18 Feb
2012
18 Feb
'12
12:44 p.m.
On Fri, Feb 17, 2012 at 07:23:12PM -0500, gene heskett wrote:
Sure, no discussion about that. But at that lenght and frequency, this
is not due to the cable's characteristic impedance but only the result
of the line amp's output impedance (600 ohm) and the the cable capacitance
forming a simple RC lowpass circuit. The C in this case is proportional
to length, while a cable's characteristic impedance is not.
At audio freqencies the cable's series resistance dominates the
impedance of its inductance, which means that the classic equation
for impedance, sqrt(L/C) is no longer valid. It still has an impedance
which turns out to be something like sqrt (R/wC). In theory you could
try and match that at both ends but since it depends on frequency that
would be very complicated.
The practical solution for long lines, and what Ma Bell does, is to
increase the cable's apparent inductance by inserting series inductors
at regular intervals. These combine with the cable's capacitance to
produce a purely resistive impedance even at audio frequencies. The
distance between the inductors determines the bandwidth. Lines used
for full bandwidth audio (e.g. broadcasting) require more inductors
per unit lenght.
The practical solution for audio connections up to a few hundred
meters is to ignore cable impedance completely, accept the capacitive
load and just drive it with a very low impedance. Anything longer than
that will be digital these days.
Ciao,
--
FA
Vor uns liegt ein weites Tal, die Sonne scheint - ein Glitzerstrahl.
The concept of
cable impedance makes sense only if the lenght
becomes a non-trivial fraction of wavelength. For audio that
means that for anything shorter than a few hundred meters it's
only capacitance that matters.
But at a 60 ohm impedance, that capacitance is considerable. I could put
20 volts P-P into a cable headed for our news dept, at say 15 kilohertz,
and 200' away, I could only see 6 or 7 volts.