% STEM content samples for LaTeX-to-HTML accessibility testing \ifdefined\HCode \documentclass[10pt,letterpaper]{article} \else \documentclass[10pt,letterpaper,twocolumn]{article} \fi \usepackage{fontspec} \setmainfont{DejaVu Serif} \setsansfont{DejaVu Sans} \setmonofont{DejaVu Sans Mono} \usepackage[english]{babel} \usepackage[ letterpaper, top=0.68in, bottom=0.72in, left=0.70in, right=0.70in ]{geometry} \usepackage{microtype} \usepackage{amsmath,amssymb,amsfonts,mathtools} \usepackage{booktabs,array,tabularx,makecell} \usepackage{graphicx} \usepackage{xcolor} \usepackage{ragged2e} \usepackage{enumitem} \usepackage{needspace} \usepackage{listings} \usepackage{caption} \usepackage{float} \usepackage{tikz} \usetikzlibrary{arrows.meta,positioning,calc} \usepackage{pgfplots} \pgfplotsset{compat=1.18} \usepackage[hidelinks]{hyperref} % ------------------------------------------------- % Two-column layout controls % ------------------------------------------------- \setlength{\columnsep}{0.26in} \setlength{\columnseprule}{0pt} \setlength{\textfloatsep}{8pt plus 2pt minus 2pt} \setlength{\floatsep}{7pt plus 2pt minus 2pt} \setlength{\intextsep}{7pt plus 2pt minus 2pt} \setlength{\dbltextfloatsep}{9pt plus 2pt minus 2pt} \setlength{\dblfloatsep}{8pt plus 2pt minus 2pt} \raggedbottom \setlength{\abovedisplayskip}{5pt plus 2pt minus 2pt} \setlength{\belowdisplayskip}{5pt plus 2pt minus 2pt} \setlength{\abovedisplayshortskip}{3pt plus 1pt minus 1pt} \setlength{\belowdisplayshortskip}{3pt plus 1pt minus 1pt} \captionsetup{ font=small, labelfont=bf, justification=raggedright, singlelinecheck=false, skip=4pt } \setlist[itemize]{leftmargin=1.45em,itemsep=2pt,topsep=3pt} \setlist[enumerate]{leftmargin=1.75em,itemsep=2pt,topsep=3pt} \newcolumntype{Y}{>{\RaggedRight\arraybackslash}X} \newcolumntype{C}{>{\Centering\arraybackslash}X} % ------------------------------------------------- % Compact code style for research-paper columns % ------------------------------------------------- \definecolor{codegray}{gray}{0.42} \lstdefinestyle{researchcode}{ basicstyle=\ttfamily\fontsize{7pt}{8pt}\selectfont, keywordstyle=\bfseries, commentstyle=\itshape, stringstyle=\ttfamily, numbers=left, numberstyle=\ttfamily\tiny\color{codegray}, stepnumber=1, numbersep=4pt, frame=tb, framerule=0.35pt, framesep=3pt, breaklines=true, breakatwhitespace=false, columns=fullflexible, keepspaces=true, showstringspaces=false, tabsize=4, xleftmargin=1.35em, framexleftmargin=1.05em, xrightmargin=0pt, linewidth=\linewidth, aboveskip=5pt, belowskip=5pt, captionpos=t, abovecaptionskip=0pt, belowcaptionskip=4pt } \lstset{style=researchcode} \hypersetup{ pdftitle={STEM Accessibility Test Document}, pdfauthor={Sohan Reddy Ganampeta}, pdfsubject={LaTeX-to-HTML accessibility testing with NVDA}, pdfkeywords={LaTeX, HTML5, MathML, MathJax, NVDA, MathCAT, accessibility} } \title{STEM Accessibility Test Document} \author{Sohan Reddy Ganampeta} \date{July 2026} \begin{document} \maketitle \begin{abstract} This document collects representative STEM content for testing LaTeX-to-HTML conversion with NVDA. It includes mathematics, code, tables, charts, diagrams, captions, and text alternatives from several academic disciplines. \end{abstract} \setcounter{tocdepth}{3} \tableofcontents \clearpage % ------------------------------------------------- % ------------------------------------------------- \section{Inline and Display Mathematics} The Pythagorean relationship is \(a^{2}+b^{2}=c^{2}\), and Newton's second law is \(F=ma\). A simple displayed equation is \begin{equation} E=mc^{2}. \end{equation} % ------------------------------------------------- \section{Fractions, Roots, Exponents, and Subscripts} The solutions of the quadratic equation \(ax^{2}+bx+c=0\), where \(a\neq0\), are \begin{equation} x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \end{equation} For a damped second-order system, the natural frequency and damping ratio may be written as \begin{align} \omega_{n} &= \sqrt{\frac{k}{m}},\\ \zeta &= \frac{c}{2\sqrt{mk}}. \end{align} % ------------------------------------------------- \section{Calculus} The derivative of a polynomial may be shown line by line: \begin{align} f(x) &= x^{3}-4x^{2}+2x-7,\\ f'(x) &= 3x^{2}-8x+2. \end{align} The definite integral of \(2x\) from zero to three is \begin{align} \int_{0}^{3}2x\,\mathrm{d}x &= \left[x^{2}\right]_{0}^{3}\\ &= 3^{2}-0^{2}\\ &= 9. \end{align} For a multivariable function \(g(x,y)\), the gradient is \begin{equation} \nabla g(x,y)= \begin{bmatrix} \dfrac{\partial g}{\partial x}\\[3pt] \dfrac{\partial g}{\partial y} \end{bmatrix}. \end{equation} % ------------------------------------------------- \section{Aligned Equations and Multi-Step Derivations} Consider the first-order differential equation \begin{equation} \dot{x}(t)+2x(t)=4, \qquad x(0)=1. \end{equation} Multiplying by the integrating factor \(e^{2t}\) gives \begin{align} e^{2t}\dot{x}(t)+2e^{2t}x(t) &=4e^{2t},\\ \frac{\mathrm{d}}{\mathrm{d}t} \left(e^{2t}x(t)\right) &=4e^{2t},\\ e^{2t}x(t)-x(0) &=2\left(e^{2t}-1\right),\\ x(t) &=2-e^{-2t}. \end{align} % ------------------------------------------------- \section{Vectors and Matrices} Let \begin{equation} \mathbf{x}= \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}, \qquad A= \begin{bmatrix} 0 & 1\\ -\dfrac{k}{m} & -\dfrac{c}{m} \end{bmatrix}, \qquad B= \begin{bmatrix} 0\\ \dfrac{1}{m} \end{bmatrix}. \end{equation} The state equation is \begin{equation} \dot{\mathbf{x}}=A\mathbf{x}+B u. \end{equation} % ------------------------------------------------- \section{Piecewise Functions and Cases} Piecewise definitions are important in control, optimization, and signal processing. A saturation function may be written as \begin{equation} \operatorname{sat}(u)= \begin{cases} u_{\max}, & u>u_{\max},\\ u, & -u_{\max}\leq u\leq u_{\max},\\ -u_{\max}, & u<-u_{\max}. \end{cases} \end{equation} % ------------------------------------------------- \section{Greek Symbols and Control Notation} Consider the Lyapunov candidate \begin{equation} V(e,\widetilde{\theta}) =\frac{1}{2}e^{\mathsf{T}}Pe +\frac{1}{2}\widetilde{\theta}^{\mathsf{T}} \Gamma^{-1}\widetilde{\theta}, \end{equation} with the derivative condition \begin{equation} \dot{V} \leq -\lambda_{\min}(Q)\lVert e\rVert^{2}. \end{equation} % ------------------------------------------------- \section{Math Inside Lists} \begin{enumerate} \item The sampling period is \(T_{s}=0.01\,\mathrm{s}\). \item The estimation error is \(e_{k}=x_{k}-\hat{x}_{k}\). \item The control command is \(u_{k}=-Kx_{k}+r_{k}\). \item The error magnitude is \(\lVert e_{k}\rVert\). \end{enumerate} % ------------------------------------------------- \section{Mathematics in a Data Table} \ifdefined\HCode\else\onecolumn\fi \noindent\begin{minipage}{\textwidth} \centering \captionsetup{width=0.84\textwidth} \captionof{table}{System parameters for a second-order example.