Conclusions and Future Work

Abstract

This chapter deals with the conclusion of the fractal properties applied in aperture coupling problems. Several future problems are also identified which can be taken as future work.

Appendices

Appendix A: Calculation of Integrals Over Triangular Domains

The area coordinates associated with a triangle are shown in Fig. 7.1. The area coordinates are defined in terms of cartesian coordinates as

$$\begin{aligned} x&= x_1\mathcal {L}_1+x_2\mathcal {L}_2+x_3\mathcal {L}_3\nonumber \\ y&= y_1\mathcal {L}_1+y_2\mathcal {L}_2+y_3\mathcal {L}_3\nonumber \\ 1&= \mathcal {L}_1+\mathcal {L}_2+\mathcal {L}_3 \end{aligned}$$

(7.1)

Thus, these coordinates are not independent, rather are related by 3rd equation. For every set of $\mathcal {L}_1,\,\mathcal {L}_2$, and $\mathcal {L}_3$, there is a unique set of cartesian coordinates. At point j, $\mathcal {L}_j=1$ and $\mathcal {L}_j=0$ at any other vertex. The linear relationship between the area coordinates and cartesian coordinates implies that the contours of $\mathcal {L}_j$ are equally spaced straight lines parallel to the side on which $\mathcal {L}_j=0$.

Fig. 7.1

Area coordinates

The area coordinate $\mathcal {L}_j$ at a point P can alternatively be defined as the ratio of area of triangle formed by P and vertices other than jth vertex to the total area of the triangle. Thus

$$\begin{aligned} L_1=\frac{\mathrm {Area~of}~\triangle P23}{\mathrm {Area~of}~ \triangle 123} \end{aligned}$$

The integrals $I_1$ and $I_2$ can be transformed into the integrals over the area coordinates by using the formula

$$\begin{aligned} I&= \int \int \limits _{{ Triangle}}f(x,y) dxdy\nonumber \\&= A_T\int \limits _0^1\int \limits _0^{1-\mathcal {L}_2} f(\mathcal {L}_2,\mathcal {L}_3)~ d\mathcal {L}_2~d\mathcal {L}_3 \end{aligned}$$

where $A_T$ denotes the area of the triangle.

Using the Gauss-Legendre quadrature formula, the above integral can be expressed as

$$\begin{aligned} I=A_T\sum _{k=1}^{K_N} W(k)f(\mathcal {L}_{2k}\mathcal {L}_{3k}) \end{aligned}$$

(7.2)

where weights $W_k$ and coefficients $\mathcal {L}_{1k},\, \mathcal {L}_{2k}$, and $\mathcal {L}_{3k}$ can be obtained from [7] for different values of $K_N$. Hence using (7.2), $I_1$ and $I_2$ can be expressed as

$$\begin{aligned} I_1=A_T\sum _{k=1}^{K_N}(x_k-x_i)\sin \left( \frac{m\pi x_k}{a}\right) \cos \left( \frac{n\pi y_k}{b}\right) \nonumber \\ I_2=A_T\sum _{k=1}^{K_N}(y_k-y_i)\cos \left( \frac{m\pi x_k}{a}\right) \sin \left( \frac{n\pi y_k}{b}\right) \end{aligned}$$

(7.3)

where,

$$\begin{aligned} x_k=x_1+(x_2-x_1)\mathcal {L}_{2k}+(x_3-x_1)\mathcal {L}_{3k}\\ y_k=y_1+(y_2-y_1)\mathcal {L}_{2k}+(y_3-y_1)\mathcal {L}_{3k}.\\ \end{aligned}$$

Appendix B: Computation of Singular Integrals

It is well known that potential integrals involving free space Green’s function suffer from singularity, which occurs when the source and observation points coincide. A number of authors have reported the evaluation of these integrals based on singularity subtraction approach [2, 3, 5, 6]. In the singularity subtraction approach, the term having the asymptotic behavior at singularity is first subtracted from the integrand, and the resulting regular integral is computed numerically. The subtracted singular term is calculated using analytical methods and the result is added back to the numerically integrated term to obtain the final result. The approach can be summarized as

$$\begin{aligned} \int \int \limits _T \mathbf{M(r) }G(r|r')ds'=\int \int \limits _T \underbrace{\mathbf{M(r) }\left[ G(r|r')-G^{asym}(r|r')\right] ds'}_\text {Numerical Integration}\\ +\int \int \limits _T \underbrace{\mathbf{M(r) }G^{asym}(r|r')ds'}_\text {Analytical Integration}\\ \end{aligned}$$

where, $\mathbf {M(r)}$ is a scalar or vector basis function. Although, the singularity subtraction approach is extensively used in the computation of singular integrals, this methods suffers from some limitations. The difference term after subtraction of the asymptotic term cannot be well approximated in the neighborhood of the singularity because of the existence of higher order derivatives. The complexity of the problem increases for complex geometries and higher order basis functions. Also, the analytical evaluation of singular term becomes complicated for complex geometries and basis functions.

