Principles of Modern Radar - Volume 2 1891121537

13.2 Polarization 597After substituting (13.5) in (13.4) and simplification, we obtainEξ 2 cos2 ψ − E ξ E η sin 2ψ + Eη 2 sin2 ψE0x2 + E ξ 2 sin2 ψ + E ξ E η sin 2ψ + Eη 2 cos2 ψE0y2 ( )Eξ 2−− E η2 sin 2ψ + 2E ξ E η cos 2ψcos δ = sin 2 δ (13.6)E 0x E 0yEquation (13.6) is the general solution for the polarization ellipse in the (ξ,η) coordinatesystem. Since the ξ- and η-axes coincide with the major and minor axes of the ellipse, thesum of the cross-product terms involving E ξ E η must vanish. ThereforeE ξ E η sin 2ψE 2 0y− E ξ E η sin 2ψE 2 0xwhich may then be solved for the tilt angle ψ using the relationtan 2ψ = 2E 0x E 0y cos δE0x 2 − E 0y2 , − π 2 ≤ ψ ≤ π 2From Figure 13-3, we may define an auxiliary angle− 2E ξ E η cos 2ψE 0x E 0ycos δ = 0 (13.7)(13.8)and rewrite the tilt angle in (13.8) asγ = tan −1 E 0yE 0x, 0 ≤ γ ≤ π 2tan 2ψ = tan 2γ cos δ, γ ≠ π 4(13.9)(13.10)If E 0x = E 0y (or γ = π/4), the tilt angle cannot be calculated from (13.8) or (13.10).Instead, it may be found from the circular polarization ratio (Chapter 3 in [3]).Next, we derive the axial ratio of the ellipse in Figure 13-3, defined as AR = e 1 /e 2 ,in terms of the ellipticity angleτ = cot −1 (∓AR), − π 4 ≤ τ ≤ π 4(13.11)The tilt angle τ is negative for right-handed and positive for left-handed polarization,respectively. Since AR is positive, it follows that the negative sign in (13.11) leads toright-handed sense, and the positive sign, to left-handed sense. It is left as an exercise toshow that the major axis e 1 and the minor axis e 2 satisfy the relationsClearly,e 2 1 = E 2 0x cos2 ψ + E 2 0y sin2 ψ + E 0x E oy sin 2ψ cos δ (13.12)e 2 2 = E 2 0x sin2 ψ + E 2 0y cos2 ψ − E 0x E oy sin 2ψ cos δ (13.13)e 1 e 2 = E 0x E oy sin δ (13.14)e 2 1 + e2 2 = E 2 0x + E 2 0y (13.15)Using the trigonometric identity sin 2τ = (2 tan τ)/(1 + tan 2 τ), it follows from(13.11), (13.14), and (13.15) thatsin 2τ = 2e 1e 2e 2 1 + e2 2= 2E 0x E 0y sin δE 2 0x + E 2 0y(13.16)