"Michael A. Terrell" wrote in news:de-
:


Ian Jackson wrote:

OK.

At 450MHz and temperature T1 centigrade, your 5000 foot of coax will
have an attenuation of A1 dB.

[The value of A1 obviously depends on the characteristics of that
particular type of coax, and on its length. For a different frequency /
temperature / length / type of coax, the value of A1 will be different.]

At temperature T2, the change of attenuation will be approximately
(T2-T1) x 0.02 x A1 dB.

At 11GHz and temperature T1, your 5 foot of coax will have an
attenuation of A2 dB.

[Again, for a different frequency / temperature / length / type of coax,
different value of A1.]

At temperature T2, the change of attenuation will be approximately
(T2-T1) x 0.02 x A2 dB.

Essentially, you don't have to know anything about the frequency /
length / type of coax. Although these do determine the amount of initial
attenuation, all you need to know is the actual value of the initial
attenuation, and the change of temperature. The higher the initial
attenuation (for whatever reason), the greater will be the change of
attenuation.


You still need to know the length. I used to have to do the
calculations for CATV system design on a four banger calculator, before
PCs were common. Levels at the highest & lowest temperature was one
limiting factor of the amplifier spacings. Some cables that were almost
identical at first glance in the catalogs could be 100 feet difference
in maximum length. At that time, pressurized, fused disk was the best
.750 cable available, but it was so fragile we wouldn't use it. Then you
had to wade through all the various foam types, the DC resistance per
100 feet, maximum certified frequency and a dozen other numbers.

A 5000 foot cable will have twice the change of a 2500 foot cable
forthe same temperature change. You still haven't stated if your .02 dB
change/degree is per fook, per 100 feet, or per mile.


Look at the examples he worked for you.

Those units are especially convenient because you do NOT need to go through
complex calculations.

The complex calculations have already been done and you are given the final
results. All you need to know is the starting temp, the final temp and the
loss at the starting temperature. Everything else has already been taken
into account.

The 5000 foot cable will start with twice the number of dB loss as the 2500
foot cable.
Therefore the change per degree will be twice as much, automatically.

As he said, you do NOT need to know the length or the frequency, all you
need to know is the dB loss and the temperature change.

The units are (dB change in loss per degree change in temperature) / (dB
loss at T1)

So, if you are given the current dB loss at T1 and multiply that by
(dB change per degree change in temperature) / (dB loss at T1)

You get dB change per degree change in temperature. You then multiply that
by degrees change and get dB change.




--
bz 73 de N5BZ k

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

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