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Antennas and radiation - some questions...

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Marko

Mon Nov 26 2007, 03:10PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Hello, all,

To this day there are too much things regarding EM theory that I too poorly understand.
I'll try to make the questions simple.

1. What are true requirements for turning electrical energy into electromagnetic?

It is said that any acceleration of charge produces electromagnetic wave.

Still, I can have a dipole antenna cut for 1/2 wavelength which has just few tens of volts on it, producing tiny electric field, and radiates extremely efficiently, or have a huge capacitor (like a tesla coil topload) with hundreds of kilovolts on it, producing massive alternating electric field and yet radiating poorly, storing most energy reactively.

Now what are really factors there? I have some sort of understanding that existence of waves is absurd at lengths smaller than their wavelength.

So, I wondered, could it be that electric/magnetic field needs to ''reach'' half of the wavelength in order to create a wave, and if source is too small compared to wavelength, field which ''reaches'' that distance is very weak and thus there is little radiation.

From QM point of view, true waves which carry the energy away irrecoverably manifest as real particles, photons.
And alternating magnetic and electric fields are virtual photons which don't truly exist, and they can't carry away the nearfield energy.

So, is a virtual photon simply one ''whose wavelength wasn't completed'', and thus it can't exist, giving the energy up back to source?

I'm too clueless on this and would really love if someone could give an explanation understandable to me.

2. All this makes me to think that there is something really weird going on with simple antennas.

People speak about ''different modes of resonance'' for various antennas and helical resonators.

The problem is, I don't have clue what modes are - I understand only one form of electrical resonance well and that is 1/2pi*sqrtLC.

Why are dipole antennas told to ''resonate'' with nothing but their length? And then there are ''resonant cavities'' of sorts?

If I just apply what I know none of these should work

3. What I find probably most fascinating, and confusing, is how between two folded pieces of wire, with only a tiny amount of capacitance between, impedance drops down by a collosal step to like 50..75 ohms.

To this day I have no idea how to properly explain that. I know about the vacuum impedance of 314 ohms which is divided by wavelength, and can assume that 'resonance' I mentioned up there plays some major role in this.

I would be very grateful to anyone who can make me understand anything out of this better.

It took me lots of learning just to be able to ask questions like this.

I hope for some good replies,

Marko

Carbon_Rod

Mon Nov 26 2007, 05:48PM

Registered Member #65 Joined: Thu Feb 09 2006, 06:43AM
Location:
Posts: 1155

1. ARRL explains such things very well. Highly suggest reading their publications (have seen PDFs in the wild too.)

2. There are "really weird" antennas, and the ARRL books cover most of them. If your library is worth anything they should have recent copies of the books.

3. They also cover Halo antennas.

Personal favorite:

WaveRider

Mon Nov 26 2007, 08:43PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Free space impedance is approx 377 ohms.

Whan talking about RF antennas, forget quantum mechanics. The sizes are too big for meaningful description.

You are asking some fairly theoretical questions. You will need to perhaps study some of the mathematical foundations to fully appreciate how antennas (or any other distribution of charges) radiate in the "classical" sense.

Balanis' book on antennas is a good place to start.

One way to envision the problem is to view the EM fields around the antanna as spherical modes (google spherical harmonics). How well the currents in the antenna overlap with the spherical harmonics that correspond to radiation fields indicates how efficient it will be as a radiator. The harmonics that correspond to evanescent modes contribute only to reactance seen at the antenna terminals. Coupling to radiation modes contributes to feedpoint resistance...

Tesla coils are generally poor radiators because their coupling to radiative modes is weak (due to their small size with respect to wavelength...however, if you tesla coil couples into a power supply lead or ground wire, it can become a very good radiator). This becomes a problem for those who dabble in HF tesla coils....

Marko

Mon Nov 26 2007, 09:22PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Free space impedance is approx 377 ohms.

Woops.. was too lazy to look into wikipedia.

Whan talking about RF antennas, forget quantum mechanics. The sizes are too big for meaningful description.

Why? They are ''microscopic'' for wavelengths they operate at.

Guys, I don't believe any of these questions require answers bigger than one post. With all the books you have stacked I'l definitely just get lost.

Pretty much the same as I got lost googling about this topic for last year or so.

