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Imagine there is an EM wave propagating along the +z direction. This wave can have either a linearly or circularly

polarization. In this exercise, we focus on the x-y projection of the E (or H) wave vector and understand its trajectories.

An obliquely linearly polarized oscillating vector (E for example) projecting onto the x/y plane follows a trajectory indicated as the red-line in the below figure. The projected Exy vector can be further decomposed into the summation of two orthogonal oscillating components (see below image), one along the x-axis, the other along the y-axis. If Exy(t)=|Exy| cos(ωt) where |Exy| is the magnitude of the E- vector on the x/y plane, then, we have

                            

where and are the unitary vector along the x/y axes, respectively. If we know the orientation of the Exy vector, i.e. angle θ, we can then write down the expression of Ex and Ey as


Ex(t)=|Exy| cos(θ) cos(ωt) = |Ex| cos(ωt)

Ey(t)=|Exy| sin(θ) cos(ωt) = |Ey| cos(ωt)


In another view, if one transmits one horizontally (x-axis) polarized EM wave and one vertically (y-axis) polarized EM wave along the same direction, the net wave is a diagonally polarized wave.

Picture

 

A generalized form of the Ex and Ey components can be written as Ex(t) = |Ex| cos(ω1t) Ey(t)= |Ey| cos(ω2t+ φ0)


Q2.1(20 pts) : Please draw the trajectory of a circularly polarized E vector on the x/y plan (similar to the red line above), and write down the expressions for Ex(t) and Ey(t), assuming |Exy|=1 and ω=2*pi;


Q2.2 (20 pts) : If ω1= ω2=1, φ0=pi/2, and |Ex|=2*|Ey|, please draw the trajectory of Exy(t)


Q2.3 (20 pts) : If ω1=2, ω2=3, φ0=pi/4, and |Ex|=|Ey|, please draw the trajectory of Exy(t)

Picture diagrama.png

Picture diagrama.png

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