EV6_HW_Imp/doc/crosstalk/main.tex

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\title{
    Crosstalk experiment\linebreak
    \large{EV6 - Hardware Implementation}
}

\author{
    van Iterson, Arne\\
    Student Nr: 1798423
    \and
    Selier, Tom\\
    Student Nr: 1808444
}

\makeindex

\begin{document}
\maketitle

\begin{abstract}
    This document describes the process of, and measurements taken during the crosstalk experiment; Part of the EV6 Hardware Implementation course at the University of applied sciences Utrecht.
\end{abstract}

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\setlist[itemize]{itemjoin=\hspace*{\fill},itemjoin*=\hspace*{\fill}}

\begin{multicols}{2}
    \section{Introduction}
        Two conductors running parallel to each other can cause crosstalk, this is when the signal on one line induces a signal on the other line. This can cause signal integrity issues and can even cause the transmission hardware to malfunction.

        This experiment will be performed on a pre-built contraption consisting of three wires above an metal plate connected to signal ground, two of which will be used. By putting a signal on one wire and measuring the resulting signal on the other, the difference between capacitive and inductive crosstalk can be observed, as well as the effect of different terminators on the signal and crosstalk.  

        \subsection{Objective}
        The purpose of the experiment is to learn the difference between capacitive and inductive crosstalk and how different terminators affect the signal and crosstalk. Using the results of the experiment the mutual self-inductance and coupling capacitance of the setup should be determined. 

    \section{Methodology}
        The lab manual does not provide a clear methodology for the experiment, requiring the students to setup a measurement plan themselves. We determined the following measurements would be required to determine the inductance and capacitance of the setup: % yes we determined this right before we did them so that counts as a plan right?
        \begin{itemize}
            \item Near and far side of the interfered conductor while the signal conductor is not terminated.
            \item Near and far side of the interfered conductor while the signal conductor is shorted to ground.
            \item Near and far side of the interfered conductor while the signal conductor is terminated characteristically. 
        \end{itemize}

        The lab manual includes Figure \ref*{fig:setup_ee} which describes the electrical properties of the setup. 

        \begin{figure}[H]
            \includegraphics[width=\linewidth]{./img/setup_ee.png}
            \caption{Measurement setup}
            \label{fig:setup_ee}
        \end{figure}

        Since the experiment setup includes a series 180 $\Omega$ resistor, the characteristic impedance would be the sum of the impedance of the resistor and the impedance of the function generator. This would make the characteristic impedance 230 $\Omega$. The setup did include a terminator of 230 $\Omega$, but it seemed to be somewhat sketchy, being hand made, and we were not sure if it was part of the setup at all. We decided to use both the 50 $\Omega$ and 230 $\Omega$ terminators to be sure and to see if there was any difference in the results.

        The experiment called for the students to determine the ideal frequency at which to measure the crosstalk. We setup the function generator to sweep the signal between 1 kHz to 40 MHz, the maximum frequency of the function generator, and observed the crosstalk on the near and using the oscilloscope. We determined the frequency at which the crosstalk was the highest, which was around 20 MHz.

        \subsection{Equipment}
        The following equipment and settings will be used during the experiment:
        \begin{itemize}[beginpenalty=10000]
            \item Rigol DG 2041A Function/Arbritrary Waveform Generator
            \begin{itemize}
                \item Setup according to the method described in the lab manual and the determined frequency:
                \begin{itemize}
                    \item Sinusoidal waveform
                    \item Amplitude of 5 $V_{pp}$
                    \item Offset of 0 $V$
                    \item Frequency 20 MHz
                    \item Output impedance 50 $\Omega$
                \end{itemize}
            \end{itemize}
    
            \item DPO 2012 Oscilloscope
            \begin{itemize}
                \item Using default settings
            \end{itemize}

            \item Oscilloscope probe with a 10:1 attenuation ratio

            \item Multiple 1 meter BNC cables with a characteristic impedance of 50 $\Omega$
            \begin{itemize}
                \item One for trigger output, one for the experiment itself
            \end{itemize}
            
            \item Various BNC accessories
            \begin{itemize}
                \item Short circuit terminator
                \item 50 $\Omega$ terminator
                \item 230 $\Omega$ terminator
                \item T-connectors
                \item Probe to BNC connector
            \end{itemize}
        \end{itemize}