} \label{tab:system-parameters} \small \begin{tabularx}{0.84\textwidth}{@{}Y c c@{}} \toprule \textbf{Parameter and meaning} & \textbf{Symbol} & \textbf{Value}\\ \midrule Mass (object inertia) & \(m\) & \(1.0\,\mathrm{kg}\)\\ Damping coefficient (velocity resistance) & \(c\) & \(0.4\,\mathrm{N\,s/m}\)\\ Spring constant (position-restoring coefficient) & \(k\) & \(2.0\,\mathrm{N/m}\)\\ Natural frequency (undamped oscillation frequency) & \(\omega_{n}\) & \(\sqrt{k/m}\)\\ Damping ratio (normalized damping measure) & \(\zeta\) & \(c/(2\sqrt{mk})\)\\ \bottomrule \end{tabularx} \end{minipage} % ------------------------------------------------- % Keep the heading and introduction with the nonfloating chart block. If the % remaining space is too small, the complete section starts on the next page. \Needspace{4.2in} \section{Chart with a Text Equivalent} \noindent\begin{minipage}{\textwidth} \noindent \begin{minipage}[t]{0.66\textwidth} \vspace{0pt} \centering \IfFileExists{figures/stability_plot.png}{% \includegraphics[ width=\linewidth, height=1.68in, keepaspectratio ]{figures/stability_plot.png}% }{% \begin{tikzpicture} \begin{axis}[ width=\linewidth, height=1.68in, xmin=0,xmax=10, ymin=0,ymax=1.1, xlabel={Time (seconds)}, ylabel={Tracking error magnitude}, grid=major, tick label style={font=\small}, label style={font=\small} ] \addplot[blue,thick,smooth] coordinates { (0,1.00) (0.5,0.60) (1,0.48) (1.5,0.55) (2,0.38) (2.5,0.25) (3,0.29) (3.5,0.20) (4,0.13) (4.5,0.15) (5,0.11) (5.5,0.07) (6,0.08) (7,0.045) (8,0.028) (9,0.017) (10,0.010) }; \end{axis} \end{tikzpicture}% } \end{minipage}\hfill \begin{minipage}[t]{0.30\textwidth} \vspace{0pt} \footnotesize \textbf{Text equivalent.} From zero to ten seconds, the tracking error starts near \(1.00\), decreases rapidly, oscillates with progressively smaller peaks, and approaches zero. This downward trend is consistent with a stable response. \medskip The visual shape is not the only source of information: selected numerical values are also provided in Table~\ref{tab:stability-data}. \end{minipage} \captionof{figure}{Tracking error decreases over time in a stable system.} \label{fig:stability-chart} \smallskip \centering \captionsetup{width=0.56\textwidth} \captionof{table}{Selected data points from the stability chart.} \label{tab:stability-data} \begin{tabular}{@{}cccccc@{}} \toprule \textbf{Time (s)} & 0 & 2 & 4 & 6 & 10\\ \textbf{Error magnitude} & 1.00 & 0.38 & 0.13 & 0.08 & 0.01\\ \bottomrule \end{tabular} \end{minipage} % ------------------------------------------------- % Prevent the block-diagram heading from being left at the bottom of a page. \Needspace{2.2in} \section{Block Diagram with a Text Equivalent} \noindent\begin{minipage}{\textwidth} \noindent \begin{minipage}[t]{0.64\textwidth} \vspace{0pt} \centering \resizebox{\linewidth}{!}{% \begin{tikzpicture}[ >={Latex}, node distance=1.25cm, block/.style={draw,minimum height=0.85cm,minimum width=2.1cm,align=center}, sum/.style={draw,circle,inner sep=1.5pt,minimum size=0.62cm}, every node/.style={font=\small} ] \node[block] (reference) {Reference}; \node[sum,right=of reference] (sum) {$\sum$}; \node[block,right=of sum] (controller) {Controller}; \node[block,right=of controller] (plant) {Plant}; \node[block,right=of plant] (output) {Output}; \draw[->,thick] (reference) -- (sum); \draw[->,thick] (sum) -- node[above]{error} (controller); \draw[->,thick] (controller) -- node[above]{control input} (plant); \draw[->,thick] (plant) -- node[above]{response} (output); \draw[->,thick] (output.south) -- ++(0,-1.0) -| node[pos=0.35,below]{feedback path} (sum.south); \node[left=2pt of sum.south,yshift=-7pt] {$-$}; \end{tikzpicture}% } \end{minipage}\hfill \begin{minipage}[t]{0.32\textwidth} \vspace{0pt} \footnotesize \textbf{Text equivalent.} The reference signal enters a summing junction. The measured output is returned through a feedback path and subtracted from the reference. The resulting error enters the controller. The controller sends a control input to the plant, and the plant produces the system output. \end{minipage} \captionof{figure}{Closed-loop control diagram with reference input, controller, plant, output, and feedback.} \label{fig:closed-loop} \end{minipage} % Resume the standard two-column research-paper layout. \ifdefined\HCode\else\twocolumn\fi % ------------------------------------------------- \section{Algorithm as Structured Text} \begin{enumerate} \item Initialize the state estimate \(\hat{\mathbf{x}}_{0}\), parameter estimate \(\hat{\theta}_{0}\), and sampling period \(T_{s}\). \item At sample \(k\), obtain the measurement \(y_{k}\). \item Compute the estimation error \[ e_{k}=y_{k}-\hat{y}_{k}. \] \item Update the state estimate using the system model and correction term. \item Update the parameter estimate according to the specified adaptive law. \item Increase \(k\) by one and repeat until the stopping condition is met. \end{enumerate} % ------------------------------------------------- \section{Mathematics and Statistics} \subsection{Linear Equation With Explicit Steps} Solve \(3x+5=20\): \begin{align} 3x+5 &=20,\\ 3x &=15,\\ x &=5. \end{align} Each row performs one operation on both sides of the equality. \subsection{Quadratic Equation by Completing the Square} For \(x^{2}-6x+5=0\), \begin{align} x^{2}-6x &=-5,\\ x^{2}-6x+9 &=4,\\ (x-3)^{2} &=4,\\ x-3 &=\pm2,\\ x &=1\quad\text{or}\quad x=5. \end{align} \subsection{Derivative With Separate Left and Right Sides} Verify that \begin{equation} \frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+3x)=2x+3. \end{equation} \textbf{Left-hand side:} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+3x) &=\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}) +\frac{\mathrm{d}}{\mathrm{d}x}(3x)\\ &=2x+3. \end{align} \textbf{Right-hand side:} \begin{equation} 2x+3. \end{equation} Therefore, the independently interpreted left- and right-hand sides agree. \subsection{Matrix Multiplication} Let \begin{equation} A=\begin{bmatrix}1&2\\3&4\end{bmatrix}, \qquad \mathbf{x}=\begin{bmatrix}5\\6\end{bmatrix}. \end{equation} Then \begin{align} A\mathbf{x} &= \begin{bmatrix} 1(5)+2(6)\\ 3(5)+4(6) \end{bmatrix}\\ &= \begin{bmatrix} 17\\ 39 \end{bmatrix}. \end{align} \subsection{Proof by Mathematical Induction} Prove that \begin{equation} 1+2+\cdots+n=\frac{n(n+1)}{2} \end{equation} for every positive integer \(n\). \textbf{Base case.} For \(n=1\), \begin{equation} 1=\frac{1(1+1)}{2}=1. \end{equation} \textbf{Inductive hypothesis.} Assume \begin{equation} 1+2+\cdots+k=\frac{k(k+1)}{2}. \end{equation} \textbf{Inductive step.