Keywords

Transcendental equation of rectangular DRA provides complete solution of propagation constants , i.e., k x , k y , and k z . The propagation constant gives rise to resonant frequency with the help of characteristic equation . The wave numbers k x , k y , and k z are in x, y, and z direction, respectively. The free space wave number is k 0. The exact solution of RDRA resonant frequency can be determined from combined solution of transcendental equation and characteristic equation. These equations have unique solution if RDRA boundary conditions are fixed. For example, top and bottom walls are PMC and rest of the four walls is PEC and vice versa, only two different transcendental equations will be developed.

To get this solution, H z fields and derivative of H z fields need to be solved. They are solved for continuous propagating fields conditions. The fields are assumed continuous at interface of RDRA. The RDRA along with eigen vectors is shown in Fig. 4.1a, b.

Fig. 4.1
figure 1

a Rectangular DRA. b Eigen currents (current vectors) versus wavelength

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FormalPara CASE#1 RDRA solution:

See Fig. 4.2.

Fig. 4.2
figure 2

RDRA under defined boundaries

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To derive transcendental equation, the fields inside the resonator and outside the resonator are required.

$$\tan \left( {k_{z} d} \right) = \frac{{k_{z} }}{{\sqrt {\left( {\epsilon_{{r}} - 1} \right) {k_{0}}^{2} - {k_{z}}^{2} } }}\left( {{\text{transcendental}}\;{\text{equation}}} \right)$$
(4.1)
$$\epsilon_{{r}} {k_{0}}^{2} = {k_{x}}^{2} +{k_{y}}^{2} + {k_{z}}^{2} \left( {{\text{wave}}\;{\text{equation}}} \right)$$
(4.2)
$$k_{x} = m\uppi/a$$
(4.3a)
$$k_{y} = n\uppi/b$$
(4.3b)
$$k_{z} = p\uppi/d$$
(4.3c)

where a, b, and d are dimensions; m, n, and p are the indices.

TEδ11, TE1δ1, and TE11δ are dominant modes .

The solution of resonant frequency can be had if solution of k y propagation constant is obtained from characteristic equation , \(\epsilon_{{r}} {k_{0}}^{2} = {k_{x}}^{2} + {k_{y}}^{2} + {k_{z}}^{2}\), and then substituted in transcendental equation to compute resonant frequency f 0.

FormalPara Boundary condition

Propagation constant, \(\gamma_{mn}^{2} = {k_{0}}^{2} + {h_{mn}}^{2}\)

\({k} = 2 \pi /\lambda = \omega \sqrt[]{\mu \epsilon} = \omega /{\text{c}};\)

$$\int \nolimits E^{2} {\text{d}}V = \int \nolimits H^{2} {\text{d}}V$$

Time average electric energy = time average magnetic energy

$$\epsilon_{{r}} {k_{0}}^{2} = {k_{x}}^{2} + {k_{y}}^{2} + {k_{z}}^{2}$$
(4.4)
$$\epsilon_{0} {k_{0}}^{2} = {k_{x}}^{2} + {k_{y}}^{2} + {{k_{z}}^{\prime}}^{2}$$
(4.5)
$$k_{z} = p\uppi/d$$

Subtracting Eq. (4.1) from Eq. (4.2), we get

$${{k_{z}}^{\prime}}^{2} - \;{k_{z}}^{2} = \epsilon_{0}{{k_{0}}^{\prime}}^{2} - \epsilon_{{r}} {k_{0}}^{2}$$
$${k_{z}}^{\prime{2}} - {k_{z}}^{2} = \epsilon_{0} \mu_{0} \omega^{2} - \epsilon_{{r}} \mu_{0} \omega^{2}$$

Taking value of \(\epsilon_{0} = 1\) and \(\mu_{0} = {\mu }\), we get

$${ k_{z}}^{\prime{2}} - {k_{z}}^{2} = \omega^{2} {\mu }({1} - \epsilon_{{r}} )$$
(4.6)

4.1 Case-1: Top and Bottom Walls as PMC and Rest of the Four Walls are PEC

See Fig. 4.3.

Fig. 4.3
figure 3

RDRA with boundaries

Full size image

Assuming that the top and bottom surface plane be at \(z = 0,d\) to be PMC

$$\therefore \quad {n} \times H = 0$$

And

$$n \cdot E = 0$$

or,

$$\begin{aligned} H_{y} = H_{x} & = 0 \\ \qquad E_{z} & = 0\end{aligned}$$

Rest of the other walls is PEC

$$\therefore \quad n \times E = 0$$

And

$$n \cdot H = 0$$

At

$$\begin{aligned} x = 0, a \quad E_{y} = E_{z} & = 0 \\ H_{x} & = 0 \end{aligned}$$

At

$$\begin{aligned} y & = 0,b \quad E_{x} = E_{z} = 0 \\ H_{y} & = 0 \end{aligned}$$

We also know

$$E_{x} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial H_{z} }}{\partial y} - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial x}} \right]$$
(4.7a)
$$E_{y} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ { - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial y} - \frac{{\partial H_{z} }}{\partial x}} \right]$$
(4.7b)
$$H_{x} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial E_{z} }}{\partial y} - \frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial x}} \right]$$
(4.7c)
$$H_{y} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial y} - \frac{{\partial E_{z} }}{\partial x}} \right]$$
(4.7d)

Now, the solution of second-order differential equation is given as follows:

$$\psi_{z} = X\left( x \right)Y\left( y \right)Z(z)$$

where

$$X\left( x \right) = A_{1} \sin k_{x} x + A_{2} \cos k_{x} x$$
(4.8a)
$$Y\left( y \right) = A_{3} \sin k_{y} y + A_{4} \cos k_{y} y$$
(4.8b)
$$Z\left( z \right) = A_{5} \sin k_{z} z + A_{6} \cos k_{z} z$$
(4.8c)

