# Attenuators in Waveguides

Let's consider for instance a rectangular waveguide with TE01 mode.

As shown here, a possible attenuator may be realized with a metal card placed so that it is tangential to the electric field. Here the explanation of this behaviour.

There are some things I do not understand:

1) Why should it be placed with electric field tangential to it? I'd say that the reason is that a tangential electric field may induce a surface current that dissipates power on the card. But why does an orthogonal field keep undisturbed? I'd say that it would induce an orthogonal current inside the card, which also dissipates power.

2) Is that card a good conductor, a very good conductor, or a bad conductor? If it is a very good conductor, I understand that an orthogonal electric field does not induce current because current does not flow internally in a perfect conductor. But if it is a very good conductor, why do we have this big dissipation of current with a tangential electric field? It induces me to think it is not a good conductor, but just a resistor.

## 1 Answer

But why does an orthogonal field keep undisturbed? I'd say that it would induce an orthogonal current inside the card, which also dissipates power.

Remember the conductive part of the pad is just a thin film coated onto the glass plate.

It has to be thin presumably to have enough resistance to not disturb the field significantly. If it did, instead of absorbing the signal it would reflect it back towards the source. This would still reduce the forward travelling signal, but might be detrimental to the operation of the source.

Since the film is so thin, there's nowhere for a significant current to flow if you apply the electric field orthogonal to its surface.

Is that card a good conductor, a very good conductor, or a bad conductor?'

It has to be only moderately conductive.

If it were a very good or ideal conductor, it would produce reflections rather than absorption when placed in the waveguide.