Abstract

Keywords

Circumferential Slot, Cylindrical Waveguide, Poynting’s Vector, Mutual Admittance

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1. Introduction

Electromagnetic radiation from an aperture waveguide is one kind of classical and practical problems which has been considered for several years. Because of cylindrical waveguide and conformal antenna applications, same as aperture waveguide radiation, electromagnetic radiation from slots on cylindrical waveguide is a prominent problem. To have a better analysis, desired pattern and also to have a proper matching in input, electromagnetic radiation and mutual coupling between slots present an equation to compute mutual admittance between apertures on cylindrical waveguide.

To compute mutual admittance, apart from aperture shape on cylindrical waveguide, there are different methods which have been used but generally they are divided into two general categories:

・ The methods which are based on reciprocity (reaction) theorem that involve modal solution [1] - [7] and surface ray or GTD-Solutions [8]

・ The methods which are based on Poynting’s vector [9] .

In modal analysis method we need to have several mode combinations to obtain an accurate approach, if the mode numbers aren’t sufficient, analysis accuracy will be decreased. To have more mode numbers, radius of cylinder must be increased, so this method is not useful for such a problem which the mode numbers are insufficient. If the radius of cylinder be one or two times bigger than wavelength units, surface ray method or GTD-Solution will be helpful. The method which is used in here is based on Poynting’s vector method due to its advantages for circumferential slot array designing.

This method has much more conformity with cylindrical waveguide to reach convergence. The method which is used in here was suggested by Hill [10] to find input impedance for dipole antenna. It also can be used to find mutual impedance between axial slots on cylindrical waveguide [11] . So Poynting’s vector method can be indicated such as following relation:

$Y_{12}=\frac{1}{V_2{V}_1{}^{\text{*}}}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{2\text{π}}{E}_2\times H_1^{\text{*}}\cdot \hat{n}{r}^2\sin \theta \text{d}\theta \text{d}\phi$ (1)

Which $Y_{12}$ is mutual impedance between two slots number 1 and 2, $E_2$ is far electrical radiated field which is emerged from $V_2$ and $H_1^{\text{*}}$ is far conjugate magnetic field which is emerged by $V_1$ in (1).

2. Computing Mutual Admittance

Far field radiation from circumferential slots [12] whereas voltage has cosine distribution can be written as follows:

$E_{i\theta }=\frac{V_i{\alpha }_i{\text{e}}^{-jk{r}_i}}{j\text{π}{r}_i\sin \left(\theta \right)}{\displaystyle {\sum }_{n=-\infty }^{\infty }\frac{j^n\cos \left(\frac{n{\alpha }_i}{2}\right){\text{e}}^{jn\left(\phi -{\phi }_i\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_i\right)}^2\right]H_n^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)}\text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)}$ (2)

$\begin{array}{c}{E}_{i\phi }=-\frac{V_i{\alpha }_i{\text{e}}^{-jk{r}_i}}{\text{π}{r}_{i}ka\sin \left(\theta \right)}\cot \left(\theta \right){\displaystyle {\sum }_{n=-\infty }^{\infty }\frac{m{j}^m\cos \left(\frac{m{\alpha }_i}{2}\right){\text{e}}^{jm\left(\phi -{\phi }_i\right)}}{\left[{\text{π}}^2-{\left(m{\alpha }_i\right)}^2\right]H_m^{\left(2\right)}{}^{\prime }\left(a\sin \left(\theta \right)\right)}}\\ \cdot \text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (3)

$H_{\phi }=\frac{E_{\theta }}{\eta },H_{\theta }=-\frac{E_{\phi }}{\eta }$ (4)

where $W=3.5\text{mm}$ and $k=\frac{2\text{π}}{\lambda }$ which λ is wavelength of propagated wave

with vacuum dielectric, a is radius of cylinder and $V_i$ is voltage which is emerged from distributed field on th slot and ${\alpha }_i$ is slot length depending on degree. In here $r_i\approx r-z_{i}k\cos \left(\theta \right)$ and ${\phi }_i$ is offset of th slot according to x axis, η is intrinsic impedance of environment and $r,\theta$ and φ are spherical coordinate (Figure 1).