Due to these limitations, a more efficient method has been proposed in [4], which is based on the singularity cancellation method. The fundamental advantage of this method is that, the singular integral can be evaluated by purely numerical quadrature approach. Here, we have followed the singularity cancelation method for the evaluation of singular integrals.

The integral to be calculated is of the form

$$\begin{aligned} I=\int \int \limits _T \mathbf M (\bar{r}')\frac{e^{-jkR}}{4\pi R}ds' \end{aligned}$$

(7.4)

where, $R=|\bar{r}-\bar{r}'|$ is the distance between the source and observation point. The geometry of the triangle over which the integration has to be performed is shown in Fig. 7.2. Here, $\bar{r}_1$, $\bar{r}_2$ and $\bar{r}_3$ are the position vectors of the vertices of triangle and $\bar{r}_0$ denotes the observation point. The original triangle is subdivided into three subtriangles by connecting the vertices of the original triangle with the observation point as shown in Fig. 7.2. The contribution of the integral defined in (7.4) is calculated for each subtriangle and then added back to obtain the final result.

The geometrical quantities for subtriangle 1 (see Fig. 7.3) are

Fig. 7.2

Subdivision of original triangle

Fig. 7.3

Local coordinate system for subtriangle 1

$$\begin{aligned} \bar{r}_1'&= \bar{r}_0 \qquad \qquad \qquad \qquad \qquad \,\, \bar{l}_1'=\bar{r}_3'-\bar{r}_2'\nonumber \\ \bar{r}_2'&= \bar{r}_2\rightarrow \bar{r}_3\rightarrow \bar{r}_1 \qquad \qquad \qquad \bar{l}_2'=\bar{r}_1'-\bar{r}_3'\nonumber \\ \bar{r}_3'&= \bar{r}_3\rightarrow \bar{r}_1\rightarrow \bar{r}_2 \qquad \qquad \qquad \bar{l}_3'=\bar{r}_2'-\bar{r}_1'\nonumber \\ \hat{n}'&= \frac{\bar{l}_1' \times \bar{l}_2'}{|\bar{l}_1' \times \bar{l}_2'|}\nonumber \\ \bar{h}_1&= \frac{2A'}{|\bar{l}_1'|^2}\bar{l}_1' \times \hat{n}'\nonumber \\ A'&= \frac{\hat{n}'.\bar{l}_1' \times \bar{l}_2'}{2} \end{aligned}$$

(7.5)

Here, the arrows indicate the parameters for subtriangles 1–3 as one goes from subtriangle 1 to subtriangle 3. The primed and unprimed parameters denote the quantities for the original triangle and subtriangle, respectively.

The integral over the subtriangle 1 can now be expressed as

$$\begin{aligned} I_1=\int \limits _0^{h1'} \int \limits _{x_L}^{x_U}\mathbf {M}(r')\frac{e^{-jkR}}{4\pi R} dx'dy' \end{aligned}$$

(7.6)

where, $R=(\sqrt{(x')^2+(y')^2})$, is the distance between the source and observation point. The limits of the integration can be expressed in terms of normalized area coordinate $\mathcal {L}_{1}'$ at node $1'$. $\mathcal {L}'_{1}$ is unity at $y'=0$ and zero at $y'=h_1'$. Hence

$$\begin{aligned} y'=(1-\mathcal {L}_1')h_1' \end{aligned}$$

(7.7)

$$\begin{aligned} x_L=\hat{n}'\;{\cdot }\;(\hat{h}_1'\times \bar{l}_2')(1-\mathcal {L}_1') \end{aligned}$$

(7.8)

$$\begin{aligned} x_U=-\hat{n}'\;{\cdot }\;(\hat{h}_1'\times \bar{l}_3')(1-\mathcal {L}_1') \end{aligned}$$

(7.9)

Now, as the source and observation points coincide, the term $\frac{1}{R}$ becomes singular. To remove this singularity, we make the substitution

$$\begin{aligned} du=\frac{dx'}{R}=\frac{dx'}{\sqrt{x'^2+y'^2}} \end{aligned}$$

(7.10)

Integrating (7.10) with respect to $x'$, we obtain

$$\begin{aligned} u(x')=\sinh ^{-1}\left( \frac{x'}{y'}\right) \end{aligned}$$

(7.11)

Hence, (7.6) can be expressed as

$$\begin{aligned} I_1=\frac{1}{4\pi }\int \limits _0^{h1'} \int \limits _{u_L}^{u_U}\mathbf {M}(r')e^{-jkR}dudy' \end{aligned}$$

(7.12)

where, $R=y'\cosh u$.