} Then \begin{align} 1+2+\cdots+k+(k+1) &=\frac{k(k+1)}{2}+(k+1)\\ &=(k+1)\left(\frac{k}{2}+1\right)\\ &=\frac{(k+1)(k+2)}{2}. \end{align} This is the required formula with \(n=k+1\); therefore, the statement holds for all positive integers. \subsection{Separate LHS and RHS Trigonometric Derivations} Verify the identity \begin{equation} \frac{1-\cos(2x)}{\sin(2x)}=\tan(x). \end{equation} \textbf{Left-hand side derivation:} \begin{align} \frac{1-\cos(2x)}{\sin(2x)} &=\frac{2\sin^{2}(x)}{2\sin(x)\cos(x)}\\ &=\frac{\sin(x)}{\cos(x)}\\ &=\tan(x). \end{align} \textbf{Right-hand side derivation:} \begin{align} \tan(x)&=\frac{\sin(x)}{\cos(x)}. \end{align} Both sides reduce to \(\sin(x)/\cos(x)\), so the identity is verified wherever the original expressions are defined. \subsection{LHS Derived to a Constant RHS} For the infinite geometric series, \begin{equation} 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2. \end{equation} Let \begin{equation} S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots. \end{equation} Multiply both sides by \(2\): \begin{equation} 2S=2+1+\frac{1}{2}+\frac{1}{4}+\cdots. \end{equation} Subtract the original series from the doubled series: \begin{align} 2S-S &=\left(2+1+\frac{1}{2}+\frac{1}{4}+\cdots\right) -\left(1+\frac{1}{2}+\frac{1}{4}+\cdots\right),\\ S&=2. \end{align} Therefore, \begin{equation} 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2. \end{equation} Here the left-hand side is derived, while the right-hand side remains the constant \(2\). \section{Computer Science} \subsection{Cross-Language Number-Analysis Program} The four listings implement the same number-analysis task in Python, R, MATLAB, and C++. \subsubsection{Python Example} \begin{lstlisting}[language=Python,caption={Python palindrome and number-analysis program},label={lst:python-palindrome}] def normalize_number(value: int) -> int: """Return the nonnegative form of an integer.""" return abs(value) def reverse_number(value: int) -> int: number = normalize_number(value) reversed_value = 0 while number > 0: digit = number % 10 reversed_value = (reversed_value * 10) + digit number //= 10 return reversed_value def digit_sum(value: int) -> int: number = normalize_number(value) total = 0 while number > 0: total += number % 10 number //= 10 return total def is_palindrome(value: int) -> bool: number = normalize_number(value) return number == reverse_number(number) def is_prime(value: int) -> bool: number = normalize_number(value) if number < 2: return False divisor = 2 while divisor * divisor <= number: if number % divisor == 0: return False divisor += 1 return True def analyze_numbers(values: list[int]) -> list[dict[str, object]]: results = [] for index, value in enumerate(values, start=1): result = { "item": index, "original": value, "reverse": reverse_number(value), "digit_sum": digit_sum(value), "palindrome": is_palindrome(value), "prime": is_prime(value), } results.append(result) return results def print_report(results: list[dict[str, object]]) -> None: print("\nNUMBER ANALYSIS REPORT") print("-" * 76) print( f"{'Item':>4} | {'Number':>10} | {'Reverse':>10} | " f"{'Digit sum':>9} | {'Palindrome':>10} | {'Prime':>5}" ) print("-" * 76) for result in results: palindrome_text = "yes" if result["palindrome"] else "no" prime_text = "yes" if result["prime"] else "no" print( f"{result['item']:>4} | {result['original']:>10} | " f"{result['reverse']:>10} | {result['digit_sum']:>9} | " f"{palindrome_text:>10} | {prime_text:>5}" ) def main() -> None: default_data = "121, 1331, 42, 97, -44, 1001" prompt = f"Enter integers separated by commas [{default_data}]: " raw_text = input(prompt).strip() if raw_text == "": raw_text = default_data try: values = [int(part.strip()) for part in raw_text.split(",")] except ValueError as error: print(f"Input error: {error}. The default list will be used.") values = [121, 1331, 42, 97, -44, 1001] results = analyze_numbers(values) print_report(results) palindrome_count = 0; prime_count = 0 for result in results: palindrome_count += int(result["palindrome"]) prime_count += int(result["prime"]) summary = { "total": len(results), "palindromes": palindrome_count, "primes": prime_count, } print( f"\nSummary: total={summary['total']}, " f"palindromes={summary['palindromes']}, " f"primes={summary['primes']}." ) if __name__ == "__main__": main() \end{lstlisting} \subsubsection{R Example} The R version performs the same task so that code navigation and punctuation pronunciation can be compared across languages. \begin{lstlisting}[language=R,caption={R palindrome and number-analysis program},label={lst:r-palindrome}] normalize_number <- function(value) { return(abs(as.integer(value))) } reverse_number <- function(value) { number <- normalize_number(value) reversed_value <- 0L while (number > 0L) { digit <- number %% 10L reversed_value <- (reversed_value * 10L) + digit number <- number %/% 10L } return(reversed_value) } digit_sum <- function(value) { number <- normalize_number(value) total <- 0L while (number > 0L) { total <- total + (number %% 10L) number <- number %/% 10L } return(total) } is_palindrome <- function(value) { number <- normalize_number(value) return(number == reverse_number(number)) } is_prime_number <- function(value) { number <- normalize_number(value) if (number < 2L) { return(FALSE) } limit <- floor(sqrt(number)) if (limit >= 2L) { for (divisor in 2L:limit) { if ((number %% divisor) == 0L) { return(FALSE) } } } return(TRUE) } analyze_numbers <- function(values) { rows <- lapply(seq_along(values), function(index) { value <- values[index] data.frame( Item = index, Number = value, Reverse = reverse_number(value), Digit.Sum = digit_sum(value), Palindrome = is_palindrome(value), Prime = is_prime_number(value), stringsAsFactors = FALSE ) }) return(do.call(rbind, rows)) } main <- function() { default_data <- "121, 1331, 42, 97, -44, 1001" prompt_text <- paste0( "Enter integers separated by commas [", default_data, "]: " ) raw_text <- trimws(readline(prompt = prompt_text)) if (identical(raw_text, "")) { raw_text <- default_data } pieces <- trimws(unlist(strsplit(raw_text, ","))) values <- suppressWarnings(as.integer(pieces)) if (length(values) == 0L || any(is.na(values))) { cat("Input error: the default list will be used.\n") values <- c(121L, 1331L, 42L, 97L, -44L, 1001L) } results <- analyze_numbers(values) cat("\nNUMBER ANALYSIS REPORT\n") print(results, row.names = FALSE) palindrome_count <- sum(results$Palindrome); prime_count <- sum(results$Prime) summary <- list( total = nrow(results), palindromes = palindrome_count, primes = prime_count ) cat(sprintf( "\nSummary: total=%d, palindromes=%d, primes=%d.\n", summary$total, summary$palindromes, summary$primes )) } main() \end{lstlisting} \subsubsection{MATLAB Example} The MATLAB version uses a script followed by local functions. It tests array notation, structure fields, colon-based ranges, formatted output, and function boundaries. \begin{lstlisting}[language=Matlab,caption={MATLAB palindrome and number-analysis program},label={lst:matlab-palindrome}] % Palindrome and number-analysis program clear; clc; defaultData = [121, 1331, 42, 97, -44, 1001]; promptText = ['Enter integers separated by commas ', ... '[121, 1331, 42, 97, -44, 1001]: ']; rawText = strtrim(input(promptText, 's')); if isempty(rawText) values = defaultData; else pieces = split(rawText, ','); values = str2double(strtrim(pieces)); values = values(:).'; if isempty(values) || any(isnan(values)) fprintf('Input error: the default list will be used.\n'); values = defaultData; end end template = struct( ... 'Item', 0, ... 'Number', 0, ... 'Reverse', 0, ... 'DigitSum', 0, ... 'Palindrome', false, ... 