Question

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C=capacitance of equivalent ckt.[7]

\begin{equation} 

C=\dfrac{\epsilon_{ef}\epsilon_{o}L_{e}W}{2     h} F

\end{equation}

where\

\begin{center}

$F=\cos ^{ - 2} ({\pi}X_{f}/L)$

\end{center}

L=inductance of equivalent ckt.[7]

 \begin{equation} 

L=\frac{1}{({2\pi}f_{r})^{2}C}

\end{equation}\

$\Delta L$=additional series inductance

\begin{equation} 

\Delta L=\frac{Z_{01}+Z_{02}}{16\pi{f_{r}}F} tan(\pi{f_{r}{L_{n}}}/C)

\end{equation}\

$Z_{01} and Z_{02}$ are the characteristics impedances of microstrip lines with width of 

$w_{1} and w_{2}$ respectively.The values

\begin{equation}

Z_{01}=120\pi/(\frac{w_{1}}{h}+1.393+0.667\ln(\frac{w_{1}}{h}+1.444))

\end{equation}\

$$ 

Z_{02}=120\pi/(\frac{w_{2}}{h}+1.393+0.667\ln(\frac{w_{2}}{h}+1.444))

$$

\

where\

$$w_{1}=w-2{P_{s}}-W_{s}$$

\

and\

$$w_{2}=2{P_{s}}-W_{s}$$

The capacitance$\Delta C$ between center wing and side wing is calculated as gap capacitance.

\begin{equation} 

\Delta C=2L_{n}\dfrac{\epsilon_{0}}{\pi}

\left[\ln\left(2\frac{1+\sqrt{k^{'}}}{1-\sqrt{k^{'}}}\right)+\ln\coth\left(\frac{{\pi}S}{4h}\right)+0.013C_{f}\dfrac{h}{s}\right]F

\end{equation}\

where\

$$K^{'}=\sqrt{1-K^{2}}$$

\

$$K^{2}=\frac{1+\frac{W_{1}}{S}+\frac{W_{2}}{S}}{\left(1+\frac{W_{1}}{S}\right)\left(1+\frac{W_{2}}{S}\right)}$$

The resonance resistance of the two resonators are given by [11]

\begin{equation}

R_{1,2}=\frac{1}{2(G_{1}+G_{12})}\cos^{2}(\frac{\pi{X_{f}}}{l})

\end{equation}

\

$$G_{1}=\frac{1}{{120}\pi^{2}}\int_{0}^{\pi}\left[\frac{\sin\left(kw\cos\frac{\theta}{2}\right)}{\cos{\theta}}\right](\sin{\theta})^{3}d{\theta}$$

\

$$G_{12}=\frac{1}{{120}\pi^{2}}\int_{0}^{\pi}\left[\frac{\sin\left(kw\cos\frac{\theta}{2}\right)}{\cos{\theta}}\right]J_{0}(kl\sin{\theta})(\sin{\theta})^{3}d{\theta}$$

\

where $k$ is the wave numbers at center wing resonance frequency $f_{r}$ and side wings resonance frequency $f_{r}^{'}$ , which are given by

$$fr=\frac{1}{\sqrt{LC}}$$

$${f}r^{'}=\frac{1}{\sqrt{L^{'}C^{'}}}$$

\

where\

 $L^{'}=L+\Delta L$ and $C^{'}={C{\Delta C}}/{(C+\Delta C)} $\\

Coupling factor $C_{P}$ between two resonator is given as

\

\begin{equation}

C_{P}={1}/{\sqrt{Q_{T}Q^{'}_{T}}}

\end{equation}

Coupling capacitance $C_{c}$ is defined as\

\begin{equation}

C_{c}= -(C+C^{'})+\sqrt{\left(\left(C+C^{'}\right)^{2}-4CC^{'}\left(1-\frac{1}{C^{2}_{P}}\right)\right)}

\end{equation}

Input impedance of ESPA

\begin{equation}

Z_{in}=jwL_{P}+\frac{Z_{1}Z_{2}}{(Z_{1}+Z_{2})}

\end{equation}

where $Z_{1}$ and $Z_{2}$ are\

$$Z_{1}=\frac{JwLR_{1}}{R_{1}-w^{2}LCR_{1}+JwL}     +\frac{1}{jwC_{c}}          $$

and\

$$Z_{2}=\frac{JwL^{'}R_{2}}{R_{2}-w^{2}L^{'}C^{'}R_{2}+JwL^{'}}$$

\

$$L_{P}=\frac{60{h}}{C}\ln(C/{0.2886\pi{f_{r}}})$$

reflection coefficient\

$$\Gamma=\dfrac{Z_{in}-50}{Z_{in}+50}$$

Return loss

\begin{equation}

return loss=20\log_{10}

\mid{\Gamma\mid}

\end{equation}

find the output

Answer 1