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\documentclass[journal]{IEEEtran}

\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{esint}
\usepackage{physics}

\DeclareMathOperator\arctanh{arctanh}

\setlength{\parindent}{0pt}

\title{EEMAGS Equation Sheet}

\newcommand{\del}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\veps}{\varepsilon}
\newcommand{\vphi}{\varphi}

\begin{document}
\maketitle

\section{Constants}
$$\varepsilon_0 = 8.854 \times 10^{-12}$$
$$\mu_0 = 1.257 \times 10^{-6}$$
$$c = 3 \times 10^8$$
$$\eta_0 = \sqrt{{\mu_0 \over \varepsilon_0}} = 377\Omega \approx 120\pi$$

\section{Fifth's Bullet Points}
Find $\veps_r$ given $\lambda$ and $f$:
$$\veps_r = {c^2 \over \mu_r \lambda^2 f^2}$$
Find $v_p$ given $\delta$, $\alpha$, and $\beta$ given a good conductor:
$$v_p = {\omega \over \beta}$$
Find $P_\text{avg}$ given $\vec{H}(t)$ in air:

Find $\Gamma$ in polar and VWSR given $Z_L$ and $Z_0$:
$$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
Find $\vec{E}(z,t)$ given $\vec{H}(z,t)$ in a lossless medium of $\mu_r$ and $\veps_r$:

Find $\lambda$, $\veps_r$, and $\vec{H}$ given $\vec{E}$:
$$\hat{H}_y = {\hat{E}_x \over \eta}$$
$$\veps_r = {\beta^2 \over \omega^2 \mu_r \mu_0 \veps_0}$$
$$\lambda = {v_p \over f}$$
Find $\Gamma^2$ and $\langle P \rangle$ from $\veps$, lossless:

Find $\beta_\text{air}$, $\beta_\text{material}$, $\hat{\Gamma}$, $\hat{T}$ given $\vec{E}^i$, $\veps_r$, $\mu_r$, $\sigma_r$:

Find $v_p$, $L$, and $Z_0$ given $\veps_r$ and $C$, lossless:
$$v_p = {1 \over \sqrt{\mu_0 \mu_r \veps_0 \veps_r}} = {\omega \over \beta}$$
$$Z_0 = \sqrt{L \over C}$$
$$L = {\mu_0 \mu_r \veps_0 \veps_r \over C}$$
Find $C$ given $Z_0$, $R_L$, $f$, and VWSR:
$$C = {L \over Z_0^2} - {G \over j\omega} + {R \over j\omega Z_0^2}$$
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
$$\Gamma = {R_L - Z_0 \over R_L + Z_0}$$
$$\omega = 2\pi f$$
Find VWSR, $\Gamma$, $Z_\text{in}$ from $\lambda$, $Z_0$, $Z_L$, length:
$$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
$$\max[Z_\text{in}] = Z_0 \cdot \text{VWSR}$$
$$\min[Z_\text{in}] = {Z_0 \over \text{VWSR}}$$
Find $Z_0$, $\alpha$, $\beta$, $\gamma$, $\lambda$ given $Z_\text{in, sc}$ and $Z_\text{in, oc}$:
$$Z_0 = \sqrt{Z_\text{in, sc} Z_\text{in, oc}}$$
$$\gamma = {1\over l} \arctanh\left(\sqrt{Z_\text{in, sc} \over Z_\text{in, oc}}\right)$$


\section{Boundary Conditions}
Electric field boundary conditions:
$$\vec{E}_{T1} = \vec{E}_{T2}$$
$$\vec{D}_{N1} = \vec{D}_{N2}$$
$$\vec{E} = \vec{E}_N + \vec{E}_T$$
$$\vec{D} = \veps \vec{E}$$

Magnetic field boundary conditions:
$$\vec{H}_{T1} = \vec{H}_{T2}$$
$$\vec{B}_{N1} = \vec{B}_{N2}$$
$$\vec{B} = \vec{B}_N + \vec{B}_T$$
$$\vec{B} = \mu \vec{H}$$

Normal vectors:
$$\vec{n} = \vec{E}_1 - \vec{E}_2$$
$$\vec{n} = \vec{H}_1 - \vec{H}_2$$
$$\hat{n} = {\vec{n} \over \|\vec{n}\|}$$



\section{Laplace and Poisson}
The Laplacian:
$$\vec{\nabla} \cdot (-\vec{\nabla}V) = -\vec{\nabla}^2 V = {\rho \over \veps_0}$$
Laplace's Equation (charge free region):
$$\vec{\nabla}^2 V = 0$$
Poisson's Equation:
$$\vec{\nabla}^2 V = -{\rho \over \veps_0}$$

\section{General Transmission Lines}
\begin{center}
\includegraphics[width=0.3\textwidth]{Transmission_line_element.png}
\end{center}
$$\begin{aligned}
\gamma &= \alpha + j\beta = \left[(R + j\omega L)(G + j\omega C)\right]^{1/2} \\ & = {1\over l} \arctanh\left( \sqrt{Z_\text{in, sc} \over Z_\text{in, oc}}\right)
\end{aligned}$$

The characteristic impedance:
$$Z_0 = \sqrt{R + j\omega L \over G + j\omega C} = \sqrt{Z_\text{in, sc} Z_\text{in, oc}}$$
$$Z_\text{in} = Z_0 {Z_L + Z_0 \tanh(\gamma l) \over Z_0 + Z_L \tanh(\gamma l)}$$

\section{Voltage Standing Wave Ratio}
$$\Gamma = {V_\text{reflected} \over V_\text{incident}}\Big|_{z' = 0} = {Z_L - Z_0 \over Z_L + Z_0}$$
$$\text{VWSR} = {|V_\text{max}| \over |V_\text{min}|} = {1 + |\Gamma| \over 1 - |\Gamma|}$$

$$v_p = {\omega \over \beta}$$

$$P_\text{avg} = {|V_0^+| \over 2 Z_0}(1 - |\Gamma_L|^2)


\section{Transmission Line Design}
Lossless lines:
$$R = G = 0$$
$$\alpha = 0$$
$$\beta = \omega \sqrt{LC}$$

Distortionless lines:
$${R \over L} = {G \over C}$$
$$\alpha = R \sqrt{C \over L}$$
$$\beta = \omega \sqrt{LC}$$

\end{document}
