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Transmission lines and Waveguides

  1. Dec 24, 2011 #1
    Ok, it's clear to me that waveguides are for sending EM waves from one place to another (e.g. TEM, TE, TM modes).

    But what about TLs? I've seen them described as carrying AC signals (V or I) but also EM waves, e.g. TEM waves in coax cable. What gives? Does one imply the existence of the other? When the coax cable's signal enters my TV, what is the "data" - the V/I waveform or the EM waveform?
     
  2. jcsd
  3. Dec 25, 2011 #2
    It is EM wave that travel down in the dielectric. The current is the consequence of the boundary condition of the H field which is perpendicular to the direction of propagation( down the coax).

    [tex] \nabla \times \vec H = \vec J +\frac {\partial \vec D}{\partial t}\;\;\hbox { and }\;\; \vec I = \int_S \vec J \cdot d\vec S[/tex]

    The velocity of electrons moving down the coax is very slow, it is not from the movement of the electrons that create the current flow. It is only because of the EM wave propagation at close to the speed of light that's why you see the apparent current flow instantaneously.
     
    Last edited: Dec 25, 2011
  4. Dec 25, 2011 #3
    In any waveguide or transmission line you can define a set of terminal at which there is an I and V. At these terminals you typically interface a circuit.

    In reality there are I and V that could be defined all along the guiding structure but they are not helpful to describe the propagation. You need the full fields for that calculation.
     
  5. Dec 25, 2011 #4
    Good point about I and V is defined at every point along the Tx line as the consequence of the E and H as you can see the differential form of Mexwell's equation is in point form. It is not that I and V are moving even though it looks like they are moving in very high speed. That's why you can measure the voltage at the end of the Tx line and get current if you put a load at the end.
     
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