**Appendix A. Calculation of a microstrip line impedance**

The equivalent scheme of **Figure 12** includes an excitation part using the formula of Eq. (6) (the upper modes are assumed evanescent) and an interface part, which is, in this case, a dipole. The global scheme is thus, after using the one of **Figure 1**, the following is:

By posing *Ys* ¼ *Zs* �1 , we obtain the relations (**Figures 26** and **27**):

$$
\begin{vmatrix} \mathbf{E}\_0 \\ \mathbf{J} \end{vmatrix} = \begin{vmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{1} & \hat{Y}\_M + \hat{Y}\_S \end{vmatrix} \begin{vmatrix} \mathbf{J}\_0 \\ \mathbf{E} \end{vmatrix} \tag{34}
$$

We can notice that *Y*^ *<sup>S</sup>* is a diagonal operator because the surface impedance is uniformly defined on the whole surface. Its annulment on the metal subdomain is indeed by the short circuit shown in **Figure 1**.

We define the scalar product as an integral over y between 0 and *b* of the product of functions:

**Figure 26.** *Equivalent scheme of the closure at* z *= 0.*

**Figure 27.** *Equivalent scheme of the structure of Figure 12.*

$$
\langle f|\mathbf{g}\rangle = \int\_0^b f\_y \mathbf{g}\_y d\mathbf{y} \text{ (}f\_x = \mathbf{g}\_x = \mathbf{0}\text{)}\tag{35}
$$

In this problem, the *E* and *J* fields have no components at *x* and are independent of *x* (Invariance by translation along *Ox* of the structure) so, the integral on the surface can be simplified into a *y*-integral. The problem is two-dimensional, leading to many simplifications: simple series in the developments, only one mode present, and a scalar formulation. We pose, in a slightly different way than in Eq. (14):

$$J\_0 = \frac{1}{a} \mathcal{H}\_0' \text{ and } \langle \left. f\_0' \right| \mathbf{E} \rangle \tag{36}$$

*f* 0 <sup>0</sup> ¼ 1 in this case, with the definition of the scalar product of Eq. (35). The unknown field in the aperture is expressed on a basis defined from 0 to *c*. We pose successively:

$$E(\mathbf{y}) = \sum\_{p=0}^{\infty} \varkappa\_p \mathbf{g}\_p(\mathbf{y}) \tag{37}$$

with *f* <sup>0</sup> <sup>0</sup> <sup>¼</sup> 1, *<sup>g</sup>*<sup>0</sup> <sup>¼</sup> <sup>1</sup>ffi *<sup>c</sup>* <sup>p</sup> , *g<sup>p</sup>* 6¼ ffiffi 2 *c* q cos *<sup>p</sup>π<sup>y</sup> c* � �. The unknowns are *<sup>g</sup>p*; by applying the Galerkin method, the first Equation is scalarly multiplied by *f* <sup>0</sup> <sup>0</sup> and the second one successively by *g*0, *g*1, *g*<sup>2</sup> … *gp*, p is the order of truncation of the test functions. This gives us:

$$V = \sum\_{p=0}^{p} \langle \left. f' \right|\_{0} \mathbf{g}\_{p} \rangle \mathbf{x}\_{p} \tag{38}$$

The other equations have a zero first member because J is zero in the virtual source domain and thus:

$$
\left\langle \mathbf{g}\_p \right| f\_0 \text{'} \rangle = \mathbf{0} = - \left\langle f\_0 \left| \mathbf{g}\_p \right\rangle \frac{I}{a} + \sum\_{q=0}^p \left\langle \mathbf{g}\_p \right| (\hat{\mathbf{Y}}\_M + \hat{\mathbf{Y}}\_s) \Big| \mathbf{g}\_q \right\rangle \mathbf{x}\_q \tag{39}
$$

*Multiscale Auxiliary Sources for Modeling Microwave Components DOI: http://dx.doi.org/10.5772/intechopen.102795*

Eliminating the unknown vector of *xp* components between Eq. (38) and Eq. (39), we find:

$$\frac{V}{I} = \frac{1}{a} A^t [Y]^{-1} A \tag{40}$$

with, *A* ¼ *f* <sup>0</sup> <sup>0</sup> � � �*g*1 � , *f* <sup>0</sup> <sup>0</sup> � � �*g*2 � , … , *f* <sup>0</sup> <sup>0</sup> � � �*gp* h iE .

