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Transverse resonance method

  1. Nov 24, 2015 #1
    The Transverse resonance method is used to determine the propagation constant of a wave in several waveguides, like the rectangular waveguide, or also dielectric waveguides.

    It takes advantage of the fact that a standing wave is present along a certain direction (transverse with respect to the main propagation direction), due to purely reactive loads at both the ends of the transmission line (which represents the wave propagation). http://www0.egr.uh.edu/courses/ece/ECE6351-5317/SectionJackson/Class%20Notes/Notes%2013%20-%20Transverse%20resonance%20method.pptx [Broken] provides and example in the slides 3-4-5.

    In such conditions, the transmission line (ended with the reactive loads) is said to be a resonant structure. It means that [itex]Z_{in}^r (x)[/itex] looking forward must be equal to minus the [itex]Z_{in}^l (x)[/itex] looking backward, so that [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] for every position [itex]x[/itex] along the line. Why is this the resonance condition? Which is the relation between this condition and the existence of a standing wave?

    I tried to consider a transmission line with total length [itex]2 \ell[/itex] centered in [itex]x = 0[/itex] and ended at both [itex]x = \ell[/itex] and [itex]x = -\ell[/itex] with a short-circuit [itex]Z_L = 0[/itex]. The input impedance in a generic position [itex]x[/itex] is actually

    [itex]Z_{in}^l (x) = j Z_0 \tan [ \beta (\ell - |x|)][/itex]
    [itex]Z_{in}^r (x) = j Z_0 \tan [ \beta (-\ell + |x|)] = -j Z_0 \tan [ \beta (\ell - |x|)] [/itex]

    so the assumption [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] seems to be verified. But what is the physical reason for this?

    In the linked slides this is derived from the continuity of voltage and current in the position [itex]x[/itex]: [itex]V^r = V^l[/itex] and [itex]I^r = -I^l[/itex], but this can be applied to every transmission line and for me has apparently nothing to do with resonance.
     
    Last edited by a moderator: May 7, 2017
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  3. Nov 29, 2015 #2

    marcusl

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    "In the linked slides this is derived from the continuity of voltage and current in the position x: Vr=Vl and Ir=−Il, but this can be applied to every transmission line and for me has apparently nothing to do with resonance."

    Are you sure? For traveling waves, i.e., when the line is terminated at each end in its characteristic impedance Z0, you see Z0 in each direction, that is Zr = Zl = Z0. It seems to me that Zr = -Zl instead produces infinite impedance at the point of interest, making the longitudinal current zero. This is consistent with a standing wave for which no power moves down the line.

    Having said that, the concept of "transverse resonance" seems odd to me.
     
  4. Dec 1, 2015 #3
    It may be a trivial question, but how can the current be zero? Can you show me the steps?

    Yes, it would.

    It seems odd to me too, but I have to dig into it :frown:
     
  5. Dec 1, 2015 #4

    marcusl

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    I take back what I said before about currents, power, etc., after looking at Pozar's book referenced in your link. Here's what I think.

    The resonance condition he talks about is transverse to the waveguide axis (you said as much in your post, but I got confused). In the transverse x direction, a rectangular waveguide looks like a parallel plate waveguide with shorts at either end, x=0 and x=a. For the fundamental mode, the waveguide width a corresponds to λ/2. We can evaluate Z looking in either direction by referring to a Smith Chart and reading the reactance off of the outer R=0 circle since you are assuming a lossless structure. For example, consider the left/right impedances at x=a/4, which is λ/8 from one short and 3λ/8 from the other. The Smith Chart shows us that the corresponding reactances are X = 1 and X = -1, so Zleft = 1i, Zright = -1i. Regardless of which x you choose, Z looking left is always the complex conjugate of the right so they always sum to zero. That's how that condition becomes equivalent to a resonant line.
     
  6. Dec 2, 2015 #5
    Thank you for having even read that book. I agree with all your considerations and (if you look back to my first post) they are almost equivalent to what I've written about a resonant line and the [itex]Z^{left} (x)[/itex], [itex]Z^{right} (x)[/itex] impedances.
    But my question was slightly different: what is the physical reason for a line, having in any position [itex]Z^{left} (x) = -Z^{right} (x)[/itex], to be called a resonant line? Why an adapted line with [itex]Z^{left} (x) = Z^{right} (x) = Z_0[/itex] is not a resonant line?
     
  7. Dec 2, 2015 #6

    marcusl

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    The condition [itex]Z^{left} (x) = Z^{right} (x) = Z_0[/itex] describes a line that is properly terminated at each end, e.g., with a generator and a load each having impedance [itex]Z_0[/itex] . Such a line supports traveling waves that carry power down the waveguide. The condition [itex]Z^{left} (x) = -Z^{right} (x)[/itex] at every point is special. It can only be set up by the impedance transforming property of specific lengths of line acting on an identical mismatch at each end of the line. The transformation applied to line lengths of x and λ/2-x produces the desired condition, furthermore, only when the mismatch at each end of the line is 0 or infinity (short or open). You can see this from the equations for that transformation, or from the Smith Chart which is a handy graphical tool that solves those equations. The final step is to realize that a transmission line that is open- or short-circuited produces 100% reflections at each end, hence supporting a standing wave mode--in effect a one-dimensional resonant cavity. The fundamental mode occurs at precisely the frequency that produces an exact λ/2 wave between terminations, which is the definition of a resonator.
     
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