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| − | a) | + | a) |
| + | |||
<math> | <math> | ||
TEM \to E_z = H_z = 0\\ | TEM \to E_z = H_z = 0\\ | ||
| Line 9: | Line 10: | ||
\right. | \right. | ||
</math> | </math> | ||
| + | assume these solutions in region between conductors. | ||
| − | + | solve wave equations: | |
| − | + | <math> | |
| − | + | \left\{ | |
\begin{array}{ll} | \begin{array}{ll} | ||
\nabla^2\bar{E} + k^2 \bar{E} =0 \hspace{1cm} \text{with BC's to find}\\ | \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} | \nabla^2\bar{H} + k^2 \bar{E} =0 \hspace{1cm} \bar{E} \text{ and } \bar{H} | ||
\end{array} | \end{array} | ||
| − | \right. | + | \right. |
| − | + | Z = \frac{|E|}{|H|} | |
| − | + | </math><br> | |
| − | + | Alternative: from transmission line theory :<br> | |
| − | Alternative: from transmission line theory : | + | <math>Z_0 = \sqrt{\frac{L}{C}} (lossless)</math><br> |
| − | + | find C by assuming same V on line (or Q) <br> | |
| − | + | find L by assuming same I on the line <br> | |
| − | + | <math> | |
| − | + | ||
| − | + | ||
| − | + | ||
\textbf{Note:}\\ | \textbf{Note:}\\ | ||
TEM\\ | TEM\\ | ||
| − | + | Z_{TEM} = \frac{E_x}{E_y}\\ | |
| − | + | \bar{H} = \frac{1}{Z_{\text{TEM}}}(\hat{z}x\bar{E}) | |
| + | </math><br> | ||
| − | b) find | + | b) |
| − | + | ||
| − | \ | + | <math> |
| − | \ | + | 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*} | \begin{align*} | ||
V_2 - V_1 &= - \int_1^2 \bar{E}\cdot dl\\ | V_2 - V_1 &= - \int_1^2 \bar{E}\cdot dl\\ | ||
| Line 43: | Line 46: | ||
&=\frac{Q}{2\pi L\epsilon}\ln\bigg(\frac{b}{a}\bigg) | &=\frac{Q}{2\pi L\epsilon}\ln\bigg(\frac{b}{a}\bigg) | ||
\end{align*} | \end{align*} | ||
| − | \[C= \frac{2\pi L\epsilon}{\ln\big(\frac{b}{a}\big)}\] | + | \[C= \frac{2\pi L\epsilon}{\ln\big(\frac{b}{a}\big)}\]\\ |
| + | </math><br> | ||
| − | find | + | <math> |
| − | \ | + | 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*} | \begin{align*} | ||
\Phi &= \int \bar{B}\cdot d\bar{s}\\ | \Phi &= \int \bar{B}\cdot d\bar{s}\\ | ||
| Line 54: | Line 59: | ||
&= \frac{\mu IL}{2\pi}\ln\bigg(\frac{b}{a}\bigg) | &= \frac{\mu IL}{2\pi}\ln\bigg(\frac{b}{a}\bigg) | ||
\end{align*} | \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}}\\ |
| + | </math><br> | ||
| − | c) | + | c) |
| + | c) still supports a TEM mode; a uniformly lossy dielectric results in a non-zero attenuation constant <math>\alpha</math> and a conductance <math>G</math> such that Ez = 0 and Hz = 0. However, if the {\bf conductors} have a large but finite conductivity (no longer PEC), then the Poynting vector must have a component normal to the conductor surface, necessitating the presence of a small axial electric field Ez != 0 (see David K. Cheng, Field and Wave Electromagnetics, 2nd Ed. page 433). | ||
Latest revision as of 23:44, 19 July 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. $ assume these solutions in region between conductors.
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. Z = \frac{|E|}{|H|} $
Alternative: from transmission line theory :
$ Z_0 = \sqrt{\frac{L}{C}} (lossless) $
find C by assuming same V on line (or Q)
find L by assuming same I on the line
$ \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) c) still supports a TEM mode; a uniformly lossy dielectric results in a non-zero attenuation constant $ \alpha $ and a conductance $ G $ such that Ez = 0 and Hz = 0. However, if the {\bf conductors} have a large but finite conductivity (no longer PEC), then the Poynting vector must have a component normal to the conductor surface, necessitating the presence of a small axial electric field Ez != 0 (see David K. Cheng, Field and Wave Electromagnetics, 2nd Ed. page 433).