'Prime', false); results = repmat(template, 1, numel(values)); for index = 1:numel(values) value = values(index); results(index).Item = index; results(index).Number = value; results(index).Reverse = reverseNumber(value); results(index).DigitSum = digitSum(value); results(index).Palindrome = isPalindrome(value); results(index).Prime = isPrimeNumber(value); end fprintf('\nNUMBER ANALYSIS REPORT\n'); fprintf('%s\n', repmat('-', 1, 76)); fprintf('%4s | %10s | %10s | %9s | %10s | %5s\n', ... 'Item', 'Number', 'Reverse', 'Digit sum', 'Palindrome', 'Prime'); fprintf('%s\n', repmat('-', 1, 76)); for index = 1:numel(results) palindromeText = yesOrNo(results(index).Palindrome); primeText = yesOrNo(results(index).Prime); fprintf('%4d | %10d | %10d | %9d | %10s | %5s\n', ... results(index).Item, ... results(index).Number, ... results(index).Reverse, ... results(index).DigitSum, ... palindromeText, primeText); end summary.total = numel(results); summary.palindromes = sum([results.Palindrome]); summary.primes = sum([results.Prime]); fprintf('\nSummary: total=%d, palindromes=%d, primes=%d.\n', ... summary.total, summary.palindromes, summary.primes); function output = normalizeNumber(value) output = abs(fix(value)); end function output = reverseNumber(value) number = normalizeNumber(value); output = 0; while number > 0 digit = mod(number, 10); output = (output * 10) + digit; number = floor(number / 10); end end function output = digitSum(value) number = normalizeNumber(value); output = 0; while number > 0 output = output + mod(number, 10); number = floor(number / 10); end end function answer = isPalindrome(value) number = normalizeNumber(value); answer = (number == reverseNumber(number)); end function answer = isPrimeNumber(value) number = normalizeNumber(value); if number < 2 answer = false; return; end answer = true; limit = floor(sqrt(number)); for divisor = 2:limit if mod(number, divisor) == 0 answer = false; break; end end end function text = yesOrNo(condition) if condition text = 'yes'; else text = 'no'; end end \end{lstlisting} \subsubsection{C++ Example} The C++ version uses a structure, vectors, functions, exception handling, stream input, formatted output, and the conditional operator. It provides dense punctuation that is useful when evaluating screen-reader code settings. \begin{lstlisting}[language=C++,caption={C++ palindrome and number-analysis program},label={lst:cpp-palindrome}] #include #include #include #include #include #include #include struct Result { std::size_t item; long long original; long long reversed; int digitSum; bool palindrome; bool prime; }; long long normalizeNumber(long long value) { return std::llabs(value); } long long reverseNumber(long long value) { long long number = normalizeNumber(value); long long reversedValue = 0; while (number > 0) { const int digit = static_cast(number % 10); reversedValue = (reversedValue * 10) + digit; number /= 10; } return reversedValue; } int calculateDigitSum(long long value) { long long number = normalizeNumber(value); int total = 0; while (number > 0) { total += static_cast(number % 10); number /= 10; } return total; } bool isPalindrome(long long value) { const long long number = normalizeNumber(value); return number == reverseNumber(number); } bool isPrimeNumber(long long value) { const long long number = normalizeNumber(value); if (number < 2) { return false; } for (long long divisor = 2; divisor * divisor <= number; ++divisor) { if (number % divisor == 0) { return false; } } return true; } std::string trim(const std::string& text) { const std::string spaces = " \t\n\r"; const std::size_t first = text.find_first_not_of(spaces); if (first == std::string::npos) { return ""; } const std::size_t last = text.find_last_not_of(spaces); return text.substr(first, last - first + 1); } std::vector parseInput(const std::string& text) { std::vector values; std::stringstream stream(text); std::string token; while (std::getline(stream, token, ',')) { token = trim(token); std::size_t usedCharacters = 0; const long long value = std::stoll(token, &usedCharacters); if (usedCharacters != token.size()) { throw std::invalid_argument("Unexpected characters in: " + token); } values.push_back(value); } if (values.empty()) { throw std::invalid_argument("No integers were entered."); } return values; } std::vector analyzeNumbers(const std::vector& values) { std::vector results; for (std::size_t index = 0; index < values.size(); ++index) { const long long value = values[index]; results.push_back(Result{ index + 1, value, reverseNumber(value), calculateDigitSum(value), isPalindrome(value), isPrimeNumber(value) }); } return results; } void printReport(const std::vector& results) { std::cout << "\nNUMBER ANALYSIS REPORT\n"; std::cout << std::string(76, '-') << '\n'; std::cout << std::setw(4) << "Item" << " | " << std::setw(10) << "Number" << " | " << std::setw(10) << "Reverse" << " | " << std::setw(9) << "Digit sum" << " | " << std::setw(10) << "Palindrome" << " | " << std::setw(5) << "Prime" << '\n'; std::cout << std::string(76, '-') << '\n'; for (const Result& result : results) { const char* palindromeText = result.palindrome ? "yes" : "no"; const char* primeText = result.prime ? "yes" : "no"; std::cout << std::setw(4) << result.item << " | " << std::setw(10) << result.original << " | " << std::setw(10) << result.reversed << " | " << std::setw(9) << result.digitSum << " | " << std::setw(10) << palindromeText << " | " << std::setw(5) << primeText << '\n'; } } int main() { const std::string defaultData = "121, 1331, 42, 97, -44, 1001"; std::cout << "Enter integers separated by commas [" << defaultData << "]: "; std::string inputText; std::getline(std::cin, inputText); inputText = trim(inputText); if (inputText.empty()) { inputText = defaultData; } std::vector values; try { values = parseInput(inputText); } catch (const std::exception& error) { std::cerr << "Input error: " << error.what() << " The default list will be used.\n"; values = {121, 1331, 42, 97, -44, 1001}; } const std::vector results = analyzeNumbers(values); printReport(results); int palindromeCount = 0; int primeCount = 0; for (const Result& result : results) { palindromeCount += result.palindrome ? 1 : 0; primeCount += result.prime ? 1 : 0; } std::cout << "\nSummary: total=" << results.size() << ", palindromes=" << palindromeCount << ", primes=" << primeCount << ".\n"; return 0; } \end{lstlisting} \section{Physics and Astronomy} \subsection{Physics Examples} \subsubsection{Measurements, Units, and Uncertainty} A measured velocity may be reported as \[ v = (3.42\pm0.05) \times10^{5} \,\mathrm{m\,s^{-1}}. \] An energy measurement with asymmetric uncertainty may be written as \[ E = 1.50^{+0.18}_{-0.12} \,\mathrm{keV}. \] \begin{table}[tbp] \centering \caption{Physics measurements with different uncertainty formats.