For TE mode \(\left( {E_{z} = 0\;{\text{and}}\;H_{z} \ne 0} \right)\)

$$\psi_{{H_{z}}} = X\left( x \right)Y\left( y \right)Z\left( z \right)$$

After putting \(E_{z} = 0,\) we get

$$E_{y} = C^{\prime } \left[ { - \frac{{\partial H_{z} }}{\partial x}} \right]$$

or,

$$E_{y} = C^{\prime}X^{\prime}\left( x \right)Y\left( y \right)Z\left( z \right)$$

Now

$$X^{\prime}\left( x \right) = A_{1} \cos k_{x} x - A_{2} \sin k_{x} x$$

But at

$$\begin{aligned} & \quad x = 0,a\quad E_{y} = 0 \\ & \therefore \quad 0 = A_{1} \cos k_{x} 0 - A_{2} \sin k_{x} 0 \\ \end{aligned}$$

or,

$$A_{1} = 0\;\;{\text{and}}\;\;k_{x} = \frac{m\pi }{a}$$

Similarly,

$$E_{x} = C^{\prime}\left[ {\frac{{\partial H_{z} }}{\partial y}} \right]$$

or,

$$E_{x} = C^{\prime}X\left( x \right)Y\left( y \right)Z(z)$$

Now

$$Y^{\prime}\left( y \right) = A_{3} \cos k_{y} y - A_{4} \sin k_{y} y$$

But at

$$\begin{aligned} y & = 0,b\quad E_{x} = 0 \\ \quad \therefore \quad 0 & = A_{3} \cos k_{y} 0 - A_{4} \sin k_{y} 0 \end{aligned}$$

or,

$$A_{3} = 0\;\; {\text{and}}\;\;k_{y} = \frac{n\pi }{b}$$

from

$$H_{x} = C^{\prime}\left[ { - \frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial x}} \right]$$

or,

$$H_{x} = C^{\prime}X^{\prime}\left( x \right)Y\left( y \right)Z^{\prime}\left( z \right)$$

Now

$$Z^{\prime}\left( z \right) = A_{5} \cos k_{z} z - A_{6} \sin k_{z} z$$

At

$$z = 0,d\quad H_{x} = 0$$
$$\begin{aligned} \therefore \quad & A_{5} \cos k_{z} 0 - A_{6} \sin k_{z} 0 = 0 \\ & A_{5} = 0\;\;{\text{and}}\;\; k_{z} = \frac{p\pi }{d} \\ \end{aligned}$$

Hence,

$$H_{z} = A_{2} A_{4} A_{6} \cos \left( {\frac{m\pi }{a}x} \right) \cos \left( {\frac{n\pi }{b}y} \right) \cos \left( {\frac{p\pi }{d}z} \right)$$
(4.9)

Using Eqs. (4.1)–(4.4), and (4.8a)–(4.8c), we get

$$H_{x} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{m\pi }{a}} \right)\left( {\frac{p\pi }{d}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.10a)
$$H_{y} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{n\pi }{b}} \right)\left( {\frac{p\pi }{d}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.10b)
$$E_{y} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{m\pi }{a}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.10c)
$$E_{x} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{n\pi }{b}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.10d)

Now, evaluate \(E_{x}\) and \(H_{z}\) at the boundary walls of the dielectric waveguide.

As we know that at the PMC wall, the tangential component of magnetic field and normal component of electric field are equal to “zero” at the interface z = 0, d.

Hence,

$$H_{x} ,H_{y} = 0$$

and

$$E_{z} = 0$$

Also, for propagation to be possible, we need two normal components of E and H.

Thus, we take \(E_{x}\) and \(H_{y} .\)

Now, the propagating wave is continuous at the interface, i.e., \(E_{x} = E^{\prime}_{x}.\)

Therefore,

$${A}\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\left( {C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} } \right) = {A}\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.11)

or,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} = C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.12)

But at z = 0, only the inside waveform exists.

Therefore,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} = 0$$

Now substituting the value of z = 0, we get

$$\begin{aligned} & C_{1} + C_{2} = 0 \\ & {\text{or}}, \quad C_{1} = - C_{2} \end{aligned}$$
(4.13)

As \(H_{z}\) is continuous at the interface z = d,

$$H_{z} = H^{\prime}_{z} \quad {\text{and}}\quad \frac{{\partial H_{z} }}{\partial x} = \frac{{\partial H^{\prime}_{z} }}{\partial x}$$

From Eq. (4.9),

$$H_{z} = B\cos \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\cos \left( {k_{z} z} \right)$$
(4.14a)

and

$$H^{\prime}_{z} = B\cos \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\cos \left( {k^{\prime}_{z} z} \right)$$
(4.14b)

Equating Eqs. (4.14a) and (4.14b), we get

$$B\cos \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\left( {C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} } \right) = B\cos \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\left( {C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}} } \right)$$

or,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} = C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.15)

From Eq. (4.15), i.e., \(C_{1} = - C_{2}\), we get, at \(z = d,\)

$$2jC_{1} \sin (k_{z} {d}) = {C}^{\prime}_{2} {{e}}^{{ - {{j}}k_{z} d}}$$
(4.16)

Now, equating the derivative of \(H_{z} ,\) we get

$$jk_{z} \left( {C_{1} {{e}}^{{{{j}}k_{z} z}} - C_{2} {{e}}^{{ - {{j}}k_{z} z}} } \right) = - jk^{\prime}_{z} C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.17)

or,

$$2k_{z} {C}_{1} {\cos}({k_z}{d}) = - k^{\prime}_{z} {C}^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}} .$$

Dividing equation (4.16) by (4.17)

$$\frac{{j\tan k_{z} d}}{{k_{z} }} = \frac{ - 1}{{k^{\prime}_{z} }}$$
(4.18a)

Squaring both sides and substituting the value of \(k_{z}^{2}\) from Eq. (4.3c),

$${k_{z}}^{{\prime }{2}} = k_{z}^{2} - \omega^{2} {\mu }(\epsilon_{{r}} - 1)$$

and substituting \(\mu = 1.\) We get,

$$\tan (k_{z} d) = \frac{{k_{z} }}{{\sqrt {k_{0}^{2} (\epsilon_{{r}} - 1) - k_{z}^{2} } }}$$
(4.18b)

The above equation is the required transcendental equation .

