# Working of antenna

1. Oct 16, 2011

### erece

We all know that current can't flow in open circuit. Then why does current flow in the wires of half wave dipole antenna ? I am confused , can't visualize.

2. Oct 16, 2011

### yungman

Current in a linear antenna is very complicated, most of the books use only approximations. I half wave antenna, people use current phasor to approximate the current along the wire. Obviously the current is zero at the end, this will create a standing wave along the body of the antenna. This is due to the current at the end has no where to go and reflected back. When the incident current phasor sum with the reflected current, it create a standing wave along the antenna.

to understand this, you have to have knowledge of EM and phasors. This is all because it is really EM wave that travel down the antenna. the current is only the consequence of the boundary condition of the metal and air boundary. and if antenna is in air, the speed is about 3EE8 meter per second. Say if you have a half wave dipole antenna, each side is quarter wave long. Say if the antenna is for 100MHz, it would take 2.5nS for the EM wave to travel from the injection point in the middle to the end of the antenna. Current distribution is not even along the antenna.

Last edited: Oct 16, 2011
3. Oct 16, 2011

### sophiecentaur

That statement is far too simplified. By that argument, no circuit involving series capacitors could ever work.
Any piece of wire can act as an antenna, wherever you connect the RF power. Some will work a lot better than others.
Consider the familiar picture of a dipole, fed in the middle. This proves that current can flow into the feed point ('cos the damn thing works). If the wire is much shorter than a wavelength, the two ends of the wire act pretty much like the plates of a very low value Capacitor. But, because they have finite length, some current travels along the wires and takes time to reach the ends - whilst tending to zero at the ends (nowhere to go). Each time the direction of the RF reverses, current sloshes back in the other direction. The phase of the current is pretty much in quadrature (90o) with the driving voltage - but not quite. The quadrature bit represents reactive energy stored in the capacity but a tiny in-phase component represents energy that is actually dissipated (lost) due to radiation into space. If the dipole is about a half wavelength long, the dipole shows zero reactance and just 'looks like' a pure resistor. It radiates the RF power very well at this resonant length.

E and B fields close to the dipole are mostly in quadrature but, at a distance, the in-phase components are the only ones that exist and they 'carry' the radiated em energy.

4. Oct 16, 2011

### gtacs

5. Oct 17, 2011

### davenn

yes a good link and also within it some out of PF site links to dipoles for dummies.
although only links for parts 1 and 2 were given, you can when on the link replace the 2 with a 3 and get part 3 then the same for part 4 :)

Now just to throw a spanner into the works .....
nowhere in that previous discussion was mentioned that there are 2 wavelengths to worry about. There is the freespace wavelength calculated by the speed of light, c / frequency
which seems to be what they were all talking about. But in the metal of an antenna or coax cable etc there can be a significant difference in velocity.

For a nominal open wire dipole the freespace wavelength needs to be multiplied by the velocity factor, in this case ~ 0.95 (ie 95%). The majority of coax cables vary between ~ 0.55 and 0.80.
This wavelength is called the "Electrical Wavelength" and becomes very critical when dealing with freq's from a few 100 MHz and upwards. Even by the time you get to 1296MHz ( one of the lower freq's of the amateur radio microwave bands) even a millimetre or 2 in difference in length of a dipole, can make a huge difference to what freq an antenna will resonate on.

And whilst talking about antenna lengths and resonance ... just something else to contemplate.....
Say you want to put youor antenna inside a plastic or fibreglass radome to protect it from the weather. You have now just increased the capacitive effect between the antenna elements and the air.
increasing capacitance = a decrease in resonant freq. That means the antenna now appears to be too long for the desired freq of use. So you have to cut the antenna elements a little shorter to bring the resonant freq back up to where its needed.

Ohh the joys of antenna construction :) I have been at it for 30 something years

cheers
Dave
VK2TDN

6. Oct 17, 2011

### yungman

I don't get the two wave length. There is only the EM wave on the surface of the metal rod, there is no EM inside the rod as EM don't penetrate into the metal. This is not like coax where the EM wave travel inside the dielectric medium between the inner and the outer conductor.

The surface current is only the consequence of the boundary condition between the perfect conductor( well say metal) and the free space ( well say air). The EM wave travel in air, I don't get the 0.95 factor.

7. Oct 18, 2011

### davenn

The velocity of a RF current in/along a wire (conductor) is less than in free space, where it is the speed of light.

I will try to dig up some references. I dont know the physics behind it, its not my field.
I just deal in the day to day facts of what happens whilst playing RF technician.

cheers
Dave

1 basic reference.... there's lots on google

http://en.wikipedia.org/wiki/Wave_propagation_speed

Tho the effect is most noticeable with insulated wire. It does become obvious at UHF and microwave frequencies for bare wire

Last edited: Oct 18, 2011
8. Oct 18, 2011

### yungman

Thanks for the reply, I looked at the article and I understand that. But what we are talking here is the EM in air along the dipole antenna, the velocity should be the velocity in free space according to the formula in the wikipedia. If the antenna is immersed in other dielectric material, then we have to use the $\epsilon$ of the dielectric to put into the velocity formula. But you have years of experience, I must be missing something. I am studying antenna.........from the books, nothing like real world experience and can never beat the real life experience.

Thanks

Alan

9. Oct 18, 2011

### erece

thank you all for your replys... I want to know that why EM waves travel. I mean what causes them to travel from one antenna to other antennas. What is the driving force behind it?

