Principles of Modern Radar - Volume 2 1891121537

608 CHAPTER 13 Introduction to Radar PolarimetryA partially polarized wave may be regarded as the sum of a completely polarized waveand a completely unpolarized wave. Thus, the Stokes vector of a partially polarized wavemay be written in normalized form as⎡ ⎤ ⎡⎤ ⎡1dḡ = ⎢ g 1 /g 0⎥⎣ g 2 /g 0⎦ = ⎢ d cos 2τ cos 2ψ⎥⎣ d cos 2τ sin 2ψ ⎦ + ⎢⎣g 3 /g 0 d sin 2τ1 − d000⎤⎥⎦ (13.48)where the first vector on the right represents the polarized part, and the second termrepresents the unpolarized part of the wave. Equation (13.48) forms the basis of polarizationdecomposition theorems that allow one to write the partially polarized scattered signal interms of a completely polarized part corresponding to the target structure and a residue ornoise term (representing, e.g., clutter, motion) [13, 14, 23].Let g be the Stokes vector corresponding to the transmit antenna polarization stateê t , and h be the Stokes vector corresponding to the receive antenna polarization state ê r .From (13.26) and (13.27), it is evident that the backscattered power received by the radaris proportional to RCS of the target and may be written asP = |V rt | 2 = |S(m,ψ,v,τ,γ )ê t (ψ a ,τ a ) · ê r (ψ b ,τ b )| 2 (13.49)where the argument of the scattering matrix consists of the five significant Huynen parametersdescribing the scattered wave from the target (see Section 13.3.2), (ψ a ,τ a ) and (ψ b ,τ b )denote the orientation and ellipticity angles, respectively, of the transmit and receive antennapolarization states. A completely polarized wave is assumed for the moment. Huynen[14] showed that (13.49) may be concisely written in terms of the Stokes vectors of thetwo antenna polarizations, as defined in (13.21), and the Kennaugh matrix defined by thetarget parameters, asP = Kg· h (13.50)The 4 × 4 real symmetric Kennaugh matrix represents power measurements and can beexpressed in terms of the five target parameters, m,ψ,v,τ,γ, as [14, 23]⎡⎤A 0 + B 0 C ψ H ψ FC ψ A 0 + B ψ E ψ G ψK = ⎢⎥ (13.51)⎣ H ψ E ψ A 0 − B ψ D ψ ⎦F G ψ D ψ −A 0 + B 0whereA 0 = Qf cos 2 2τB 0 = Q ( 1 + cos 2 2γ − f cos 2 2τ )B ψ = B cos 4ψ − E sin 4ψC ψ = C cos 2ψD ψ = G sin 2ψ + D cos 2ψE ψ = B sin 4ψ + E cos 4ψF = 2Q cos 2γ sin 2τ