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I am trying to solve the following problem:

enter image description here

I know that in the +z direction:

linear polarization requires the field vector to have:

  1. one component OR
  2. two orthogonal linear components that are in phase or 180 degrees (or multiples of 180) out of phase.

Circular polarization requires:

  1. field must have two orthogonal linear components AND
  2. the two components must have the same magnitude AND
  3. the phase difference must be odd multiples of 90 degrees.

Polarization loss factor:

\$ PLF = |\hat{\rho_w} \cdot \hat{\rho_a}| ^{2}\$

In the +z direction

for linear:

\$\Delta \phi = \phi_y - \phi_x = n\pi, n = 0,1,2,3...\$

for RHCP:

\$|E_x| = |E_y|\$

\$\Delta \phi = \phi_y - \phi_x = -(\frac{1}{2} + n) \pi, n = 0,1,2,3...\$

for LHCP:

\$|E_x| = |E_y|\$

\$\Delta \phi = \phi_y - \phi_x = (\frac{1}{2} + n) \pi, n = 0,1,2,3...\$

Therefore:

For linear polarization: \$\phi = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\$

For RHCP: \$\phi = \frac{3\pi}{2}\$

For LHCP: \$\phi = \frac{\pi}{2}\$

For PLF:

\$\hat{\rho_a} = \frac{1}{\sqrt{2}}(\hat{a_x} + \hat{a_y})\$

\$\hat{\rho_w} = \frac{1}{\sqrt{sin^2(\phi) + cos^2(\phi)}}(\hat{a_x}sin(\phi) + \hat{a_y}jcos(\phi)) = (\hat{a_x}sin(\phi) + \hat{a_y}jcos(\phi))\$

\$ PLF = |\hat{\rho_w} \cdot \hat{\rho_a}| ^{2} = |\frac{1}{\sqrt{2}}(\hat{a_x} + \hat{a_y}) \cdot (\hat{a_x}sin(\phi) + \hat{a_y}jcos(\phi))|^{2}\$

And im not sure where to go from there for the PLF.

Are any of these answers correct? What am I doing right or wrong?

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1 Answer 1

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I think you have been mixing the following:

  • phase difference between the two field components, and
  • the angle \$ \phi \$ in the given wave expression

As you know, to have a linear polarization, we must have only one component of the field, or two components with 0 or 180 degrees phase shift.

The two given components are: \$ \sin(\phi) \$ & \$ j \cos(\phi) \$. Because of the existence of \$(j)\$, we can NOT find an angle \$\phi\$ to make the phase shift 180. This will require to generate an imaginary value for \$\sin(\phi)\$ or \$\cos(\phi)\$. Then, we have only the choice of making the field of one component. That is it, we seek \$\sin(\phi)\$ or \$\cos(\phi)\$ equal to zero. This happens at \$\phi = 0,\pi/2,\pi,3\pi/2\$. Notice that the angle \$\phi\$ is limited in \$[0,2\pi]\$.

For circular polarization, either left or right handed, you need to solve \$\sin(\phi)\$ \$= \pm \$ \$\cos(\phi)\$ to have the circular polarization condition. This occurs at \$ \phi = \pi/4,5\pi/4 \$ for left hand CP and at \$ \phi = 3\pi/4, 7\pi/4 \$ for right hand CP.

For the second point in your question, the polarization loss factor will be found at all \$ \phi \$: $$PLF = |\hat{\rho}_w .\hat{\rho}_a|^2 $$ where $$ \hat{\rho}_w = \sin(\phi)\hat{x} + j\cos{\phi}\hat{y} $$ $$ \hat{\rho}_a = \frac{1}{\sqrt{2}}(\hat{x}+\hat{y})$$ Substituting gives: $$PLF = \bigg|\frac{1}{\sqrt{2}}\sin(\phi)+j\frac{1}{\sqrt{2}}\cos(\phi)\bigg|^2$$ Then, $$PLF = \frac{1}{2}\sin^2(\phi)+\frac{1}{2}\cos^2(\phi) = \frac{1}{2}$$ So, at all \$\phi\$, the polarization loss factor will always be equal to 1/2. This is reasonable since the antenna is aligned at mid angle between the \$x\$ & \$y\$ axes, while the incoming wave has two components that are always orthogonal in space.

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