$H_m^{\left(2\right)}{}^{\prime }\left(\text{.}\right)$ is derivative of second type of Henkel function. In above relations time-dependent function ( ${\text{e}}^{j\omega t}$ ) is neglected. Now, above relations must be considered for finding mutual impedance so according to (1) the result of $E_2\times H_1^{\text{*}}$ can be presented as follows:

$E_2=E_{2\theta }{\hat{a}}_{\theta }+E_{2\phi }{\hat{a}}_{\phi }$ (5)

$H_1=H_{1\theta }{\hat{a}}_{\theta }+H_{1\phi }{\hat{a}}_{\phi }$ (6)

$\begin{array}{c}{E}_2\times H_1^{\text{*}}=\left(E_{2\theta }{\hat{a}}_{\theta }+E_{2\phi }{\hat{a}}_{\phi }\right)\times {\left(H_{1\theta }{\hat{a}}_{\theta }+H_{1\phi }{\hat{a}}_{\phi }\right)}^{\text{*}}\\ =\left(E_{2\theta }{H}_{1\phi }^{\text{*}}-E_{2\phi }{H}_{1\theta }^{\text{*}}\right){\hat{a}}_r=\left(E_{2\theta }\frac{E_{1\theta }^{\text{*}}}{\eta }+E_{2\phi }\frac{E_{1\phi }^{\text{*}}}{\eta }\right){\hat{a}}_r\end{array}$ (7)

Integration path C can be shown as Figure 2:

Suppose that in cylindrical coordination system, slot position of I and II are $\left(a,{\phi }_2,z_2\right)$ and $\left(a,{\phi }_1,z_1\right)$ .

According to above coordination system, electric and magnetic far field radiation is:

$E_{1\theta }=\frac{V_1{\alpha }_1{\text{e}}^{-jk{r}_1}}{j\text{π}{r}_1\sin \left(\theta \right)}{\displaystyle {\sum }_{n=-\infty }^{\infty }\frac{j^n\cos \left(\frac{n{\alpha }_1}{2}\right){\text{e}}^{jn\left(\phi -{\phi }_1\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]H_n^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)}\text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)}$ (8)

Which $r_1\approx r-z_{1}k\cos \left(\theta \right)$ . For $E_{2\phi }$ we have:

$\begin{array}{c}{E}_{2\phi }=-\frac{V_2{\alpha }_2{\text{e}}^{-jk{r}_2}}{\text{π}{r}_{2}ka\sin \left(\theta \right)}\cot \left(\theta \right){\displaystyle {\sum }_{m=-\infty }^{\infty }\frac{m{j}^m\cos \left(\frac{m{\alpha }_2}{2}\right){\text{e}}^{jm\left(\phi -{\phi }_2\right)}}{\left[{\text{π}}^2-{\left(m{\alpha }_1\right)}^2\right]H_m^{\left(2\right)}{}^{\prime }\left(a\sin \left(\theta \right)\right)}}\\ \cdot \text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (9)

That $r_2\approx r-z_{2}k\cos \left(\theta \right)$ .

Now the result of $E_2\times H_1^{*}$ must be defined, so first of all the term of $E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }$ must be computed as follow:

$E_{2\theta }=\frac{V_2{\alpha }_2{\text{e}}^{-jk{r}_2}}{j\text{π}{r}_2\sin \left(\theta \right)}{\displaystyle {\sum }_{n=-\infty }^{\infty }\frac{j^n\cos \left(\frac{n{\alpha }_2}{2}\right){\text{e}}^{jn\left(\phi -{\phi }_2\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]H_n^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)}\sin c\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)}$ (10)

$E_{1\theta }^{\ast }=\frac{V_1{}^{*}{\alpha }_1{\text{e}}^{jk{r}_1}}{-j\text{π}{r}_1\sin \left(\theta \right)}{\displaystyle {\sum }_{n=-\infty }^{\infty }\frac{{\left(-j\right)}^n\cos \left(\frac{n{\alpha }_1}{2}\right){\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}\text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)}$ (11)

$\begin{array}{c}\frac{E_{2\theta }{E}_{1\theta }^{\ast }}{\eta }=\frac{V_2{V}_1{}^{*}{\alpha }_1{\alpha }_2{\text{e}}^{-jk\left(r_2-r_1\right)}}{\eta {\pi }^2{r}_1{r}_2{\sin }^2\left(\theta \right)}\cdot \underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{j^{m-n}\cos \left(\frac{m{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_m^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}\cdot {\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (12)

Then for $E_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }$ we have:

$\begin{array}{c}{E}_{1\phi }^{\ast }=-\frac{V_1{}^{*}{\alpha }_1{\text{e}}^{jk{r}_1}}{\text{π}{r}_{1}ka\sin \left(\theta \right)}\cot \left(\theta \right)\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{n{\left(-j\right)}^n\cos \left(\frac{n{\alpha }_1}{2}\right){\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]H_n^{\left(2\right)}{{}^{\prime }}^{*}\left(ka\sin \left(\theta \right)\right)}\\ \cdot \text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (13)

$\begin{array}{c}{E}_{2\phi }=-\frac{V_2{\alpha }_2{\text{e}}^{-jk{r}_2}}{\text{π}{r}_{2}ka\sin \left(\theta \right)}\cot \left(\theta \right)\underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{m{j}^m\cos \left(\frac{m{\alpha }_2}{2}\right){\text{e}}^{jm\left(\phi -{\phi }_2\right)}}{\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)}\\ \cdot \text{sinc}\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (14)

$\begin{array}{l}{E}_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }=\frac{V_1{}^{*}{V}_2{\alpha }_2{\alpha }_1{\text{e}}^{-jk\left(r_2-r_1\right)}}{{\text{π}}^2{r}_1{r}_2{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\\ \cdot \underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{mn{j}^{m-n}\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{m{\alpha }_2}{2}\right){\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{{}^{\prime }}^{*}\left(ka\sin \left(\theta \right)\right)}\\ \cdot {\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (15)

To simplify integral computation, it must be divided in to two parts then compute any part separately and finally the results must be added:

$\begin{array}{l}{Y}_{12}=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}{E}_2\times H_1^{*}\cdot \hat{n}{r}^2\sin \theta \text{d}\theta \text{d}\phi \\ =\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }+E_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }\right){\hat{a}}_r\cdot {\hat{a}}_r{r}^2\sin \theta \text{d}\theta \text{d}\phi \\ =\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi \\ +\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi =Y_{12}^1+Y_{12}^2\end{array}$ (16)

$Y_{12}^1=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi$ (17)

$Y_{12}^2=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi$ (18)

In another hand to compute $Y_{12}^1$ we have:

$\begin{array}{c}{Y}_{12}^1=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi \\ =\frac{1}{V_2{V}_1{}^{\text{*}}}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{V_2{V}_1{}^{\text{*}}{\alpha }_1{\alpha }_2{\text{e}}^{-jk\left(r_2-r_1\right)}}{\eta {\text{π}}^2{r}_1{r}_2{\sin }^2\left(\theta \right)}{\displaystyle \underset{m=-\infty }{\overset{\infty }{\sum }}}{\displaystyle \underset{n=-\infty }{\overset{\infty }{\sum }}}\frac{j^{m-n}\cos \left(\frac{m{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_m^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)r^2\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (19)

According to asymmetry of integration path and term of $Y_{12}^1$ , so integration path can be assumed as:

So:

$\begin{array}{c}{Y}_{12}^1=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\theta }\frac{E_{1\theta }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi \\ =\frac{2{\alpha }_1{\alpha }_2}{\eta {\text{π}}^2{r}_1{r}_2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\left(r_2-r_1\right)}}{{\sin }^2\left(\theta \right)}{\displaystyle \underset{m=-\infty }{\overset{\infty }{\sum }}}{\displaystyle \underset{n=-\infty }{\overset{\infty }{\sum }}}\frac{j^{m-n}\cos \left(\frac{m{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_m^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)r^2\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (20)

After substitution of $r_1$ and $r_2$ in (20) and according to even symmetry with respect to φ, integral can be simplify as:

$\begin{array}{c}{Y}_{12}^1=\frac{2{\alpha }_1{\alpha }_2}{\eta {\text{π}}^2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\sin }^2\left(\theta \right)}\sin \theta {\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\epsilon }_m{\epsilon }_n{j}^{m-n}\cos \left(\frac{m{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{\cos \left(m\left(\phi -{\phi }_2\right)\right)\cos \left(n\left(\phi -{\phi }_1\right)\right)}{H_m^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\text{d}\theta \text{d}\phi \end{array}$ (21)

Which:

${\epsilon }_m=\{\begin{array}{l}1m=0\\ 2m\ne 0\end{array}$

if $m=n$ , term of $\cos \left(m\left(\phi -{\phi }_2\right)\right)\cos \left(n\left(\phi -{\phi }_1\right)\right)$ is an orthogonal function and it has definite values.