} \label{tab:physics-uncertainty} \small \begin{tabular}{@{}lll@{}} \toprule Quantity & Measured value & Unit\\ \midrule Velocity & \(3.42\pm0.05\) & \(10^{5}\,\mathrm{m\,s^{-1}}\)\\ Energy & \(1.50^{+0.18}_{-0.12}\) & \(\mathrm{keV}\)\\ Temperature & \(295.4\pm0.6\) & \(\mathrm{K}\)\\ \bottomrule \end{tabular} \end{table} % ------------------------------------------------- \subsubsection{Motion with Units and Vector Notation} Consider an object with mass \(m=2\,\mathrm{kg}\) and acceleration \[ \mathbf{a} = \begin{bmatrix} 3\\ -1\\ 2 \end{bmatrix} \,\mathrm{m\,s^{-2}}. \] Newton's second law gives \begin{align} \mathbf{F} &=m\mathbf{a}\\ &=2 \begin{bmatrix} 3\\ -1\\ 2 \end{bmatrix}\\ &= \begin{bmatrix} 6\\ -2\\ 4 \end{bmatrix} \,\mathrm{N}. \end{align} The magnitude of the force is \begin{align} \lVert\mathbf{F}\rVert &= \sqrt{6^{2}+(-2)^{2}+4^{2}}\\ &= \sqrt{56}\\ &= 2\sqrt{14}\,\mathrm{N}. \end{align} % ------------------------------------------------- \subsubsection{Electromagnetic Field Equations} Gauss's law for the electric field is \[ \nabla\cdot\mathbf{E} = \frac{\rho}{\varepsilon_{0}}. \] The Ampere--Maxwell equation is \begin{align} \nabla\times\mathbf{B} &= \mu_{0}\mathbf{J} \notag\\ &\quad+ \mu_{0}\varepsilon_{0} \frac{\partial\mathbf{E}}{\partial t}. \end{align} The Lorentz force on a particle is \[ \mathbf{F} = q \left( \mathbf{E} + \mathbf{v}\times\mathbf{B} \right). \] % ------------------------------------------------- \subsubsection{Quantum-Mechanical Notation} The time-dependent Schrödinger equation may be written as \[ i\hbar \frac{\partial}{\partial t} \lvert\psi(t)\rangle = \hat{H} \lvert\psi(t)\rangle. \] Normalization requires \begin{align} \langle\psi\mid\psi\rangle &= \int_{-\infty}^{\infty} \lvert\psi(x)\rvert^{2}\,dx\\ &=1. \end{align} The expected value of position is \[ \langle x\rangle = \langle\psi\lvert \hat{x} \rvert\psi\rangle. \] % ------------------------------------------------- \subsubsection{Tensor and General-Relativity Notation} Einstein's field equation is \[ G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu}. \] An infinitesimal spacetime interval may be written as \[ ds^{2} = g_{\mu\nu} \,dx^{\mu}dx^{\nu}. \] % ------------------------------------------------- \subsubsection{Experimental Measurements and Fitted Model} The measured response is compared with the model \[ y(t) = e^{-0.40t} \cos(1.80t). \] \begin{figure}[tbp] \centering \begin{tikzpicture} \begin{axis}[ width=0.98\columnwidth, height=0.68\columnwidth, xlabel={Time \(t\) (\(\mathrm{s}\))}, ylabel={Response \(y(t)\)}, xmin=0, xmax=3.6, ymin=-0.45, ymax=1.15, grid=major, legend style={ font=\scriptsize, draw=none, at={(0.98,0.97)}, anchor=north east }, tick label style={font=\scriptsize}, label style={font=\small} ] \addplot[ black, thick ] coordinates { (0.0,1.00) (0.5,0.68) (1.0,0.20) (1.5,-0.17) (2.0,-0.31) (2.5,-0.20) (3.0,0.02) (3.5,0.14) }; \addlegendentry{Fitted model} \addplot[ only marks, mark=*, mark size=1.7pt, black, error bars/.cd, y dir=both, y explicit ] table[ x=t, y=value, y error=error, row sep=\\ ] { t value error\\ 0.0 1.03 0.04\\ 0.5 0.69 0.04\\ 1.0 0.18 0.03\\ 1.5 -0.20 0.04\\ 2.0 -0.31 0.03\\ 2.5 -0.17 0.03\\ 3.0 0.03 0.04\\ 3.5 0.15 0.04\\ }; \addlegendentry{Measurements} \end{axis} \end{tikzpicture} \caption{Measured oscillatory response with uncertainty bars and a fitted model. Filled circles represent measurements; the solid line represents the model.} \label{fig:physics-experiment} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} Measured points follow a damped oscillatory model. \textbf{Long description.} Time runs from zero to \(3.5\) seconds. The response begins near \(1.0\), decreases through zero, reaches approximately \(-0.31\) at two seconds, and then increases toward zero. Each measured point has a vertical uncertainty bar. The measured points remain close to the fitted curve. \end{minipage} \end{figure} \begin{table}[tbp] \centering \caption{Data represented in the experimental-response graph.} \label{tab:physics-experiment-data} \small \begin{tabular}{@{}rrrr@{}} \toprule \(t\) (\(\mathrm{s}\)) & Measured & Uncertainty & Model\\ \midrule 0.0 & 1.03 & 0.04 & 1.00\\ 0.5 & 0.69 & 0.04 & 0.68\\ 1.0 & 0.18 & 0.03 & 0.20\\ 1.5 & -0.20 & 0.04 & -0.17\\ 2.0 & -0.31 & 0.03 & -0.31\\ 2.5 & -0.17 & 0.03 & -0.20\\ 3.0 & 0.03 & 0.04 & 0.02\\ 3.5 & 0.15 & 0.04 & 0.14\\ \bottomrule \end{tabular} \end{table} % ------------------------------------------------- \subsubsection{Experimental-Apparatus Diagram} \begin{figure}[tbp] \centering \begin{tikzpicture}[ component/.style={ draw, rounded corners, minimum width=1.05cm, minimum height=0.65cm, align=center, font=\scriptsize }, flow/.style={ -{Stealth[length=2mm]}, thick }, node distance=4mm ] \node[component] (laser) {Laser}; \node[component, right=of laser] (lens) {Lens}; \node[component, right=of lens] (sample) {Sample}; \node[component, right=of sample] (detector) {Detector}; \node[component, below=6mm of detector] (computer) {Computer}; \draw[flow] (laser) -- node[above,font=\tiny] {beam} (lens); \draw[flow] (lens) -- (sample); \draw[flow] (sample) -- (detector); \draw[flow] (detector) -- node[right,font=\tiny] {signal} (computer); \end{tikzpicture} \caption{Simplified optical measurement apparatus. Arrows show the direction of the optical beam and recorded signal.} \label{fig:physics-apparatus} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} A laser beam passes through a lens and sample before reaching a detector. \textbf{Long description.} From left to right, the laser sends a beam to the lens, the lens directs the beam through the sample, and the transmitted beam reaches the detector. The detector sends an electrical signal downward to a computer. \end{minipage} \end{figure} % ================================================= \subsection{Astronomy Examples} \subsubsection{Orbital Period and Received Flux} For two bodies with masses \(M\) and \(m\), the orbital period of a circular orbit with radius \(a\) is \[ T = 2\pi \sqrt{ \frac{a^{3}}{G(M+m)} }. \] The received flux from a source with luminosity \(L\) at distance \(d\) is \[ F = \frac{L}{4\pi d^{2}}. \] If the distance doubles while luminosity remains constant, then \begin{align} F_{\mathrm{new}} &= \frac{L}{4\pi(2d)^{2}}\\ &= \frac{L}{16\pi d^{2}}\\ &= \frac{1}{4}F. \end{align} % ------------------------------------------------- \subsubsection{Distance Modulus and Redshift} The distance modulus is \[ m-M = 5\log_{10} \left( \frac{d}{10\,\mathrm{pc}} \right). \] Astronomical redshift is \[ z = \frac{ \lambda_{\mathrm{obs}}-\lambda_{0} }{ \lambda_{0} }. \] For \(\lambda_{\mathrm{obs}}=660.0\,\mathrm{nm}\) and \(\lambda_{0}=656.3\,\mathrm{nm}\), \begin{align} z &= \frac{660.0-656.3}{656.3}\\ &\approx 5.64\times10^{-3}. \end{align} % ------------------------------------------------- \subsubsection{Celestial Coordinates} A source position may be written as \begin{align} \alpha &= 12^{\mathrm{h}} 34^{\mathrm{m}} 56.7^{\mathrm{s}},\\ \delta &= -45^{\circ} 18^{\prime} 22^{\prime\prime}. \end{align} Here, \(\alpha\) is right ascension and \(\delta\) is declination. \begin{table}[tbp] \centering \caption{Example celestial coordinates.