Richard Clark wrote in message . ..

The scenario begins:

"A 50-Ohm line is terminated with a load of 200+j0 ohms.
The normal attenuation of the line is 2.00 decibels.
What is the loss of the line?"

Having stated no more, the implication is that the source is matched
to the line (source Z = 50+j0 Ohms). This is a half step towards the
full blown implementation such that those who are comfortable to this
point (and is in fact common experience) will observe their answer and
this answer a

"A = 1.27 + 2.00 = 3.27dB"

Interesting. I'd first have asked if the line was really 50 ohms,
completely nonreactive. If so, L/C=R/G and I'd have said A=3.266dB.
If the line was just 50 ohms nominal, then I can think of at least one
scenario in which A=0.60dB. And I can think of another in which A=
5.2dB. The definition I have used for loss in those cases is
A=-10*log10(P1/P2), where P2 is the power delivered to the (line+load)
and P1 is the power delivered to the load, with steady-state
excitation. The answer, given that definition, never depends on
source impedance of the driving source. Of course it could with a
different definition, for example involving the maximum available
power from the source, but that just confuses the issue by lumping
"(source) mismatch loss" with acutal line loss, as has been pointed
out before.

Cheers,
Tom