|MadSci Network: Engineering|
I cannot answer this question adequately. A good answer can be obtained from Mr. Rod Raburn at firstname.lastname@example.org. My best answer is simply that the filter is chosen to propagate a narrow band of microwave energy at a certain frequency. In some cases, a filter (very low power)will consist of a certain length, shape, etc. of highly conductive metal strip that forms the waveguide. From what I remember, the dielectric that the board rests on can be a thin film to relatively thick separation between either air or another surface of metal. The dielectric is chosen to have as little loss as possible so that the impedance is not attenuated. In the case of the metal strip that has no metal on the other side of the dielectric that it is mounted on, the air/dielectric discontinuity forms a reflective (but not perfect)boundry. The thickness of the dielectric and the size of the dielectric constant define the natural frequency of microwave energy that can travel along the waveguide. For example, if the frequency one wanted to filter was 2.45 GHz, in air, its wavelength would be (3E8 m/s)/(2.45E9) = 0.12m or roughly 12 cm. Since the wavelength is a function of the square root of the dielectric constant, choosing a dielectric with a dielectric constant of 100 would reduce the thickness of the dielectric by a factor of ten. Therefore, to filter out 2.45 GHz, the dielectric thickness would have to be a fraction of 1.2 cm. Generally, if I remember correctly, this technique is a quater wave waveguide, so the board thickness would be reduced to 1.2/4 cm or roughly 0.3 cm. If you need a more specific or clearer answer than that, which I suspect you do, then please consult the person mentioned above. I worked with engineers at Los Alamos labs briefly who were experts in the design of microwave filters and gleaned a little information from them on this subject.
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