**3. Determination of the Green's function**

This function used to form the integral equation Eq.(2), thereby it represents the inverse of Laplacian operator Eq.(3), where the point ð Þ *x*, *y* is said field point created by a unit charge 1ð Þ *C* at the point *x*0, *y*<sup>0</sup> � � said source point.

$$G\_0(\mathbf{x}, \boldsymbol{y} | \mathbf{x}\_0, \boldsymbol{y}\_0) = \left(\frac{\partial^2}{\partial \mathbf{x}^2} + \frac{\partial^2}{\partial \boldsymbol{y}^2}\right)^{-1} \tag{3}$$

$$\left(\frac{\partial^2}{\partial \mathbf{x}^2} + \frac{\partial^2}{\partial \mathbf{y}^2}\right) G\_0(\mathbf{x}, \mathbf{y} | \mathbf{x}\_0, \mathbf{y}\_0) = -\delta(\mathbf{x} - \mathbf{x}\_0) \left(\mathbf{y} - \mathbf{y}\_0\right) \tag{4}$$

With *δ* is the Dirac function, so the Green's function *G*<sup>0</sup> is written in the following form:

$$G\_0(\mathbf{x}, \mathbf{y} | \mathbf{x}\_0, \mathbf{y}\_0) = -\frac{1}{2\pi} \ln \sqrt{(\mathbf{x} - \mathbf{x}\_0)^2 + \left(\mathbf{y} - \mathbf{y}\_0\right)^2} + cte \tag{5}$$

To calculate the Green's function corresponds to the coplanar line without the central strip shown in **Figure 2a**, first we calculate the Green's function of the electrical potential created by a distribution of charges between two infinite ground planes shown in **Figure 2b** using the multiple image method [2], given by [3] as follow:

$$G\_0\left(\mathbf{x}, \boldsymbol{\mathcal{y}} | \mathbf{x}\_0, \boldsymbol{\mathcal{y}}\_0\right) = -\frac{1}{4\pi} \sum\_{n=-\infty}^{n=+\infty} \ln \left( \frac{\left(\boldsymbol{\mathcal{y}} - \boldsymbol{\mathcal{y}}\_0\right)^2 + \left(\boldsymbol{\mathcal{x}} - \boldsymbol{\mathcal{x}}\_0 - 2na\right)^2}{\left(\boldsymbol{\mathcal{y}} - \boldsymbol{\mathcal{y}}\_0\right)^2 + \left(\boldsymbol{\mathcal{x}} + \boldsymbol{\mathcal{x}}\_0 - 2na\right)^2} \right) \tag{6}$$

**4. Numerical characterization of the integral equation**

1 *ε* ð*w=*<sup>2</sup> �*w=*2

*ψ*ð Þ¼ *x*

method is divided into two stages, firstly by developing *ρ x*0, *y*<sup>0</sup>

*ρ <sup>j</sup>*ð Þ¼ *x*<sup>0</sup>

X *N*

*i*¼1

*α j*

8 < :

*ψi*ð Þ¼ *x*<sup>0</sup>

1 *ε* ð*w=*<sup>2</sup> �*w=*2

*ψ*ð Þ¼ *x*, *y*

*DOI: http://dx.doi.org/10.5772/intechopen.95142*

*f j* ð Þ¼ *x*<sup>0</sup>

X *N*

*i*¼1

system Eq. (14):

**Figure 4.**

**85**

*Discretization of the central strip.*

The linear charge density *ρ* on the metal strip shown in **Figure 4** is related to the

*Analysis and Two-Dimensional Modeling of Directional Coupler Based on Two Coplanar Lines*

*G x*, *y*j*x*0, *y*<sup>0</sup>

The central strip conductor is considered infinitely thin, that allows us to write:

Assuming that the central strip is submitted to a unit potential, so the equation Eq. (11) is convenient to solve numerically using the method of moments. This

functions Eq. (12) in the form of rectangular pulses Eq. (13), secondly by using the Galerkin procedure which allows to write equation Eq. (11) as a linear equations

> X *N*

> > *j*¼1

0 *alsewhere*

X *N*

*j*¼1

� � <sup>¼</sup> *Mij* � ��<sup>1</sup>

linear equations system Eq. (14), defined on the central strip of the structure

<sup>1</sup> *if x*<sup>∈</sup> *<sup>x</sup> <sup>j</sup>* � *<sup>h</sup>*

*α j εe*

So, the linear equations system (14) can be written in the following matrix form:

With *Mij* � � is the matrix of the Green's function obtained by the inversion of the

ð*<sup>x</sup> <sup>j</sup>*þ*<sup>h</sup>* 2 *<sup>x</sup> <sup>j</sup>*�*<sup>h</sup>* 2

*: ψ <sup>j</sup>*

*α <sup>j</sup> f <sup>j</sup>*

2 ; *x <sup>j</sup>* þ *h* 2

� �

*G xi*j*x <sup>j</sup>*

� �*:<sup>ρ</sup> <sup>x</sup>*0, *<sup>y</sup>*<sup>0</sup>

� �*dl*<sup>0</sup> (10)

