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How/Why does antenna voltage reflect?

  1. Nov 19, 2013 #1
    I'm trying to understand antenna theory again, and i'm again stumped by the concept of VSW (voltage standing waves). I understand standing waves, I remember these from a level physics, however I do not really understand how voltage can be a wave, and how it reflects.

    From my understanding of voltage it is really hard to understand how it can reflect.

    a voltage is the difference in electrical potential energy (per unit charge) between two places. This electrical potential energy (per unit charge) difference describes the force on that unit charge (due to the electric fields), and the distance this charge will travel.

    For example, with an electrical potential energy (per unit charge) difference of 10V or 10J/Q, an electron at the higher potential will travel to the lower potential. If the distance between these two potentials is 2 meters, the force on the electron will be 5N, as it travels from one potential to the other. (If the electric field is constant between the two potentials. If it is not, the integral of the force across the 2 meters will still equal 10, however the force will vary at different distances)

    Voltage is just a made up concept to help explain the forces that charged particles experience. It isn't a real thing that can be reflected. So what is being reflected? I'm guessing it is the electric field, as this has an actual propagation speed, and even though the electric field is also a made up concept to describe the force a charged particle will experience at that point in the field, it is easier to imagine this being reflected.

    So can someone please explain to me how this all ties together? How does an electric field propagate through a conductor? Do the vectors in this vector field change direction when there is a bend in the wire, and why do they do this? Why wouldn't the electric field escape outside of the conductor? And how does an electric field (or voltage) reflect at a short?

    Thanks for reading.
  2. jcsd
  3. Nov 19, 2013 #2
    Oops, I've posted this in the wrong place. Can someone please move it to Engineering - Electrical Engineering.
  4. Nov 20, 2013 #3
    Moving charge (current) produces magnetic field perpendicular to the charge's direction. Electricity and magnetism are the same thing called electromagnetism. http://en.wikipedia.org/wiki/Ampère's_circuital_law
  5. Jan 2, 2014 #4


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  6. Jan 3, 2014 #5


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    I'll try to answer this to the best of my information.
    1. In antenna, a current flows through the cables even though there is an open circuit at the other end. This happens due to the transit time effect, wherein the signal changes more rapidly than the speed at which the signal travels.
    2. since there is no load and the line is assumed to be lossless, there can be no net transfer of power. Therefore, in a certain sense, the power that goes from the supply must end up coming back to it in some form or the other.

    The only way to account for both these phenomenon is that the distribution must be that of a standing wave, that carries no power. In turn, the standing wave can be viewed as the sum of two waves travelling in opposite directions. So from this perspective, reflecting wave seems more of a mathematical quantity used to account for the formation of standing waves

    There's a physical explanation too: As long as a wave is travelling inside ur transmission line, it 'sees' only a characteristic impedance. However, when it comes at the end, it sees a drastic change in impedance and it is unable to transfer all of the power it has. So only a portion of the voltage/current can be given to the 'load', the other must return back to the source, thus forming the reflected wave.

    Reflected waves are more to do with power than simply the current or voltage.

    Just my two cents :)
  7. Jan 3, 2014 #6


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    The classic video:
    Last edited by a moderator: Sep 25, 2014
  8. Jan 3, 2014 #7
    For Standing waves I think of more in terms of Conservation of energy -and it may be helpful to first think of a pulse or step wave - the step travels down the antenna and reaches the "end". From a Cons of E standpoint consider what happens when the end is A) Shorted, B) open - these two boundary conditions tell a lot. Considering of course that the antenna has a finite capacitance, a charge can be induced on the conductors, as the charge is induced, we have applied a voltage, and a current flows, therefore a small amount of energy (power) is present.
  9. Jan 3, 2014 #8


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    Brilliant. Playing with real toys beats computer simulations every time. And the lecturer obviously LOVES playing with those toys!
    Last edited by a moderator: Sep 25, 2014
  10. Jan 3, 2014 #9


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    You could think of it in terms of the old Maximum Power Theorem from schooldays. If you have a source of power with an internal resistance of R, then the load that will accept the maximum power will have a resistance of R. When there is a reactive component, the maximum power is transferred when the load has the complex conjugate impedance of the generator. When the load is matched to the line, the maximum power will pass through the junction.
    If the power is not all matched into the load then it has to go somewhere so it will be reflected. This applies whether the signal is a sine wave, a short pulse or a step function. Energy is conserved at the interface, as has been said before. (Although an antenna can have a Capacitance, Inductance or pure resistance - depending on the design and the frequency of the arriving wave.)
  11. Jan 3, 2014 #10


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    Sophie pretty much nailed it. Somehow transmission lines introduce confusion. The load at the end of the transmission line still behaves the same way RC and RL networks do in the early school days of AC theory.
    I recall in my early days of HAM radio the various views from different hams on SWR. Some said you HAVE to have the SWR down or you will wreck the transmitter. Others said that this is non-sense. I didn't really know which to believe until it was put in this way:
    It is strange how lowering the frequency to audio and driving speakers and resistors can make it more clear but that is exactly what happened to me. Take a class B amplifier driving a resistor. Take a look at when the power dissipated in the output transistors and we find that when the transistor is outputting the highest voltage it is also outputting the highest current due to the resistive load. Voltage and current are in phase. Also note here that when the most voltage is dropped across the output transistors (zero output) there is NO current flowing. Now lets drive a pure capacitive load where the current is leading the voltage by 90 degrees. This would be about the worst SWR you can have on a transmission line/antenna system. Now when the output transistors are outputting zero volts they are passing maximum current. They are dissipating a lot more power than when driving a resistive load. When put to me in this manner it all made sense. Hope it helps someone.
  12. Jan 3, 2014 #11
    This. Is. Awesome.

    He is an example of another Faraday. Science through experimentation.
    Last edited by a moderator: Sep 25, 2014
  13. Jan 3, 2014 #12
    There is an easy way to make a wave machine like that yourself with some coffee stirrers and masking tape. Google "wave tape physics" for some other construction examples.
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