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Principles behind how Waveguides work.

  1. Aug 15, 2012 #1
    So I am studying about waveguides and I am going through all this mathematics and then I realized that I do not understand the general concept of how wave guides even work. I can go through the math all I want but I don't understand one thing..

    If a wave is radiated in free space and is unrestricted, then it travels and attenuates as an inverse square law. Why is it that once we put the radiation source in a wave guide, it does not decay and can travel long distances?

    In the beginning of the mathematics, the book starts with the wave equations for E and H derived from Maxwells Equations. How ever it is the wave equations for a lossless condition. Why can we assume a waveguide is lossless? Shouldn't they attenuate as an inverse square law just as if they were radiated in free space un-restricted?
     
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  3. Aug 15, 2012 #2

    Drakkith

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    The wave attenuates because it spreads out into space. In a waveguide, if it works the way I think it does, the wave cannot spread out so you would lose no intensity.
     
  4. Aug 15, 2012 #3

    marcusl

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    It is not true that a wave traveling in free space is attenuated. If you integrate the power flux density through a sphere of large radius, you will find that all of the power from the source is still there.

    If you measure the field at a given point, however, it will diminish as 1/R^2 (assuming that you are in the far field) because the wave spreads as it travels out. Remember that field intensity is proportional to the density of flux lines, and the flux density falls with distance.

    An ideal waveguide prevents any spread (obviously the fields stay in the box) so the peak E (or H) field strength is the same everywhere along its length. In practice, copper's conductivity is high but not infinite so there is a small loss down a waveguide. I seem to recall that the loss is computed using an approximation or expansion, if you are interested in actual values. I'd have to look it up to be sure. Jackson's book has the calculation, also books on waveguides like those by Ragan or Marcuvitz.
     
  5. Aug 16, 2012 #4
    Thank you for your reply, I have gained some insight. But I still do not understand one thing. Why is it the field does not diminish in the wave guide as it does in free space? Is it because of the reflections on the conductive material? Does the wave constructively interfere with itself or something?

    I will be purchasing Jacksons book soon for my next course in E&M.
     
  6. Aug 16, 2012 #5

    marcusl

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    Yes, you can think of a wave (TE mode, e.g.) propagating down a waveguide as the superposition of two plane waves that zig-zag down the length by bouncing off of the walls. If the walls are perfectly conductive, then there is no loss.

    This decomposition used to be commonly shown in early texts on waveguides (Sarbacher and Edson, 1943 covers it very nicely in sec. 5.15), but it isn't seen much any more. That's a shame, since it also gives insight into phase and group velocity, and what happens physically in the waveguide at cutoff. BTW, these authors credit Leon Brillouin for first looking at waveguide propagation this way.

    For a more modern reference (but briefer treatment), see Collin, Foundations for Microwave Engineering (1966), on p. 105. (You'll also find the waveguide attenuation calculation on p. 101.) Either of these two books may help you intuitively understand waveguide propagation.
     
    Last edited: Aug 16, 2012
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