½ � *<sup>Y</sup>* �<sup>1</sup> is the inverse of the matrix and *<sup>Y</sup>* is built on the operator *<sup>Y</sup>*^*<sup>M</sup>* <sup>þ</sup> *<sup>Y</sup>*^*s*. The general term of *Y* is written, in this definition:

$$\left[\left[Y\right]\_{pq} = \left<\mathbf{g}\_p\right| \left(\hat{Y}\_M + \hat{Y}\_s\right) \middle| \mathbf{g}\_q\right>\tag{41}$$

with as in Eq. (6):

$$\hat{Y}\_M = \sum\_{n>0} Y\_{Mn} \hat{P}\_n = \sum\_{n>0} \langle f\_n | Y\_{Mn} | f\_n \rangle \tag{42}$$

We can find Eq. (40) with the help of auxiliary sources by representing the problem in **Figure 12** differently: at the level of *P*, we consider a set of sources for each mode *g<sup>p</sup>* defined for *y* <*c*. The multipole *Q* represents the discontinuity between the guide of height b and that of height *c*; the sources *g<sup>p</sup>* will then be closed on the impedance *Zs*. A classical modal connection studies the multipole Q. The conditions at *z* = 0 are written:

$$\begin{cases} \mathbf{E} = \mathbf{E}' 
with \mathbf{0} < \mathbf{y} < \mathbf{b} \\ \mathbf{J} + \mathbf{J}' = \mathbf{0} 
with \mathbf{0} < \mathbf{y} < \mathbf{c} \end{cases} \tag{43}$$

We deduce that, by scalar multiplication by *f <sup>n</sup>* of the first Equation and by *g<sup>p</sup>* of the second, the generalized transformer relations:

$$\begin{cases} \mathbf{E}\_n = \sum\_{q=0}^p \langle f\_n | \mathbf{g}\_q \rangle \mathbf{E}'\_q \\\\ \mathbf{J}'\_p = -\sum\_{m=0}^\infty \langle \mathbf{g}\_p | f\_m \rangle \mathbf{J}\_m \end{cases} \tag{44}$$

Using the closure relation on the mode admittances of evanescent modes *f <sup>m</sup>*, (*m* > 0), the second relation of Eq. (44) combined with the first gives:

$$\begin{cases} \mathbf{J}'\_{p} = \left< \mathbf{g}\_{p} \right| f\_{0} \rangle I\_{0} + \sum\_{m>0}^{\infty} \left< \mathbf{g}\_{p} \right| f\_{m} \rangle Y\_{Mn} \left< f\_{m} \middle| \mathbf{g}\_{p} \right> \mathbf{E}\_{q}^{-P} \\\\ \mathbf{E}\_{0} = \sum\_{q=0}^{P} \left< f\_{0} \middle| \mathbf{g}\_{q} \right> \mathbf{E}'\_{q} \end{cases} \tag{45}$$

We notice the similarity between Eq. (44) and Eq. (38), Eq. (41). The difference comes from the definition of I using a function *f* <sup>0</sup> <sup>0</sup> Eq. (38), whereas here, we use the fundamental mode of the guide.

Now posing the closure relation:

$$\mathbf{J}' = -\mathbf{Y}\_\prime \mathbf{E}'\_p \tag{46}$$

We find the expression of the impedance seen from the source:

$$E\_0 = B^t \left[ \hat{Y}\_M + \hat{Y}\_S \right] B \mathbf{J}\_0 \tag{47}$$

With

$$\mathcal{B}^t = \left| \langle f\_0 | \mathbf{g}\_1 \rangle, \langle f\_0 | \mathbf{g}\_2 \rangle, \dots, \dots, \langle f\_0 | \mathbf{g}\_p \rangle \right| \tag{48}$$

Considering that *<sup>f</sup>* <sup>0</sup> <sup>¼</sup> <sup>1</sup>ffiffiffiffi *ab* <sup>p</sup> , (normed fundamental mode), that the scalar product of Eq. (48) concerns the integral in the whole right section, therefore is equal to that Eq. (35) multiplied by *a*. Finally, *V* ¼ *E*0*b* and *I* ¼ *J*0*a*, we find exactly the expression of Eq. (40). In terms of auxiliary sources, we can say that there is an identity of treatment between the direct approach and the passage through the auxiliary sources provided that these are sufficiently numerous, each having the field distribution corresponding to the functions *gp*. The formula of Eq. (40) is simple in the case of a single test function *g*0. We can establish the expression of the input admittance and examine to what extent the introduction of a second test function modifies the result.