4.2 Case-2

For transcendental equation, we need to compute the fields inside the resonator and outside it.

$$\tan \left( {k_{z} d} \right) = \frac{{k_{z} }}{{\sqrt {\left( {\epsilon_{{{r}}} - 1} \right)k_{0}^{2} - k_{z}^{2} } }}$$
(4.19)

where \(\epsilon_{{r}} k_{0}^{2} = k_{x}^{2} + k_{y}^{2} + k_{z}^{2}\) (characteristic wave equation)

$$k_{x} = \frac{m\pi }{a}$$
(4.20a)
$$k_{y} = \frac{n\pi }{b}$$
(4.20b)
$$k_{z} = \frac{p\pi }{d}$$
(4.20c)

where a, b, and d are dimensions; m, n, and p are modes.

TEδ11, TE1δ1, and TE11δ are dominant modes.

Boundary condition

Propagation constant, \(\gamma_{mn}^{2} = k_{0}^{2} + {hmn}^{2}\) where \(k = \frac{2\pi }{\lambda } = \omega \sqrt {\mu \epsilon } = \frac{\omega }{c}.\)

From the energy conservation principle,

$$\int {E^{2} {\text{d}}V} = \int {H^{2} {\text{d}}V}.$$

i.e., time average electric energy = time average magnetic energy.

When top and bottom walls are PMC, rest of the other walls is PEC

Assuming that the top and bottom surface plane be at \(z = 0,d\)

$$\therefore \quad {n} \times H = 0$$

And

$$n \cdot E = 0$$

or,

$$\begin{aligned} H_{y} & = H_{x} = 0 \\ E_{z} & = 0 \\ \end{aligned}$$

Rest of the other walls is PEC

$$\therefore \quad {n} \times E = 0$$

And

$$n \cdot H = 0$$

At

$$\begin{aligned} x = 0,{a}\quad E_{y} & = E_{z} = 0 \\ H_{x} & = 0 \end{aligned}$$

At

$$\begin{aligned} y = 0,{b}\quad E_{x} & = E_{z} = 0 \\ H_{y} & = 0 \\ \end{aligned}$$

We also know

$$E_{x} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial H_{z} }}{\partial y} - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial x}} \right]$$
(4.21a)
$$E_{y} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ { - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial y} - \frac{{\partial H_{z} }}{\partial x}} \right]$$
(4.21b)
$$H_{x} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial E_{z} }}{\partial y} - \frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial x}} \right]$$
(4.21c)
$$H_{y} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial y} - \frac{{\partial E_{z} }}{\partial x}} \right]$$
(4.21d)

Now, the solution of second-order differential equation is given as follows:

$$\psi_{z} = X\left( x \right)Y\left( y \right)Z(z)$$
(4.22)

where

$$\begin{aligned} X\left( x \right) & = A_{1} \sin k_{x} x + A_{2} \cos k_{x} x \\ Y\left( y \right) & = A_{3} \sin k_{y} y + A_{4} \cos k_{y} y \\ Z\left( z \right) & = A_{5} \sin k_{z} z + A_{6} \cos k_{z} z \\ \end{aligned}$$

For TE mode \((E_{z} = 0\;\;{\text{and}}\;\;H_{z} \ne 0)\)

$$\begin{aligned} \psi_{H_{z}} & = X\left( x \right)Y\left( y \right)Z(z) \\ E_{z} & = 0 \end{aligned}$$

we get

$$E_{y} = C^{\prime}\left[ { - \frac{{\partial H_{z} }}{\partial x}} \right]$$

or,

$$E_{y} = C^{\prime}X^{\prime}\left( x \right)Y\left( y \right)Z\left( z \right)$$

Now

$$X^{\prime}\left( x \right) = A_{1} \cos k_{x} x - A_{2} \sin k_{x} x$$

But at

$$\begin{aligned} x & = 0,a\;\; E_{y} = 0 \\ \therefore \quad 0 & = A_{1} \cos k_{x} 0 - A_{2} \sin k_{x} 0 \end{aligned}$$

or,

$$A_{1} = 0\;\;{\text{and}}\;\;k_{x} = \frac{m\pi }{a}$$

Similarly,

$$E_{x} = C^{\prime}\left[ {\frac{{\partial H_{z} }}{\partial y}} \right]$$

or,

$$E_{x} = C^{\prime}X\left( x \right)Y^{\prime}\left( y \right)Z\left( z \right)$$

Now

$$Y^{\prime}\left( y \right) = A_{3} \cos k_{y} y - A_{4} \sin k_{y} y$$

But at

$$\begin{aligned} y & = 0,b\;\;E_{x} = 0 \\ \therefore \quad 0 & = A_{3} \cos k_{y} 0 - A_{4} \sin k_{y} 0 \end{aligned}$$

or,

$$A_{3} = 0\;\;{\text{and}}\;\;k_{y} = \frac{n\pi }{b}$$
$$H_{x} = C^{\prime}\left[ { - \frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial x}} \right]$$

or,

$$H_{x} = C^{\prime}X^{\prime}\left( x \right)Y\left( y \right)Z^{\prime}\left( z \right)$$

