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Subject: Re: Who thinks this?
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
NNTP-Posting-Date: Sat, 02 Nov 2002 06:43:36 GMT
Organization: AT&T Broadband
Date: Sat, 02 Nov 2002 06:43:36 GMT
First of all, the capacitive portion of the transimission line does store
charge just like an ideal capacitor. The difference is how it behaves while
it's being charged or discharged. When applying the 5V step, a capacitor
with series R will draw exponentially decreasing current over time. The
transmissionline with series R will draw a constant current for a fixed
period (2 * tp) and then abruptly quit.
You mention that the source voltage is 2.5V, but this is only temporary
until the reflected wave arrives back at the source end at which point the
whole system will settle with both ends (indeed, the entire transmission
line) charged to 5V.
If you folks want to try a SPICE-like simulator that actually shows you the
wave propagation INSIDE of a transmission line, check out a program called
Travis 2.0 CE at http://www.eyeseeware.com/. The example discussed in this
thread is fairly trivial--Travis will let you see far more complicated
scenarios. The full version of the program (called simply Travis 2.0) also
includes diodes for those of you wanting to see how they affect propagating
waves. You can download a fully functioning demo of Travis 2.0 CE for free.
(Yeah, I'm one of the authors. This is a shameless plug.)
"Christopher R. Carlen" wrote in message
> I find many people have the following conception:
> A transmission line is like a capacitor. If you have a long line, and
> feed it a digital logic transition, the output signal at the other end
> will look like a RC exponential response. The slowness of that response
> will be related, for the most part proportionally, to the length of the
> cable and the output impedance of the driver.
> Furthermore, that putting a resistor in series with the output of a
> driver, and the input of the cable, will somehow slow the response even
> more, which seems logical if you think the line is capacitive.
> Of course all of these notions are very incorrect, assuming one is
> talking about an almost ideal line with a purely real complex
> propagation constant, and thus a purely real characteristic impedance,
> and a non-dispersing, non-distorting line, such as typical controlled
> impedance cables that we use every day.
> What do you think?
> Fun questions are then derived from these considerations like:
> You have a driver generating a very stiff 5V output step, and you
> connect it to a 50R line with a 50R series resistor.
> Why does the edge at the unterminated output end of the line snap to 5V
> after the propagation delay, with the same risetime as if the driver
> were driving a simple 50R resistor, no matter how long the line (again
> assuming that the line length is such that the line is very close to
> ideally non-distorting)? And why then is the input voltage to the line
> only 2.5V when the output voltage steps to 5V?
> Or doesn't it?
> (I'd say it does.)
> Good day!
> Christopher R. Carlen
> Principal Laser/Optical Technologist
> Sandia National Laboratories CA USA
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