4
\$\begingroup\$

I want to design a horn antenna to detect the 21 cm hydrogen line. I know how to optimize the antenna dimensions by taking into account some functions that approximate the phase errors and efficiencies. (The method I'm following is proposed by Leandro de Paula Santos Pereira and Marco Antonio Brasil Terada in New Method for Optimum Design of Pyramidal Horn Antennas). They end up with a slightly different horn antenna equation (which is a 4th order polynomial) which has to be numerically solved to get the value of the aperture width B. However, that equation contains the waveguide dimensions a and b, which are assumed to be known.

That leads me to my question: how exactly do I find out what these waveguide dimensions are? Looking around on the internet, I found out about how rectangular waveguides work, but all those formulae are based on the fact that the waveguide width is known. Nowhere is the waveguide height taken into account, and I understand why it isn't, but there's a huge difference between not taking it into account in theory and actually making the waveguide section for the horn antenna.

Can't I just go "Since I want to detect 1.42 GHz, I'll make sure that my cutoff frequency for the waveguide is some x less than 1.42GHz, and I'll make sure that 1.42GHz falls in the middle of the 'operating range' (the average of 125% and 189% - as a side note, why are these the given range of operating frequencies for a rectangular waveguide? It seems a bit arbitrary...). I can find out the cutoff frequency by dividing 1.42GHz by 1.57, and I know that the TE10 is the dominant mode in the waveguide, so I can calculate the waveguide width." How do I calculate the height? If I divide the width by 2, why? Is there any text that explains this? Is there some other way to calculate the optimal height and waveguide thickness?

Also, the folks over at microwavetools.com have the dimensions I'm looking for as a fuction of strictly the required wavelength, and their values are close to the waveguide in horn antennas that others built - but why do their dimensions work? Why do they have seemingly arbitrary multipliers for the wavelength?

Any help would be appreciated, maybe a link to a textbook chapter or a research paper that explains how to calculate waveguide dimensions in detail, as I'm at my wits' end trying to make sense of this. As a side note, since I'm inserting a pin into the waveguide to excite the horn antenna, how long should I make the waveguide? I've heard values from 3\$\lambda\$ (somewhere around 62 centimeters) to 'the length is arbitrary'....

\$\endgroup\$
5
  • \$\begingroup\$ What are your goals for Beamwidth 15%? 10%? and Return Loss? 15 dB 20 dB? \$\endgroup\$ Feb 12, 2022 at 17:36
  • 1
    \$\begingroup\$ I'm flexible on both these values (mostly because I don't know what's 'good' values are....) but I can guesstimate 50 degrees for the beamwidth, a return loss of ~19dB at 1.42GHz \$\endgroup\$
    – requiemman
    Feb 12, 2022 at 18:09
  • \$\begingroup\$ a & b apertures affect the XY gain for EH fields . Narrow apertures lead to wide beamwidth and low gain. Beamwidth is inverse to gain. so if you want high gain, then it must be narrow beamwidth \$\endgroup\$ Feb 12, 2022 at 18:26
  • 1
    \$\begingroup\$ Right, I understand what you're saying, but that still doesn't explain why b is a/2 for rectangular waveguides...is it literally just 'a/2 is a good enough length for higher frequency waves to be attenuated'? \$\endgroup\$
    – requiemman
    Feb 12, 2022 at 18:39
  • 1
    \$\begingroup\$ Rectangular WG's don't have to be 2:1. They can be any ratio but beam and bandwidth are affected. but That ratio must have desirable properties for a reasonable BW. But it could be square or round . bing.com/videos/… \$\endgroup\$ Feb 12, 2022 at 21:40

2 Answers 2

1
\$\begingroup\$

There is a simpler way to approach this problem: see how others have solved it before.

I'm not too experienced with microwave design and hardware myself, so let's go for a search:
Hydrogen line - Wikipedia
I already know from physics (and the question :) )this is 21cm / 1420MHz, but I also know from experience, there is a defined system of microwave bands. Aha, it's in the L band. Let's follow that link.
L band - Wikipedia
Dang, nothing about waveguide. Let's search "L band waveguide". A broad article pops up,
Waveguide (radio frequency) - Wikipedia
search for "L band", we find a hit down in the #Waveguides_in_practice section. Aha, WR650 (and two other standards) waveguide lands right in the middle of the band of interest. Bingo!

As for geometry -- again, this will be a rough explanation, and probably a bit inaccurate -- the 1/2 ratio comes from a combination of reasonable parameters: to get an octave usable range between cutoff and the next mode(s); the wave impedance is reasonable (both for purposes of losses and power limit, and for ease of coupling from external circuitry); uh probably things about dispersion (phase velocity exceeds group velocity!), though that will be in general a characteristic of waveguide, but the particular value will depend on geometry; etc.

Note also that you want to make the flanges a specific size and shape: a thin space with a groove around it acts as a 1/4 wave transmission line, which is open on one side (outside) so reflects the opposite (a short circuit) at the inside. The alternative is cramming EMI gasket into the joint, which might not be very repeatable, or even seal as well. These are all part of the respective standards, so follow the drawings or buy standard parts, and you're good to go.

\$\endgroup\$
1
  • \$\begingroup\$ Wow, this question takes me back. I ended up solving the problem myself using essentially the same method you posted. \$\endgroup\$
    – requiemman
    Oct 23, 2023 at 1:15
0
\$\begingroup\$

Try this explanation from RF cafe: -

enter image description here

See also this answer on this site: -

The cut-off point should be easily understood but to understand that a waveguide can pass a bunch of frequencies needs a little leap of faith (without going into the math): -

enter image description here

The picture above is a drawing of a waveguide and shows two "busbars" - these can be regarded as the electrical connections in or out. If the busbar is thin, dimension a/2 is half the physical width of the waveguide.

This defines the lower frequency limit of the waveguide.

Now, imagine that for higher frequencies the busbar got fatter. Yes, I know there isn't a busbar inside the waveguide because it's one solid piece of metal shaped into a tube but, the busbar analogy is to help realize the concept that higher frequencies will work...

If the busbar gets fatter then a/2 reduces. Whatever value a/2 is, it will be a quarter wave at some frequency and this will be an open circuit to the electrical signal applied.

Thus, waveguides can pass a range of frequencies and are not limited to exactly one spot frequency.

\$\endgroup\$
6
  • \$\begingroup\$ I understand what RF Cafe's saying, but it doesn't explain why the height is half the width, or am I missing something? I understood how I'd calculate the width, though \$\endgroup\$
    – requiemman
    Feb 12, 2022 at 18:42
  • \$\begingroup\$ @requiemman the formula includes dimension b - did you not see it? \$\endgroup\$
    – Andy aka
    Feb 12, 2022 at 18:52
  • 2
    \$\begingroup\$ Yes, I did, but that's jsut the formula for the cutoff frequency and wavelength, isn't it? I'm asking how I'd determine the dimensions a and b for a waveguide if all I knew was my operating frequency. The cutoff frequency is a single equation with two unknowns... \$\endgroup\$
    – requiemman
    Feb 12, 2022 at 18:54
  • 1
    \$\begingroup\$ I forgot to add - I can get a from my cutoff frequency but I'm specifically asking why I'm limited to b=a/2 and not, say, b=0.75*a or b=0.45*a \$\endgroup\$
    – requiemman
    Feb 12, 2022 at 19:07
  • 1
    \$\begingroup\$ Maybe this can help: web.mit.edu/6.013_book/www/chapter13/13.4.html - it also includes info on the probing pin you mentioned. \$\endgroup\$
    – Andy aka
    Feb 12, 2022 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.