Now

$$Z^{\prime}\left( z \right) = A_{5} \cos k_{z} z - A_{6} \sin k_{z} z$$

At

$$\begin{aligned} z & = 0,d\;\;H_{x} = 0 \\ \therefore \quad & A_{5} \cos k_{z} 0 - A_{6} \sin k_{z} 0 = 0 \\ & A_{5} = 0\;\;{\text{and}}\;\;k_{z} = \frac{p\pi }{d} \end{aligned}$$

Hence,

$$H_{z} = A_{2} A_{4} A_{6} \cos \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.23)

Using Eqs. (4.1)–(4.4), and (4.8a)–(4.8c), we get

$$H_{x} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{m\pi }{a}} \right)\left( {\frac{p\pi }{d}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.24a)
$$H_{y} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{n\pi }{b}} \right)\left( {\frac{p\pi }{d}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.24b)
$$E_{y} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{m\pi }{a}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right) \cos \left( {\frac{p\pi }{d}z} \right)$$
(4.24c)
$$E_{x} = C^{\prime\prime}A_{2} A_{4} A_{6} \left( {\frac{n\pi }{b}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.24d)

Above equations can also be written as follows:

$$\begin{aligned} H_{x} & = \frac{{k_{x} k_{z} }}{{j\omega \mu_{0} }}\sin (k_{x} x) \cos (k_{y} y) \sin (k_{z} z) \\ H_{y} & = \frac{{k_{y} k_{z} }}{{j\omega \mu_{0} }}\cos (k_{x} x) \sin (k_{y} y) \sin (k_{z} z) \\ E_{y} & = - k_{x} \sin (k_{x} x)\cos (k_{y} y)\cos (k_{z} z) \\ E_{x} & = k_{y} \cos (k_{x} x) \sin (k_{y} y) \cos (k_{z} z) \\ H_{z} & = \frac{{k_{x}^{2} + k_{y}^{2} }}{{j\omega \mu_{0} }}\cos (k_{x} x) \cos (k_{y} y) \cos (k_{z} z) \\ \end{aligned}$$

Since \(H_{z}\) is continuous, i.e., \(\frac{{{\text{d}}H_{z} }}{{{\text{d}}z}} \ne 0 ,\)

$$H^{\prime}_{z} = \frac{{k_{x}^{2} + k_{y}^{2} }}{{j\omega \mu_{0} }}\cos (k_{x} x) \cos (k_{y} y) \cos ({k}^{\prime}_{z} z)$$

Now, \(H_{y}\) can be written as follows:

$$H_{y} = \frac{{k_{y} k_{z} }}{{j\omega \mu_{0} }}\cos (k_{x} x) \sin (k_{y} y)(C_{1} {{e}}^{{{{j}}k_{z} d}} - C_{2} {{e}}^{{ - {{j}}k_{z} d}} )$$

But

$$H_{y} = 0\;{\text{at}}\;d = 0$$
$$C_{1} - C_{2} = 0$$

or,

$$C_{1} = C_{2}$$
$$\frac{{{\text{d}}H_{y} }}{{{\text{d}}z}} =A^{\prime}jk_{z} \cos (k_{x} x) \sin (k_{y} y)(C_{1}{{e}}^{{{{j}}k_{z} d}} + C_{2} {{e}}^{{ - {{j}}k_{z} d}} )$$

or,

$$\frac{{{\text{d}}H_{y} }}{{\text{d}}z} = C_{1} jk_{z} \cos(k_{x} x) \sin (k_{y} y)({{e}}^{{{{j}}k_{z} d}} +{{e}}^{{ - {{j}}k_{z} d}} )$$

or,

$$\begin{aligned}\frac{{{\text{d}}H_{y} }}{{{\text{d}}z}} & = C_{1} 2jk_{z} \cos(k_{x} x) \sin (k_{y} y) \cos (k_{z} d) \\ H^{\prime}_{y} & =C^{\prime}_{1} \cos (k_{x} x) \sin (k_{y} y) {{e}}^{{ -{{j}}k^{\prime}_{z} d}} {\text{ outside the cavity }} \end{aligned}$$

For \(H_{z}\) to be continuous,

$$\frac{{{\text{d}}H_{y} }}{{{\text{d}}z}} = \frac{{{\text{d}}H^{\prime}_{y} }}{{{\text{d}}z}}$$

or,

$$C_{1} 2jk_{z} \cos (k_{x} x) \sin (k_{y} y) \cos (k_{z} d) = - jk^{\prime}_{z} C^{\prime}_{1} \cos (k_{x} x) \sin (k_{y} y){{e}}^{{ - {{j}}k^{\prime}_{z} d}}$$

or,

$$2C_{1} k_{z} \cos (k_{z} d) = - k^{\prime}_{z} C^{\prime}_{1} {{e}}^{{ - {{j}}k^{\prime}_{z} d}}$$
(4.25)

From above equations, we have

$$E_{x} = k_{y} \cos \left( {k_{x} x} \right)\sin \left( {k_{y} y} \right)\left( {C_{1} {{e}}^{{{{j}}k_{z} d}} + C_{2} {{e}}^{{ - {{j}}k_{z} d}} } \right)$$

