Question 3 of 8: Two Crossed Plane Waves — Planes of Circular Polarization

Question 3: Two Crossed Plane Waves — Planes of Circular Polarization (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two equal-power plane waves at $f=10$ GHz, $S=10$ W/m$^2$ each, both propagating horizontally: wave 1 at 30° east of north, $\mathbf{E}_1$ horizontal (in the horizontal plane); wave 2 at 30° west of north, $\mathbf{E}_2$ vertical.

Find. The spacing between the vertical north–south planes where the resultant field is circularly polarized, and the field amplitude in those planes.

Approach. Set up east–west/north/up axes $(x,y,z)$. Because $\mathbf{E}_1$ (horizontal) and $\mathbf{E}_2$ (vertical) are always mutually perpendicular, the pair is circularly polarized wherever they have equal amplitude and a $90^\circ$ relative phase; find how that relative phase varies with position.

1. Propagation unit vectors. With $x=$East, $y=$North, $z=$Up: $$\hat{k}_1=(\sin30^\circ,\cos30^\circ,0),\qquad \hat{k}_2=(-\sin30^\circ,\cos30^\circ,0).$$
2. Relative phase versus position. The phase of wave $i$ at point $\mathbf{r}$ is $-k\hat{k}_i\cdot\mathbf{r}$ (common $k=2\pi/\lambda$, same frequency). The phase of wave 2 relative to wave 1 is $$\delta(\mathbf{r})=k(\hat{k}_1-\hat{k}_2)\cdot\mathbf{r}=k(1,0,0)\cdot\mathbf{r}=kx,$$ since $\hat{k}_1-\hat{k}_2=(1,0,0)$ — the north components cancel and the phase difference depends on the east–west coordinate $x$ alone. $\mathbf{E}_1$ (horizontal, perpendicular to $\hat k_1$) and $\mathbf{E}_2=\hat z E_0\cos(\omega t-k\hat k_2\cdot\mathbf r)$ are always orthogonal, so wherever $\delta(\mathbf r)=\pm\pi/2+n\pi$ the two equal-amplitude, orthogonal, quadrature components trace a circle.
3. Plane spacing. Circular polarization requires $kx=\pi/2+n\pi$, i.e. $x=\lambda/4+n\lambda/2$: a family of vertical N–S planes spaced $$\Delta x=\frac{\lambda}{2},\qquad \lambda=\frac{c}{f}=\frac{3\times10^8}{10\times10^9}=0.03\ \text{m}=3\ \text{cm}.$$ $$\boxed{\Delta x=1.5\ \text{cm}}.$$
4. Field amplitude in those planes. Both waves carry the same $S$, so $$E_0=\sqrt{2\eta_0 S}=\sqrt{2\times377\times10}=86.8\ \text{V/m (peak)}.$$ For two orthogonal components of equal amplitude $E_0$ in exact phase quadrature, the resultant is a rotating vector of CONSTANT magnitude $E_0$ (not $\sqrt2\,E_0$): $\mathbf E(t)=E_0[\hat e_1\cos\omega t+\hat e_2\sin\omega t]$, and $|\mathbf E(t)|=E_0\sqrt{\cos^2\omega t+\sin^2\omega t}=E_0$ for all $t$. So $$\boxed{E_{circ}=86.8\ \text{V/m (peak)}=61.4\ \text{V/m (rms)}}.$$