Lab 1 – Time Domain Reflectometry
Autor
charles lee doyle
Letzte Aktualisierung
vor 8 Jahren
Lizenz
Creative Commons CC BY 4.0
Abstrakt
Time Domain Reflectometry DIT
Time Domain Reflectometry DIT
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{textcomp}
\usepackage{graphicx,psfrag,epsf}
\usepackage{enumerate}
\usepackage{natbib}
\usepackage{float}
\usepackage{wrapfig}
\newcommand{\blind}{0}
\addtolength{\oddsidemargin}{-.75in}%
\addtolength{\evensidemargin}{-.75in}%
\addtolength{\textwidth}{1.5in}%
\addtolength{\textheight}{1.3in}%
\addtolength{\topmargin}{-.8in}%
\begin{document}
%\bibliographystyle{natbib}
\def\spacingset#1{\renewcommand{\baselinestretch}%
{#1}\small\normalsize} \spacingset{1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\if0\blind
{
\title{\bf Lab 1 – Time Domain Reflectometry}
\author{Charles Lee Doyle \\ c12763425}
\maketitle
} \fi
\if1\blind
{
\bigskip
\bigskip
\bigskip
\begin{center}
{\LARGE\bf Title}
\end{center}
\medskip
} \fi
\bigskip
%\noindent%
%{\it Keywords:} 7 or fewer keywords
\spacingset{1.45}
\section{Introduction}
How do you connect a receiver or transmitter to an antenna? Simple - a transmission line. You are no doubt familiar with transmission lines (sometimes abbreviated as tx lines). If you plug any electric device into a wall outlet, the wires that connect the wall outlet to the device is a transmission line.
However, transmission lines behave very oddly at high frequencies. In traditional (low-frequency) circuit theory, wires connect devices, but have zero resistance. There is no phase delay across wires; and a short-circuited line always yields zero resistance.
For high-frequency transmission lines, things behave quite differently. For instance, short-circuits can actually have an infinite impedance; open-circuits can behave like short-circuited wires. The impedance of some load (ZL=XL+jYL) can be transformed at the terminals of the transmission line to an impedance much different than ZL. The goal of this Lab report is to understand transmission lines and the reasons for their odd effects.
Also for instance reflected signals in a network that has an open circuit will create excess traffic, bandwidth goes down and the network collapses.
In our experiment, the transmission line were tested through nine different conditions, simulated using pSpice and Matlab.
Each of these conditions is determined by finding the reflection coefficient which is calculated by using the equation:
\bigskip
\begin{equation}
\Gamma = \frac{Vref}{Vinc}
\end{equation}
\bigskip
which is written as:
\begin{equation}
\Gamma = \frac{ZL-ZO}{ZL+ZO}
\end{equation}
\bigskip
Where:
\bigskip
ZL is the impedance of the load.
\bigskip
Z0 is the impedance of the line.
\bigskip
$\Gamma$ is the reflection coefficient.
%\label{sec:intro}
\section{objective}
\label{sec:meth}
\begin{itemize}
\item To determine the behaviour of a transmission line under the nine different conditions listed above,
\item Obtain waveforms from pSpice simulations then validate each of our results using Matlab\textcopyright,
\item mathematical analysis, to verify results.
\end{itemize}
\section{Required tests:}
\label{sec:verify}
using Pspice the following 10 conditions were built and simulated determining what waveform's were created by each condition, using ZO=50 $\Omega$
\begin{enumerate}
\item Unstable system Rs = short circuit and RL= circuit.
\item Stable system, using a pulse waveform where will RS be short circuit and RL will be open circuit. With a max step size of $0.1nS$
\item RS = $50 \Omega$ and RL will be open circuit.
\item matched condition where RS = 50$\Omega$ and RL = 50$\Omega$.
\item A 1nF capacitor connected in series to a matched RS.
\item A 1nF capacitor connected in parallel to a matched RS.
\item A 10μH inductor connected in parallel to a matched RS of 50$\Omega$.
\item A 10μH inductor connected in series to a matched RS of 50$\Omega$.
