Home exercise 6.1.2: An elliptically polarised wave towards -x

$a_k=-a_x$. With $a_x\times a_y=a_z$ and $a_x\times a_z=-a_y$,

At $x=0$, $E(0,t)=\mathrm{Re}\{(4a_y-\mathrm{j}3a_z){\mathrm{e}}^{\mathrm{j}\omega t}\}=4\cos \omega t{a}_y+3\sin \omega t{a}_{z}mV/m$. The tip runs on the ellipse $(E_y/4{)}^2+(E_z/3{)}^2=1$, from $4a_y$ ($\omega t=0$) to $3a_z$ ($\omega t=\pi /2$).

The rotation from $a_y$ to $a_z$ is about $a_y\times a_z=+a_x=-a_k$: against the right hand, left-handed.

The components are in time quadrature and lie along the axes, so the axes of the ellipse are the coordinate axes: $a=4$, $b=3mV/m$, $\mathrm{AR}=4/3$, that is $20{\log }_{10}(4/3)=2.50\mathrm{dB}$, with the major axis along $y$.

Take $u=a_z$, $v=a_y$, for which $a_z\times a_y=-a_x=a_k$. Then $E_0=-\mathrm{j}3u+4v=E_R(u-\mathrm{j}v)+E_L(u+\mathrm{j}v)$ gives $E_R+E_L=-3\mathrm{j}$ and $\mathrm{j}(E_L-E_R)=4$, so

$\mathrm{AR}=(3.5+0.5)/(3.5-0.5)=4/3$, and $|E_L|>|E_R|$ confirms the left hand. In power, $3.5^2/(3.5^2+0.5^2)=98\%$ of the wave is LHCP.