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Ohmic losses in straight waveguide

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Hi,

I am simulating ( RF module) a straight WR42 hollow waveguide of rectangular section with Copper walls. I used Impedence Buondary Conditions to consider the Ohmic losses. On the input Port i inserted two Port Nodes ( TE10 with wave exc ON and TE20 with wave exc OFF), the same on the Output Port ( TE10 with wave exc OFF and TE20 with wave exc OFF), because I also want to simulate the TE10->TE20 mode conversion. As I expect, the mode conversion is very low, since I am simulating a frequency range for which the WG is optimized. The TE10, however, is not fully transmitted ( the corresponding S-parameter is close to 1 but not exactly one), since it is attenuated by Ohmic losses given by the Copper. When Ohmic losses are present, one can associate to each mode an attenuation coefficient, which says the characteristic length over which each mode is attenuated by Ohmic losses (P=Po*exp(-2(att_coeff)z). How can this coefficient be extracted from the results? The only similar coeff that I can extract is the propagation coefficient beta. From the results I have four propagation coefficients (emw.beta_1/2/3/4), what is their physical meaning if I have just two modes (TE10 and TE20)?


5 Replies Last Post Dec 5, 2023, 12:22 p.m. EST
Robert Koslover Certified Consultant

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Posted: 11 months ago Dec 1, 2023, 9:28 a.m. EST
Updated: 11 months ago Dec 1, 2023, 9:36 a.m. EST

I'll respond to this only in part. If ohmic lossess in the waveguide walls are small, don't try to use the computed S parameters (which are computed only using the fields at the ports) to extract them, since numerical noise (due to the finite discretization) can wash out such small changes relative to unity. Instead, compute the thermal loss all along the path by integrating the surface losses on the walls. You can do this in either post-processing (Derived values -> integration -> surface integration) or by setting up a surface integration probe. Either way, integrate the quantity emw.Qsh over the lossy surfaces. Compare that to your input power (typically set to 1W, if you use the default value) and you'll immediately derive the conductive losses, and with greater confidence. I realize that this does not distinguish between losses in each mode present, if more than one mode is present. But I'll leave that question for others here who may wish to offer additional advice. Good luck.

p.s. You can also compute the losses in a straight rectangular waveguide, by mode, analytically. Refer to any decent EM theory / waveguide textbook. Many good ones have been written, and many still sit covered with dust on library shelves, going back to (roughly) the 1930s, although I mostly prefer the post-WW2 books. :-).

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
I'll respond to this only *in part*. If ohmic lossess in the waveguide walls are *small*, don't try to use the computed S parameters (which are computed only using the fields at the ports) to extract them, since numerical noise (due to the finite discretization) can wash out such small changes relative to unity. Instead, compute the thermal loss all along the path by integrating the surface losses on the walls. You can do this in either post-processing (Derived values -> integration -> surface integration) or by setting up a surface integration probe. Either way, integrate the quantity emw.Qsh over the lossy surfaces. Compare that to your input power (typically set to 1W, if you use the default value) and you'll immediately derive the conductive losses, and with greater confidence. I realize that this does not distinguish between losses in each mode present, if more than one mode is present. But I'll leave that question for others here who may wish to offer additional advice. Good luck. p.s. You can also compute the losses in a straight rectangular waveguide, by mode, analytically. Refer to any decent EM theory / waveguide textbook. Many good ones have been written, and many still sit covered with dust on library shelves, going back to (roughly) the 1930s, although I mostly prefer the post-WW2 books. :-).

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Posted: 11 months ago Dec 1, 2023, 11:39 a.m. EST
Updated: 11 months ago Dec 1, 2023, 11:29 a.m. EST

You would expect mode conversion only if there is some sort of disturbance in the waveguide, say a post or some other geometric change. Any TE20 that appears will probably change in amplitude if the mesh size, etc. is varied.

I agree that a reliable value for attenuation can't be calculated from Comsol S parameters, at least for highly conducting walls.

My favorite reference for these things is an old copy of Ramo Whinnery and Van Duzer obtained from a former colleague. For WWII era- the MIT Radiation Lab series can be found as a free download.

You would expect mode conversion only if there is some sort of disturbance in the waveguide, say a post or some other geometric change. Any TE20 that appears will probably change in amplitude if the mesh size, etc. is varied. I agree that a reliable value for attenuation can't be calculated from Comsol S parameters, at least for highly conducting walls. My favorite reference for these things is an old copy of Ramo Whinnery and Van Duzer obtained from a former colleague. For WWII era- the MIT Radiation Lab series can be found as a free download.

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Posted: 11 months ago Dec 5, 2023, 7:49 a.m. EST

Thanks,

Do you think that bending the waveguide could also accentuate Ohmic losses, or just excite mode conversion?

Thanks, Do you think that bending the waveguide could also accentuate Ohmic losses, or just excite mode conversion?

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Posted: 11 months ago Dec 5, 2023, 12:20 p.m. EST
Updated: 11 months ago Dec 5, 2023, 12:08 p.m. EST

Thanks,

Do you think that bending the waveguide could also accentuate Ohmic losses, or just excite mode conversion?

My intuition is that a bend will have additional ohmic loss (intuition easily checked by simulation). And that the bend will excite some higher modes.

By the way- higher order modes may be above cutoff. If so- they will be exicited by a disturbance but they decay exponentially with distance (evanescent modes) and are unlikely to be observed unless a port is quite close.

>Thanks, > >Do you think that bending the waveguide could also accentuate Ohmic losses, or just excite mode conversion? My intuition is that a bend will have additional ohmic loss (intuition easily checked by simulation). And that the bend will excite some higher modes. By the way- higher order modes may be above cutoff. If so- they will be exicited by a disturbance but they decay exponentially with distance (evanescent modes) and are unlikely to be observed unless a port is quite close.

Robert Koslover Certified Consultant

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Posted: 11 months ago Dec 5, 2023, 12:22 p.m. EST
Updated: 11 months ago Dec 5, 2023, 12:53 p.m. EST

There exist published papers that address the subject of losses in waveguides with bends. You might want to do a search at IEEE Explore -- see ieeexplore.ieee.org/Xplore/home.jsp , for phrases such as "loss in waveguide bend".

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
There exist published papers that address the subject of losses in waveguides with bends. You might want to do a search at IEEE Explore -- see ieeexplore.ieee.org/Xplore/home.jsp , for phrases such as "loss in waveguide bend".

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