At

$$d = 0,\;E_{x} = 0,$$

so,

$$C_{1} + C_{2} = 0$$

or,

$$\begin{aligned} C_{1} & = - C_{2} \\ \therefore \quad E_{x} & = k_{y} \cos \left( {k_{x} x} \right)\sin \left( {k_{y} y} \right)C_{1} \left( {{{e}}^{{{{j}}k_{z} d}} - {{e}}^{{ - {{j}}k_{z} d}} } \right) \end{aligned}$$

or,

$$E_{x} = 2jC_{1} k_{y} \cos \left( {k_{x} x} \right)\sin \left( {k_{y} y} \right)\sin \left( {k_{z} d} \right)$$

Also

$$E_{x}^{\prime } = k_{y} \cos \left( {k_{x} x} \right)\sin \left( {k_{y} y} \right)\cos \left( {k_{z}^{\prime } z} \right)$$

or,

$$E_{x}^{\prime } = C_{1}^{\prime } k_{y} \cos \left( {k_{x} x} \right) \sin \left( {k_{y} y} \right){{e}}^{{ - {{j}}k_{z} d}}$$

For \(H_{z}\) to be continuous,

$$E_{x} = E^{\prime}_{x}$$

or,

$$jC_{1} k_{y} \cos (k_{x} x) \sin (k_{y} y) \sin (k_{z} d) = C_{1}^{\prime} k_{y} \cos (k_{x} x) \sin (k_{y} y) {{e}}^{{ - {{j}}k_{z} d}}$$

or,

$$2jC_{1} \sin (k_{z} d) = C^{\prime}_{1} {{e}}^{{ - {{j}}k_{z} d}}$$
(4.26)

Dividing Eq. (4.16) by Eq. (4.25), we get

$$\frac{{2C_{1} j\sin (k_{z} d)}}{{2C_{1} k_{z} \cos (k_{z} d)}} = \frac{{ - C^{\prime}_{1} {{e}}^{{ - {{j}}k_{z} d}} }}{{k^{\prime}_{z} C^{\prime}_{1} {{e}}^{{ - {{j}}k_{z}^{\prime} d}} }}$$

or,

$$\frac{{j\tan k_{z} d}}{{k_{z} }} = \frac{ - 1}{{k^{\prime}_{z} }}$$

or,

$${j} \tan k_{z} d = - \frac{{k_{z} }}{{k^{\prime}_{z} }}$$

On squaring and putting \({k_{z}^{\prime}}^{2} = k_{z}^{2} + \omega^{2} \mu_{0} (1 - \epsilon_{{r}} )\)

$$\tan^{2} k_{z} d = - \frac{{k_{z}^{2} }}{{k_{z}^{2} + \omega^{2} \mu_{0} (1 - \epsilon_{{r}} )}}$$

or,

$$\tan^{2} k_{z} d = \frac{{k_{z}^{2} }}{{\omega^{2} \mu_{0} \left( {\epsilon_{{r}} - 1} \right) - k_{z}^{2} }}$$
$$\tan k_{z} d = \frac{k_{z}}{ \sqrt {\left( {\epsilon_{{r}} - 1} \right)k_{0}^{2} - k_{z}^{2}}}$$
(4.27)

With the help of transcendental equation , we can find the propagation factor. Also with the help of this equation, we can obtain resonant frequency .

CASE#3

For transcendental equation, we need to compute the fields inside the resonator and outside it.

$$\begin{aligned} k_{z} \tan (k_{z} {d}) &= \sqrt {\left( { {\epsilon_r} - 1} \right)k_{0}^{2} - k_{z}^{2} } ; \\ \epsilon_{{r}} k_{0}^{2} & = k_{x}^{2} + k_{y}^{2} + k_{z}^{2}; \end{aligned}$$

and

$$k_{x} = {m}\uppi/{a}$$
(4.28a)
$$k_{y} = {n}\uppi/{b}$$
(4.28b)
$$k_{z} = {p}\uppi/{d}$$
(4.28c)

where a, b, and d are dimensions; m, n, and p are the indices.

TEδ11, TE1δ1, TE11δ are dominant modes.

Boundary conditions

Propagation constant, \(\gamma_{mn}^{2} = k_{0}^{2} + h_{mn}^{2}\)

\({k } = 2\uppi{\kern 1pt} /{\kern 1pt} \lambda = \omega \sqrt[2]{\mu \epsilon } = \omega /{c;}\)

$$\int E^{2} {\text{d}}V = \int H^{2} {\text{d}}V$$

Time average electric energy = time average magnetic energy

$$\epsilon_{0} \epsilon_{{r}} k_{0}^{2} = k_{x}^{2} + k_{y}^{2} + k_{z}^{2}$$
(4.29a)
$$\epsilon_{0} k_{0}^{2} = k_{x}^{2} + k_{y}^{2} + k_{z}^{{\prime}{2}}$$
(4.29b)
$$k^{\prime}_{z} \ne {p}\uppi/{d}$$

Subtracting Eq. (4.1) from Eq. (4.2), we get

$$\begin{aligned} k_{z}^{\prime{2}} - k_{z}^{2} &= \epsilon_{0} k_{0}^{\prime{2}} - \epsilon_{0} \epsilon_{{r}} k_{0}^{2} \\ k_{z}^{\prime{2}} - k_{z}^{2} &= \epsilon_{0} \mu_{0} \omega^{2} - \epsilon_{0} \epsilon_{{r}} \mu_{0} \omega^{2} \end{aligned}$$

Taking the value of \(\epsilon_{0}=1\) and \(\mu_{0} =\upmu , {\text{we get}}\)

$$k_{z}^{\prime{2}} - k_{z}^{2} = \omega^{ 2}\upmu\epsilon_{0} \;({1} - \epsilon_{{r}} )$$
(4.30)

When top and bottom walls are PEC, rest of the other walls is PMC.

