## 3 Results
### Sampling of Continuous Apertures
#### 3.1.1 Example A.1  
We produce an array that is a linear \(x\)-axis array with linear \(y\)-axis arrays as elements. The final planar array should be square and should use the same space taper in both the \(x\)- and in \(y\)-axes. With such an array, the 2D design is reduced to a 1D design.  
The limited-field-of-view indices are \(SLL_{\text{L}}<-20\) dB, \(HPBW_{\text{L}}\)-\(1.2^{\circ}\), and \(D_{\text{L}}>48\) dBi with \(\theta_{0}\)-\(8^{\circ}\). Let us now follow the steps of the proposed design procedure:  
Step 1: The scaling factor is equal to 7.  
Step 2: Using Table 1, the respective FFoV indices are \(SLL_{\text{F}}<-20\) dB, \(HPBW_{\text{F}}\)-\(8.4^{\circ}\), \(D_{\text{F}}>31\) dBi.  
Step 3: In this step, the aperture distribution and the size of the aperture are determined from the FFoV indices. We use a (cos\(\theta\))\({}^{n}\) distribution for the linear array, and from the \(SLL_{\text{F}}\) we determine the exponent \(n\). This is done as follows: From Table 1.1 of [4], we copy that for \(SLL=[-13.2,\,-23,\,-32]\) and the respective \(n\) takes the values \(n=[0,\,1,\,2]\). Using interpolation for \(SLL_{\text{F}}=-20\) dB, we find \(n\)-\(0.694\). Then, from \(HPBW_{\text{F}}\), we determine the length of the linear aperture of the FFoV linear array. Indeed, again from Table 1.1 of [1], we copy that for \(n=[0,\,1,\,2]\), which is \(HPBW=[50.8,\,68.8,\,83.2]/(\text{aperture length in wavelengths})\). Thus, using interpolation for \(n\)-\(0.694\) (the previously determined exponent), we obtain \(HPBW_{\text{F}}\)-\(63.29/(\text{aperture length in wavelengths})\). Thus, solving for the aperture length of the FFoV array, we find that it is approximately equal to \(7.53\lambda\).  
Step 4: In order to determine the sampling procedure, we need to determine the number of elements and choose a proper array element placement procedure. We start from the determination of the number and size of the elements. If the nearest distance between the array elements [21] is \(\varepsilon\), then the noise-like pattern [22] (\(\sim\)-inversely proportional to the number of the elements of the array), starts from the first grating lobe. The formula is [23]:  
\[|u|=\frac{3.8}{k\cdot\varepsilon}, \tag{19}\]  
where \(k\) is equal to \(2\pi\) and \(\varepsilon\) is measured in wavelengths. By equating \(u\) as the sine of the field of view we get:  
\[\sin(FoV)=\frac{3.8}{k\cdot\varepsilon} \tag{20}\]  
From (20), the minimum distance between the elements can be determined. We use elements that cover all the available space [24], since their patterns should approximate the brick-wall function (Equation (11)). From the above, we can proceed to the determination of the element directivity, assuming square, uniformly illuminated elements:  
\[D_{e}=4\pi S^{2}\Rightarrow D_{e}=4\pi(2\cdot\varepsilon)^{2}. \tag{21}\]  
Then from [1]:  
\[D_{L}=N_{e}\cdot D_{e}, \tag{22}\]  
we can find the number of elements. Taking into account that the square of the number of elements should be an integer (since we aim at a square array), the result is approximately equal to \(N_{\text{e}}=1600\).  
Thus, now that we have determined the number of elements, we can use the procedure described in [23] (p. 582) to design the linear array.  
Step 5. Last, we scale the design to produce the final array. Although we have determined the size of the elements from the previous step, we must avoid any possible overlapping between the elements. Thus, the side-length \(s\) takes the minimum value (multiplied, of course, by the scaling factor) between \(\varepsilon\) and the minimum inter-distance of adjacent elements. The final array has 1600 elements of \(s=1.86\lambda\). The final array layout and power pattern are given in Figure 4. The achieved directivity is equal to \(D=48.50\) dBi.