## 3 Near Field Models of Reflectarray and Transmitarray Antennas
### Near Field as Superposition of Far Field Contributions  
To develop this model, we have to consider each unit cell in the array as a unit of radiation and calculate its far field. In such a case, the total near field will be the superposition of the far fields radiated by each element:  
\[\vec{E}_{\text{NF}}(\vec{r}) =\sum_{i=1}^{N}\vec{E}_{\text{FF},i}(\vec{r}) \tag{23a}\] \[\vec{H}_{\text{NF}}(\vec{r}) =\sum_{i=1}^{N}\vec{H}_{\text{FF},i}(\vec{r}), \tag{23b}\]  
where \(\vec{E}_{\text{FF},i}\) and \(\vec{H}_{\text{FF},i}\) are calculated as the far field radiated by an aperture [26]. The electric far field is:  
\[\vec{E}_{\text{FF},i}=E_{\theta,i}\hat{\theta}+E_{\varphi,i}\hat{\varphi}, \tag{24}\]  
and its components according to [PERSON]'s equivalent principle:  
\[E_{\theta,i} =\frac{jk_{0}e^{-jk_{0}R_{i}}}{4\pi R_{i}}\big{[}P_{x,i}\cos\varphi_ {i}+P_{y,i}\sin\varphi_{i}-\eta\cos\theta_{i}\big{(}Q_{x,i}\sin\varphi_{i}-Q_{y,i}\cos\varphi_{i}\big{)}\big{]}, \tag{25a}\] \[E_{\varphi,i} =-\frac{jk_{0}e^{-jk_{0}R_{i}}}{4\pi R_{i}}\big{[}\eta\big{(}Q_{x,i} \cos\varphi_{i}+Q_{y,i}\sin\varphi_{i}\big{)}+\cos\theta_{i}\big{(}P_{x,i}\sin \varphi_{i}-P_{y,i}\cos\varphi_{i}\big{)}\big{]}, \tag{25b}\]  
where the angles \(\theta_{i}\) and \(\varphi_{i}\) are defined in Figure 2.  
\(P\) and \(Q\) are known as the spectrum functions, and they are the Fourier transform of the field at the aperture:  
\[\begin{split} P_{x/y,i}&=\iint_{S_{i}}E_{\text{ap},x/y,i}(x,y)\,e^{\text{i}k_{0}(ux+\gamma y)}\,dx\,dy,\\ Q_{x/y,i}&=\iint_{S_{i}}H_{\text{ap},x/y,i}(x,y)\,e^{ \text{i}k_{0}(ux+\gamma y)}\,dx\,dy,\end{split} \tag{26}\]  
where \(u=\sin\theta_{i}\cos\varphi_{i}\) and \(v=\sin\theta_{i}\sin\varphi_{i}\). Since the aperture is element \(i\) of the array, we consider a constant field, which comes out of the integral, yielding:  
\[P_{x/y,i} =E_{\text{ap},x/y,i}\iint_{S_{i}}e^{\text{i}k_{0}(ux+\gamma y)}\, dx\,dy, \tag{27a}\] \[Q_{x/y,i} =H_{\text{ap},x/y,i}\iint_{S_{i}}e^{\text{i}k_{0}(ux+\gamma y)}\, dx\,dy. \tag{27b}\]  
The double integral in (27) can be solved analytically yielding [28]:  
\[\iint_{S_{i}}e^{\text{i}k_{0}(ux+\gamma y)}\,dx\,dy=p_{x}p_{y}\,\text{sinc} \bigg{(}\frac{k_{0}up_{x}}{2}\bigg{)}\,\text{sinc}\bigg{(}\frac{k_{0}vp_{y}}{ 2}\bigg{)}, \tag{28}\]  
where \(\text{sinc}\,x=\sin x/x\) is the unnormalized sinc function.  
Once the electric far field has been obtained, the magnetic far field may be obtained using the plane wave relation:  
\[\vec{H}_{\text{FF},i}=\frac{\hat{r}\times\vec{E}_{\text{FF},i}}{\eta}=\frac{E_ {\varphi,i}\hat{\theta}+E_{\theta,i}\hat{\varphi}}{\eta}. \tag{29}\]  
The far field of an array element obtained with (24) and (29) is expressed in spherical coordinates, while we are interested in obtaining the near field in a Cartesian basis. Thus, to perform the superposition of (23), the far field of an array element must be expressed as:  
\[\vec{E}_{\text{FF},i} =E_{\text{FF},x,i}\hat{x}_{a}+E_{\text{FF},y,i}\hat{y}_{a}+E_{\text {FF},z,i}\hat{z}_{a}, \tag{30a}\] \[\vec{H}_{\text{FF},i} =H_{\text{FF},x,i}\hat{x}_{a}+H_{\text{FF},y,i}\hat{y}_{a}+H_{ \text{FF},z,i}\hat{z}_{a}, \tag{30b}\]  
Figure 2: The definition of the angles \(\theta_{i}\) and \(\varphi_{i}\) for the computations of the far field radiated by an array element are defined in the element coordinate system \((\hat{x}^{\prime},\hat{y}^{\prime},\hat{z}^{\prime})\), which is the same as the ACS but translated by \(\vec{\tau}_{i}^{\prime}\).  
where the Cartesian components are found with:  
\[\left(\begin{array}{c}E_{\mathrm{FF},x,i}\\ E_{\mathrm{FF},y,i}\\ E_{\mathrm{FF},z,i}\end{array}\right)=\left(\begin{array}{cc}\cos\theta_{i} \cos\varphi_{i}&-\sin\varphi_{i}\\ \cos\theta_{i}\sin\varphi_{i}&\cos\varphi_{i}\\ -\sin\theta&0\end{array}\right)\left(\begin{array}{c}E_{\theta}\\ E_{\varphi}\end{array}\right). \tag{31}\]  
The same procedure is applied to the \(\vec{H}_{\mathrm{FF}}\) field.