logbook
Electrical
- Sep 8, 2003
- 764
I am trying to work out how to calibrate a return loss measurement jig in waveguide. The system is just a CW power source, two directional couplers and two power meters. I thought that I could put on a short-circuit piece and an open circuit piece and say that both should give 0dB return loss. But then I realised that a waveguide "open-circuit" probably has no meaning. Just leaving the guide open presumably allows a large amount of field to escape. What would the return loss of an open waveguide be?
In order to get the traditional open and short circuit return loss values, does one simply use a sliding short or perhaps a short and a quarter lambda spacer?
In order to get the traditional open and short circuit return loss values, does one simply use a sliding short or perhaps a short and a quarter lambda spacer?