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From: Ian Walker
Subject: Re: How to check for a 50 or 75 ohm connector
Date: Tue, 24 Dec 2002 22:30:18 +0000
NNTP-Posting-Date: Tue, 24 Dec 2002 23:00:52 +0000 (UTC)
User-Agent: Turnpike/6.02-U ()
In article , John Woodgate
>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?
Try a little thought experiment (or a real one if you have a VNA)
connect a length of 50 ohm co-axial cable to a co-axial 50 ohm
termination and now measure the impedance presented to the open end of
the co-axial line; for all frequencies below cut-off you will obtain a
value of 50 +j0 ohm (or 50 +o degrees), if you perform the experiment
then the readings obtained will depend on the uncertainties in the
system. It can only present a non-reactive value if for all values of w
if XL = -XC. If it did not then it could only be used a spot frequency
with a specific length, if at all. Having demonstrated that finite
analysis fails I regret to advise you the you will have to use the
mathematics of the infinitely small.
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