What is the impedance looking into a length of coaxial cable that is short-circuited at the far end and a half wave long? I have been told that this the the basis of what is known as a wave guide "Choke Joint" can someone explain the usefulness of having such a device in a waveguide, of all places, which is meant to transport signals from point a to point b in the most efficient manner possible?
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That will form a short circuit. – Andy aka Aug 14 '20 at 07:30
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Since the joint of 2 waveguides must be created with two 1/4λ lengths, the shorted end will create an open circuit at the joint so that even metal conduction is unnecessary as the current is choked. Therefore an airtight seal gasket can be used.
More useful is the mm waveguide with both shorts 1/4λ apart then at right angles 1/4λ deep on one piece against the connecting flange. It can be rectangular or circular.
Tony Stewart EE75
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