Waveguide multihole coupler design

[elusquinos]

Waveguide coupler design

Hi everyone, i will need to design a waveguide coupler for WR-34 and the center frequency in 28GHz. Coupling required=10dB, Directivity>30dB.

Can you help me?

Thank you in advance

 

[sergio mariotti]

Re: Waveguide coupler design

Yes i may help you.
There are mainly 2 types of couplers, namely:
Moreno Cross: small, crossed waveguide, max directivity 20dB. Coup=15 to 30dB
Side coupled: larger, highest directivity (typ 35 dB), Coup=6 to 40+ dB

I've a software that can calculate the side coupled type. It's an old dos styled and it is not frindly, so i may do the calculation for you.
Tell me if you want the design.

The mechanical build is easy, just have enough rectangular waveguide, two flanges, a drill machine, a vice, an appropriate iron, SnPb wire and a good fluxant.

 

[elusquinos]

Re: Waveguide coupler design

Thank you very much Sergio, i have been trying to apply Multihole coupler method but the results are not satisfactories. In Pozar`s book(MW Engineering) you have an example but when you apply it for 28 GHz then results are not coherent. Could you say me where can i study yours proposed methods?.

Our coupler waveguide would have the following specifications:

C=10 dB
D=>20dB
Waveguide==>WR-34
Frequency==>28.5GHz


Thank you again.

 

[HOHOH]

Re: Waveguide coupler design

In case of Pozar`s book, only explained a concept and basic theory by domain mode.

If you simulate example in it, then you have a different result.

So, you must consider higher mode for holes in waveguide.

But you need a little change of parameters, for example, offset length, hole size, distance between holes.

If you have ansof hfss or cst microwave studio, you have a good results.

I upload source code by Octave code(similar matlab, www.octave.org), please search this board.

 

[sergio mariotti]

Re: Waveguide coupler design

Thank you very much Sergio, i have been trying to apply Multihole coupler method but the results are not satisfactories. In Pozar`s book(MW Engineering) you have an example but when you apply it for 28 GHz then results are not coherent. Could you say me where can i study yours proposed methods?.

Our coupler waveguide would have the following specifications:

C=10 dB
D=>20dB
Waveguide==>WR-34
Frequency==>28.5GHz


Thank you again.

I'm sorry but the cad i have only allow coupling <-13 dB, so i cannot compute coupling of 10 dB.

 

[elusquinos]

Re: Waveguide coupler design

Thank you to both. I have a little doubts about Multihole Coupler method explained in Pozar´s book.

Firstly, s parameter is the aperture offset form the sidewall of the guide. In Pozar`s book example(Page 331 in 3rd Edition) s=a/4. But it is a fixed value or you can choose another value.


On the other hand, we have opened a 10dB directional coupler in WR-28 and we have observed two rows of 20 holes each one. We are thinking about developing chebishev polinomial for 20 holes in WR-34 waveguide in order to calculate radii holes(and distance between holes=λg/4) but this seems very complex(i am going to try solve it with Mathematica but i do not know if it will be right. Are there easier solutions?

Finally, a question for HOHOH, your source code is for calculate the radii holes? I didn´t find it yet, could you give more details about how find it?

Thank you again, you are a big help for me.

 

[elusquinos]

Re: Waveguide coupler design

Hello,

Finally i think we have found a possible solution to our problem. Radius holes have been developed by binomial response. But we have trunked the method because distance between center holes is lower than hole diameter, therefore we have thought to make the 4 central holes equal to his neighobour holes. For example if the hole number is 10 ,then radius hole are [a b c c c c c c b a] where each letter is the radius of each hole. With this trunk we loose maximally flat in the frequency response but at least we think it is realisable.

Another important thing is number of rows. If you have two rows of holes then

Coupling Factor C( of two rows holes)=Coupling Factor(single row holes)-6dB


Therefore i.e. if you want a total C=10dB then you can make two rows holes whose coupling C=16dB each one. This last is easier to built than a single row because the radius holes are smaller and phisically they can be realised. I think this one can be a good technique for small coupling factors required.

We are going to implement this solution as soon as possible when this happens i am going to comment the results.

I am sorry if it is too heavy.