For radiating structures or scatterers, the far field quantities are of primary interest. [[EM.Cube]]'s [[FDTD Module]] can calculate the far field radiation patterns of an antenna or the radar cross section (RCS) of a target. In general, by far fields we mean the electric fields evaluated in the far zone of a physical structure, which satisfies the following condition:
:<math>r << \frac{2D^2}{\lambda_0}</math><!In the FDTD method, the far fields are calculated using a near-field-to-far-field transformation of the field quantities on a given closed surface. [[Image:FDTD79EM.pngCube]]-->uses rectangular boxes to define these closed surfaces. You can use [[EM.Cube]]'s default radiation box or define your own. Normally, the radiation box should enclose the entire FDTD structure. In this case, the calculated radiation pattern corresponds to the entire radiating structure. The radiation box may also contain only parts of a structure, which results in partial radiation patterns.
where r is the distance between the observation and source points, λ<sub>0</sub> is the free space wavelength and D is the largest dimension of the radiating structure. In [[EM.Cube]], the far-zone electric fields '''E<sup>ff</sup>'''(θ, φ) are functions of the spherical observation angles only and are defined as
:<math> \begin{align} & \mathbf{E(r) = E} (r,\theta,\phi) = \frac{e^{-jk_0 r}}{r} \mathbf{E^{ff}}(\theta,\phi) \\ & \mathbf{H(r) = H} (r,\theta,\phi) = \frac{1}{\eta_0} \mathbf{ \hat{r} \times E^{ff}(r) } \end{align} \quad k_0 r >> 1 </math><!--[[Image:FDTD104(1).png]]--> where k<sub>0</sub> = 2π/λ<sub>0</sub> and η<sub>0</sub> = 120π Ω is the intrinsic impedance of the free space. In the FDTD method, the far fields are calculated using a near-field-More details pertaining to-far-field transformation of the field quantities on a given closed surface. [[EM.Cube]] uses rectangular boxes to define these closed surfaces. You can use [[EM.Cube]]Tempo's default radiation box or define your own. The far-zone electric field can be written as: [[Image:FDTD105.png]] where '''J '''and '''M''' Far Field calumniation are the equivalent electric and magnetic surface currents on the surface of the enclosing box. '''G<sub>A,ff</sub>''' is the asymptotic form of the dyadic Green's function associated with the magnetic vector potential '''A''' and '''G<sub>EM,ff</sub>''' is the asymptotic form of the dyadic Green's function of the electric field due to a magnetic current. In most FDTD problems, the background medium of your physical structure is the free space and these functions reduce to the much simpler and familiar free-space Green's function: exp(-jk<sub>0</sub>r)/(4πr). In that case, one can define a pair of electric and magnetic radiation integrals: :<math> \begin{align} & \mathbf{N(r)} = \iint_S \mathbf{J(r')} e^{ -jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \\ & \mathbf{L(r)} = \iint_S \mathbf{M(r')} e^{ -jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \\ \end{align} </math><!--described at [[Image:FDTD107.png]]--> where :<math> \mathbf{\hat{r}} = \sin\theta \cos\phi \mathbf{\hat{x}} + \sin\theta \sin\phi \mathbf{\hat{y}} + \cos\theta \mathbf{\hat{z}} </math><!--[[Image:FDTD108.png]]--> In that case, the θ and φ components of the far fields can be computed from the following relationships: :<math> \begin{align} & E_{\theta}^{ff}(\theta, \phi) = -\frac{jk_0}{4\pi} (L_{\phi} + \eta_0 N_{\theta}) \\ & E_{\phi}^{ff}(\theta, \phi) = \frac{jk_0}{4\pi} (L_{\theta} + \eta_0 N_{\phi}) \end{align} </math><!--[[Image:FDTD106.png]]--> where the θ and φ components of the radiation integrals are given by: :<math> \begin{align} & N_{\theta}(\theta,\phi) = \iint_S [J_x\cos\theta\cos\phi + J_y\cos\theta\sin\phi - J_z\sin\theta] e^{ jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \\ & N_{\phi}(\theta,\phi) = \iint_S [-J_x \sin\phi + J_y\cos\phi] e^{ jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \end{align} </math> :<math> \begin{align} & L_{\theta}(\theta,\phi) = \iint_S [M_x\cos\theta\cos\phi + M_y\cos\theta\sin\phi - M_z\sin\theta] e^{ jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \\ & L_{\phi}(\theta,\phi) = \iint_S [-M_x \sin\phi + M_y\cos\phi] e^{ jk_0 \mathbf{\hat{r} \cdot r'} } \, ds' \end{align} </math><!--[[Image:FDTD109.png]]--> Normally, the radiation box should enclose the entire FDTD structure. In this case, the calculated radiation pattern corresponds to the entire radiating structure. The radiation box may contain only parts of a structure, which results Farfield Calculations in partial radiation patternsEM. In calculating the far field quantities, using Poynting's theorem, one can define the radiated power density as: :<math> \mathbf{W} = \frac{1}{2} \text{Re}(\mathbf{E \times H^*}) = \frac{\mathbf{\hat{k}}}{2\eta_0} |\mathbf{E}(r,\theta,\phi)|^2 \, |_{r \to \infty} </math><!--[[Image:FDTD110.pngTempo]]--> To eliminate the dependency on r, a normalized quantity called "Radiation Intensity" in the following way: :<math> S(\theta,\phi) = \lim_{r \to \infty} r^2 |\mathbf{W}| = \frac{1}{2\eta_0} | \mathbf{E^{ff}}(\theta,\phi)|^2 </math><!--[[Image:FDTD111.png]]--> The total radiated power can now be calculated as: :<math> P_{rad} = \int\limits_0^{2\pi} d\phi \int\limits_0^{\pi} d\theta \, S(\theta,\phi) \sin\theta = \frac{1}{2\eta_0} \int\limits_0^{2\pi} \int\limits_0^{\pi} |\mathbf{E^{ff}}(\theta,\phi)|^2 \sin\theta \, d\theta \, d\phi </math><!--[[Image:FDTD112.png]]-->
===Defining The Far Field Box===