} \label{tab:celestial-coordinates} \scriptsize \begin{tabular}{@{}lll@{}} \toprule Source & Right ascension & Declination\\ \midrule A & 12:34:56.7 & \(-45\):18:22\\ B & 12:35:04.1 & \(-45\):17:49\\ C & 12:35:15.8 & \(-45\):19:03\\ \bottomrule \end{tabular} \end{table} % ------------------------------------------------- \subsubsection{Multi-Panel Astronomical Image} \begin{figure*}[tbp] \centering \begin{tikzpicture}[font=\small] % Panel A \begin{scope} \fill[black!6] (0,0) rectangle (5.1,3.5); \draw (0,0) rectangle (5.1,3.5); \node[anchor=north west,font=\bfseries] at (0.10,3.40) {(A) Optical}; \foreach \x/\y/\r in { 0.45/0.55/0.04, 0.85/2.55/0.06, 1.25/1.35/0.04, 1.75/2.85/0.05, 2.05/0.60/0.04, 2.55/1.75/0.09, 3.15/2.55/0.05, 3.65/0.85/0.04, 4.10/2.90/0.06, 4.55/1.45/0.05, 4.75/0.45/0.04 }{ \fill (\x,\y) circle (\r); } \draw[very thick] (2.55,1.75) circle (0.22); \draw[-{Stealth[length=2mm]}] (0.45,2.65) -- (0.45,3.15); \node[above,font=\scriptsize] at (0.45,3.15) {N}; \draw[-{Stealth[length=2mm]}] (0.45,2.65) -- (0.95,2.65); \node[right,font=\scriptsize] at (0.95,2.65) {E}; \draw[thick] (3.70,0.35) -- (4.55,0.35); \node[above,font=\scriptsize] at (4.12,0.35) {\(10^{\prime\prime}\)}; \end{scope} % Panel B \begin{scope}[xshift=5.55cm] \fill[black!14] (0,0) rectangle (5.1,3.5); \draw (0,0) rectangle (5.1,3.5); \node[anchor=north west,font=\bfseries] at (0.10,3.40) {(B) Infrared}; \foreach \x/\y/\r in { 0.50/0.55/0.07, 0.90/2.60/0.08, 1.30/1.40/0.07, 1.80/2.80/0.08, 2.10/0.65/0.07, 2.55/1.75/0.20, 3.20/2.50/0.08, 3.65/0.90/0.07, 4.10/2.85/0.10, 4.55/1.45/0.08 }{ \fill[black!75] (\x,\y) circle (\r); } \fill[black!35] (2.55,1.75) circle (0.48); \draw[very thick] (2.55,1.75) circle (0.22); \draw[thick] (3.70,0.35) -- (4.55,0.35); \node[above,font=\scriptsize] at (4.12,0.35) {\(10^{\prime\prime}\)}; \end{scope} % Panel C \begin{scope}[xshift=11.10cm] \fill[black!9] (0,0) rectangle (5.1,3.5); \draw (0,0) rectangle (5.1,3.5); \node[anchor=north west,font=\bfseries] at (0.10,3.40) {(C) Composite}; \foreach \x/\y/\r in { 0.50/0.55/0.05, 0.90/2.60/0.06, 1.30/1.40/0.05, 1.80/2.80/0.06, 2.10/0.65/0.05, 2.55/1.75/0.11, 3.20/2.50/0.06, 3.65/0.90/0.05, 4.10/2.85/0.07, 4.55/1.45/0.06 }{ \fill (\x,\y) circle (\r); } \draw[thick] (2.55,1.75) ellipse (0.45 and 0.30); \draw[thick,dashed] (2.55,1.75) ellipse (0.75 and 0.52); \draw[thick,dotted] (2.55,1.75) ellipse (1.05 and 0.75); \draw[very thick] (2.55,1.75) circle (0.20); \draw[thick] (3.70,0.35) -- (4.55,0.35); \node[above,font=\scriptsize] at (4.12,0.35) {\(10^{\prime\prime}\)}; \end{scope} \end{tikzpicture} \caption{Controlled multi-panel astronomy image. Panel A represents an optical observation, Panel B represents an infrared observation, and Panel C combines the source positions with three contour levels. Circles identify the target; solid, dashed, and dotted contours represent increasing spatial extent.} \label{fig:astronomy-panels} \begin{minipage}{0.96\textwidth} \small \textbf{Short text alternative.} Three observations show the same astronomical field in optical, infrared, and composite representations. \textbf{Long description.} All three panels show the same rectangular sky region. The target is near the center and is marked by a thick circle. Panel A contains several compact optical sources. Panel B shows the central target as a broader infrared source. Panel C overlays solid, dashed, and dotted elliptical contours around the central target. North points upward, east points right, and the scale bar represents ten arcseconds. The panel labels, contour line styles, target marker, orientation arrows, and scale bar provide information without relying only on color. \end{minipage} \end{figure*} % ------------------------------------------------- \subsubsection{Astronomical Light Curve} \begin{figure}[tbp] \centering \begin{tikzpicture} \begin{axis}[ width=0.98\columnwidth, height=0.68\columnwidth, xlabel={Time (\(\mathrm{h}\))}, ylabel={Relative flux}, xmin=0, xmax=8, ymin=0.955, ymax=1.012, grid=major, legend style={ font=\scriptsize, draw=none, at={(0.03,0.03)}, anchor=south west }, tick label style={font=\scriptsize}, label style={font=\small}, yticklabel style={ /pgf/number format/fixed, /pgf/number format/precision=2 } ] \addplot[ black, dashed, thick ] coordinates { (0,1.000) (1,1.000) (2,0.998) (3,0.985) (4,0.965) (5,0.985) (6,0.998) (7,1.000) (8,1.000) }; \addlegendentry{Transit model} \addplot[ only marks, mark=*, mark size=1.7pt, black, error bars/.cd, y dir=both, y explicit ] table[ x=time, y=flux, y error=error, row sep=\\ ] { time flux error\\ 0 1.002 0.004\\ 1 0.999 0.004\\ 2 0.997 0.004\\ 3 0.983 0.004\\ 4 0.964 0.005\\ 5 0.986 0.004\\ 6 0.999 0.004\\ 7 1.001 0.004\\ 8 1.000 0.004\\ }; \addlegendentry{Observed flux} \end{axis} \end{tikzpicture} \caption{Example transit light curve. Filled circles with error bars represent observed flux; the dashed line represents the fitted transit model.} \label{fig:light-curve} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} Relative flux temporarily decreases near four hours. \textbf{Long description.} Relative flux remains near \(1.00\) before decreasing to approximately \(0.964\) at four hours. It then returns to approximately \(1.00\). This symmetric decrease represents a possible transit. Vertical error bars show measurement uncertainty. \end{minipage} \end{figure} \begin{table}[tbp] \centering \caption{Representative light-curve data.} \label{tab:light-curve-data} \small \begin{tabular}{@{}rrr@{}} \toprule Time (\(\mathrm{h}\)) & Relative flux & Uncertainty\\ \midrule 0 & 1.002 & 0.004\\ 1 & 0.999 & 0.004\\ 2 & 0.997 & 0.004\\ 3 & 0.983 & 0.004\\ 4 & 0.964 & 0.005\\ 5 & 0.986 & 0.004\\ 6 & 0.999 & 0.004\\ 7 & 1.001 & 0.004\\ 8 & 1.000 & 0.004\\ \bottomrule \end{tabular} \end{table} % ------------------------------------------------- \subsubsection{Stellar Spectrum} \begin{figure}[tbp] \centering \begin{tikzpicture} \begin{axis}[ width=0.98\columnwidth, height=0.68\columnwidth, xlabel={Wavelength (\(\text{\AA}\))}, ylabel={Normalized flux}, xmin=6400, xmax=6700, ymin=0.35, ymax=1.65, grid=major, tick label style={font=\scriptsize}, label style={font=\small}, scaled x ticks=false ] \addplot[ black, thick ] coordinates { (6400,1.02) (6440,1.00) (6470,0.97) (6495,0.52) (6520,0.98) (6545,1.00) (6563,1.52) (6580,1.02) (6620,0.99) (6660,1.01) (6700,1.00) }; \draw[dashed] (axis cs:6495,0.35) -- (axis cs:6495,0.75); \node[ font=\scriptsize, anchor=south ] at (axis cs:6495,0.75) {absorption}; \draw[dotted] (axis cs:6563,1.20) -- (axis cs:6563,1.62); \node[ font=\scriptsize, anchor=south ] at (axis cs:6563,1.53) {emission}; \end{axis} \end{tikzpicture} \caption{Example stellar spectrum containing an absorption feature near \(6495\) angstroms and an emission feature near \(6563\) angstroms.} \label{fig:stellar-spectrum} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} The spectrum contains one strong absorption feature and one strong emission feature. \textbf{Long description.} Normalized flux is close to \(1.0\) across most wavelengths. It decreases to approximately \(0.52\) near \(6495\) angstroms, producing an absorption line. It increases to approximately \(1.52\) near \(6563\) angstroms, producing an emission line. \end{minipage} \end{figure} \begin{table}[tbp] \centering \caption{Representative spectrum values.