With mathematics too.

One way to envision the problem is to view the EM fields around the antanna as spherical modes (google spherical harmonics). How well the currents in the antenna overlap with the spherical harmonics that correspond to radiation fields indicates how efficient it will be as a radiator.

I looked after it and stll have no clue what it means.
I remember those vertically overlapping magnetic and electric field lines creating each other endlessly, and can't figure much more.

Still I haven't got even my first and simplest question answered, why does a dipole radiate so efficiently, and TC topload producing thousands times stronger electric field doesn't?

Only answer I can give is that topload is much smaller than wavelength, but that's closely not enough.

Ignoring my assumptions, there should be no current flowing into halfwave dipole apart from tiny capacitive current firstly at all.

But somehow we can put few tens of V on it at right frequency and have amps flowing through mid air into nothing.

Marko

WaveRider

Mon Nov 26 2007, 09:32PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Marko,
I would like to write the magic words that allow you to understand in 5 minutes what took me (and other students of RF engineering) years of study to fully understand. It is just not that simple! The whys and hows require effort on your part that go beyond this forum, if you are truly interested. I am happy to encourage you and offer information where I can..

Antennas are not popular study topics because truly understanding them requires dedicated work... Building simple ones by trial and error can be satisfying...but the understanding requires almost religious devotion... The reference to quantum mechanics indicates that the understanding is just not there.

The dipole radiates so well because it couples very well to radiation modes (as I have stated). It cannot be well described by simple circuit models that you may be used to. It is up to you to look into the details.... This is what universities are for!

Chris Russell

Mon Nov 26 2007, 10:07PM

... not Russel!
Registered Member #1 Joined: Thu Jan 26 2006, 12:18AM
Location: Tempe, Arizona
Posts: 1052

I hope I can be of some help. I'm going to try to stick to general concepts for now, as the math can (and has) take up entire books.

1. What are true requirements for turning electrical energy into electromagnetic?

It is said that any acceleration of charge produces electromagnetic wave.

Still, I can have a dipole antenna cut for 1/2 wavelength which has just few tens of volts on it, producing tiny electric field, and radiates extremely efficiently, or have a huge capacitor (like a tesla coil topload) with hundreds of kilovolts on it, producing massive alternating electric field and yet radiating poorly, storing most energy reactively.

In order for an antenna to radiate, the basic requirement is this: alternating current must flow a distance through a conductor. The more current and the longer the conductor, the more the antenna radiates. Let's look at your halfwave dipole example a little more closely. Imagine what happens along the length of your dipole antenna, if a 10V sine wave is applied at the feed point. When one end is at 10V, the other end of the dipole is at 0V. They, of course, switch off rapidly, setting up a standing wave. The ends of the antenna are swinging up and down in potential, while at the feedpoint, there is almost no change in voltage at all. These large voltage potentials cause RF current to flow through the antenna. Current flows from one half of the antenna to the other, reaching maximum current right at the feedpoint. This current does the job of radiating. The total amount of current that actually flows is determined by your antenna's feedpoint impedance. In this case, roughly 73 ohms, purely resistive.

Your Tesla coil radiates poorly simply because the distance that the current flows is very small in most cases, especially considering the wavelength at which the Tesla coil operates -- hundreds upon hundreds of meters long. A halfwave dipole at 300kHz would be 500 meters long. Compared to a TC that is, perhaps, two meters from topload to ground rod, it starts to become apparent why the radiation efficiency is very very poor.

Tom540

Mon Nov 26 2007, 10:37PM

Banned on 3/17/2009.
Registered Member #487 Joined: Sun Jul 09 2006, 01:22AM
Location:
Posts: 617

"Still I haven't got even my first and simplest question answered, why does a dipole radiate so efficiently, and TC topload producing thousands times stronger electric field doesn't?"

Well as my old boss told me. the longer the wave length the bigger the antenna because the wave needs more room to propagate.
For example an antenna thats too short trying to tx a low frequency the wave gets clipped easier by objects like the ground etc. I know this is why they use higher 900MHz and up frequencies for wireless net because it can radiate much easier and penetrate walls etc. The wave carrries less energy though. This is also why some big long range transmitters use low frequencies and huge antennas. So basically that Tesla coil running at a few hundred KHz needs massive surface area at least 1/4 of the wavelength length.