So integral can be simplified such as following relations:

$\begin{array}{c}{Y}_{12}^1=\frac{2{\alpha }_1{\alpha }_2}{\eta {\text{π}}^2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\sin }^2\left(\theta \right)}\sin \theta {\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\epsilon }_n{\epsilon }_n\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{\cos \left(n\left(\phi -{\phi }_2\right)\right)\cos \left(n\left(\phi -{\phi }_1\right)\right)}{H_n^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{}^{*}\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\text{d}\theta \text{d}\phi \end{array}$ (22)

Above relations can be simplified more, if integral be found in term of following relations:

$\begin{array}{c}{Y}_{12}^1=\frac{4{\alpha }_1{\alpha }_2}{\eta \text{π}}{{{\displaystyle \int }}^{\text{}}}_c\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\sin }^2\left(\theta \right)}\sin \theta \cdot {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\epsilon }_n\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left|H_n^{\left(2\right)}\left(ka\sin \left(\theta \right)\right)\right|}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\text{d}\theta \end{array}$ (23)

The integral can be divided to imaginary and real parts.

For real part we have:

$k_z=k\cos \left(\theta \right),\text{d}{k}_z=-k\sin \theta \text{d}\theta ,\theta =\left[0,\frac{\text{π}}{2}\right]\to k_z=\left[k,0\right]$

So result can be shown as follows:

$\begin{array}{c}{Y}_{12}^1=\frac{4{\alpha }_1{\alpha }_{2}k}{\eta \text{π}}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{{\text{e}}^{-j{k}_z\left(z_2-z_1\right)}}{k^2-k_z^2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\epsilon }_n\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left|H_n^{\left(2\right)}\left(a\sqrt{k^2-k_z^2}\right)\right|}{\text{sinc}}^2\left(\frac{W{k}_z}{2\text{π}}\right)\text{d}{k}_z\end{array}$ (24)

Now we consider imaginary part, for new path can be shown:

$\begin{array}{c}{k}_z=k\cos \left(\theta \right)=k\cos \left(\mathrm{Re}\left(\theta \right)+j\mathrm{Im}\left(\theta \right)\right)\\ =k\left[\cos \left(\mathrm{Re}\left(\theta \right)\right)\cosh \left(\mathrm{Im}\left(\theta \right)\right)-\sin \left(\mathrm{Re}\left(\theta \right)\right)\sinh \left(\mathrm{Im}\left(\theta \right)\right)\right]\end{array}$

$\mathrm{Re}\left(\theta \right)=\frac{\text{π}}{2},\mathrm{Im}\left(\theta \right)=\left[0,\infty \right]\Rightarrow k_z=\left[0,-j\infty \right]$

It must be mentioned that $k_z=-j{k^{\prime }}_z,{k^{\prime }}_z=\left[0,\infty \right]$ so:

$\begin{array}{c}{Y}_{12}^1=\frac{j4{\alpha }_1{\alpha }_{2}k}{\eta \text{π}}{{{\displaystyle \int }}^{\text{}}}_0^{\infty }\frac{{\text{e}}^{-{k^{\prime }}_z\left(z_2-z_1\right)}}{k^2+{k^{\prime }}_z^2}\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}\frac{{\epsilon }_n\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(\frac{n{\alpha }_1}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}\\ \cdot \frac{\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{{\left|H_n^{\left(2\right)}\left(a\sqrt{k^2+{k^{\prime }}_z^2}\right)\right|}^2}{\text{sinhc}}^2\left(\frac{W{k^{\prime }}_z}{2\text{π}}\right)\text{d}{k^{\prime }}_z\end{array}$ (25)

where:

$\text{sinch}\left(\frac{W{k^{\prime }}_z}{2\text{π}}\right)=\frac{\sinh \left(\frac{W{k^{\prime }}_z}{2}\right)}{\frac{W{k^{\prime }}_z}{2}}$

Same as $Y_{12}$ we try to find $Y_{12}^2$ so:

$Y_{12}^2=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\left(E_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }\right)r^2\sin \theta \text{d}\theta \text{d}\phi$ (26)

$\begin{array}{c}{E}_{2\phi }\frac{E_{1\phi }^{\ast }}{\eta }=\frac{V_1{}^{*}{V}_2{\alpha }_2{\alpha }_1{\text{e}}^{-jk\left(r_2-r_1\right)}}{\eta {\text{π}}^2{r}_1{r}_2{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{mn{j}^{m-n}\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{m{\alpha }_2}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_n^{\left(2\right)}{{}^{\prime }}^{\ast }\left(ka\sin \left(\theta \right)\right)H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\end{array}$ (27)