� � in series of N basic

(13)

ð Þ *x*<sup>0</sup> (12)

� �*dx*<sup>0</sup> (14)

h i (15)

*G x*ð Þ j*x*<sup>0</sup> *:ρ*ð Þ *x*<sup>0</sup> *dx*<sup>0</sup> (11)

distribution potential *ψ* in the cross-section of the coplanar line, and it's zero elsewhere. For that reason, the integral equation Eq. (2) can be written as:

#### **Figure 2.**

*(a) Micro-coplanar line without central conductor; (b) structure with two infinite ground planes.*

With *a* is the gap between the two infinite metal plate **Figure 2a**, its sum given by [6] as:

$$G\_0\left(\mathbf{x}, \boldsymbol{\mathcal{y}} | \mathbf{x}\_0, \boldsymbol{\mathcal{y}}\_0\right) = -\frac{1}{4\pi} \ln \left( \frac{\sin^2 \frac{\pi}{2a} (\mathbf{x} - \mathbf{x}\_0) + sh^2 \frac{\pi}{2a} (\mathbf{y} - \mathbf{y}\_0)}{\sin^2 \frac{\pi}{2a} (\mathbf{x} + \mathbf{x}\_0) + sh^2 \frac{\pi}{2a} (\mathbf{y} - \mathbf{y}\_0)} \right) \tag{7}$$

The next step consists of applying a suitable conformal mapping which makes it possible to transform the Green's function into the given structure in **Figure 2a**, it's given as:

$$G(\mathbf{x},\boldsymbol{\eta}|\mathbf{x}\_{0},\boldsymbol{\eta}\_{0}) = -\frac{1}{4\pi} \ln \left( \frac{\sin^{2}\left(\frac{1}{2}\left(\cos^{-1}\left(\frac{\mathbf{x}}{\mathbf{b}\cdot\mathbf{z}}\right) - \cos^{-1}\left(\frac{\mathbf{x}\_{0}}{\mathbf{b}\cdot\mathbf{z}\_{0}}\right)\right)\right) + sh^{2}\left(\frac{1}{2}\left(\boldsymbol{ch}^{-1}(\mathbf{z}) - \boldsymbol{ch}^{-1}(\mathbf{z}\_{0})\right)\right)}{\sin^{2}\left(\frac{1}{2}\left(\cos^{-1}\left(\frac{\mathbf{x}}{\mathbf{b}\cdot\mathbf{z}}\right) + \cos^{-1}\left(\frac{\mathbf{x}\_{0}}{\mathbf{b}\cdot\mathbf{z}\_{0}}\right)\right)\right) + sh^{2}\left(\frac{1}{2}\left(\boldsymbol{ch}^{-1}(\mathbf{z}) - \boldsymbol{ch}^{-1}(\mathbf{z}\_{0})\right)\right)}\right) \tag{8}$$

$$\text{With:} z = \left(\frac{a+\beta}{2}\right)^{1/2}; z\_0 = \left(\frac{a\_0+\beta\_0}{2}\right)^{1/2}; \text{ and } a = \left(\frac{x}{b}\right)^2 + \left(\frac{y}{b}\right)^2; \beta = \left(a^2 - 4\left(\frac{x}{b}\right)^2\right).$$

The Green's function Eq. (8) is also a reciprocal function [7], and it's can be expressed as the superposition of two functions for non-homogeneous [8] middle, and for isotropic or anisotropic [9–10] middles.

So, for a non-homogeneous middle shown in the **Figure 3**, the Green's function is written as [3]:

$$G(\mathbf{x}, y | \mathbf{x}\_0, y\_0) = -\frac{1}{4\pi} \left( \ln \left( \frac{\sin^2(\mathbf{R}^-) + sh^2(T^-)}{\sin^2(\mathbf{R}^+) + sh^2(T^-)} \right) + R.\ln \left( \frac{\sin^2(\mathbf{R}^-) + sh^2(T^+)}{\sin^2(\mathbf{R}^+) + sh^2(T^+)} \right) \right) \tag{9}$$

With: *<sup>R</sup>*� <sup>¼</sup> <sup>1</sup> <sup>2</sup> *cos* �<sup>1</sup> *<sup>x</sup> b:z* � � � *cos* �<sup>1</sup> *<sup>x</sup>*<sup>0</sup> *b:z*<sup>0</sup> � � � � ; *<sup>T</sup>*� <sup>¼</sup> <sup>1</sup> <sup>2</sup> *ch*�<sup>1</sup> ð Þ� *<sup>z</sup> ch*�<sup>1</sup> ð Þ *z*0*:* � � And: *<sup>R</sup>* <sup>¼</sup> *<sup>ε</sup>*1�*ε*<sup>2</sup> *ε*1þ*ε*<sup>2</sup> .

**Figure 3.** *Micro-coplanar (non-homogeneous middle).*

*Analysis and Two-Dimensional Modeling of Directional Coupler Based on Two Coplanar Lines DOI: http://dx.doi.org/10.5772/intechopen.95142*
