Question

The electric field of a uniform plane wave travelling along the negative z direction is given by the following equation:

${\overrightarrow{E}}_w^i=({\hat{a}}_x+j{\hat{a}}_y)E_0{e}^{jkz}$

This wave is incident upon a receiving antenna placed at the orgin and whose radiated electric field towards the incident wave is given by the following equation:

${\overrightarrow{E}}_a=({\hat{a}}_x+2{\hat{a}}_y)E_t\frac{1}{r}{e}^{-jkr}$

The polarization of the incident wave, the polarization of the antenna and losses due to the polarization mismatch are, respectively,

Answer 1

${\overrightarrow{E}}_w^i=({\hat{a}}_x+j{\hat{a}}_y)E_0{e}^{jkz}$

$\Rightarrow$ Wave contains two orthogonal components and Y component leads X component by ${90}^{\circ }$ and also wave is travelling in negative z-direction.

${\overrightarrow{E}}_a=({\hat{a}}_x+2{\hat{a}}_y)E_t\frac{1}{r}{e}^{-jkr}$

$\Rightarrow$ Wave contains two orthogonal components with unequal amplitudes and both are in-phase.

$\Rightarrow$ Linear polarization.

Polarization Loss Factor, PLF = ${|{\hat{E}}_T\cdot {\hat{E}}_R|}^2$

Where, $E_T=\frac{{\hat{a}}_x+j{\hat{a}}_y}{\sqrt{2}}$

$E_R=\frac{{\hat{a}}_x+2{\hat{a}}_y}{\sqrt{5}}$

PLF = ${\left|\frac{1+j2}{\sqrt{10}}\right|}^2=\frac{5}{10}=\frac{1}{2}$


$\Rightarrow PLF\left(dB\right)=10log\frac{1}{2}=-3dB$