What Are the Key Factors Influencing Waveguide Cut-Off Frequency?

AI Thread Summary
The discussion focuses on the factors influencing waveguide cut-off frequency, addressing several key points. Cut-off frequency applies to specific modes of wave propagation, which are determined by the waveguide's dimensions and shape, impacting boundary conditions. While theoretically, no electromagnetic waves should propagate below the cut-off frequency, practical scenarios may allow for some leakage, typically resulting in significant power loss. The dimensions of the waveguide, particularly its major and minor axes, dictate the cut-off frequencies for different modes, with smaller waveguides suitable for higher frequencies. Overall, understanding these principles is essential for designing effective waveguides for specific frequency applications.
rokper
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Hi,

For experts among you it might be trivial, but I am not exactly "at home" here :confused:
so, here it goes:

1.Does the "cut off frequency" in waveguides apply to all incident angles of electromagnetic waves propagation through the waveguide or not.

2. Does the shape of cross-section (circular or rectangular) plays any role in this?

3.Does this apply also to waves propagating parallel to the walls (or main axis) of waveguides (usually this is direction "z"): so no waves reflecting from walls due to incident angle.

4.Is "cutoff" 100% or do some electromagnetic waves "leak out" at that frequency (or lower)
at the open end.

5.Has length anything to do with it: how long must the ratio (length/width) be, for those "rules" to be valid. Is there such thing, as minimum

Many thanks,

Rok
 
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I'm not an expert either, but we're all here to learn.

1. Certainly there will be a loss when you insert a wave with oblique incidence into a waveguide, but if you just consider a wave already propagating inside a rectangular or circular waveguide, the wave will propagate in set modes dependent on the source frequency and dimensions of the waveguide.

2. The shape will change the solution of the wave equation by changing the boundary conditions, and thus change the cutoff frequencies.

3. This shouldn't matter, if I understand the question.

4. Cutoff is due to the propagation constant becoming imaginary below a certain cutoff wavenumber (or frequency), therefore the wave will exponentially decay with no propagation component. Theoretically this is 100%, but in reality we might see -50 dB of power transmission.
If you have an open end of the waveguide with no termination into a load, you will have an impedance mismatch with some reflection.

5. The major (a) and minor (b) dimension of the waveguide determines the cutoff frequencies of each mode. I suppose there is no such thing as a minimum waveguide size, as smaller ones are intended for higher and higher frequencies and you will find that the cutoff frequencies rise to optical levels if you get ridiculously small - not good for RF work.

Electromagnetics isn't an easy subject, so I recommend you start at the beginning. I like the Microwave Engineering book by Pozar.
 
Thanks, PatrickEE...

Yes, Electromagnetics isn't an easy subject, but interesting one. Unfortunately, I don't have time to dig dipper into it. There are some good lectures on internet, though: http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-17/

So, let's say, I want to have some kind of selective frequency waveguide, something like one you can see in attachment. So the question is, if I have a standing wave on the opening, do the blue lines get reflected as well as red ones? Presuming a is approx. 1/2 wavelength. So the question is, does the cutoff frequency apply also for blue, as well as red ones. I think, it does, although, it seem a bit counter intuitive to me (-50dB seems quite close to perfect to me)..
I want higher frequencies to be able to pass...
 

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