} \label{tab:spectrum-data} \small \begin{tabular}{@{}rrl@{}} \toprule Wavelength (\(\text{\AA}\)) & Normalized flux & Interpretation\\ \midrule 6440 & 1.00 & Continuum\\ 6470 & 0.97 & Continuum\\ 6495 & 0.52 & Absorption\\ 6520 & 0.98 & Continuum\\ 6563 & 1.52 & Emission\\ 6580 & 1.02 & Continuum\\ 6620 & 0.99 & Continuum\\ \bottomrule \end{tabular} \end{table} % ------------------------------------------------- \subsubsection{Hertzsprung--Russell Diagram} \begin{figure}[tbp] \centering \begin{tikzpicture} \begin{axis}[ width=0.98\columnwidth, height=0.74\columnwidth, xlabel={Temperature (\(\mathrm{K}\))}, ylabel={Absolute magnitude}, xmin=3500, xmax=14000, ymin=-5, ymax=12, x dir=reverse, y dir=reverse, grid=major, tick label style={font=\scriptsize}, label style={font=\small}, legend style={ font=\scriptsize, draw=none, at={(0.02,0.02)}, anchor=south west } ] \addplot[ only marks, mark=*, mark size=1.8pt, black ] coordinates { (12000,-2) (10000,0) (8000,2) (6500,4) (5500,5.5) (4500,8) (3800,11) }; \addlegendentry{Main sequence} \addplot[ only marks, mark=square*, mark size=2pt, black ] coordinates { (5200,-1.0) (4600,-2.2) (4000,-3.0) }; \addlegendentry{Giants} \addplot[ only marks, mark=triangle*, mark size=2.2pt, black ] coordinates { (12500,9.5) (11000,10.5) (9500,11.5) }; \addlegendentry{White dwarfs} \end{axis} \end{tikzpicture} \caption{Controlled Hertzsprung--Russell diagram. Circles represent main-sequence stars, squares represent giants, and triangles represent white dwarfs.} \label{fig:hr-diagram} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} Three stellar groups occupy different regions of a temperature--magnitude diagram. \textbf{Long description.} Temperature decreases from left to right. Absolute magnitude is also displayed in reverse order, with brighter negative magnitudes near the top. Main-sequence stars form a diagonal pattern from hot and bright at upper left to cool and faint at lower right. Giants appear near the upper-right region. White dwarfs appear near the lower-left region. Marker shape, rather than color alone, identifies each group. \end{minipage} \end{figure} % ------------------------------------------------- \subsubsection{Astronomical Source Catalog} \begin{table*}[tbp] \centering \caption{Example astronomical source catalog containing uncertainties, limits, and unavailable values.} \label{tab:astronomy-catalog} \small \begin{tabular}{@{}lcccccl@{}} \toprule Source & Right ascension & Declination & \(g\) magnitude & Flux (\(\mu\mathrm{Jy}\)) & Redshift & Note\\ \midrule SRC-001 & 12:34:56.7 & \(-45\):18:22 & \(18.42\pm0.03\) & \(3.21\pm0.14\) & 0.125 & Detected\\ SRC-002 & 12:35:04.1 & \(-45\):17:49 & \(>22.10\) & \(<0.18\) & \(\ldots\) & Upper limit\\ SRC-003 & 12:35:15.8 & \(-45\):19:03 & \(19.07\pm0.05\) & \(2.48\pm0.17\) & 0.138 & Detected\\ SRC-004 & 12:35:22.4 & \(+02\):14:08 & \(20.31\pm0.08\) & \(0.91\pm0.12\) & 0.207 & Low signal\\ \bottomrule \end{tabular} \begin{minipage}{0.96\textwidth} \small \textit{Table note.} A greater-than sign in the magnitude column indicates that the source is fainter than the stated limiting magnitude. A less-than sign in the flux column indicates an upper limit. An ellipsis indicates that no redshift is available; it does not mean zero. \end{minipage} \end{table*} % ================================================= \section{Engineering} \subsection{Electrical Circuit and Kirchhoff's Current Law} Consider a voltage divider with source voltage \(V_s\), resistors \(R_1\) and \(R_2\), and output voltage \(V_o\). \begin{figure}[tbp] \centering \begin{tikzpicture}[ component/.style={ draw, minimum width=1.05cm, minimum height=0.48cm, font=\small }, flow/.style={ -{Stealth[length=2mm]}, thick }, font=\small ] \node[draw,circle,minimum size=0.85cm] (source) at (0,1.35) {}; \node at (-0.13,1.55) {\scriptsize \(+\)}; \node at (-0.13,1.15) {\scriptsize \(-\)}; \node[left=2mm of source] {\(V_s\)}; \node[component] (r1) at (1.65,2.55) {\(R_1\)}; \node[component] (r2) at (3.30,1.25) {\(R_2\)}; \draw[thick] (source.north) -- (0,2.55) -- (r1.west); \draw[thick] (r1.east) -- (3.30,2.55) -- (r2.north); \draw[thick] (r2.south) -- (3.30,0.15) -- (0,0.15) -- (source.south); \fill (3.30,2.55) circle (1.8pt); \node[right] at (3.30,2.55) {\(V_o\)}; \draw[flow] (0.65,2.90) -- (2.15,2.90); \node[above] at (1.40,2.90) {\(i\)}; \draw[thick] (1.65,0.15) -- (1.65,-0.05); \draw[thick] (1.30,-0.05) -- (2.00,-0.05); \draw[thick] (1.40,-0.18) -- (1.90,-0.18); \draw[thick] (1.50,-0.31) -- (1.80,-0.31); \node[below] at (1.65,-0.31) {ground}; \end{tikzpicture} \caption{Voltage-divider circuit. Current \(i\) flows from the source through \(R_1\). The output node connects \(R_1\) to \(R_2\), and \(R_2\) connects the output node to ground.} \label{fig:voltage-divider} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} A voltage source and two series resistors form a voltage divider. \textbf{Long description.} The positive terminal of source \(V_s\) connects to \(R_1\). The other terminal of \(R_1\) is the output node \(V_o\). Resistor \(R_2\) connects the output node to ground. The negative terminal of the source also connects to ground. Current \(i\) flows from the source through \(R_1\) toward the output node. \end{minipage} \end{figure} Kirchhoff's current law at the output node gives \begin{align} \frac{V_s-V_o}{R_1} &= \frac{V_o}{R_2},\\ R_2(V_s-V_o) &= R_1V_o,\\ R_2V_s-R_2V_o &= R_1V_o,\\ R_2V_s &= (R_1+R_2)V_o,\\ V_o &= \frac{R_2}{R_1+R_2}V_s. \end{align} For \(V_s=12\,\mathrm{V}\), \(R_1=2\,\mathrm{k\Omega}\), and \(R_2=1\,\mathrm{k\Omega}\), \begin{align} V_o &= \frac{1}{2+1}(12)\\ &= 4\,\mathrm{V}. \end{align} \begin{table}[tbp] \centering \caption{Voltage-divider component connections.} \label{tab:voltage-divider-connections} \small \begin{tabular}{@{}llll@{}} \toprule Component & From & To & Value\\ \midrule Source & Ground & Input node & \(12\,\mathrm{V}\)\\ \(R_1\) & Input node & Output node & \(2\,\mathrm{k\Omega}\)\\ \(R_2\) & Output node & Ground & \(1\,\mathrm{k\Omega}\)\\ \bottomrule \end{tabular} \end{table} % ================================================= \subsection{State-Space and Feedback-Control Models} A continuous-time linear system may be written as \begin{align} \dot{\mathbf{x}}(t) &= A\mathbf{x}(t)+B\mathbf{u}(t),\\ \mathbf{y}(t) &= C\mathbf{x}(t)+D\mathbf{u}(t), \end{align} where \begin{align} A&= \begin{bmatrix} 0 & 1\\ -2 & -3 \end{bmatrix}, & B&= \begin{bmatrix} 0\\ 1 \end{bmatrix},\\ C&= \begin{bmatrix} 1 & 0 \end{bmatrix}, & D&=0. \end{align} The transfer function can be derived as \begin{align} G(s) &= C(sI-A)^{-1}B+D,\\ sI-A &= \begin{bmatrix} s & -1\\ 2 & s+3 \end{bmatrix},\\ (sI-A)^{-1} &= \frac{1}{s^{2}+3s+2} \begin{bmatrix} s+3 & 1\\ -2 & s \end{bmatrix},\\ G(s) &= \frac{1}{s^{2}+3s+2}. \end{align} A closed-loop control system is shown in Figure~\ref{fig:closed-loop-control}. \begin{figure}[tbp] \centering \begin{tikzpicture}[ block/.style={ draw, minimum width=1.05cm, minimum height=0.58cm, align=center, font=\scriptsize }, sum/.style={ draw, circle, minimum size=0.52cm, inner sep=0pt }, flow/.style={ -{Stealth[length=2mm]}, thick }, font=\scriptsize ] \node[sum] (sum) at (0,0) {\(\sum\)}; \node[block] (controller) at (1.45,0) {Controller\\\(K(s)\)}; \node[block] (plant) at (3.05,0) {Plant\\\(G(s)\)}; \node[block] (sensor) at (2.25,-1.25) {Sensor\\\(H(s)\)}; \draw[flow] (-1.05,0) -- (sum.west) node[midway,above] {\(R(s)\)}; \draw[flow] (sum.east) -- (controller.west) node[midway,above] {\(E(s)\)}; \draw[flow] (controller.east) -- (plant.west) node[midway,above] {\(U(s)\)}; \draw[flow] (plant.east) -- (4.25,0) node[midway,above] {\(Y(s)\)}; \fill (4.05,0) circle (1.6pt); \draw[flow] (4.05,0) -- (4.05,-1.25) -- (sensor.east); \draw[flow] (sensor.west) -- (0,-1.25) -- (sum.south); \node at (-0.18,0.20) {\(+\)}; \node at (-0.18,-0.22) {\(-\)}; \node[below] at (3.45,-1.25) {feedback}; \end{tikzpicture} \caption{Closed-loop control system with negative feedback.