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Circularly polarized waves

Consider a time-harmonic E field, given as:

\[\mathbf{E}(\mathbf{r},t) = \text{Re}\Big[\mathbf{E}(\mathbf{r})\text{e}^{j\omega t}\Big]\]

For a plane wave, the \(\mathbf{E}\) field phasor is

\[\mathbf{E}(\mathbf{r}) = \hat{n}~E_{0}~\text{e}^{-j~\mathbf{k}\cdot \mathbf{r}}\]

Now, if we consider a plane wave with \(\mathbf{E}\) field of amplitude \(E_{0}\), pointing in a direction \(45^{circ}\) with respect to the \(x\)-axis and traveling in the \(+z\) direction, the \(\mathbf{E}\) field phasor is

\[\mathbf{E}(\mathbf{r}) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{e}^{-j~k~z}\]

The \(\mathbf{E}\) field in this wave always points along the line defined by the diagonal \(\hat{\mathbf{y}} + \hat{\mathbf{z}}\). It turns out that we still have a linearly polarized wave, rotated by \(45^{\circ}\) and with an amplitude divided by \(\sqrt{2}\).

\[\mathbf{E}(\mathbf{r}) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{e}^{-j~k~z} \; \Rightarrow \; \mathbf{E}(\mathbf{r},t) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{cos}~(\omega t - kz)\]

If an observer sits at \(z=0\), along the direction of propagation, he sees an oscillating \(\mathbf{E}\) field given by

\[\mathbf{E}(\mathbf{r},t)\Big|_{z=0} = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{cos}~(\omega t)\]

Now, if we consider a plane wave given by the phasor

\[\mathbf{E}(\mathbf{r}) = (\hat{x} - j~\hat{y})~E_{0}~\text{e}^{-j~k~z}\]

we notice that it has a complex polarization vector.

Finding the time-dependent \(\mathbf{E}\) field:

\[\mathbf{E}(\mathbf{r},t) = \text{Re}\Big[\mathbf{E}(\mathbf{r})\text{e}^{j\omega t}\Big] = E_{0}~\big[\hat{x}~\text{cos}(\omega t - kz) + \hat{y}~\text{sin}(\omega t -kz)\big]\]

The important thing to notice here is that the \(x\)- and \(y\)-components of the \(\mathbf{E}\) field have the same amplitude, but are \(\pi/2\) out of phase.

Now, an observer sitting at \(z=0\), will see the \(\mathbf{E}\) field vector rotating in a circle

\[\mathbf{E}(\mathbf{r}) = (\hat{x}-j~\hat{y})~E_{0}~\text{e}^{-j~k~z} \; \Rightarrow \; \mathbf{E}(\mathbf{r},t) = E_{0} \big[\hat{x}~\text{cos}(\omega t - kz) + \hat{y}~\text{sin}(\omega t - kz)\big]\]

\[\mathbf{E}(\mathbf{r},t)\Big|_{z=0} = E_{0}~\big[\hat{x}~\text{cos}(\omega t) + \hat{y}~\text{sin}(\omega t)\big]\]