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let's consider a waveguide excited with a coaxial cable antenna: enter image description here

You may see that this structure is sized with a distance between the antenna and the plate equal to \$\lambda/4\$. My professor has told me the reasons of this choice are that:

1) The antenna irradiates waves both at right and left. But we want all the wave at left. Therefore we put a very good conductor plate at left, in order to reflect the wave towards right.

2) The distance between the antenna and the previous plate is sized to be \$\lambda/4\$ because in this way it acts as a quarter wave transformer with a short circuit load (represented by the plate), and so the short circuit is seen as an open circuit by the antenna.

So, it is a narrow band system, designed to work ideally for only a single frequency. If for instance we excite the antenna with a frequency such that the distance antenna - plate becomes \$\lambda/2\$, the antenna will see a short circuit.

Now my question is: why is it a problem if the source (antenna) sees a short circuit? From a circuital point of view, I'd say that it is a voltage source in parallel with a short circuit, which is dangerous because it means infinite current which may damage the circuit. But this is not a circuit. So which is the disadvantage/danger of this condition?

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My waveguide knowledge is a bit sparse. But here's the way I think it works.

What you've shown on the left of your drawing is a classic quarter wave stub. This creates a 180 deg phase shift (90 deg out, 90 deg back) in the signal that's returned to the "T" junction.

But, the waveguide section on the left is terminated in a short. When the RF reflects off that short, another 180 deg shift is introduced. So what's returned to the "T" junction has been shifted 180 deg (out and back) + 180 deg (reflection off yhe short), or 360 deg. That is to say the reflected signal comes back and reinforces the signal propagating to the right, out to the open end of the waveguide.

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In high frequency electronics, a short circuit has the effect of total signal reflection. (It can be compared to what a mirror does with a light beam). So this short circuit will reflect all signal power back into the antenna.

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  • \$\begingroup\$ But in theory this will appen also with an open circuit, correct? \$\endgroup\$ – Kinka-Byo Mar 15 at 18:09
  • \$\begingroup\$ Yes, but your circuit is not open because the wave can propagate to the right. \$\endgroup\$ – Stefan Wyss Mar 15 at 18:14

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