    \subsection{Setup}
    All measurements will be taken using the setup in Figure \ref*{fig:setup}, switching between the near and far side of the setup where applicable.
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/setup.png}
        \caption{Measurement setup}
        \label{fig:setup}
    \end{figure}

    \section{Expected results}
    The results expected for the different types of crosstalk are described in the following sections.\\
    
    \textbf{Capacitive crosstalk}\\
    Capacitive crosstalk is caused by the electric field of the signal conductor inducing a voltage on the interfered conductor. This electric field is generated by the presence of a voltage on the signal conductor. Therefore, when the signal conductor is terminated by a short terminator, the voltage on the signal conductor is practically zero, making the capacitive crosstalk minimal. When the signal conductor is not terminated, the crosstalk should be at its maximum. Capacitive crosstalk does not have a phase shift. \\ 

    \textbf{Inductive crosstalk}\\
    Inductive crosstalk is caused by the magnetic field of the signal conductor inducing a voltage on the interfered conductor. The magnetic field is generated by the current flowing through the signal conductor. When the signal conductor is terminated, the crosstalk should be at its maximum since the current is as high as it can be. When the signal conductor is not terminated, the crosstalk should be minimal. The magnetic field generates a current in the interfered conductor that is opposite to the signal conductor, creating a 180$\degree$ phase shift\\

    When the signal conductor is terminated with a characteristic terminator, the resulting will be a combination of the capacitive and inductive crosstalk since the voltage is not shorted completely, creating a resistor divider, and there is some current flowing through the signal conductor; Causing both the electric and magnetic field to induce a voltage on the interfered conductor.

    \section{Results}
    This section will show the measurement results of the experiment. The results will be presented in the form of graphs which include peak-to-peak voltage, frequency and phase shift within the legend. In the next section, the results will be analysed and compared to the expected results.

    \subsection{Measurements}
    \subsubsection{Open termination}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - Capacative - Near end.png}
        \caption{Open termination near end}
        \label{fig:graph_cap_near}
    \end{figure}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - Capacative - Far end.png}
        \caption{Open termination far end}
        \label{fig:graph_cap_far}
    \end{figure}

    \subsubsection{Short termination}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - Inductive - Near end.png}
        \caption{Short termination near end}
        \label{fig:graph_ind_near}
    \end{figure}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - Inductive - Far end.png}
        \caption{Short termination far end}
        \label{fig:graph_ind_far}
    \end{figure}

    \subsubsection{Characteristic termination}
    \textbf{50 Ohm terminator}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - 50 Ohm - Near end.png}
        \caption{50 Ohm termination near end}
        \label{fig:graph_char_50_near}
    \end{figure}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - 50 Ohm - Far end.png}
        \caption{50 Ohm termination far end}
        \label{fig:graph_char_50_far}
    \end{figure}

    \textbf{230 Ohm terminator}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - 230 Ohm - Near end.png}
        \caption{230 Ohm termination near end}
        \label{fig:graph_char_230_near}
    \end{figure}
    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Graph - 230 Ohm - Far end.png}
        \caption{230 Ohm termination far end}
        \label{fig:graph_char_230_far}
    \end{figure}

    \subsection{Analysis}
    In this section the results will be compared with the theory given in the reader. 
 
    \subsubsection{Capacative}
    The formula for capacitive crosstalk is given by:
    
    \begin{equation} \label{eq:capacitive crosstalk}
        \begin{aligned}
        U &= Z_0 \frac{j \omega C_{1,2} \cdot U_g}{2} \\
        \end{aligned}
    \end{equation}

    Solving equation \ref{eq:capacitive crosstalk} for $C_{1,2}$ gives
    \begin{equation}
        \begin{aligned}
        U &= Z_0 \frac{j \omega C_{1,2} \cdot U_g}{2} \\
        2U &= Z_0 \cdot j\omega C_{1,2} \cdot U_g \\
        \frac{2U}{C_{1,2}} &= Z_0 \cdot j\omega \cdot U_g \\
        C_{1,2} &= \frac{2U}{Z_0 \cdot j\omega \cdot U_g}
        \end{aligned}
    \end{equation}

    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 $Z_0 = 220\ \Omega$ 

    \begin{equation}
        \begin{aligned}
        C_{1,2} &= \frac{2 \cdot 0.1\ V}{220\ \Omega \cdot j20\cdot10^6 Hz \cdot 2\ V} \\
        &\approx j2.27\cdot10^{-11} F \\
        &\approx j0.027\ nF
        \end{aligned}
    \end{equation}

    Simulating the circuit to crosscheck if $0.027\ nF$ is a realistic answer gives a promising result.