} \label{fig:closed-loop-control} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} A controller and plant form a forward path with sensor-based negative feedback. \textbf{Long description.} The reference \(R(s)\) enters the positive input of a summing junction. The error \(E(s)\) passes through controller \(K(s)\), producing control signal \(U(s)\). The plant \(G(s)\) produces output \(Y(s)\). The output also passes through sensor \(H(s)\) and returns to the negative input of the summing junction. \end{minipage} \end{figure} The closed-loop transfer function is obtained from \begin{align} E(s) &= R(s)-H(s)Y(s),\\ Y(s) &= G(s)K(s)E(s),\\ Y(s) &= G(s)K(s) \left[ R(s)-H(s)Y(s) \right],\\ \left[ 1+G(s)K(s)H(s) \right]Y(s) &= G(s)K(s)R(s),\\ \frac{Y(s)}{R(s)} &= \frac{G(s)K(s)} {1+G(s)K(s)H(s)}. \end{align} An actuator magnitude limit \(u_{\max}>0\) may be represented by \[ \operatorname{sat}(v) = \begin{cases} -u_{\max}, & v<-u_{\max},\\ v, & -u_{\max}\leq v\leq u_{\max},\\ u_{\max}, & v>u_{\max}. \end{cases} \] The implemented control input is \[ u(t) = \operatorname{sat} \left( -K\mathbf{x}(t) \right). \] % ================================================= \subsection{Bode Magnitude and Phase Response} For the first-order system \[ G(s)=\frac{1}{s+1}, \] the frequency response is \begin{align} \left|G(j\omega)\right| &= \frac{1}{\sqrt{1+\omega^{2}}},\\ M(\omega) &= 20\log_{10} \left|G(j\omega)\right|,\\ \phi(\omega) &= -\tan^{-1}(\omega). \end{align} \begin{figure}[tbp] \centering \begin{tikzpicture} \begin{axis}[ width=0.97\columnwidth, height=0.52\columnwidth, xmode=log, xlabel={Angular frequency \(\omega\) (\(\mathrm{rad\,s^{-1}}\))}, ylabel={Magnitude (\(\mathrm{dB}\))}, xmin=0.1, xmax=10, ymin=-22, ymax=2, grid=major, tick label style={font=\scriptsize}, label style={font=\scriptsize} ] \addplot[ black, thick, mark=* ] coordinates { (0.1,-0.04) (0.2,-0.17) (0.5,-0.97) (1.0,-3.01) (2.0,-6.99) (5.0,-14.15) (10.0,-20.04) }; \draw[dashed] (axis cs:1,-22) -- (axis cs:1,2); \node[ font=\scriptsize, anchor=south west ] at (axis cs:1,-18) {\(\omega_c=1\)}; \end{axis} \end{tikzpicture} \vspace{2mm} \begin{tikzpicture} \begin{axis}[ width=0.97\columnwidth, height=0.52\columnwidth, xmode=log, xlabel={Angular frequency \(\omega\) (\(\mathrm{rad\,s^{-1}}\))}, ylabel={Phase (\(^{\circ}\))}, xmin=0.1, xmax=10, ymin=-95, ymax=5, ytick={-90,-60,-45,-30,0}, grid=major, tick label style={font=\scriptsize}, label style={font=\scriptsize} ] \addplot[ black, thick, mark=square* ] coordinates { (0.1,-5.7) (0.2,-11.3) (0.5,-26.6) (1.0,-45.0) (2.0,-63.4) (5.0,-78.7) (10.0,-84.3) }; \draw[dashed] (axis cs:1,-95) -- (axis cs:1,5); \end{axis} \end{tikzpicture} \caption{Bode response of \(G(s)=1/(s+1)\). The upper panel shows magnitude in decibels, and the lower panel shows phase in degrees. Both panels use a logarithmic frequency axis.} \label{fig:bode-response} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} Magnitude decreases and phase approaches negative \(90\) degrees as frequency increases. \textbf{Long description.} Frequency increases logarithmically from \(0.1\) to \(10\) radians per second. Magnitude begins near zero decibels, equals approximately negative \(3.01\) decibels at one radian per second, and decreases to approximately negative \(20.04\) decibels. Phase decreases from approximately negative \(5.7\) degrees to negative \(84.3\) degrees. The cutoff frequency is one radian per second. \end{minipage} \end{figure} \begin{table}[tbp] \centering \caption{Numerical values represented in the Bode plots.} \label{tab:bode-data} \small \begin{tabular}{@{}rrr@{}} \toprule \(\omega\) & Magnitude (\(\mathrm{dB}\)) & Phase (\(^{\circ}\))\\ \midrule 0.1 & \(-0.04\) & \(-5.7\)\\ 0.2 & \(-0.17\) & \(-11.3\)\\ 0.5 & \(-0.97\) & \(-26.6\)\\ 1.0 & \(-3.01\) & \(-45.0\)\\ 2.0 & \(-6.99\) & \(-63.4\)\\ 5.0 & \(-14.15\) & \(-78.7\)\\ 10.0 & \(-20.04\) & \(-84.3\)\\ \bottomrule \end{tabular} \end{table} % ================================================= \subsection{Beam Loading and Free-Body Diagram} Consider a simply supported beam of length \(L\) with a downward point load \(P\) at its midpoint. \begin{figure}[tbp] \centering \begin{tikzpicture}[ force/.style={ -{Stealth[length=2.5mm]}, very thick }, dimension/.style={ {Stealth[length=1.8mm]}-{Stealth[length=1.8mm]}, thin }, font=\small ] \draw[very thick] (0,1.15) -- (5,1.15); % Pin support \draw[thick] (0,1.10) -- (-0.32,0.55) -- (0.32,0.55) -- cycle; \draw[thick] (-0.45,0.48) -- (0.45,0.48); % Roller support \draw[thick] (5,1.10) -- (4.68,0.62) -- (5.32,0.62) -- cycle; \draw[thick] (4.82,0.50) circle (0.10); \draw[thick] (5.18,0.50) circle (0.10); \draw[thick] (4.60,0.35) -- (5.40,0.35); % Applied load \draw[force] (2.5,2.20) -- (2.5,1.25); \node[above] at (2.5,2.20) {\(P\)}; % Reactions \draw[force] (0,0.10) -- (0,1.00); \node[left] at (0,0.55) {\(R_A\)}; \draw[force] (5,0.10) -- (5,1.00); \node[right] at (5,0.55) {\(R_B\)}; % Coordinate \draw[-{Stealth[length=2mm]}] (0,1.45) -- (1.15,1.45); \node[above] at (1.15,1.45) {\(x\)}; % Dimensions \draw[dimension] (0,-0.25) -- (2.5,-0.25); \draw[dimension] (2.5,-0.25) -- (5,-0.25); \node[below] at (1.25,-0.25) {\(L/2\)}; \node[below] at (3.75,-0.25) {\(L/2\)}; \draw[thin] (0,0.05) -- (0,-0.40); \draw[thin] (2.5,1.05) -- (2.5,-0.40); \draw[thin] (5,0.05) -- (5,-0.40); \node[below] at (0,-0.48) {support A}; \node[below] at (5,-0.48) {support B}; \end{tikzpicture} \caption{Free-body diagram of a simply supported beam with a midpoint load.} \label{fig:beam-free-body} \begin{minipage}{0.96\columnwidth} \small \textbf{Short text alternative.} A midpoint load acts downward on a beam with upward reactions at both ends. \textbf{Long description.} A horizontal beam extends from pin support A at \(x=0\) to roller support B at \(x=L\). A downward force \(P\) acts at \(x=L/2\). Reactions \(R_A\) and \(R_B\) act upward at the two supports. The positive \(x\)-direction points from support A toward support B. \end{minipage} \end{figure} Force and moment equilibrium give \begin{align} \sum F_y &= R_A+R_B-P=0,\\ \sum M_A &= R_BL-P\frac{L}{2}=0,\\ R_A &= R_B=\frac{P}{2}. \end{align} The internal shear force is \[ V(x) = \begin{cases} \dfrac{P}{2}, & 0