What causes the unexpected annihilation point in the Magic-Tee configuration?

  • #1
Leopold89
14
2
I have some problems understanding the magic-tee. There is a configuration for the E and H arm, where the signal output is blocked. As far as I understand you should be able to set one arm to 0 and the other to 1/4 of a wavelength, so the reflected wave's phase will be shifted by pi compared to the incoming and they will annihilate.
But now I found a setup where the annihilation happens at 3/8 of a wavelength. As far as I understand there are two possibilities:
1. is the impedance at the reflecting arm not matched, leading to a different phase of the reflecting wave?
2. maybe a phase shift from optics (reflection at a medium with n2>n1)?

Could you explain this really weird phenomenon to me, please?
 
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  • #2
Is it possible to send a diagram so we are certain which ports we are discussing?
 
  • #3
220px-MagicTee.jpg

Here is the image. I used port 1 as input, port 2 as output, blocked port 3 and 4. Then I set the height of port 4 to 0mm and of port 3 to 3/8 of the wavelength.
 
  • #4
I suspect the disparity is caused by the wavelength in the waveguide.
 
  • #5
If you mean this conversion from vacuum wave length to waveguide wavelength, then unfortunately no, I have considered this.
 
  • #6
Is Port 2 terminated in a matched load?
 
  • #8
tech99 said:
Is Port 2 terminated in a matched load?
Yes, the load of around 523Ohm (even though I would have expected around 533.9Ohm) is the same at port 1 and 2.

tech99 said:
If you terminate arm 4 with a non contacting variable short circuit (https://www.ainfoinc.com/waveguide-...sliding-short-plates-4-9-7-05-ghz-fdp58-udr58) then the zero setting is lambda/4 not zero. This may give the offset you are observing.
I do not understand why. If the wave in arm 3 gets phase shifted by pi and the wave that would go into arm 4 instead is unchanged, I would expect destructive interference. But if both are phase shifted by ##\pi##, then I would expect constructive interference.
Also the offset is ##\phi = 2\pi\frac{2\Delta x_i}{\lambda_w}##, right? So if I set arm 4 to 0, I get an offset of 0.
 
  • #9
So, I now set both arms, 3 and 4, to ##\frac{\lambda_w}{4}\approx 43##mm length, but get the minima for dampening at ##\approx 2.4##GHz with -70dB and ##\approx 2.6##GHz with -50dB, referring to the frequency of the signal generator, while I would have expected 2.45GHz.
 
  • #10
That is only a 2% error. Maybe the offset of the movable short circuits is not so precise.
 
  • #11
Even air has sufficient relative permittivity to slow radio waves a little, as I found when I passed them through a bell jar for a demonstration. When I pumped the air out of the bell jar I found the jar was slightly defocusing the beam.
 
  • #12
Usually I would agree, but in this case I have a computer simulation, so everything is perfect. Perfect vacuum, precise control, perfect conductor and so on.
 
  • #13
Leopold89 said:
Usually I would agree, but in this case I have a computer simulation, so everything is perfect. Perfect vacuum, precise control, perfect conductor and so on.
And a perfect mesh generator. I am so jealous. I wish I had one of those.
 
  • #14
I am not sure if you have done the experiment or are using a simulation. If the latter, how can we discuss errors arising from it, in view of its being perfect?
 
  • #15
It is a simulation, CST to be precise.
tech99 said:
how can we discuss errors arising from it, in view of its being perfect?
Here is a misunderstanding. I was not suggesting that the setup/simulation was wrong, but that I may have missed something in my calculation of the arm position. Maybe the corners add a weird contribution. Maybe I was reading the wrong books, because I thought the S matrix would have entries dependent on something like ##e^{2\pi i\frac{2\Delta x}{\lambda_w}}##.

But if you say you block the signal at ##\frac{\lambda_w}{4}##, then I have to try another simulation software and perform an experiment.
 
  • #16
Could it be that ##\frac{\lambda_w}{4}## only works, if you have ##\cos\alpha=\frac{\lambda}{\lambda_w}=\frac{1}{\sqrt{2}}##? Because otherwise you have a different ##\lambda_w## in the arms than at the input?
 

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