> I hate to start something here, but where does the "0.18 dB
> additional loss come from with any length of line"?
> Mismatch in and of itself does not necessarily represent loss.
The "additional loss" is a function of line VSWR and is added to the matched
line loss. For example, consider a 100-foot length of transmission line
that according to the manufacturer's spec sheet exhibits 1.0 dB of loss when
perfectly matched (e.g., 1.0 dB per 100-ft @ 100 Mhz).
When that same transmission line terminates into a mismatched load that
results in a VSWR of 3:1, an *additional loss* of 0.5 dB is produced from
the mismatch alone. Accordingly, for that 100-feet of line terminated into
a load that results in a 3:1 VSWR, the total transmission line loss is (1.0
dB + 0.5 dB = 1.5 dB).
Now, increase the mismatch at the load to a 10:1 VSWR and the loss due only
to the mismatch is now up to 1.7 dB. Added to the 1.0 dB loss of 100-ft of
line when perfectly matched, results in a total loss of 2.7 dB.
For ARRL subscribers, there's an excellent article that presents this
two-dimentional loss curve at Fig 2:
http://www.arrl.org/tis/info/pdf/7901019.pdf
A mistake that is commonly made occurs when cable loss is referenced on the
manufacturers spec sheet and an assumption is drawn that the loss depicted
in their line length versus frequency table computes into total transmission
line loss. In the example above, the table showed that 100-ft of line has a
loss of 1.0 dB @ 100 MHz. But what you get from that computation is just
half the answer when we're dealing with total line loss. You need to know
the line VSWR in order to get to total transmission line loss. The only
time the spec sheet loss can be used in its entirety, without correction, is
when the line is terminated into its characteristic impedance.
Paul, W9AC
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