Worked example 6.1.3: TV aerial and satellite LNB

$\lambda =c/f=3\times 10^8/6\times 10^8=0.50\mathrm{m}$, so a half-wave element is about $25\mathrm{cm}$ long, and

Tilted by $30^{\circ }$: $20{\log }_{10}\cos 30^{\circ }=-1.25\mathrm{dB}$, hardly noticeable. Vertical ($\psi =90^{\circ }$): only the cross-polar leakage is left, $-20\mathrm{dB}$, that is $35.8\mathrm{dB}\mu V/m$, a hundred times less power.

$\Delta \lambda =42.0^{\circ }-30.5^{\circ }=11.5^{\circ }$:

Left upright, the wanted polarisation loses only $20{\log }_{10}\cos 13.5^{\circ }=-0.24\mathrm{dB}$, but the other one leaks in at $20{\log }_{10}\sin 13.5^{\circ }=-12.6\mathrm{dB}$ instead of the $\approx 30\mathrm{dB}$ isolation that the H/V frequency plan relies on. The LNB is turned by $13.5^{\circ }$, anticlockwise seen from behind the dish, since the satellite is east of the site.

An exact calculation of the skew angle from the satellite and site positions (the line of sight and the Earth's axis projected on the plane normal to it) gives $13.5^{\circ }$ as well, so the simple formula is good to a tenth of a degree here.