10. Oct 18, 2011

### sophiecentaur

The different speeds of waves on the antenna and in space will affect the radiation pattern of the antenna. All antennae are 'slow wave structures'. If you want n extreme example of this, look at a Beverage Antenna, which consists of a long wire, suspended on poles along the ground. They can be hundreds of metres long. The antenna receives medium and low frequency signals and relies on the forward tilt of a (VP) ground wave as it goes over ground with finite resistance. An emf is induced from the ground wave onto the tethered wave on the wire and gradually builds up over the length of the wire - giving you Gain in the direction of the wire. The wave on the wire goes a bit slower than the wave it's receiving so, after a certain length (governed by wire height ground characteristics) the two waves get out of step and the pickup drops.
Another example of a 'slow wave' antenna is the 'normal mode helix' which is common on walkie talkies. It looks like a spring covered in plastic (which is what it is). You use about 1/4 wavelength of wire, coiled up into about ten cm of plastic sleeving. They work very well, 'considering'.

11. Oct 19, 2011

### yungman

I am just studying antenna, this is my understanding and if anyone think I am wrong, please join in. I am using a very short dipole in this example as it is easier and I only consider FAR FIELD. Let the antenna rod in +z direction:

1) Voltage at the center feed of the dipole launch a EM wave into the air along the surface of the antenna rod. The boundary condition cause a current standing wave on the antenna rod.

2) The varying current on the surface create traveling in +z direction generate a circling magnetic field with center at z and parallel with xy plane with the right hand rule ( your thumb is +z, your 4 fingers is the direction of the magnetic field). The formulas for this is:

$$\vec A = -\frac{\mu_0I(t_r)}{4\pi}\left(\frac {e^{-j\beta R}}{R}\right)\sin\theta\;\;\hbox { and magnetic field }\;\; \vec B=\nabla \times \vec A$$

Where R is the distance from the origin to the point you are measuring in spherical coordinates. $\theta \;$ is the same as in spherical coordinates. The magnetic field is in B. Don't worry too much about the formulas, just know that the current distribution along the antenna rod producing an Magnetic field $\vec B$ that is circling around the rod and travel out.

3) In varying magnetic field, there is always an $\vec E$ accompanied along with it with the formula:

$$\vec E = \frac 1 {j\omega \epsilon_0} \nabla \times \vec H$$

Again, don't worry about the formula, just understand the current distribution on the rod cause a circling magnetic field around the rod which cause an electric field that propagate out into the air. This is the nature of the EM wave, once you launch it, it will propagate out into the space.

This is my understanding, I hope someone can have a better answer. What I presented is for a very short dipole antenna, a half wave antenna can be approximated as infinite number of these short dipoles stack on top end to end and the field is integral of each small dipole.

12. Oct 19, 2011

### sophiecentaur

A simple explanation of why a wave is actually propagated:

When the field at the antenna changes, it takes time for that change to reach a point at some distance away. By the time that new value of field has been reached at that distance, the field at the antenna has changed once more. So the changes at a distance never quite keep up with the changes at the antenna. This forms a wave, which propagates outwards from the antenna and varies in time, as the frequency of the transmitted signal and the wavelength (distance between peaks) will be equal to the speed (c) divided by the frequency.
You can make a simple model of this process by waggling one end of a rope and seeing how the up-down motion of the rope lags behind your hand movements and the motion of the rope forms a wave, moving away from your hand. Notice that the faster you waggle your hand, the shorter is the wavelength.

13. Oct 19, 2011

### yungman

Thanks for your reply, I think this sounds like traveling wave antenna that I am not there yet!!! I'll read through that first. I am still working on this!!!........Better yet, I have been on strike for the last 3 weeks doing some guitar electronics design, need to get back to the studying soon.

14. Oct 20, 2011

### sophiecentaur

I was just making the point that the speeds are different - so that makes the wavelengths different - one in free space and one on / around the metal. "Travelling wave" is a term used for antennae when the difference in the two wavelengths becomes significant.

15. Oct 20, 2011

### yungman

Thanks, I have not learn this yet. I am sure I'll have more question by that time.

16. Oct 21, 2011

### erece

still not clear ??

17. Oct 21, 2011

### sophiecentaur

The change takes time to reach a distant place. If you plotted the value of the field from antenna to that point, you would get a sinewave that has the wavelength of the radiation and, obviously, the same frequency as the radiation. If things are not instantaneous, you must have a wave, as the effect moves outwards.
I could ask what alternative picture can you invent in your head that satisfies the basic requirements? When you find you can't come up with one, the wave explanation will seem more appealing, perhaps.

18. Oct 21, 2011

### cmb

I think the one aspect not well drawn out in this, and the other recent thread, is why more electric current feeds into the antenna. It's fine to picture the electrons sloshing around back and forth, but why don't they just oscillate endlessly, rather than pumping the antenna with more power?

This is because energy is decopled away from the antenna. Why is this? Well, if you consider the very first pulse of current passing up along an antenna element, it passes through a region of space in which there were no previous currents, and thus no magnetic fields. However, as it stirs the electrons at the end of the element and 'washes' back, it will meet a stream of current flowing towards it.

This new pulse of current leaves a magnetic field behind in its 'wake', which the first current pulse passes through.

It is this combination of current in the antenna and magnetic field from the next cycle that causes coupling of the energy of that current into free space. They intermodulate and can then propagate through space.

It is not necessary to have such long antenna structures. It is also possible to apply an alternating electric current to an antenna strucutre and artificially, by magnetic means, apply an alternating magnetic field around it to 'simulate' what would happen in a much longer antenna. Efficient EM propagation and coupling to free space can then be achieved for structures much smaller than wavelength.

Many years ago in school I imagined this solution and I dreamed of trying this out when I grew up. When I did, I found out someone else had also thought it up, and tried it out, apparently successfully, before me!!

I should add that there is still some question marks over the performance of an antenna made like this, though.