big math
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\end{figure}
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\subsection{Analysis}
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\subsubsection{Characteristic termination}
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In this section the results will be compared with the theory given in the reader.
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\subsubsection{Capacative}
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The formula for capacitive crosstalk is given by:
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\begin{equation} \label{eq:capacitive crosstalk}
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\begin{aligned}
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U &= Z_0 \frac{j \omega C_{1,2} \cdot U_g}{2} \\
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\end{aligned}
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\end{equation}
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Solving equation \ref{eq:capacitive crosstalk} for $C_{1,2}$ gives
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\begin{equation}
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\begin{aligned}
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U &= Z_0 \frac{j \omega C_{1,2} \cdot U_g}{2} \\
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2U &= Z_0 \cdot j\omega C_{1,2} \cdot U_g \\
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\frac{2U}{C_{1,2}} &= Z_0 \cdot j\omega \cdot U_g \\
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C_{1,2} &= \frac{2U}{Z_0 \cdot j\omega \cdot U_g}
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\end{aligned}
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\end{equation}
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Taking the results from figure \ref{fig:graph_cap_far},\newline $U_g = 2\ V$, $U = 0.14\ V$, $\omega = 20 \cdot 10^6\ Hz$ and $Z_0 = 220\ \Omega$
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\begin{equation}
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\begin{aligned}
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C_{1,2} &= \frac{2 \cdot 0.1\ V}{220\ \Omega \cdot j20\cdot10^6 Hz \cdot 2\ V} \\
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&\approx j2.27\cdot10^{-11} F \\
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&\approx j0.027\ nF
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\end{aligned}
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\end{equation}
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Simulating the circuit to crosscheck if $0.027\ nF$ is a realistic answer gives a promising result.
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\begin{figure}[H]
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\includegraphics[width=\linewidth]{./img/Spice - Capacative Schematic.png}
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\caption{LTSpice schematic}
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\label{fig:schematic_capacitive}
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\end{figure}
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\begin{figure}[H]
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\includegraphics[width=\linewidth]{./img/Spice - Capacative Simulation.png}
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\caption{LTSpice Simulation, signal conductor(green) and interefered conductor(blue)}
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\label{fig:simulation_capacitive}
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\end{figure}
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The amplitude of the interfered conductor is roughly 0.2V, which is in the ballpark of the actual signal shown in figure \ref{fig:graph_cap_near} and figure \ref{fig:graph_cap_far}.
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\subsubsection{Inductive}
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The formula of inductive crosstalk is given by
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\begin{equation} \label{eq:inductive crosstalk}
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U = j \omega L_{M} \frac{U_{g}}{{2}\cdot Z_{0}}
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\end{equation}
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Solving equation \ref{eq:inductive crosstalk} for $L_M$ gives
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\begin{equation}
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\begin{aligned}
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U &= \frac{U_{g} \cdot j \omega L_{M}}{2\cdot Z_{0}} \\
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U_{g} \cdot j \omega L_{M} &= 2U \cdot Z_{0} \\
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L_{M} &= \frac{2U \cdot Z_{0}}{U_{g} \cdot j \omega}
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\end{aligned}
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\end{equation}
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Taking the results from figure \ref{fig:graph_ind_far},\newline $U_g = 2.2\ V$, $U = 0.14\ V$, $\omega = 20 \cdot 10^6\ Hz$ and $Z_0 = 220\ \Omega$
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\begin{equation}
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\begin{aligned}
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L_{M} &= \frac{2\cdot 0.14\ V \cdot 220\ \Omega}{2.2\ V \cdot j 20 \cdot 10^{6}} \\
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&= j1.4 \cdot 10^{-6} H \\
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&= j1.4\ \mu H
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\end{aligned}
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\end{equation}
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A simulation for inductive crosstalk was not done, since this was way more complicated than the capacative crosstalk. However, the method of solving was the same, thus the result must simular to the capacative crosstalk.
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\subsubsection{Characteristic}
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From the measurements, it becomes clear that the correct characteristic termination is, in fact, 230 $\Omega$. The 50 $\Omega$ terminator does dampen the crosstalk on the far end of the conductor (Figure \ref*{fig:graph_char_50_far}), while the 230 $\Omega$ terminator effectively eliminates it (Figure \ref*{fig:graph_char_230_far}).
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\paragraph{Far end}
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For the voltage at the far end, let the equation be
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\begin{equation}
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U_{fe} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(\frac{C_{12}}{C_{l}}- \frac{L_{M}}{L_l})
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\end{equation}
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According to the reader and the measurements, when $Z_0$ = $R_L$ the circuit is terminated characteristically. Thus, $\frac{C_{12}}{C_{l}}= \frac{L_{M}}{L_l}$ has to be true. This makes the equation for the far end
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\begin{equation}
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\begin{aligned}
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U_{fe} &= j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(0) \\
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U_{fe} &= 0\ V
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\end{aligned}
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\end{equation}
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\paragraph{Near end}
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According to the reader the voltage at the near end is
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\begin{equation}
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U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(\frac{C_{12}}{C_{l}}+ \frac{L_{M}}{L_l})
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\end{equation}
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When the circuit is terminated characteristically and $\frac{C_{12}}{C_{l}}= \frac{L_{M}}{L_l}$ this equation is non-zero. Since these fractions are the same, the equation can be written using a constant, $K$.
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\begin{equation}
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U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(2K)
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\end{equation}
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Substituting $K$ with $\frac{L_M}{L_l}$ gives
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\begin{equation}
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U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}\cdot \frac{2L_M}{L_l}
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\end{equation}
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Simplifying the equation gives
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\begin{equation}
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\begin{aligned}
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U_{ne} &= j \omega L_{l} \frac{2 \cdot U_{g} \cdot L_{M}}{4 \cdot Z_{0} \cdot L_{l}} \\
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&= \frac{2 \cdot U_{g} \cdot L_{M} \cdot j \omega L_{l}}{4 \cdot Z_{0} \cdot L_{l}} \\
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&= \frac{2 \cdot U_{g} \cdot L_{M} \cdot j \omega}{4 \cdot Z_{0}}
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\end{aligned}
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\end{equation}
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Losing the dependance of the unknown variables $L_l$ and $C_l$ makes it possible to crosscheck this equation against one of the results. Figure \ref{fig:graph_char_230_near} gives the values: $U_g = 2.04\ V$, $\omega = 20 \cdot 10^6\ Hz$ and $Z_0 = 220\ \Omega$
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\begin{equation}
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\begin{aligned}
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U_{ne} &= \frac{2 \cdot 2.04\ V \cdot 1.4 \cdot 10^{-6}\ H \cdot j 20 \cdot 10^6\ Hz}{4 \cdot 220\ \Omega} \\
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&= 0.129 \overline{81}\ V \\
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&\approx 0.13\ V
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\end{aligned}
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\end{equation}
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$0.13\ V$ is equal to the amplitude measured in Figure \ref{fig:graph_char_230_near}, concluding that $230 \Omega$ is the characterstic termination.
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\section{Conclusion}
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\end{multicols}
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