Document 4 - Basics in beamer (13)

\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[english, greek]{babel}

\title{Document 4 - Basics in Beamer}
\author{Thomas Boufikos}
\date{05/10/2021}
\usetheme{Boadilla}
\usecolortheme{whale}

\begin{document}

    \selectlanguage{english}
    \begin{frame}{Telecommunication Channel}
        \[
            v(t) = \sum _{n=-\infty} ^ {\infty} a_ng_T(t-nT) \\
        \]
         $
            h(t) = c(t) \textasteriskcentered  g_T(t) \\
            x(t) = h(t) \textasteriskcentered  g_R(t) \\
            y(t) = \sum _{n=-\infty} ^ {\infty} a_nx(t-nT) + v(t) \\
            y(mT+t_0) = a_mx(t_0) + \sum _{n \neq m} a_nx((m-n)T + t_0) + v_m \\
            v_m = v(mT + t_0) \\
        $
         \\ \\
        \selectlanguage{greek}
        Όλα τα πάνω ισχύουν, εφόσον η δειγματοληψία γίνει τις χρονικές στιγμές
        $ mT + T_0 $, διαφορετικά μιλάμε για μία τυχαία στιγμή $ t $.
    \end{frame}


    \selectlanguage{english}
    \begin{frame}{Telecommunication Channel - Continue}
        \selectlanguage{greek}
        Αν θέλουμε να στείλουμε τα $ a_n $, στέλνουμε τα $ b_n $, σύμφωνα με τη σχέση: $ b_n = 2a_n - 1$, επομένως τα μπιτ πληροφορίας $ a_n $ γίνονται πλάτη $ b_n $. Για παράδειγμα: \\ \\
        \phantom{ssssssss} Πληροφορία:
        \begin{tabular}{|c|c|c|c|c|c|c|c|}
            \hline
             a_0 &  a_1 &  a_2 &  a_3 &  a_4 &  a_5 &  a_6 &  a_7 \\
             \hline
             1 & 0 & 0 & 0 & 1 & 1 & 0 & 1 \\
             \hline
        \end{tabular}

        Μεταδιδόμενη ακολουθία:
        \begin{tabular}{|c|c|c|c|c|c|c|c|}
            \hline
             b_0 &  b_1 &  b_2 &  b_3 &  b_4 &  b_5 &  b_6 &  b_7 \\
             \hline
             1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\
             \hline
        \end{tabular}

    \end{frame}


    \begin{frame}{Basic Math Stuff}
        Here is some text followed by an equation: \\
        \[
            p(x) = ax^2 + bx + c = 0
        \] \\
        and the polynomial roots are: \\
        \begin{align*}
            x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}
        \end{align*}, where:
        $ \Delta = b^2 - 4ac $
    \end{frame}


    \begin{frame}{Columns}
        This is the text above the 2 columns. I will leave a vertical space of 5mm and then
        I will write text in the 2 columns.
        \vspace{5mm}

        \begin{columns}
            \column{0.48 \linewidth}
            This is the 1st column. Here is some additional text to ensure that there is enough space to fill out the width of slide. Placement: left!
            \column{0.48 \linewidth}
            This is the 2nd column. Here is some additional text to ensure that there is enough space to fill out the width of slide. Placement: right! \\
            I will write down more text to enlight the difference.
        \end{columns}

        \vspace{5mm}
        That's all about the columns. End of text!
    \end{frame}

    \begin{frame}{Pausing text}
        Here is some text. \\
        \pause
        Here is some more text. \\
        \pause
        More, more, more text! \\
        \pause
        Next step is the final one. \\
        \pause
        Final text. \\
    \end{frame}


    \begin{frame}{Specification Overlays}
        \begin{itemize}
            \item<1->  This appears on all frames. \\
            \item<2>   This appears only on frame 2. \\
            \item<2,4> This appears on frames 2 and 4. \\
            \item<1-3> This appears on frames 1-3. \\
            \item<2->  This appears on frames 2-4. \\
            \item<6> This appears only on last frame (6th). \\
        \end{itemize}
    \end{frame}

    \begin{frame}{Specification Overlays}
        Frame \only<1>{1} \only<2>{2}
        \begin{itemize}
            \item This will be \textbf<2>{bold} text.\\
            \item This will be \textit<2>{italic} text.\\
            \item This will be \textcolor<2>{blue}{blue} text.\\
            \item This will be \textcolor<2>{pink}{pink} text.\\
        \end{itemize}
    \end{frame}

    \begin{frame}{Blocks}
        Blocks are like theorem environments.
        \begin{block}{Block 1}
            This is block 1. \\
        \end{block}
        \begin{block}{Block 2}
            This is block 2. \\
            It is under block 1 of course. \\
        \end{block}

        \newtheorem{thm}{Thomas Theorem}
        \begin{thm}[Pythagorean Theorem]
            In a triangle with angles $ a = 90, b < 90, c < 90 $, if C, A, B are the opposite sides of these angles, there must be: \\
            \[
                C^ 2 = A^2 + B^2
            \]
        \end{thm}
        \begin{proof}
            There is no supported proof right now!
        \end{proof}
    \end{frame}




\end{document}