Now,

Assuming that the top and bottom surface plane be at \(z = 0,d\)

$$\therefore \quad {{n}} \times E = 0$$

and

$$n \cdot H = 0$$

or,

$$E_{y} = E_{x} = 0$$
$$H_{z} = 0$$

Rest of the other walls is PMC

$$\therefore \quad {{n}} \times H = 0$$

And

$$n \cdot E = 0$$

At

$$\begin{aligned} x & = 0,{a}\quad H_{y} = H_{z} = 0 \\ E_{x} & = 0 \\ \end{aligned}$$

At

$$\begin{aligned} y &= 0,{b}\quad H_{x} = H_{z} = 0 \\ E_{y} & = 0 \end{aligned}$$

We also know

$$E_{x} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial H_{z} }}{\partial y} - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial x}} \right]$$
(4.31a)
$$E_{y} = \frac{1}{{j\omega \epsilon \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ { - \frac{1}{j\omega \mu }\frac{{\partial^{2} E_{z} }}{\partial z\partial y} - \frac{{\partial H_{z} }}{\partial x}} \right]$$
(4.31b)
$$H_{x} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{{\partial E_{z} }}{\partial y} - \frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial x}} \right]$$
(4.31c)
$$H_{y} = \frac{ - 1}{{j\omega \mu \left( {1 + \frac{{\gamma^{2} }}{{k^{2} }}} \right)}}\left[ {\frac{1}{j\omega \epsilon }\frac{{\partial^{2} H_{z} }}{\partial z\partial y} - \frac{{\partial E_{z} }}{\partial x}} \right]$$
(4.31d)

Now, the solution of second-order differential equation is given as follows:

$$\psi_{z} = X\left( x \right)Y\left( y \right)Z\left( z \right)$$

where

$$\begin{aligned} X\left( x \right) & = A_{1} \sin k_{x} x + A_{2} \cos k_{x} x \\ Y\left( y \right) & = A_{3} \sin k_{y} y + A_{4} \cos k_{y} y \\ Z\left( z \right) & = A_{5} \sin k_{z} z + A_{6} \cos k_{z} z \end{aligned}$$
  1. (i)

    For TE mode \(\left( E_{z} = 0 \;\; {\text{and}}\;\; H_{z} \ne 0 \right)\)

    $$\psi_{H_{z}} = X\left( x \right)Y\left( y \right)Z\left( z \right)$$

At

$$x = 0, a \quad H_{z} = 0,$$

or,

$$\begin{aligned} & A_{1} \sin k_{x} 0 + A_{2} \cos k_{x} 0 = 0 \\ & \therefore \quad A_{2} = 0\,\;{\text{and}}\,\;k_{x} = \frac{m\pi }{a} \end{aligned}$$

Also at

$$y = 0, b \quad H_{z} = 0$$

or,

$$\begin{aligned} & A_{3} \sin k_{y} 0 + A_{4} \cos k_{y} \,0 = 0 \\ & \therefore \quad A_{4} = 0\;\; {\text{and}}\;\; k_{y} = \frac{n\pi }{b} \end{aligned}$$

At

$$z = 0,d \quad H_{z} = 0$$
$$\begin{aligned} & \therefore \quad A_{5} \sin k_{z} 0 + A_{6} \cos k_{z} 0 = 0 \\ & A_{6} = 0\quad {\text{and}}\;\;k_{z} = \frac{p\pi }{d} \end{aligned}$$

Hence,

$${H}_{z} = {A}_{1} {A}_{3} {A}_{5} \;{\sin}\left( {\frac{m\pi }{a}{x}} \right)\;{\sin}\left( {\frac{n\pi }{b}{y}} \right)\;{\sin}\left( {\frac{p\pi }{d}{z}} \right)$$
(4.32)

Using Eqs. (4.1)–(4.4), and (4.8a)–(4.8c), we get

$$H_{x} = C^{\prime\prime}A_{1} A_{3} A_{5} \left( {\frac{m\pi }{a}} \right)\left( {\frac{p\pi }{d}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.33a)
$$H_{y} = C^{\prime\prime}A_{1} A_{3} A_{5} \left( {\frac{n\pi }{b}} \right)\left( {\frac{p\pi }{d}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\cos \left( {\frac{p\pi }{d}z} \right)$$
(4.33b)
$$E_{y} = C^{\prime\prime}A_{1} A_{3} A_{5} \left( {\frac{m\pi }{a}} \right)\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.33c)
$$E_{x} = C^{\prime\prime}A_{1} A_{3} A_{5} \left( {\frac{n\pi }{b}} \right)\sin \left( {\frac{m\pi }{a}x} \right)\cos \left( {\frac{n\pi }{b}y} \right)\sin \left( {\frac{p\pi }{d}z} \right)$$
(4.33d)

Now, evaluate \(H_{x}\) and \(H_{z}\) at the boundary walls of the dielectric waveguide.

As we know that at the PEC wall, the tangential component of electric field and normal component of magnetic field is equal to “zero” at the interface \(z = 0,{d}.\)

Hence,

$$E_{x} ,E_{y} = 0$$

and

$$H_{z} = 0$$

Also, for propagation to be possible, we need two normal components of E and H.

Thus we take \(E_{y}\) and \(H_{x} .\)

Now, the propagating wave is continuous at the interface, i.e., \(H_{x} = H^{\prime}_{x}.\)

Therefore,

$${A}\cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)(C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} ) = {A} \cos \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$

or,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} = C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.34)

But at \(z = 0\), only the inside waveform exists.