\item RLC circuit
\end{enumerate}
\bigskip
\bigskip
\bigskip
We were asked to set up a pulse wave using the following settings in Pspice:
\begin{itemize}
\item TD = 19.6n
\item V1 = 0
\item V2 = 1
\item TD = 0
\item TR = 10n
\item TF = 20n
\item PW = 10
\item PER = 20
\end{itemize}
\subsection{Unstable system, Rs is a short circuit and RL an open circuit.}
\label{sec:conc}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{1st.JPG}
\caption{\label{fig:Circuit} Circuit diagram}
\end{figure}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{1st_2.JPG}
\caption{\label{fig:Circuit} Waveform}
\end{figure}
As in figure 2 we can see that the output looks somewhat like a chirp signal. however It is inherently unstable. Later this will be rectified by using alternating sampling steps in pSpice.
\subsection{Stable system, using a pulse waveform where will RS be short circuit and RL will be open circuit. With a max step size of 0.1nS}
\label{sec:conc}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{2nd.JPG}
\caption{\label{fig:Circuit} Waveform 1}
\end{figure}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{2nd_2.JPG}
\caption{\label{fig:Circuit} Waveform 2}
\end{figure}
In Fig 3 we can see the classic chirp signal. After we changed the sampling step size in pSpice to 1nS. So that the waveform is now stablised compared to figure 2.
To calculate the reflection coefficient:
\begin{equation}
\Gamma = \frac{Vref}{Vinc}
\end{equation}
which is written as:
\begin{equation}
\Gamma = \frac{ZL-ZO}{ZL+ZO}
\end{equation}
\begin{equation}
\Gamma = \frac{ \infty - 50}{\infty + 50} = 1 \angle 0^{\circ}
\end{equation}
In Fig 4 one can see that the output voltage is 2 volts, even though our input voltage was 1V. Which is explained as follows:
\begin{equation}
Vref = \Gamma Vinc
\end{equation}
\begin{equation}
Vref = Vinc
\end{equation}
\begin{equation}
Vout=Vinc+Vinc
\end{equation}
\begin{equation}
1+1=2V
\end{equation}
We get 2v output as our reflection coefficient is equal to 1V, where our supply voltage 1v. The reason for this phenomena is because our supply voltage is reflected through the line as our signal is tranmitted at different impedances, causing the voltage to appear double.
\subsection{RS = 50$\Omega$ and RL will be open circuit.}
\label{sec:conc}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{3rd.JPG}
\caption{\label{fig:Circuit} Circuit}
\end{figure}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{3rd_2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
The reflection coefficient is:
\begin{equation}
\Gamma = \frac{Vref}{Vinc}
\end{equation}
\begin{equation}
\Gamma = \frac{ZL-Z0}{ZL + Z0}
\end{equation}
\begin{equation}
\Gamma L = \frac{100k-50}{100k+50} = 1
\end{equation}
\begin{equation}
\Gamma s = \frac{50-50}{50+50} = 0
\end{equation}
In the above waveform fig6, the voltage from source to load is 0.5v. This $\frac{1}{2} V$ doubles in as the value of $\Gamma L$
is equal to $1v$.
\subsection{matched condition where RS = 50$\Omega$ and RL = 50$\Omega$.}
\label{sec:conc}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{4th.JPG}
\caption{\label{fig:Circuit}circuit}
\end{figure}
This is called a matched condition as RS and RL and both equal to 50$\Omega$
There is no reflection back through the line as both impedance's are equal
This reflection coefficient can be proven below:
\begin{equation}
\Gamma = \frac{ZL - Z0}{ZL + Z0}
\end{equation}
\begin{equation}
\Gamma = \frac{50 − 50}{50 + 50} = 0
\end{equation}
As seen in figure 8, the signal from the source is being completely
absorbed by the load, therefore, there is no reflection through the line.
\begin{equation}
\Gamma = \frac{ZL-Z0}{ZL + Z0}
\end{equation}
\begin{equation}
\Gamma = \frac{50-50}{50+50} = 0
\end{equation}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{4th_2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
\bigskip
\bigskip
\bigskip
\subsection{A 1nF capacitor connected in parallel to a matched RS.}
\label{sec:conc}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{5th.JPG}
\caption{\label{fig:Circuit} circuit}
\end{figure}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{5th2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
for conditon 5 we derived a set of calculations to determine the step reponse before put the values into Matlab\textcopyright using the step function to get the step response of the system.
here below is how the step response of the system was derived:
since this circuit has a matched condition,
\begin{equation}
Rs = RL.