By taking:

$$\mathbf{g}\_0 = \frac{1}{\sqrt{c}} ; \mathbf{g}\_p \neq \sqrt{\frac{2}{c}} \cos \left( \frac{p \pi y}{c} \right) \tag{49}$$

$$Y\_{Mn} = \frac{jao\varepsilon\_0}{\chi\_n}; \gamma\_n = \sqrt{\frac{n^2 \pi^2}{b^2} - k\_0^2} \tag{50}$$

The excited evanescent modes are of type *TM*0*<sup>n</sup>*. Hence, the expression of Eq. (44) is as follows.

$$Y = \frac{I}{V} = a \frac{\sum\_{n>0}^{j\_{\text{low}}} \left| \left< \mathbf{g}\_0 \middle| \, f\_n \right> \right|^2}{\left| \left< \mathbf{g}\_0 \middle| \, f\_n' \right> \right|^2} + Y\_s \frac{a}{c} \tag{51}$$

takes the form after development:

$$Y = 2j a e \frac{a}{b} \sum \frac{1}{\sqrt{\frac{n^2 \pi^2}{b^2} - k\_0^2}} \left( \frac{\sin\left(\frac{nc}{b}\right)}{\frac{nc}{b}} \right)^2 + Y\_s \frac{a}{c} \tag{52}$$

Assuming that the impedance is localized, b representing a quantity of the order of dimension of the box (lower however than the half-wavelength in the vacuum), we can admit that b is very large in front of *c*. By posing:

$$\infty = n \frac{c}{b}; d\infty = \frac{c}{b} \tag{53}$$

*Multiscale Auxiliary Sources for Modeling Microwave Components DOI: http://dx.doi.org/10.5772/intechopen.102795*

*x* is a practically continuous variable, and the series that appears in Eq. (52) turns into an integral. Using the approximations in Eq. (53), we have:

$$Y = Y\_s \frac{a}{c} + 2j \sqrt{\frac{\varepsilon}{\mu}} \int\_{\frac{\varepsilon}{5}}^{\infty} \frac{ak\_0}{\sqrt{\pi^2 \varkappa^2 - K\_0^2 c^2}} \frac{\sin^2(\varkappa)}{\varkappa^2} d\varkappa \tag{54}$$

When *c* tends to zero, we see that the integral of Eq. (54) diverges as log ð Þ *K*0*c* , while the purely surface admittance part can be kept constant by making *Ys* tend to zero proportionally to *c*; the capacitive part placed in parallel tends towards infinity, which forbids to consider an impedance of zero dimension.

If we desire to consider the edge effects, it is necessary to introduce other test functions *gp*. This gives an idea of the precision of the auxiliary source concept. For example, considering two test functions, we find for the input admittance [according to Eq. (40)]:

$$Y' = a \frac{Y\_{11} \left(1 - \frac{Y\_{11}^2}{Y\_{11} Y} \right)}{\left| \left< \mathbf{g} \right| f\_0' \right> )^2} + Y\_s \frac{a}{c} \tag{55}$$

The relative degree of accuracy of the capacitive part is given by the term:

$$\frac{\partial C}{C} = \frac{-Y\_{12}^2}{Y\_{11}Y\_{22}} \text{ With} \begin{cases} Y\_{11} = \sum\_{n>0} \frac{joe\varepsilon}{Y\_n} \left| \langle \mathbf{g}\_0 | f\_n \rangle \right|^2 \\\\ Y\_{22} = \sum\_{n>0} \frac{joe\varepsilon}{Y\_n} \left| \langle \mathbf{g}\_1 | f\_n \rangle \right|^2 \\\\ Y\_{11} = \sum\_{n>0} \frac{joe\varepsilon}{Y\_n} \langle \mathbf{g}\_0 | f\_n \rangle \langle f\_n | \mathbf{g}\_0 \rangle \end{cases} \tag{56}$$

From these expressions, we can see that the accuracy depends essentially on the ratio *<sup>c</sup>=b*, as a first approximation, the relative error is proportional to ð Þ *<sup>c</sup>=<sup>b</sup>* <sup>2</sup> ; we can therefore admit that the auxiliary sources are a good approximation for dimensions between one-tenth and one-hundredth of the dimensions of the case.

*Recent Microwave Technologies*
