Topic: Electromagnetics
The electric field of a linearly polarized electromagnetic wave is given by $E_i = \hat{a}_x E_0(x,y) e^{-jkz}$ is incident upon a linearly polarized antenna whose electric field polarization is expressed as $E_a = (\hat{a}_x + \hat{a}_y) E(r, \theta, \phi)$. Find the polarization loss factor.
- A. 1/2
- B. 3/2
- C. 2/3
- D. 1/4
Answer: A. 1/2
Explanation: The Polarization Loss Factor (PLF) is calculated using the normalized dot product of the incident wave and antenna polarization vectors: $PLF = \frac{|\mathbf{E}_{inc} \cdot \mathbf{E}_{ant}|^2}{|\mathbf{E}_{inc}|^2 |\mathbf{E}_{ant}|^2}$. Here, $\mathbf{E}_{inc}$ is in the $\hat{a}_x$ direction and $\mathbf{E}_{ant}$ is in the $(\hat{a}_x + \hat{a}_y)$ direction. Taking the dot product gives $1$. The squared magnitudes are $|\hat{a}_x|^2 = 1$ and $|\hat{a}_x + \hat{a}_y|^2 = 1^2 + 1^2 = 2$. Therefore, $PLF = \frac{1^2}{1 \times 2} = \frac{1}{2}$.