Therefore,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} = 0$$

Now, substituting the value of \(z = 0,\) we get

$$C_{1} + C_{2} = 0$$

or,

$$C_{1} = - C_{2}$$
(4.35)

As \(H_{z}\) is continuous at the interface \(z = {d} .\)

Therefore,

$$H_{z} = H^{\prime}_{z} \;\; {\text{and}}\;\;\frac{{\partial H_{z} }}{\partial x} = \frac{{\partial H^{\prime}_{z} }}{\partial x}$$
$${H}_{z} = {B}\;{\sin}\left( {\frac{m\pi }{a}{x}} \right){\sin}\left( {\frac{n\pi }{b}{y}} \right){\sin}\left( {{k}_{z} {z}} \right)$$
(4.36a)

and

$${H^{\prime}}_{z} = {B}\;{\sin}\left( {\frac{m\pi }{a}{x}} \right)\;{\sin}\left( {\frac{n\pi }{b}{y}} \right)\;{\sin}\left( {{k^{\prime}}_{z} {z}} \right)$$
(4.36b)

Equating Eqs. (i) and (ii), we get

$$B\sin \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)(C_{1} {{e}}^{{{{j}}k_{z} z}} - C_{2} {{e}}^{{ - {{j}}k_{z} z}} ) = B\sin \left( {\frac{m\pi }{a}x} \right)\sin \left( {\frac{n\pi }{b}y} \right)C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$

or,

$$C_{1} {{e}}^{{{{j}}k_{z} z}} - C_{2} {{e}}^{{ - {{j}}k_{z} z}} = C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$

From Eq. (1b), i.e., \(C_{1} = - C_{2} ,\) we get, at \({z} = {d} ,\)

$${2C}_{1} \cos (k_{z} {d}) = { C}^{\prime}_{2} {{e}}^{{ - {{j}}k_{z} d}}$$
(4.37)

Now, equating the derivative of \(H_{z} ,\) we get

$${j}k_{z} (C_{1} {{e}}^{{{{j}}k_{z} z}} + C_{2} {{e}}^{{ - {{j}}k_{z} z}} ) = - {j}k^{\prime}_{z} C^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$

or,

$$2jk_{z} {C}_{1} \sin (k_{z} {d}) = - k^{\prime}_{z} {C}^{\prime}_{2} {{e}}^{{ - {{j}}k^{\prime}_{z} z}}$$
(4.38)

Dividing equation (iv) by (iii), we get

$$jk_{z} \tan k_{z} d = - k^{\prime}_{z}$$

Squaring both sides and substituting the value of \(k_{z}^{{\prime}{2}}\) from Eq. (4.3c), we get

$$k^{\prime}_{z} = k_{z}^{2}- \omega^{2} {\mu }(\epsilon_{{r}} - 1)$$

and substituting \(\mu = 1\), we get isolated DRA case as:

$$k_{z} \tan (k_{z} {d}) = \sqrt {\left( { \epsilon_{{r}} - 1} \right)k_{0}^{2} - k_{z}^{2} }$$
(4.39a)

DRA with ground plane case as:

$$k_{z} \tan (k_{z} {d/2}) = \sqrt {\left( { \epsilon_{{r}} - 1} \right)k_{0}^{2} - k_{z}^{2} }$$
(4.39b)

Hence, the solution of transcendental equation is completely obtained.

4.3 MATLAB Simulation Results

The same can be seen if MATLAB simulation is obtained as given below:

Relationship between delta distance and its impact on resonant frequency is shown in Fig.  4.4.

Fig. 4.4
figure 4

Frequency versus delta distance

Full size image

The resonant frequency is increasing as the delta length is increasing as shown in Fig. 4.4. Also, radiation lobe is increasing as the number of resonant mode is increasing as shown in Fig. 4.5.

Fig. 4.5
figure 5

Radiation lobes of radiation pattern in RDRA

Full size image

Radiation Lobes:

RDRA dimensions are given to compute resonant modes using MATLAB.

MATLAB Program for Ez field

MATLAB program for transcendental equation and resonant frequency of RDRA:

The MATLAB-simulated resonant modes in Figs. 4.6, 4.7, 4.8, 4.9, 4.10, 4.11 and 4.12 have been drawn, and resonant frequency using transcendental equation is placed in table form.

Fig. 4.6
figure 6

Resonant modes in xy plane

Full size image
Fig. 4.7
figure 7

Resonant modes in xy plane

Full size image
Fig. 4.8
figure 8

Resonant modes in RDRA in xy plane

Full size image
Fig. 4.9
figure 9

Resonant modes in RDRA in xy plane

Full size image
Fig. 4.10
figure 10

Resonant modes 3D in RDRA in x y z plane

Full size image
Fig. 4.11
figure 11

TE 341 resonant modes

Full size image
Fig. 4.12
figure 12

TE 323 resonant modes

Full size image
Table 4.1 RDRA materials, permittivity, dimensions, and computed resonant frequency
Full size table
Table 4.2 Fringing effect along “b” dimensions increased effective along y-direction of RDRA
Full size table

Solved examples of RDRA resonant frequency:

Example 1

Calculate the dimension of “d” in RDRA:

For \({\text{TE}}_{111}\) mode when

$$\begin{aligned} & \epsilon_{{r}} = 100 \\ & {a } = 10\,{\text{mm}} \\ & {b } = 10\, {\text{mm}} \\ & f_{{r}} = 7.97\,{\text{GHz}} \\ \end{aligned}$$

Solution

Resonant frequency:

$$\begin{aligned} f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{p}{d}^{2} } \\ 7.97 \times 10^{9} & = \frac{{3 \times 10^{8} }}{{2\sqrt {100} }}\sqrt {100^{2} + 100^{2} + \frac{1}{d}^{2} } \\ 531.33 & = \sqrt {20000 + \frac{1}{d}^{2} } \\ \frac{1}{d} & = 512.167 \\ {d} & = 1.95\,{\text{mm}} \end{aligned}$$

