just an example) you would need to add a top loaded capacity (a ball, plate, slight increase of the antenna length, etc.) of 20pF to the antenna.

Then assuming the feedline is matched to Zin, this should radiate most of the energy away neglecting internal wire losses. Does this seem plausible?

[E Kafeman]

What I wrote above was just a (over-)simplified way to understand space impedance relative antenna impedance, but it is not how space sees the antenna. Impedance of waves with origin from loop or monopoles, in fare field do they all have 377 Ohm radiation impedance. My comparisons above are only correct as a transformation between an electrical transmission line and free space at long distance.
For antenna in nearfield is impedance given by E/H (compare U/I=R). A resonant electric monopole in near field can be several kOhm in radiation impedance. Nearfield is a space where full transformation not yet have fully occurred.
Antenna, how it is seen from space, is not a result of a feeding impedance, feeding impedance is of interest for best electrical matching. Assume a resonant dipole, it can be feed in the middle with a matched impedance of 72 Ohm, but it can also be feed somewhere else along its length and do then have a different feeding impedance but from radiation view have nothing changed.
From space seen is an antenna a structure with aperture area, not a point with a certain impedance. Even an isotropic antenna have an aperture area.
However there are relations between space impedance and dipole radiation resistance, one example:
Ra/Rs=Pin/(〖Sd*4*π〗^2*d^2 )
where Ra=Antenna radiating resistance, Rs=Space impedance, Pin=Power feed to antenna, Sd=power intensity, d=distance to antenna.

Again, compare with a soundwave that travels in space, it do have an impedance This impedance backward relation with a speaker electrical impedance is a bit complex. even if it was the speaker that was origin and translated electrical signals to the soundwaves with a certain power.
Even if it can be complex is it relative easy to measure that it is a coupling between these impedance in both directions. Measure electrical impedance and then put the speaker in a lake, and its electrical impedance for sound-frequencies (not DC resistance) will change due to the change in acoustical impedance for water compared to air.
Similar as for EM antennas according to formula above.

[Nojodes]

I am not sure about the "full transformation" view of radiation fields. The power leaving an antenna in the far field I thought would be the same power leaving in the near field (just much weaker in the far field due to the spread). This is what I've somewhat observed so far from experimenting, please bare with me though.

The difference I've seen in the near field appears to be due to electrical standing waves (or reactive power, better known as reflected power) being present in the antenna which is superimposed over the power leaving the antenna in electromagnetic radiation. If I measure the phase angle between the E and H in the near field it will not be zero, it will have some angle. The better the wave impedance of the antenna is matched to free space (the less reflections), the less that angle should be. Based on what I've read, the wave impedance of an antenna is what they call the "characteristic impedance". I'm not sure why they had to call wave impedance another name - I guess to make it more confusing for people like myself. The standing wave field drops off faster than the radiated field so once you get far enough away the phase angle will be zero. Its like all that real and reactive power stuff they teach in school but with a distance dependence.

I'm working with helical "non-radiating" resonators which have very small radiation resistances and generate nothing but a standing waves. If I measure the E and H in the near field (or about 10 inches from the coil), you get a phase angle of about 89 degrees. This correlates directly with the 0.7 ohm radiation resistance I see from the electrical input side (after I account for all the dissipated power, the only real power left is what is radiated). The radiated power using the measured phase angle is Pin*cos(89), but that Pin takes into consideration the total so it accounts for dissipation losses.

I guess what I thought I was observing in my coils was this superposition of real and reactive power in space. I have yet to be able to measure the far field only because the operating frequency is 2.4MHz and to get 2 wavelengths away I'd need a distance of about 820ft and I can't do that in my 40ft square lab.

In my case for those helical coils the phase angle between E and H is almost 90 degrees because they really don't radiate well. I thought if you had an antenna that radiated say 90%, then the magnitude of the reactive field would be much smaller than the radiated field in the near-zone and you would have a phase angle somewhere closer to 25 degrees. If you could radiate 100%, then there would be no reflection and no standing wave would exist since no power would get reflected and you would see a phase angle of 0 degrees both in the near and far zones.

Does this all seem plausible or am I missing something? Also Kafeman, thank you a ton for taking time to respond to all my questions and ramblings. It really helps!

[E Kafeman]

Yes total amount of radiated power is constant, independent of distance.
It is the ratio of E/H that changes from near field to far field.

For a small loop in near field do magnetic field dominate.
A such antenna can still be perfect matched and causing minimal reflections. Less efficiency then a half-wave dipole or a full wave loop but still useful as it is a simple antennas to make and tune, when space is limited. No need of ground-plane. 10-15% efficiency for a lambda/10 loop antenna is about what I normally reach in frequency range 400-900 MHz.

Resistive and reactive mismatch do both cause reflections.

Assume a dipole with characteristic impedance of 73 Ohm.
Now do we redesign it to a double fold dipole. Characteristic impedance is increased *4 but wave impedance have not changed.
http://www.antenna-theory.com/definitions/intrinsicimpedance.php

Not sure what "standing wave fields" you is thinking about. Half wave dipole, is also known as a standing wave antenna.

E and H fields are 90 degree apart and both are perpendicular to direction of a planar wave in far field. E and H field are in phase in a planar wave.
Planar wave can not exist in near field but some trials have been done with phased arrays to simulate near field planar wave.
It is not possible to make an antenna that directly is emitting a almost perfect planar wave. It would be against too many physical laws.
But it may change according to quantum-people.