written 8.5 years ago by | • modified 8.5 years ago |
This question appears in Mumbai University > Satellite Communication and Network subject
Marks: 8 M, 10 M
Year: Dec 2014, May 2015
written 8.5 years ago by | • modified 8.5 years ago |
This question appears in Mumbai University > Satellite Communication and Network subject
Marks: 8 M, 10 M
Year: Dec 2014, May 2015
written 8.5 years ago by |
With circular polarization faraday shift simply adds to the overall rotation and does not affect the cross polar components of the electric field. Thus the depolarizing effects the faradays rotation are counted faradays rotation is therefore of no concern with circulars polarized waves. However, if linear polarization is to be used, polarization-tracking equipment may be installed at the antenna
Depolarization is caused by rain and ice.
Rain Depolarisation:
Larger raindrops can be considered to be oblate spheroids with some flattening underneath as a result of resistance. Also in a realistic situation, aerodynamic forces will cause some tilting of the drops, thus there will be certain randomness in the angle of tilt
A linear polarized wave can be resolved into two component waves, one vertically polarized and the other horizontally polarized. Consider a wave with its electric vector at some angle T relative to the major axis of the raindrop. Here the major axis of the rain drop is shown to be horizontal. The vertical component of the electric field lies parallel to the minor axis of the raindrop and therefore encounters less water than the horizontal component. Hence there will be a difference in the attenuation and phase shift experienced by each of the electric field components. These differences are termed as differential attenuation and differential phase shift and they result in depolarization of wave.
Expression for XPD in dB associated with rain
$XPD=U-V log A$
Where U and V are empirically determined coefficients and A is the rain attenuation. U, V and A are in dB.
$V=20; for 8 \lt f \lt 15 GHz$ $=23; for 15 \lt f \lt 35 GHz$ And $U+ 30 log f – 10 log (0.5-0.4697 cos4T) - 40 log (cosEl)$
Where,
f: frequency in GHz
El: angel of elevation of the propagation path at the earth station.
T: tilt angel of polarization relative to horizontal
T: 45° for circular polarization
When the electric field is parallel to the ground [horizontal], T = 0°
With the electric field vector in the reference plane containing the direction of propagation and the local vertical, T = 90° – El [all angles in degrees]
Ice Depolarisation:
Ice layer that is present over the rain region causes depolarization. Ice which is comparatively a good dielectric component leads to more loss. Shape of ice crystals is like a needle or like a plate and if they are all aligned; their effect of depolarization is much more. Sudden increase in the XPD coinciding with lightening can cause severe damage to the signals. Generally a fixed value in decibels of XPD is calculated for account of ice depolarization.
[Dielectric: it is an electric insulator that can be polarizes by an applied electric field.]