Polarization of Two Orthogonal Plane Waves: 17-Phys-A3 Electromagnetics, December 2016, Q4

Question 4: Polarization of Two Orthogonal Plane Waves (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 10 GHz plane waves in free space, equal power density $S=1\ \text{W/m}^2$ each; wave 1 travels due north with $\mathbf E_1$ horizontal (east–west); wave 2 travels due east with $\mathbf E_2$ vertical; at point A the two fields are in phase (given: total field linear).

Find. (i) RMS amplitude of the total field at A; (ii) the nearest point to A where the total field is circularly polarized.

Wave 1's $\mathbf E$ (east–west) and wave 2's $\mathbf E$ (vertical) are mutually perpendicular at A; moving diagonally (NW/SE) shifts their relative phase fastest.

Approach. Get each wave's RMS field from $S=E_{rms}^2/\eta_0$; because $\mathbf E_1\perp\mathbf E_2$ and they are in phase at A, the resultant magnitude is the vector sum of two equal, orthogonal, in-phase components. For part (ii), each wave's phase depends only on distance travelled along its OWN propagation direction, so move away from A in the direction that changes the two waves' relative phase fastest — the diagonal (NW/SE) direction — until that phase difference reaches $90^\circ$.

RMS field of one wave. $$E_0=\sqrt{S\,\eta_0}=\sqrt{(1)(376.8)}=19.41\ \text{V/m}.$$
Total field at A. $\mathbf E_1$ and $\mathbf E_2$ are equal in magnitude, mutually perpendicular (east–west vs. vertical), and in phase at A, so they add as perpendicular vectors: $$E_{rms}=\sqrt{E_1^2+E_2^2}=E_0\sqrt2=\boxed{27.45\ \text{V/m}}.$$
Phase gradient toward circular polarization. With north $=\hat x$, east $=\hat y$, wave 1's phase is $kx$ and wave 2's is $ky$ ($k=2\pi/\lambda$), so their relative phase is $\Delta\phi=k(x-y)$. This changes fastest per unit distance moved along the diagonal $(\hat x-\hat y)/\sqrt2$ (NW), at rate $k\sqrt2$ per metre. Circular polarization needs $|\Delta\phi|=90^\circ=\pi/2$ relative to A (which starts at $\Delta\phi=0$, linear): $$d=\frac{\pi/2}{k\sqrt2}=\frac{\lambda}{4\sqrt2}.$$
Evaluate. $\lambda=c/f=(3\times10^8)/(10\times10^9)=0.03$ m, so $$d=\frac{0.03}{4\sqrt2}=\boxed{5.30\ \text{mm}}.$$ This distance applies in EITHER diagonal sense: $5.30$ mm to the north-west of A gives $\Delta\phi=+90^\circ$ (one circular sense), and $5.30$ mm to the south-east gives $\Delta\phi=-90^\circ$ (the opposite sense) — both are equally the "nearest" point.

Field at A and nearest circular-polarization point
Quantity	Result
$E_{rms}$ at A	27.45 V/m
Wavelength	30 mm
Nearest circular-pol. distance	5.30 mm (NW or SE of A)

Topic: Superposition of two orthogonally-polarized, orthogonally-propagating plane waves.

Key relations: $S=E_{rms}^2/\eta_0$; two equal, perpendicular, in-phase fields sum to magnitude $E_0\sqrt2$ (linear, at $45^\circ$); a $90^\circ$ relative phase between equal orthogonal components gives circular polarization.

Why this works: each wave's phase advances only along its own propagation direction, so relative phase is a genuinely two-dimensional (spatial) quantity; the fastest way to change it per unit distance travelled is to move along the gradient of $(x-y)$, i.e. diagonally between the two propagation directions, not along either axis alone.

Common pitfall: moving along north or east only and expecting the polarization to change — moving purely along one wave's own propagation direction changes ONLY that wave's phase but at a rate $k$, not the combined rate $k\sqrt2$ of the true steepest-ascent direction, and gives the wrong (too large) distance.

Source: Sadiku, Elements of Electromagnetics, Ch. 10 (wave polarization); Balanis, Antenna Theory, Ch. 4.

Question 4 of 8: Polarization of Two Orthogonal Plane Waves

Question 4: Polarization of Two Orthogonal Plane Waves (equal value)