Actually, there is something 'magic' about 50 ohms. An air-dielectric
co-axial cable has minimum loss per metre when its characteristic impedance is
76.7 ohms

I presume that the 76.7-0hm figure comes from a trade-off beween RF current and
conductor resistance. In other words, increasing the impedance value, the RF
current would become lower (for a given RF power), but the inner conductor
resistance would become higher because of the lower diameter needed to obtain
the higher impedance value (for a given outer diameter cable). And viceversa.


and the relative permittivity of polythene is 2.26 so a polythene-dielectric
co-axial cable has lowest loss when its characteristic impedance is
76.7/SQRT(2.26) = 51 ohms, which is most often rounded down to 50.

Under the assumption that dielectric loss is negligible, a permittivity 2.26
time higher than that of air results in a lower inner conductor diameter, for a
given outer diameter cable and a given impedance. Probably, lowering impedance
from 75 to about 50 ohm, the loss advantage one experiences thanks to the higher
inner conductor diameter needed for the lower impedance value is higher than the
loss disadvantage caused by the higher RF current (for a given RF power).

Maximum power handling, for a polythene-dielectric cable, occurs at a much
lower impedance: 30/SQRT(2.26) = 20 ohms.

I do not succeed to understand that statement. Maximum power handling is bound
to maximum temperature which is in turn bound to dissipated power. If 50 ohm is
the impedance at which minimum loss occurs (for a given RF power), why lowering
impedance to 20 ohm should result in a loss reduction. In the equation
30/SQRT(2.26) = 20 ohms, which is meaning of the figure 30?

I wonder whether you could indicate us a reference where all those trade-offs
are mathematically discussed.

73

Tony I0JX