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Need clarification in understanding how an e-field detaches from a dipole antenna

  1. Apr 9, 2010 #1
    In my electromagnetic engineering class we were talking about Hertzian dipole antennas with an oscillating current source (transmission line) driving the antenna. As part of the description of the behavior as the two ends of the antenna have opposite charges an electric field is created from the positive charge to the negative charge. Since the current is oscillating through the antenna at one point in the middle the charges become neutralized and the e-field has already propagated some distance into space so it detaches but remains continuous. I'm having difficulty in conceptualizing how this e-field (and h-field) can exist in a state where they are neither bound by charges nor propagating as a plane wave. Could another engineer or a kind physicist help a student understand this phenomena better?
     
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  3. Apr 9, 2010 #2

    sophiecentaur

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    It's a really hard concept (I reckon so, anyway).
    For a start, the current and voltage on the dipole are in Quadrature but the E and H fields in the radiated wave are In Phase. Actually, the I and V on the radiator are not exactly in quadrature; there is a resistive (in phase) component which is referred to as the radiation resistance. There is a region where radiated and resonant fields exist (the near field) and this is all a bit of a blur to me! The Maths looks convincing but I have never followed it through thoroughly.
    My Dad, years ago, gave me a way of looking it at it and that is- the field at a distance takes time to change as the antenna currents / volts change, being limited by the speed of light. The fields will, therefore be retarded wrt the radiator. The energy in the field at a distance 'would' flow inwards and outwards BUT the fields between this distant point and the radiator will interfere with an inward flow and promote an outward flow (because of the slight phase difference). Not very rigorous but my arm waving may help a bit, I hope.
     
  4. Apr 9, 2010 #3
    Have you ever done a physics experiment where you make strings vibrate? For certain frequencies, you'll get a standing wave. Specifically, if the string is 1/2 of the frequency's wavelegth, you'll get a simple standing wave and the whole string moves up and down in unison.

    Something similar happens in an organ pipe. The pipe has a resonant frequency too and, if the pipe length is 1/2 the wavelength, then the air molecules will compress and decompress inside the pipe in unified wave-like pattern.

    The antenna is just like the organ pipe. If you give the antenna one of it's resonant frequencies (n*wave-length*1/2; for n > 0), the electrons will vibrate up and down in a unified pattern.

    You might ask, what happens if we send a signal to the antenna that is not a resonant frequency? Theoretically, nothing will happen. The wave will go into the atenna, down to the end, bounce back, and return to the source it came from. At any give point and at some instance in time, the non-resonant vibrations will tend to be random so the fields outside the antenna will tend to cancel out.

    In the real world, you will get a little bit of transmission at any frequency but the dipole will be most efficient at its theoretical resonant frequencies. The reason is that any real dipole is coupled to a complex environment which complicates its possible resonant modes.

    Does that help?
     
  5. Apr 9, 2010 #4

    sophiecentaur

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    That just has to be an oversimplification. "Theoretically" is not the correct term to use here because the 'theory' says just the opposite. Any bit of wire will radiate. A resonant dipole, even, has a very broad resonance because it is damped much more than an organ pipe or vibrating string. The resonance is not the aspect that causes radiation - it just increases the efficiency. You can discuss the whole matter of energy radiation by considering short dipoles or small loops which are very inefficient.
     
  6. Apr 9, 2010 #5
    It's a simplification but it is correct. If you have a forward travelling wave and a reflected backward travelling wave that are not in phase (non-resonance), then they will tend to cancel each other out on average. I was using this analogy to describe how the electrons will literally move in a pattern that creates the E field and H field but it does require the right conditions.

    Sure, but we're talking about 1/2 dipoles.

    That's a simplification too. If it's in a balanced transmission line (untwisted wire pair for example) then the fields in each wire will balance and cancel each other out. If you're very far from any kind of balanced transmission line then the EM wave will be negligible. You can unbalance a transmission line with a balun device but that means the wave will be reflected back at some point unless you have perfect impedance matching at the end point (ie, a resonant antenna).

    If you hook a dipole up to a VNA(Vector Network Analyzer), you'll see that the reflection magnitude is extremely close to unity for frequencies below the first resonant frequency. That means it's transmitting next to nothing. It's gets close to unity between the resonant frequencies too.
     
  7. Apr 10, 2010 #6

    sophiecentaur

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    Any structure will radiate. Some more than others- even the proverbial 'dead sheep'. A dipole (thin wire) has a 3dB bandwidth of about 5%. Once you make it fatter or conical, you can increase the bandwidth to almost an octave. You just can't say that "nothing happens" off frequency. The efficiency goes down when not tuned but the real part of VI still produces radiated EM waves.
    It struck me that this the meat of the OP's question - how does the energy actually get radiated? Resonance is an added factor which improves performance and adds complication.
     
  8. Apr 10, 2010 #7
    Your right, I got away from the op's question. How about this? Think of an electron as a person holding the end of a rope in their hand. The rope is an analogy for electric field lines that extend out from an electron. If the person waves their hand up and down then waves will travel outward along the rope. In an antenna, the electrons are literally vibrating up an down an sending waves out. It's a simplified analogy because it doesn't bring up the magnetic field but I think it's pretty good.

    You know, early pioneers in EM like Faraday believed that field lines were literal physical things similar to this rope analogy.
     
  9. Apr 10, 2010 #8

    sophiecentaur

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    Except it's not a matter of just the electrons moving up and down. You do know how 'far' an electron will move during one cycle of 10MHz, don't you - the peak velocity will be a few mm per second, so, in a fraction of a microsecond . . . . . . Adding the electrons to the explanation just makes it more complicated. I know kids are taught that everything is particles, these days, but it really hasn't done them any favours when trying to understand simple classical phenomena.
     
  10. Apr 12, 2010 #9
    It's similar to a drop hitting a pond. The waves will eventully get far away nd straighten out into plane waves. Near the source though there is a lot of curvature and a lot of "reactive" energy. That means near the antenna the energy doesn't have a well defined direction in space. Some of it is reentering the antenna too. This complexity is called the near field. It gets weaker with distance until only the outward going waves
    remain.
     
  11. Jun 2, 2010 #10
    I'd like to thank everyone who replied to this thread. I apologize for not replying sooner, finals and the end of the semester got the better of me. All of your explanations have been very helpful. I still find the phenomena somewhat difficult to conceptualize, but it could be in part that I am very inclined to view the phenomena from a particle standpoint. The analogy of the rope makes a lot of sense although it does not completely describe the phenomena at hand. I guess that's the nature of any good analogy, it has its limits.
    Thank you again for your replies.
     
  12. Jun 11, 2010 #11

    sophiecentaur

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    The problem is that trying to view the phenomenon from a particle standpoint opens up a whole new can of worms. How big would they be, for a start?

    If you consider the history of Scientists' views of light you can see that they, reasonably enough, used the 'corpuscular theory', initially, based on vaguely defined tiny bullets that came from a source and bounced into our eyes (that's after they had discarded the 'tactile theory'. in which we had sort of feelers which came out of our eyes and touched things, in some way, to make us aware of them!).

    Once the phenomenon of diffraction was noted, the wave nature needed to be considered. This works very well and explains more or less everything we experience - including light pressure. Quantum Theory re-introduced the idea of packets of energy but, in itself, doesn't really involve 'particles' at all, despite what is said. It just involves discrete bundles of energy.
    If you want to use a particle mode, how would you envisage the 'particles' radiated from a 1500m radio transmitting antenna, for instance? Would they be microscopic or 1500m radius?
     
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