Let us look at the loss of a short, open-ended, lossy transmission line stub by
examining the energy in the system. I will ignore dielectric loss for now and
look at only conductor loss. The conductor loss is (I^2)R. The stored energy is
(1/2)(V^2)C. Apply an AC voltage and run through the calculations. Now let us
double the line length. Apply the same AC voltage. R has doubled, I has
doubled, and C has doubled. The conductor loss has increased by a factor of
eight. The stored energy has increased by a factor of two. The ratio of stored
energy to dissipated energy has decreased by a factor of four. We can say then
that the Q is proportional to 1/(l^2), where l is the length.
Now let us look at a VERY long length of lossy transmission line. The energy
reflected from the open end (far, far away) is virtually zero. The line appears
as virtually a resistor having a resistance equal to the line characteristic
impedance. This is certainly a very, very low Q component.
How about examining it from a line loss standpoint using the published line
loss charts? Say a line has a loss, at a particular frequency, of 3 dB/100
meters. Let's drive it and leave the end open. The return loss is -6 dB. That
is, of the incident energy, 1/4 returns to the input of the line. Double the
line length to 200 meters. The one-way loss is 6 dB. The return loss is 12 dB.
Now 1/16 of the incident energy returns to the input. The amount of energy
stored in the line has doubled while the loss has quadrupled. The Q has
decreased by a factor of 2.
Dave WX7G
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