\end{equation}
we can assume that:
\begin{equation}
Rs = RL = R.
\end{equation}
this is a parallel capacitor and we are looking step response so we use the laplace transfrom.
\begin{equation}
Zc = \frac{1}{sc} = \frac{1}{jwc}
\end{equation}
\begin{equation}
Vo = \frac{ZRC}{ZRC + 50} \enspace \enspace \enspace where \enspace Zrc = \frac{\frac{1}{sc}}{\frac{1}{sc} + R}
\end{equation}
$Thus \enspace Zrc \enspace can \enspace be \enspace further \enspace simplified \enspace as:$
\begin{equation}
\frac{R}{scr + 1}
\end{equation}
$So:$
\begin{equation}
Vo = \left (\frac{R}{R+(scr+1)50} \right)*Vin
\end{equation}
\begin{equation}
Vo = \left ( \frac{50}{s(1*10^{-8})50^{2}+100}\right)*Vin
\end{equation}
\begin{equation}
Vo = \left ( \frac{1}{sRC+2}\right)*Vin
\end{equation}
So our equation to calculate the step response is:
\begin{equation}
H = \left ( \frac{1}{sRC+2}\right)*Vin
\end{equation}
\begin{equation}
H = \left ( \frac{1}{(50)(1*10^{-8})s+2}\right)*Vin
\end{equation}
\begin{equation}
H = \left ( \frac{1}{(50*10^{-8})s+2}\right)*Vin
\end{equation}
We inputted this equation was into Matlab\copyright \enspace using the step function and gained the plot shown below:
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{5th3.JPG}
\caption{\label{fig:Circuit} matlab step response}
\end{figure}
to calculate the reflection coefficient of this system we use:
\begin{equation}
\Gamma = \frac{ZL - Z0}{ZL + Z0}
\end{equation}
\begin{equation}
\Gamma = \frac{50 - 50}{50 +50} = 0
\end{equation}
There is no reflection back throught the transmission line so to calculate the transmission in the load we use:
\begin{equation}
\Gamma L = \frac{0 - 50}{0 +50} = -1
\end{equation}
\subsection{A 1nF capacitor connected in series to a matched RS.}
\label{sec:conc}
In figure 13. We can see that at the start of the simulation there is no reflection until 24ns. At 0.5volts though The capacitor
acts like an open circuit, so current is not passed through the capacitor. At this point the output voltage will equal the input voltage
as seen in Matlab\copyright \enspace output.
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{6th.JPG}
\caption{\label{fig:Circuit} circuit}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{6th2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
\bigskip
Calculations for step response as entered into Matlab\copyright
\begin{equation}
H = \frac{Vo}{Vin}
\end{equation}
\begin{equation}
H = \frac{\frac{1}{sC}+R+R}{\frac{1}{sC}+2R}
\end{equation}
Which can be simplified as:
\begin{equation}
H= \frac{sRC+1}{2sRC+1}
\end{equation}
\begin{equation}
H= \frac{(50)(1*10^{-9})s+1}{2(50)(1*10^{-9})s+1}
\end{equation}
simplified as:
\begin{equation}
H= \frac{(5*10^{-8})s+1}{100*10^{-9})s+1}
\end{equation}
Step response graph from Matlab\copyright:
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{6th3.JPG}
\caption{\label{fig:Circuit} matlab step response}
\end{figure}
\bigskip
\subsection{A 10μH inductor connected in parallel to a matched RS of 50$\Omega$.}
\label{sec:conc}
for this circuit we have an inductor in series, so for this circuit there will be a different step response even though the calculations for matlab\copyright will be similar to those of having a capacitor connected in parallel
this matched where RS and RL are equal :
\begin{equation}
Rs = RL = R
\end{equation}
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{7th.JPG}
\caption{\label{fig:Circuit} circuit}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{7th2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
Calculations required for MAtlab\copyright:
\begin{equation}
H= \frac{Vo}{Vin}
\end{equation}
\begin{equation}
\frac{ZRL\left | \right | L}{ZRL\left | \right | L +R}
\end{equation}
\begin{equation}
\frac{Vo}{Vin}= \frac{\frac{sLR}{sL+R}}{\frac{sLR}{sL+R}+R}
\end{equation}
Which is simplified to:
\bigskip
\begin{equation}
\frac{sLR}{sLR+sLR+R^{2}}
\end{equation}
So our step response can be calculated as:
\begin{equation}
H= \frac{sL}{2sL+R}
\end{equation}
\begin{equation}
H= \frac{(1*10^{-5})s}{2(1*10^{-5})s+50}
\end{equation}
simplified as:
\begin{equation}
H= \frac{(1*10^{-5})s+1}{(2*10^{-5})s+50}
\end{equation}
Below is the graph produced by matlab\copyright
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{7th3.JPG}
\caption{\label{fig:Circuit} matlab step response}
\end{figure}
\subsection{A 10μH inductor connected in series to a matched RS of 50$\Omega$.}
\label{sec:conc}
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{8th.JPG}
\caption{\label{fig:Circuit} circuit}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{8th2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
At lower frequencies, the inductor acts as an open
circuit as can be seen in figure 19, the voltage starts to drop. At 0.5v it acts like a short.