I could be way off on this so you any of you Ham guys feel free to correct me.

Dave Marshall

Tue Nov 27 2007, 12:17AM

Registered Member #16 Joined: Thu Feb 02 2006, 02:22PM
Location: New Wilmington, PA
Posts: 554

Actually Tom, the higher the frequency, the more easily RF is influenced by surrounding objects.

Just as efficient antennas get smaller as frequencies increase, the size of efficient reflectors and parasitic elements shrink as well. This is the reason that weather and aviation radar operate at and above 3GHz (9cm wavelength).

At these frequencies objects the size of small aircraft are highly reflective, and objects like dense foliage and heavy precipitation may as well be a highly polished mirror. While 9cm is appreciably larger than a single raindrop or leaf, the shear number and coverage ensures a significant RF return.

You can witness this effect for yourself by standing between your wifi router and PC while watching the signal strength meter. While operating on the ham bands, I can easily reach a receiver 10 or 15 miles away with only 500mW at 50MHz. 500mW at 2.4GHz with a decent dipole antenna would be half a mile under perfect circumstances. This is both due to scattering by objects nearby, and atmospheric attenuation. The reason freqs above 2GHz are used for 802.11a/b/g has more to do with bandwidth, but thats a topic for a different thread.

Don't confuse feed point impedance/radiation angle with poor radiation efficiency. For antennas to behave 'ideally' (radiating near the horizon, where the RF is effective for LOS communications), they need to be at least 1/2 wave above the ground. Below this height, interaction with the ground can adversely affect radiation pattern as well as feed point impedance. Given a proper impedance match between transmitter and antenna though, a similar amount of power will be radiated. Its just going to tend to be radiated straight up.

Also, the higher the frequency at a given amplitude, the GREATER the energy. Big broadcasting stations use low frequencies because of the desirable propagation characteristics (i.e. ground wave or skywave), as well as the reduced influence of small objects like foliage and buildings which could prevent reception.

Dave

Dr. Slack

Tue Nov 27 2007, 08:34AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Guys, I don't believe any of these questions require answers bigger than one post.

Oh, if only that were true ... but I'll try to add one to the others

There's an important distinction between near-field and far-field. Near field consists of static and evanescent modes, which are due to just electric or just magnetic effects, they do not propagate, they carry reactive power. They fall off faster than r^2, so are strong near the antenna and insignificant more than a wavelength away. Far-field transports energy and consists of both electric and magnetic field at right angles in phase quadrature (wikipedia for Poynting vector). They fall of at r^2, so actually have the chracteristic of a fixed an=mount of power being reduced in intensity by being spread out on the surface of an expanding sphere surface.

What Chris said, about the current in a dipole being signficiant, is true enough for an electric dipole, which is the type we meet overwhelmingly. But it is incomplete. You can also get loop aerials. A small loop will act like a magnetic dipole and also be a poor transmitter. Although there is current flowing, a small loop will not have the inductance to support significant voltage differences, and so still be a poor radiator, just as a short electric dipole will not have the capacitance to draw significant feedpoint current. It is only when the structure is big enough and both electric and magnetic fields are significant, that it becomes a good radiator.

What I am trying to point at is that for the radiation of power, that is real energy transport, you need both voltage AND current, and it is their product that launches the far field. The simple magnitude of either stores energy locally in the near field.

You can transport energy using the near field, but only with a physical receiver. For instance two induction loops coupled to act as a transformer. The real current drawn in the primary is as a result of currents in the secondary coupled into it. In a radiating antenna, real primary current is as a result of the radiation emitted, regardless of what it strikes.

WaveRider

Tue Nov 27 2007, 08:43AM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Hi Marko,
Sorry if I sounded a bit harsh in my last post... I did a little searching on the web and found this page that has a movie of the fields radiating from a dipole.

The key to radiation is that the "time retardation" or phase of the currents need to match closely the time retardation (phase) of a corresponding EM field that radiates well.

Also, notions of voltage on radiators in antennas have little meaning since the electric scalar potential cannot be defined over lengths that correspond to a wavelength.... Currents, on the other hand can be uniquely defined and all wire antenna models solve for currents...

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