$\begin{array}{l}{Y}_{12}^2=\frac{1}{V_2{V}_1^{\ast }}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{V_1{}^{*}{V}_2{\alpha }_2{\alpha }_1{\text{e}}^{-jk\left(r_2-r_1\right)}}{{\text{π}}^2{r}_1{r}_2{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\\ \cdot \underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{mn{j}^{m-n}\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{m{\alpha }_2}{2}\right){\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{{}^{\prime }}^{*}\left(ka\sin \left(\theta \right)\right)}\\ \cdot {\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)r^2\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (28)

$\begin{array}{l}\Rightarrow Y_{12}^2=\frac{{\alpha }_2{\alpha }_1}{{\text{π}}^2{r}_1{r}_2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\left(r_2-r_1\right)}}{{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\underset{m=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\underset{n=-\infty }{\overset{\infty }{{\displaystyle \sum }}}\frac{mn{j}^{m-n}\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{m{\alpha }_2}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_n^{\left(2\right)}{{}^{\prime }}^{*}\left(ka\sin \left(\theta \right)\right)H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)r^2\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (29)

After substitution and by following above steps, the result will be as (29):

$\begin{array}{l}{Y}_{12}^2=\frac{2{\alpha }_2{\alpha }_1}{\eta {\text{π}}^2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\underset{m=1}{\overset{\infty }{{\displaystyle \sum }}}\underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{mn{j}^{m-n}\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{m{\alpha }_2}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(m{\alpha }_2\right)}^2\right]}\\ \cdot \frac{{\text{e}}^{jm\left(\phi -{\phi }_2\right)}{\text{e}}^{-jn\left(\phi -{\phi }_1\right)}}{H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)H_n^{\left(2\right)}{{}^{\prime }}^{\ast }\left(ka\sin \left(\theta \right)\right)}{\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)r^2\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (30)

According to orthogonality property, “m” is equal to “n” so:

$\begin{array}{c}{Y}_{12}^2=\frac{4{\alpha }_2{\alpha }_1}{\eta {\text{π}}^2}{{{\displaystyle \int }}^{\text{}}}_c{{{\displaystyle \int }}^{\text{}}}_0^{\text{2π}}\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left(\phi -{\phi }_2\right)\right)\cos \left(n\left(\phi -{\phi }_1\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)\right|}^2}\\ \cdot {\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (31)

Modified integral path is shown in Figure 3.

The integration in term of φ will result in (34):

$\begin{array}{c}{Y}_{12}^2=\frac{8{\alpha }_2{\alpha }_1}{\eta \text{π}}{{{\displaystyle \int }}^{\text{}}}_c\frac{{\text{e}}^{-jk\cos \left(\theta \right)\left(z_2-z_1\right)}}{{\left(ka\sin \left(\theta \right)\right)}^2}{\left(\cot \left(\theta \right)\right)}^2\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_m^{\left(2\right)}{}^{\prime }\left(ka\sin \left(\theta \right)\right)\right|}^2}\\ \cdot {\text{sinc}}^2\left(\frac{Wk\cos \left(\theta \right)}{2\text{π}}\right)\sin \theta \text{d}\theta \text{d}\phi \end{array}$ (32)

The integral must be divided in to real and imaginary part so for real part we have:

$k_z=k\cos \left(\theta \right),\text{d}{k}_z=-k\sin \theta \text{d}\theta ,\theta =\left[0,\frac{\text{π}}{2}\right]\to k_z=\left[k,0\right]$

$\begin{array}{l}{Y}_{12}^2=\frac{8{\alpha }_2{\alpha }_1}{\eta \text{π}{a}^{2}k}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{{\text{e}}^{-j{k}_z\left(z_2-z_1\right)}}{k^2-k_z^2}\frac{k_z^2}{k^2-k_z^2}\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_m^{\left(2\right)}{}^{\prime }\left(a\sqrt{k^2-k_z^2}\right)\right|}^2}{\text{sinc}}^2\left(\frac{W{k}_z}{2\text{π}}\right)\text{d}{k}_z\end{array}$ (33)

Now imaginary part of path must be computed. At first of hand, path must be modified i.e. for new path we have:

$\begin{array}{c}{k}_z=k\cos \left(\theta \right)=k\cos \left(\mathrm{Re}\left(\theta \right)+j\mathrm{Im}\left(\theta \right)\right)\\ =k\left[\cos \left(\mathrm{Re}\left(\theta \right)\right)\cosh \left(\mathrm{Im}\left(\theta \right)\right)-j\sin \left(\mathrm{Re}\left(\theta \right)\right)\sinh \left(\mathrm{Im}\left(\theta \right)\right)\right]\end{array}$ (34)

