﻿ Polarization | Dragon Notes

# Dragon Notes

UNDER CONSTRUCTION
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## POLARIZATION

[Elliptical]: Superposition of waves polarized in varying directions and out of phase in time.
[Counterclockwise]: If $$\vec{E}$$ is a superposition of two linearly polarized waves – one polarized in the $$x$$-direction, the oher in the $$y$$-direction and lagging $$90^o$$ in time phase,
$$\ds \vplup \bvec{E}=\uvec{x}E_1(z)+\uvec{y}E_2(z)=\uvec{x}E_1e^{-jkz}+\uvec{y}jE_2e^{-jkz}\rightarrow \\ \bvec{E}(z,t)=\uvec{x}E_1 \cos(\omega t-kz)+\uvec{y}E_2 \sin(\omega t -kz -\pi/2)$$

Setting $$z=0$$ to examine the direction change of $$\bvec{E}$$ with $$t$$ at a given point,
$$\ds \boxed{\bvec{E}(0,t) = \uvec{x}E_1 \cos(\omega t) + \uvec{y}E_2 \sin(\omega t)} \vpl$$
which is a vector rotating counterclockwise with increasing $$t$$. (Right-hand polarized)
[Clockwise]: Changing the second wave so that it leads the first by $$90^o$$,
$$\ds \vplup \bvec{E} =\uvec{x}E_1 e^{-jkz} - \uvec{y} jE_2 e^{-jkz} \rightarrow \\ \bvec{E}(z,t) =\uvec{x}E_1 \cos(\omega t-kz)+\uvec{y}E_2 \cos(\omega t -kz+\pi/2)$$
At $$z=0$$,
$$\ds \vplup \boxed{\bvec{E}(0,t) = \uvec{x}E_1 \cos(\omega t) - \uvec{y}E_2 \sin(\omega t)}$$
$$\vplup$$which is a vector rotating clockwise with increasing $$t$$. (Left-hand polarized)
Note that if $$E_1 \neq E_2$$, the vectors trace an elliptical path. Else, the wave is circularly polarized.
[Linear]: Superposition of waves polarized in varying directions and in phase in time. If $$\bvec{E}$$ is a superposition of two linearly polarized waves – one polarized in the $$x$$-direction, the oher in the $$y$$-direction and in phase,
$$\ds \vplup \bvec{E}=\uvec{x}E_1(z)+\uvec{y}E_2(z)=\uvec{x}E_1e^{-jkz}+\uvec{y}jE_2e^{-jkz}\rightarrow \\ \bvec{E}(z,t)=\uvec{x}E_1 \cos(\omega t-kz)+\uvec{y}E_2 \cos(\omega t -kz)$$
which is a vector oscillating back and forth along a line that makes an angle $$\tan^{-1}(E_2/E_1)\vplup$$ with the $$x$$-axis. A linearly polarized plane wave can be resolved into a right-hand circularly polarized wave and a left-hand circularly polarized wave of equal amplitude:
$\bvec{E}(z)=\uvec{x}E_0e^{-jkz} \\ \Rightarrow \bvec{E}(z)=\bvec{E}_{rc}(z)+\bvec{E}_{lc}(z)\rightarrow \\ \boxed{\bvec{E}_{rc}(z)=\frac{E_0}{2}(\uvec{x}-j\uvec{y})e^{-jkz}} \quad \boxed{\bvec{E}_{lc}(z)=\frac{E_0}{2}(\uvec{x}+j\uvec{y})e^{-jkz}}$
A plane wave polarized in the $$x$$-direction was assumed without loss of generality.
Counterclockwise

Clockwise

Linear