The step response of the system is calculated as below:
\begin{equation}
H= \frac{Vo}{Vin} = \frac{sL+R}{sL+R+R}
\end{equation}
So our step response is:
\begin{equation}
H= \frac{sL+R}{sL+2R} =\frac{(10*10^{-6})s+50}{(10*10^{-6})s+2(50)} = \frac{(10*10^{-6}s+50)}{(10*10^{-6})s+100}
\end{equation}
Using the step function in Matlab\copyright we gained the following graph:
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{8th3.JPG}
\caption{\label{fig:Circuit} matlab step response}
\end{figure}
\newpage
\subsection{RLC circuit}
\label{sec:conc}
\bigskip
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{9th.JPG}
\caption{\label{fig:Circuit} Applying 6.0v to the the regulator}
\end{figure}
\bigskip
\begin{figure}[!ht]
\centering
\includegraphics[width= 0.50\textwidth]{9th2.JPG}
\caption{\label{fig:Circuit} waveform}
\end{figure}
To calculate the step response of the RLC circuit we use the equations
below:
\bigskip
\begin{equation}
ZRLC= sL + \frac{R}{sRC+1}
\end{equation}
$Where:$
\begin{equation}
H= \frac{ZRLC}{ZRLC+R}
\end{equation}
$So:$
\begin{equation}
\frac{sL + \frac{R}{sRC+1}}{sL + \frac{R}{sRC+1} + R}
\end{equation}
$Simplfied \enspace to:$
\begin{equation}
\frac{sL(sRc+1)+R}{sL(sRC+1)+R+R(sRC+1)}
\end{equation}
\begin{equation}
\frac{s^{2}+\frac{s}{RC}+\frac{1}{LC}}{s^{2}+s \left( \frac{1}{RC}+\frac{R}{LC} \right)+ \frac{2}{LC}}
\end{equation}
\begin{equation}
H= \frac{s^{2}+s(20*10^{6})+10^{14}}{s^{2}+s(10^{14})+(2*10^{14})}
\end{equation}
\begin{figure}[!h]
\centering
\includegraphics[width= 0.50\textwidth]{9th3.JPG}
\caption{\label{fig:Circuit} matlab step response}
\end{figure}
%\bigskip
%\begin{center}
%{\large\bf SUPPLEMENTAL MATERIALS}
%\end{center}
%\begin{description}
%\item[Title:] Brief description. (file type)
%\item[R-package for MYNEW routine:] R-package ÒMYNEWÓ containing code to perform the diagnostic methods described in the article. The package also contains all datasets used as examples in the article. (GNU zipped tar file)
%\item[HIV data set:] Data set used in the illustration of MYNEW method in Section~ 3.2. (.txt file)
%\end{description}
%\begin{thebibliography}{}
%\bibitem[Azzalini(2005)]{azza:05}
%Azzalini, A. (2005).
%\newblock The skew-normal distribution and related multivariate families.
%\newblock \emph{Scandinavian Journal of Statistics} \textbf{32}, 159--188.
%\end{thebibliography}{}
\end{document}