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Microwave cavity resonators as filters

  1. Mar 8, 2015 #1
    I'm not sure if I should be asking this in the EE section, but here goes. I'm trying to understand more about resonant cavities and how they act to filter out frequencies other than the resonant frequency.
    In the case where a cavity resonator is coupled between two transmission lines, how does the resonator "filter" the non resonant frequencies? Does the cavity present an impedance mismatch at the input for non resonant frequencies and therefor causing the non resonant frequencies to be reflected? Or, is the input impedance into to cavity the same regardless of frequency, and the non resonant frequencies are simply attenuated?
    Just trying to have a better understanding. Thanks
     
  2. jcsd
  3. Mar 9, 2015 #2
    A resonant cavity is pretty much the same as any other L C circuit. Excite it and it resonates.
     
  4. Mar 9, 2015 #3

    tech99

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    The filter works by creating mismatch at the input and reflecting power. It does this by creating either a very high impedance or a very low one, so there are two types.
     
  5. Mar 9, 2015 #4

    Baluncore

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  6. Mar 9, 2015 #5
    Thanks tech99
    So at the resonant frequency, the input impedance (and i suppose output impedance) will match the characteristic impedance of the transmission line or coaxial cable at the input and output?
    If i look at the input of the cavity with a VNA, and plot S11 on a smith chart for a range of frequencies, should the point for the resonant frequency be in the center at 1, and the off resonant frequencies should appear more on the reactive sides of the scale?
    Oh, also does that mean that the wave impedance inside the resonator is the same as the input and output?
     
  7. Mar 10, 2015 #6

    tech99

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    I think you are correct items for 1 and 2. For item 3, I am not sure you can assume that, because the filter relies on having standing waves inside it.
     
  8. Mar 10, 2015 #7
    I thought about point #3 right after I posted, and what I said didn't make sense for the reason you said. But I'm wondering if the longitudinal placement of the coupling probe or loop matters as the impedance varies from 0 to infinity along the cavity.
     
  9. Mar 11, 2015 #8
    A rectangular microwave waveguide or coaxial transmission line or even discrete capacitors and inductors all work the same way fundamentally. Physical geometry leads to capacitance and inductance directly from Maxwell's equations. The most basic forms are parallel plates giving capacitance and current carrying wires (especially in loops) giving inductance.

    For example, in a rectangular waveguide, the longer sides act as capacitive elements with a positive voltage on one face and a negative voltage on the opposite face; during the next half of the cycle, current flows between the two larger sides via the shorter sides which act like an inductive element. This ends up with the classic 'distributed element' picture of a transmission line.

    So fundamentally, a resonant chamber is just like any other LC based filter circuit tuned to either short out a certain frequency (zero impedance) or not allow a certain frequency to pass(infinite impedance). The structures do not have to be physically connected; remember that the leads of any capacitor are not actually physically connected, just two separate sheets close to each other and rolled up.
     
  10. Mar 11, 2015 #9

    tech99

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    Agree. Remember that any filter can, in effect, be fitted with transformers at input and output to obtain any desired impedance. Regarding a coupling loop, it is usual to place it at a current node, to obtain maximum coupling and to minimise electric field coupling. The coupling factor can be varied by altering the size and orientation of the loop[.
     
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