    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Spice - Capacative Schematic.png}
        \caption{LTSpice schematic}
        \label{fig:schematic_capacitive}
    \end{figure}

    \begin{figure}[H]
        \includegraphics[width=\linewidth]{./img/Spice - Capacative Simulation.png}
        \caption{LTSpice Simulation, signal conductor(green) and interefered conductor(blue)}
        \label{fig:simulation_capacitive}
    \end{figure}

    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}.

    \subsubsection{Inductive}
    The formula of inductive crosstalk is given by 

    \begin{equation} \label{eq:inductive crosstalk}
        U = j \omega L_{M} \frac{U_{g}}{{2}\cdot Z_{0}}
    \end{equation}

    Solving equation \ref{eq:inductive crosstalk} for $L_M$ gives

    \begin{equation}
        \begin{aligned}
            U &= \frac{U_{g} \cdot j \omega L_{M}}{2\cdot Z_{0}} \\
            U_{g} \cdot j \omega L_{M} &= 2U \cdot Z_{0} \\
            L_{M} &= \frac{2U \cdot Z_{0}}{U_{g} \cdot j \omega}
        \end{aligned}
    \end{equation}

    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 $Z_0 = 220\ \Omega$ 

    \begin{equation}
        \begin{aligned}
            L_{M} &= \frac{2\cdot 0.14\ V \cdot 220\ \Omega}{2.2\ V \cdot j 20 \cdot 10^{6}} \\
            &= j1.4 \cdot 10^{-6} H \\
            &= j1.4\ \mu H
        \end{aligned}
    \end{equation}

    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.

    \subsubsection{Characteristic}
    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}). 

    \paragraph{Far end}
    For the voltage at the far end, let the equation be 
    \begin{equation}
        U_{fe} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(\frac{C_{12}}{C_{l}}- \frac{L_{M}}{L_l})
    \end{equation}

    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 

    \begin{equation}
        \begin{aligned}
        U_{fe} &= j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(0) \\
        U_{fe} &= 0\ V
        \end{aligned}
    \end{equation}

    \paragraph{Near end}
    According to the reader the voltage at the near end is
    \begin{equation}
        U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(\frac{C_{12}}{C_{l}}+ \frac{L_{M}}{L_l})
    \end{equation}

    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, $K$.

    \begin{equation}
        U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}(2K)
    \end{equation}

    Substituting $K$ with $\frac{L_M}{L_l}$ gives

    \begin{equation}
        U_{ne} = j \omega L_{l} \frac{U_{g}}{4 \cdot Z_{0}}\cdot \frac{2L_M}{L_l}
    \end{equation}

    Simplifying the equation gives

    \begin{equation}
        \begin{aligned}
            U_{ne} &= j \omega L_{l} \frac{2 \cdot U_{g} \cdot L_{M}}{4 \cdot Z_{0} \cdot L_{l}} \\
            &= \frac{2 \cdot U_{g} \cdot L_{M} \cdot j \omega L_{l}}{4 \cdot Z_{0} \cdot L_{l}} \\
            &= \frac{2 \cdot U_{g} \cdot L_{M} \cdot j \omega}{4 \cdot Z_{0}}
            \end{aligned}
    \end{equation}

    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 $Z_0 = 220\ \Omega$ 

    \begin{equation}
        \begin{aligned}
        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} \\
        &= 0.129 \overline{81}\ V \\
        &\approx 0.13\ V
        \end{aligned}
    \end{equation}

    $0.13\ V$ is equal to the amplitude measured in Figure \ref{fig:graph_char_230_near}, concluding that $230 \Omega$ is the characterstic termination.

    \section{Conclusion}

\end{multicols}
\end{document}