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Inside a waveguide there is a piece (rotary) of dissipative material (not metallic). The Electric field and the piece of metallic material form an angle phi.
The value phi is the variable of the attenuator.
The component of the E field parallel to the piece o material (phi=0) is dissipated; at the output of the waveguide I'll find only the orthogonal one (phi=90).
I would add some personal opinions about the question.
I agree with respect the marcomdd's answer.
Perhaps, FD357 woul'd also know
Why one shoul'd use the rotary attenuator (instead the more cheaper types) ?
The anser is:
The Rotary Vane attenuator is a metrology standard attenuator. It means that the accuracy is very high.
It's needed to calibrate other attenuators.
Also it's attenuation level is stable across frequency.
Of course, due to its funcuion, it is an expensive instruments.
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