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Quality Factor of Cavity Resonator

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the quality factor of a rectangular cavity resonator which is filled with a lossy dielectric but has perfectly conducting walls, is given by

    [tex]Q = \frac{1}{\tan\delta}[/tex]

    where [itex]\tan\delta[/itex] is the loss tangent, i.e. the ratio of the imaginary part of the dielectric constant to the real part of the dielectric constant.

    2. Relevant equations

    3. The attempt at a solution

    [tex]Q = \omega\frac{W}{P_{loss}}[/tex]

    I have no idea what to do here, but I don't understand why textbooks attribute [itex]P_{loss}[/itex] only to the conducting wall loss. Or at least I haven't encountered a textbook where the dielectric loss -- not just the attenuation due to a lossy dielectric -- is calculated.

    Also, it seems that this expression is true for both TE and TM Modes?

    Please help.

    Thanks in advance.
     
  2. jcsd
  3. Apr 1, 2009 #2
  4. Nov 30, 2010 #3
    but how?
     
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