Example 2

RDRA with following data:

$$\begin{aligned} \epsilon_{{r}} & = 35 \\ {a} & = 18\,{\text{mm}} \\ {b} & = 18\,{\text{mm}} \\ f_{{r}} & = 2.45\,{\text{GHz}} \end{aligned}$$

Solution

$$\begin{aligned} f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{p}{d}^{2} } \\ 2.45 \times 10^{9} & = \frac{{3 \times 10^{8} }}{{2\sqrt {35} }}\sqrt {\frac{1000}{18}^{2} + \frac{1000}{18}^{2} + \frac{1}{d}^{2} } \\ 9337.222 & = \sqrt {2\left(\frac{1000}{18}^{2} \right) + \frac{1}{d}^{2} }\\ \frac{1}{d} & = 56.252 \\ d & = 17.77\,{\text{mm}} \\ \end{aligned}$$

Example 3

Calculate the resonant frequency for

\({\text{TE}}_{111}\) mode using the given data of RDRA:

$$\begin{aligned} \epsilon_{{r}} & = 10 \\ {a} & = 14\,{\text{mm}} \\ {b} & = 8\,{\text{mm}} \\ {d} & = 8\,{\text{mm}} \end{aligned}$$

Solution

$$\begin{aligned}f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{p}{d}^{2} } \\ f_{{r}} & = \frac{{3 \times 10^{8} }}{{2\sqrt {10} }}\sqrt {\frac{1000}{14}^{2} + \frac{1000}{8}^{2} + \frac{1000}{8}^{2} } \\ f_{{r}} & = 9.04 \,{\text{GHz}} \end{aligned}$$

Example 4

$$\begin{aligned} \epsilon_{{r}} & = 10 \\ {a} & = 14\,{\text{mm}} \\ {b} & = 8\,{\text{mm}} \\ {d} & = 16\,{\text{mm}} \end{aligned}$$

Solution

$$\begin{aligned} f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{p}{d}^{2} } \\ f_{{r}} & = \frac{{3 \times 10^{8} }}{{2\sqrt {10} }}\sqrt {\frac{1000}{14}^{2} + \frac{1000}{8}^{2} + \frac{1000}{16}^{2} } \\ f_{{r}} & = 7.44\, {\text{GHz}} \end{aligned}$$

Example 5

Calculate the resonant frequency for the \({\text{TE}}_{11\delta }\) mode using the given data:

$$\begin{aligned} \epsilon_{{r}} & = 10 \\ {a} & = 14\,{\text{mm}} \\ {b} & = 8\,{\text{mm}} \\ {d} & = 8\,{\text{mm}} \end{aligned}$$

Solution

$$\begin{aligned} f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{\delta }{d}^{2} } \\ f_{{r}} & = \frac{{3 \times 10^{8} }}{{2\sqrt {10} }}\sqrt {\frac{1000}{14}^{2} + \frac{1000}{8}^{2} + 0} \\ f_{{r}} & = 6.82\,{\text{GHz}} \end{aligned}$$

Example 6

$$\begin{aligned}\epsilon_{{r}} & = 10 \\ {a} & = 14\, {\text{mm}} \\ {b} & = 8\,{\text{mm}} \\ {d} & = 16\,{\text{mm}} \end{aligned}$$

Solution

$$\begin{aligned} f_{{r}} & = \frac{c}{{2\sqrt {\epsilon_{{r}} } }}\sqrt {\frac{m}{a}^{2} + \frac{n}{b}^{2} + \frac{p}{d}^{2} } \\ f_{{r}} & = \frac{{3 \times 10^{8} }}{{2\sqrt {10} }}\sqrt {\frac{1000}{14}^{2} + \frac{1000}{8}^{2} + 0} \\ f_{{r}} & = 6.82\, {\text{GHz}} \end{aligned}$$

4.4 Resonant Frequency of RDRA for Experimentations

The RDRAs can be prototyped with various materials and sizes as per the requirements.

Table 4.1 consists of list of RDRA materials, permittivity , dimensions, and computed resonant frequency.

Example 7

Compute resonant frequency when RDRA dimensions are 10 × 10 × 10 mm3 and dielectric constant of material used is 10.

$$\left( {f_{{r}}} \right)m,n,p = \frac{c}{{2{\uppi} \sqrt[]{\epsilon \mu }}}\sqrt {\left( {\frac{m\pi }{a}} \right)^{2} + \left( {\frac{n\pi }{b}} \right)^{2} + \left( {\frac{p\pi }{d}} \right)^{2} }$$

Resonant frequencies in isolated case are 49.7 and 25.8 GHz with ground plane (Table 4.2).

MATLAB program and simulation effective length due to fringing effect:

Effective increased length computations due to fringing effect:

Program 1

Results:

Program 2

Results:

Program 3

MATLAB programs taking parameters a, b, d same and comparing frequency using:

Program 1: Characteristic equation

Output:

Program 4

Transcendental equation for same dimensions:

Output:

Program 5

MATLAB programs taking parameters a, b, d same and comparing frequency using:

Characteristic equation

Output:

Program 6

Transcendental equation

Program 7

MATLAB programs taking parameters a,b,d same and comparing frequency using:

Characteristic equation

Output:

Program 8

Transcendental equation

Output:

Program 9

Program 10

  1. Q.No. 1

    Develop transcendental equation for moat-shaped RDRA.

  2. Q.No. 2

    Compute propagation constants in x-, y-, and z-directed propagated RDRAs, when feed probe is given. Compute its resonant frequency when RDRA dimensions are 5 × 5× 3 mm3 and dielectric constant used is 20.