>Following Autek's instructions, I tuned for minimum Z. At 28 Mhz,
>according to their formula, my loss was 1.3 db and at 7 Mhz it was 0.7
>db! I used the formula 0.17 x Z = loss in db (for 50 ohm coax). This
>seems awfully high, or does it?
Well, doing some back calculating, this almost sounds correct, except you
didn't include the Z you measured.
There is a rule of thumb that Insertion Loss is roughly equal to one half
the Return Loss that you would measure with an open (or shorted) line.
If for example, you measured a 10 ohm Z (10+j0), it would have a return loss
of 3.54 dB. Half of that would be 1.77 dB, which work work out to the
same answer as Autek's 0.17 x Z = 1.7 dB.
If you measured 2 ohms, it would have a return loss of ~0.72 dB, or an
insertion loss of 0.362 dB. Using 0.17 * 2 = 0.34 dB is pretty darn
close, eh?
One thing that I cannot answer is if the "tuning for minimum Z" all works
out. What one would assume is that if you tune for minimum Z, that you
are operating at j=0. If so, then the above calculations work out.
You just cannot pick a frequency, measure the Z, and convert it to loss
by multiplying by 0.17. It would appear that you have to choose a freq,
such that the coax would be a half wave, such that you are at the
Zreal < 50 and j=0. As you move away from that frequency that the coax
is a half wavelength, you will get more inductive (as you go higher in freq)
and the Z will increase, even though the Return Loss (and thus insertion loss)
remains relatively constant.
>It is old coax, but only 20 ft long. Does it really deteriorate like
>that being indoors the whole time? Or is that formula not correct?
>73 Barry
That I can't answer.
---
Chad Kurszewski, WE9V e-mail: WE9V@qth.com
The Official "Sultans of Shwing" Web Site: http://www.QTH.com/sos
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