From the above equation it is clear that the integrand is analytic in $u$ and $y'$ and the integral can be evaluated using repeated Gauss-Legendre quadrature method with weights $w_i$ and coefficients $\mathcal {L}_i$ in the normalized interval (0, 1) as given in [1]. Using this approach

$$\begin{aligned} I_1=\frac{1}{4\pi }\sum _{i=1}^{K}\sum _{j=1}^{N}w_iw_jh_{1}'(u_U^{(j)}-u_L^{(j)})\mathbf {M}(r')e^{-jkR^{i,j}} \end{aligned}$$

(7.13)

where, the superscripts $(i)$ or $(j)$ denotes the ith or jth sampled values of corresponding variable and $R^{(i,j)}=y_j'^{(j)} \cosh u_i^{(i)}$.

The sampled values in the $(u,y)$ domain can be expressed in terms of the area coordinates of the original triangle as follows:

• 1. $y'$ samples are calculated from (7.7) as

  $$\begin{aligned} y'^{(j)}=h_1'(1-\mathcal {L}_j) \end{aligned}$$

  (7.14)

• 2. $x_l$ and $x_U$ are calculated using (7.8) and (7.9).

• 3. Once these are calculated, we can use (7.11) to calculate $u_L$ and $u_U$.

• 4. $u^{(i,j)}$ can be calculated using

  $$\begin{aligned} u^{(i,j)}= u_l^{(j)}(1-\mathcal {L}_i)+u_U^{(j)}\mathcal {L}_i \end{aligned}$$

  (7.15)

• 5. Once the u samples are calculated, corresponding x samples can be obtained from (7.11) as

  $$\begin{aligned} x'^{(i,j)}=y'^{(j)} \sinh u^{(i,j)} \end{aligned}$$

  (7.16)

• 6. Next, we can calculate the remaining area coordinates as

  $$\begin{aligned} \mathcal {L}_3'^{(i,j)}=\frac{\hat{n'}\;{\cdot }\;\bar{l}_3' \times (\hat{h}_1'y'^{(j)}-\hat{l}_1x'^{(i,j)})}{2A'} \end{aligned}$$

  (7.17)
  $$\begin{aligned} \mathcal {L}_2'^{(i,j)}=1-\mathcal {L}_3'^{(i,j)}-\mathcal {L}_1'^{(i,j)} \end{aligned}$$

  (7.18)

• 7. Finally, the area coordinates of the subtriangles are transformed to that of original triangle as

  $$\begin{aligned} \left[ \begin{matrix} {\mathcal {L}}_1^{(j)}\\ {\mathcal {L}}_2^{(j)}\\ {\mathcal {L}}_3^{(j)}\\ \end{matrix}\right] =[T]\left[ \begin{matrix}{\mathcal {L}}_1'^{(j)}\\ {\mathcal {L}}_2'^{(j)}\\ {\mathcal {L}}_3'^{(j)}\\ \end{matrix}\right] \end{aligned}$$

  (7.19)

  where,

  $$\begin{aligned}{}[T]=\left[ \begin{matrix} {\mathcal {L}}_1^0 &{} 0 &{} 0 \\ {\mathcal {L}}_2^0 &{} 1 &{} 0 \\ {\mathcal {L}}_3^0 &{} 0 &{} 1 \\ \end{matrix}\right] \rightarrow \left[ \begin{matrix} {\mathcal {L}}_1^0 &{} 0 &{} 1 \\ {\mathcal {L}}_2^0 &{} 0 &{} 0 \\ {\mathcal {L}}_3^0 &{} 1 &{} 0 \\ \end{matrix}\right] \rightarrow \left[ \begin{matrix} {\mathcal {L}}_1^0 &{} 1 &{} 0 \\ {\mathcal {L}}_2^0 &{} 0 &{} 1 \\ {\mathcal {L}}_3^0 &{} 0 &{} 0 \\ \end{matrix}\right] \end{aligned}$$

  (7.20)

  where, $(\mathcal {L}_1^0,\,\mathcal {L}_2^0,\,\mathcal {L}_3^0)$ denote the area coordinates at the observation point. The arrow indicates the value of matrix as we go from subtriangle 1 to subtriangle 3.

Thus, the integral can finally be expressed as a sum of contribution of three subtriangles as

$$\begin{aligned} I\approx 2A\sum _k W_k \mathbf {M}(\bar{r}')\frac{e^{-jkR^{(k)}}}{4\pi R^{(k)}} \end{aligned}$$

(7.21)

where, $W_k$ are the weights corresponding to the sample points of area coordinates of original triangle of area A. From (7.13), we can write

$$\begin{aligned} W_k=(\hat{n}'\;{\cdot }\;\hat{n})\frac{w_i w_j h_1'(u_U^{(j)}-u_L^{(j)})R^{(i,j)}}{2A}. \end{aligned}$$

(7.22)