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E-field and H-field in a waveguide

  1. May 2, 2010 #1
    I find it a bit hard to visualize E field and H field inside a waveguide, specially in modes other than TEM.
    I know E and H are perpendicular to each other and to the direction of propagation. But I need to see some examples or something like a right hand rule.
    Basically, I am confused.
  2. jcsd
  3. May 2, 2010 #2
    If you've got access to an engineering library there are a few textbooks that depict field lines in waveguides for varous modes. It's a rare thing though, you'll have to look through a lot of books.

    P.s. They are all mutually perpendicular for TEM modes only. This isn't the case for other modes. Maybe that helps?
  4. May 2, 2010 #3


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    That is the direction of GUIDED propagation. Unless you have the source of the waves included inside your waveguide, then you are always going to have waves that are TEM to the true direction of propagation (assuming the usual caveats, source-free isotropic homogeneous). People get lazy when it comes to talking about the direction of propagation because there are two in this case. The direction that the wave is actually traveling and the direction of guided propagation.

    What actually happens is that a wave must enter and propagate through most waveguides at an angle. The waves come in at an angle and bounce off of the sides of the waveguide. This is done so that the waves satisfy the proper boundary conditions. As the wavelength changes, assuming that we are still in the same propagating mode, the angle of incidence with the sides changes to keep the wave satisfying the boundary conditions. Even though the wave is bouncing around, it maintains a net displacement along the direction of propagation. Since the direction of propagation is often the only thing that you care about in the end, people refer to TE, TM, and TEM modes of a waveguide in terms of being transverse to the direction of guided propagation, not the actual direction of propagation.

    Look at slide two in this lecture: http://www.amanogawa.com/archive/docs/EM15.pdf

    The direction of actual propagation is the black vector. You will notice that the H field on the left and the E field on the right are still transverse to the k-vector, but not transverse to the direction of guided propagation. As the wave travels, indicated by the thin red line, the wave bounces back and forth. Take a look at the rest of the slides too.

    For complicated waveguides, like Antiphon stated, you may be able to find plots of field distributions in textbooks, like Balanis'.
  5. May 2, 2010 #4
    Hello Born2wire

    I seem to recall that a rectangular wave guide can be modeled as an infinite array of waveguides with the short walls side by side. Number them sequencially ...-2,-1,0,1,2,3... The signal in every other guide is out of phase. When the current in the walls of the even numbed guides is upward at any given cross section, the current in the odd numbered guides is equal but downward. The currents cancel, so the walls are emiminated, and the boundry conditions are still met.

    If we happened to have a guide 1/2 lambda wide the fields inside the guide would be represented by two planar waves propagating +/-30 degrees from the axis. Is this still a valid model for other modes?
  6. May 3, 2010 #5


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    I am not familiar about that model but I think I may understand what they are talking about.

    Taking a parallel plate waveguide of separation a along the x direction and assuming that the guided direction is in the z direction, then for the TE mode we find that the wave equation becomes:

    [tex]i\omega\mu\left(\frac{\partial}{\partial z} H_x - \frac{\partial}{\partial x} H_z \right) = \omega^2\mu\epsilon E_y[/tex]

    Since the boundary conditions dictate that the tangential electric field be zero when x=0,a, then assuming plane wave solutions,

    [tex] E_y = E_0 \sin \left(k_x x\right) e^{ik_zz} = -\frac{iE_0}{2} \left( e^{ik_xx}-e^{-ik_xx} \right) e^{ik_zz} [/tex]

    So what we can see in the parenthesis is that we have a combination of upward and downward traveling waves (taking x as the vertical direction). An infinite sheet of current produces an electromagnetic plane wave whose electric field is polarized along the direction of the current. Thus, we probably could think of your infinitesimally thin currents as, alternatingly, creating the upward traveling wave and downward traveling wave. These are then scaled by a plane wave travelling in the z direction. So what we have is a superposition of two plane waves. One is travelling upwards at an angle of \theta from the normal of the top conductor and a second wave is travelling downwads with an angle of \theta from the normal of the bottom conductor. The resulting dispersion relation for the two plane waves is

    [tex] \omega^2\mu\epsilon = k_x^2+k_z^2 [/tex]

    where, due to the boundary conditions,

    [tex] k_xa = m\pi[/tex]

    So that the sine function is zero when x=0,a. The m here is the mode number. Using our angle theta again,

    [tex] \mathbf{k} = k_x\hat{x} = k_y\hat{y} = k\cos\theta\hat{x}+k\sin\theta\hat{y} [/tex]


    [tex] k_x = k\cos\theta = \frac{m\pi}{a} [/tex]
    [tex] \theta = \cos^{-1} \left( \frac{m\pi}{\omega\sqrt{\mu\epsilon} a} \right) [/tex]

    So we can see that for a given mode m, as we adjust the frequency of the wave then the angle that the wave propagates at must adjust accordingly to preserve the boundary conditions.

    So the idea that we have two waves propagating at +/- \theta from the axis is valid. It's just that the \theta is dependent upon the frequency, mode, and the dimensions of the waveguide.
  7. May 3, 2010 #6
    Guys, how do you draw H field, given E-field.
  8. May 3, 2010 #7
    I'm startled at how quickly you managed the derivation. The half can I follow makes sense. I solved it graphically, but only for m=1.

    It roughly models an array of stub antennas, alternating 180 degrees in phase. So it models something like a diffraction grating at well. So it would makes sense that theta depends upon frequency.

    I think the wave velocity is had from theta:

    [tex]w = c/sin(\theta)[/tex]
  9. May 6, 2010 #8
  10. May 6, 2010 #9
    You'll have to be a little more specific--what sort of guide, and what sort of mode.
  11. May 8, 2010 #10
    Rectangular waveguide - TEM and TE10 mode.
  12. May 8, 2010 #11
    Wikipedia says that a rectangular guide does not support TEM, which makes sense, but others seem to have a different definition of TEM. You should be able to construct and obtain field values for TE10 by a superposition of planar electromagnetic waves via the construction born2bwire and I were discussing above.
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