From: John Woodgate
Subject: Re: How to check for a 50 or 75 ohm connector
Date: Tue, 24 Dec 2002 20:00:53 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Tue, 24 Dec 2002 20:44:05 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Ian Walker
wrote (in ) about 'How to check
for a 50 or 75 ohm connector', on Tue, 24 Dec 2002:
>In article , John Woodgate
>>I read in sci.electronics.design that Ian Walker
>>wrote (in <$SRMgh+9L2B+EwOu@newbrain.demon.co.uk>) about 'How to check
>>for a 50 or 75 ohm connector', on Mon, 23 Dec 2002:
>>>In the special case of a
>>>transmission line XL = -XC for all frequencies of use.
>>The mind boggles.
>Why? Have you done your calculus for this?
Yes, but I'm not sure that differential calculus is necessary to explain
the boggle. A finite element method seems adequate.
A uniform transmission line has a fixed inductance L per unit length
(which can be as short a length as you like) and a fixed capacitance C
per the same unit length. XL = wL, XC = 1/(wC) where w is the angular
frequency. So how can XL = -XC for all w?
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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