$\mathrm{Re}\left(\theta \right)=\frac{\text{π}}{2},\mathrm{Im}\left(\theta \right)=\left[0,\infty \right]\Rightarrow k_z=\left[0,-j\infty \right]$

Same as above relation: $k_z=-j{k^{\prime }}_z,{k^{\prime }}_z=\left[0,\infty \right]$ so we have:

$\begin{array}{c}{Y}_{12}^2=\frac{j8{\alpha }_2{\alpha }_1}{\eta \text{π}{a}^{2}k}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{{\text{e}}^{-{k^{\prime }}_z\left(z_2-z_1\right)}}{k^2+{k^{\prime }}_z^2}\frac{{k^{\prime }}_z^2}{k^2+{k^{\prime }}_z^2}\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_n^{\left(2\right)}{}^{\prime }\left(a\sqrt{k^2+{k^{\prime }}_z^2}\right)\right|}^2}{\text{sinch}}^2\left(\frac{W{k^{\prime }}_z}{2\text{π}}\right)\text{d}{k^{\prime }}_z\end{array}$ (35)

As a result for totality of $Y_{12}^2$ :

$\begin{array}{c}{Y}_{12}^2=\frac{8{\alpha }_2{\alpha }_1}{\eta \text{π}{a}^{2}k}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{{\text{e}}^{-k_z\left(z_2-z_1\right)}}{k^2-k_z^2}\frac{k_z^2}{k^2-k_z^2}\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_n^{\left(2\right)}{}^{\prime }\left(a\sqrt{k^2+{k^{\prime }}_z^2}\right)\right|}^2}{\text{sinc}}^2\left(\frac{W{k}_z}{2\text{π}}\right)\text{d}{k}_z\\ +\frac{j8{\alpha }_2{\alpha }_1}{\eta \text{π}{a}^{2}k}{{{\displaystyle \int }}^{\text{}}}_0^{\infty }\frac{{\text{e}}^{-{k^{\prime }}_z\left(z_2-z_1\right)}}{k^2+{k^{\prime }}_z^2}\frac{{k^{\prime }}_z^2}{k^2+{k^{\prime }}_z^2}\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_n^{\left(2\right)}{}^{\prime }\left(a\sqrt{k^2+{k^{\prime }}_z^2}\right)\right|}^2}{\text{sinch}}^2\left(\frac{W{k^{\prime }}_z}{2\text{π}}\right)\text{d}{k^{\prime }}_z\end{array}$ (36)

The totality of $Y_{12}$ is sum of $Y_{12}^1$ and $Y_{12}^2$

$Y_{12}=Y_{12}^1+Y_{12}^2$ (37)

$Y_{12}$ Can be divided in to real and imaginary part:

$\begin{array}{c}{G}_{12}=\frac{4{\alpha }_1{\alpha }_2}{\eta \text{π}}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{\cos k_z\left(z_2-z_1\right)}{k^2-k_z^2}\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}\frac{{\epsilon }_n\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]}^2}\\ \cdot \frac{\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left|H_n^{\left(2\right)}\left(a\sqrt{k^2-k_z^2}\right)\right|}{\text{sinc}}^2\left(\frac{W{k}_z}{2\text{π}}\right)\text{d}{k}_z\\ +\frac{8{\alpha }_2{\alpha }_1}{\eta \text{π}{a}^{2}k}{{{\displaystyle \int }}^{\text{}}}_0^k\frac{\cos k_z\left(z_2-z_1\right)}{k^2-k_z^2}\frac{k_z^2}{k^2-k_z^2}\\ \cdot \underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}\frac{n^2\cos \left(\frac{n{\alpha }_1}{2}\right)\cos \left(\frac{n{\alpha }_2}{2}\right)\cos \left(n\left({\phi }_1-{\phi }_2\right)\right)}{\left[{\text{π}}^2-{\left(n{\alpha }_1\right)}^2\right]\left[{\text{π}}^2-{\left(n{\alpha }_2\right)}^2\right]{\left|H_n^{\left(2\right)}{}^{\prime }\left(a\sqrt{k^2-k_z^2}\right)\right|}^2}{\text{sinc}}^2\left(\frac{W{k}_z}{2\text{π}}\right)\text{d}{k}_z\end{array}$ (38)