Subject: Re: Who thinks this?
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Fri, 01 Nov 2002 11:57:51 GMT
NNTP-Posting-Date: Fri, 01 Nov 2002 12:57:51 MET
Well, if the line is open higher frequency waves get reflected there and
travel back. The reflection is in phase and adds the 2 signals at this end.
since the other side is terminated the reflected wave gets dissipated in the
resistor and doesn't travel back.
The longer the line is, the lower the reflected frequencies. This is
compensated by the increasing capacity. So the pulse might look exactly the
same until resistive components skin-effect etc. come into play.
Now if you terminate the open end as well, the pulse is only 2.5V but
equally perfect in risetime.
electronic hardware designer
"Christopher R. Carlen" schrieb im Newsbeitrag
> 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