(Created page with "a) $TEM \to E_z = H_z = 0$\\ $\left\{ \begin{array}{ll} \bar{E} = E(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \gamma = \alpha + j\beta\\ \bar{H} = H(x,y)e^{-\gamma z}e^{j\om...") |
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| − | a) | + | a) |
| − | + | <math> | |
| + | TEM \to E_z = H_z = 0\\ | ||
| + | \left\{ | ||
\begin{array}{ll} | \begin{array}{ll} | ||
\bar{E} = E(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \gamma = \alpha + j\beta\\ | \bar{E} = E(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \gamma = \alpha + j\beta\\ | ||
\bar{H} = H(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \beta = \omega\sqrt{\mu \epsilon} | \bar{H} = H(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \beta = \omega\sqrt{\mu \epsilon} | ||
\end{array} | \end{array} | ||
| − | \right. | + | \right. |
| + | </math> | ||
\begin{itemize} | \begin{itemize} | ||
Revision as of 19:50, 3 June 2017
a) $ TEM \to E_z = H_z = 0\\ \left\{ \begin{array}{ll} \bar{E} = E(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \gamma = \alpha + j\beta\\ \bar{H} = H(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \beta = \omega\sqrt{\mu \epsilon} \end{array} \right. $
\begin{itemize} \item assume these solutions in region between conductors. \item solve wave equations: $\left\{ \begin{array}{ll} \nabla^2\bar{E} + k^2 \bar{E} =0 \hspace{1cm} \text{with BC's to find}\\ \nabla^2\bar{H} + k^2 \bar{E} =0 \hspace{1cm} \bar{E} \text{ and } \bar{H} \end{array} \right.$ \item $Z = \frac{|E|}{|H|}$ \end{itemize}
Alternative: from transmission line theory : $Z_0 = \sqrt{\frac{L}{C}}$ (lossless) \begin{itemize} \item find $C$ by assuming same $V$ on line (or $Q$) \item find $L$ by assuming same $I$ on the line \end{itemize}
\textbf{Note:}\\
TEM\\
$Z_{TEM} = \frac{E_x}{E_y}$\\
$\bar{H} = \frac{1}{Z_{\text{TEM}}}(\hat{z}x\bar{E})$
b) find $C: C= \frac{Q}{V} $ \[\oint \bar{D}\cdot d\bar{s} = Q\] \[\int_0^L \int_0^{2\pi}\epsilon E_r(rd\phi dz) = Q\] \[\epsilon E_r(2\pi r)L = Q\] \[\bar{E} = \frac{Q}{2\pi r\epsilon(L)}\hat{r}\] \begin{align*} V_2 - V_1 &= - \int_1^2 \bar{E}\cdot dl\\ &=-\int_b^a \frac{Q}{2\pi L\epsilon}\bigg(\frac{1}{r}\bigg)dr\\ &=\frac{Q}{2\pi L\epsilon}\ln\bigg(\frac{b}{a}\bigg) \end{align*} \[C= \frac{2\pi L\epsilon}{\ln\big(\frac{b}{a}\big)}\]
find $L: L = \frac{\Phi}{NI}$ \[\oint \bar{H}\cdot d\bar{l} = I_{enc}\] \[\int_0^{2\pi} H_\phi(rd\phi) = I\] \[\bar{H} = \frac{I}{2\pi r}\hat{\phi}\] \begin{align*} \Phi &= \int \bar{B}\cdot d\bar{s}\\ &=\int_0^L\int_a^b \frac{\mu I}{2\pi r}drdz\\ &= \frac{\mu IL}{2\pi}\ln\bigg(\frac{b}{a}\bigg) \end{align*} \[dl = dr\hat{r} + rd\phi\hat{\phi} +dz\hat{z}\] \[ds_{\phi} = drdz\] \[L = \frac{\mu L \ln \big(b/a\big)}{2\pi}\] \[Z_o = \sqrt{\frac{L}{C}} = \sqrt{\frac{\frac{\mu L\ln(b/a)}{2\pi}}{{\frac{2\pi L\epsilon}{\ln(b/a)}}}} = \sqrt{\frac{\mu}{\epsilon}\bigg(\frac{\ln(b/a)}{2\pi}\bigg)^2}=\frac{\ln(b/a)}{2\pi}\sqrt{\frac{\mu}{\epsilon}}\]
c) no longer supports TEM mode; loss will cause $\bar{E}_z \ne 0$ and/or $\bar{H}_z \ne 0$ due to variation of fields along $z$ (attenuation)
