Section 6.1 · Worked examples · 1 of 5
Worked example 6.1.1: GPS L1: field strength and polarisation
Five short calculations tied to the applications. They use only , , the polarisation vectors, Malus' law and the two-way Doppler shift, and no new physics. Numerical constants: , . The polarisation match of a wave and an antenna, with unit polarisation vectors written for the same , is , with for RHCP and LHCP along . A good way to use them: read the given quantities, produce the formula from the corresponding application or problem yourself, and only then put the numbers in.
GPS L1: field strength and polarisation
#Given. GPS L1, , RHCP, at the ground. An echo from a building arrives weaker and LHCP. The RHCP patch antenna rejects LHCP by .
Find. (a) and , (b) the amplitudes of and , (c) the fraction of the power received by an RHCP, a linear and an LHCP antenna, (d) the echo-to-direct power ratio for a linear antenna and for the RHCP patch.
A CP wave has two components of amplitude , so and . The tip of keeps the constant length .
With : an RHCP antenna, , receives (). A linear antenna, at any angle , receives (). An LHCP antenna receives .
Linear antenna: direct , echo , ratio (echo amplitude of the direct one). RHCP patch: direct , echo , ratio (amplitude ).
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Microvolts per metre, below the receiver noise: GPS works thanks to its spreading code, not to a strong field. Circular polarisation costs the direct ray nothing and cuts the echo amplitude ten-fold, from to , which is what matters for the range measurement.
cf. Application 6.1.2