\documentclass{article}

\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{listings}
\usepackage{caption}
\usepackage{subcaption}
\usepackage{float}
\usepackage[margin=1in]{geometry}


\title{}

\author{Aidan Sharpe \& Elise Heim}

\DeclareMathOperator{\sinc}{sinc}

\begin{document}
\begin{titlepage}
\maketitle
\end{titlepage}

\begin{abstract}
\end{abstract}

\section{Introduction}

\section{Results \& Discussion}

\subsection{Uniform Convergence}
Consider the discrete time signal, $x[n] = a^n u[n]$, where $u[n]$ is the unit-step and $a=0.9$. Let $X(e^{j\omega})$ be the discrete time Fourier transform (DTFT) of $x[n]$. For the DTFT of a signal to exist, it must be absolutely summable over all $n \in \mathbb{Z}$. If the signal is absolutely summable, then it must also be bounded, meaning $|x[n]| < \infty, \forall n \in \mathbb{Z}$. In other words, $x[n]$ must have a finite maximum value.
\\
\\
In this case, $x[n]$ is bounded because $a^n$ grows with smaller values of $n$, and $u[n]$ is zero when $n<0$. Therefore, the maximum value of $x[n] = a^n u[n]$ is $x[0] = a$.
\\
\\
Additionally, $x[n]$ takes the form of a geometric series, so its sum is given by
\begin{equation}
	\sum_{n=m}^\infty s^n = {s^m \over 1-s}, |s| < 1.
\end{equation}
In this case, $m=0$, because the signal has no value for $n<0$. Since $a=0.9$, the sum evaluates to is ${1 \over 1-0.9} = 10$. Considering that $x[n]$ is always a positive real number, each term is its own absolute value, so the sum and the absolute sum are equivalent. By taking the absolute sum of the first 200 terms of $x[n]$, it becomes clear that it approaches 10 in the limit, seen in figure \ref{fig:abs_sum_anun}.

\begin{figure}[h]
	\center
	\includegraphics[width=\textwidth]{abs_sum_anun.png}
	\caption{The signal value of $x[n]=a^n u[n]$ and its absolute sum}
	\label{fig:abs_sum_anun}
\end{figure}



\section{Conclusions}

\end{document}