From b01f69fec101e78cdb168994952c4aae46f361bf Mon Sep 17 00:00:00 2001 From: Claudio Maggioni Date: Fri, 19 Oct 2018 16:05:36 +0200 Subject: [PATCH] Added work up to now --- Homework 5.tex | 185 +++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 185 insertions(+) create mode 100644 Homework 5.tex diff --git a/Homework 5.tex b/Homework 5.tex new file mode 100644 index 0000000..4dd4d0d --- /dev/null +++ b/Homework 5.tex @@ -0,0 +1,185 @@ +\documentclass[12pt]{article} + +\usepackage{karnaugh-map} +\usepackage[utf8]{inputenc} +\usepackage[margin=2cm]{geometry} + +\title{Howework 5 -- Computer Architecture} +\author{Claudio Maggioni \and Tommaso Rodolfo Masera} + +\newcommand{\bn}[1]{ + \overline{#1} +} + +\begin{document} +\maketitle +\section{Question 1} +\subsection{\label{sec:11}Sub-question 1} +\paragraph{Starting with:} +$f(A, B, C, D) = \bn{A} \times \bn{B} \times \bn{C} \times \bn{D} + +\bn{A} \times \bn{B} \times C \times \bn{D} + \bn{A} \times B \times C \times \bn{D} + +A \times \bn{B} \times \bn{C} \times \bn{D} + A \times \bn{B} \times \bn{C} \times D + +A \times \bn{B} \times C \times \bn{D} + A \times B \times C \times D + +\bn{A} \times \bn{B} \times \bn{C} \times D + \bn{A} \times B \times C \times D =$ + +\paragraph{apply the distributive law:} +$\bn{A} \times (\bn{B} \times \bn{C} \times \bn{D} + \bn{B} \times C \times \bn{D} + +B \times C \times \bn{D} + \bn{B} \times \bn{C} \times D + B \times C \times D) + +A \times (\bn{B} \times \bn{C} \times \bn{D} + \bn{B} \times \bn{C} \times D + +\bn{B} \times C \times \bn{D} + B \times C \times D) =$ + +\paragraph{apply it again:} +$\bn{A} \times (\bn{B} \times (\bn{C} \times \bn{D} + C \times \bn{D} + \bn{C} \times D) ++ B \times C \times (\bn{D} + D)) + A \times (\bn{B} \times (\bn{C} \times \bn{D} + \bn{C} \times D + +C \times \bn{D}) + B \times C \times D) = $ + + +\subsubsection{\label{sec:cd}Solving $\bn{C} \times \bn{D} + C \times \bn{D} + \bn{C} \times D$} + +\paragraph{Apply the distributive law:} +$\bn{C} \times (\bn{D} + D) + C \times \bn{D} = $i + +\paragraph{then apply the inverse law:} +$\bn{C} \times 1 + C \times \bn{D} =$ + +\paragraph{then apply the identity law:} +$\bn{C} + C \times \bn{D} =$ + +\paragraph{then apply De Morgan's law:} +$\bn{C \times \bn{C \times \bn{D}}} =$ + +\paragraph{then apply it again:} +$\bn{C \times (\bn{C} + D)} =$ + +\paragraph{then apply the distributive law:} +$\bn{C \times \bn{C} + C \times D} =$ + +\paragraph{then apply the inverse law:} +$\bn{0 + C \times D} =$ + +\paragraph{then, finally, apply the identity law, obtaining:} +$\bn{C \times D}$ + +\subsubsection{Back to the main function} +\paragraph{by \ref{sec:cd} and the inverse law, we continue this way:} +$\bn{A} \times (\bn{B} \times \bn{C \times D} + B \times C \times 1) + +A \times (\bn{B} \times \bn{C \times D}) + B \times C \times D) =$ + +\paragraph{then apply the identity law:} +$\bn{A} \times (\bn{B} \times \bn{C \times D} + B \times C) + +A \times (\bn{B} \times \bn{C \times D}) + B \times C \times D) =$ + +\paragraph{then apply the distributive law:} +$\bn{A} \times \bn{B} \times \bn{C \times D}) + \bn{A} \times B \times C + +A \times \bn{B} \times \bn{C \times D} + A \times B \times C \times D =$ + +\paragraph{then apply it again:} +$\bn{B} \times \bn{C \times D} \times (\bn{A} + A) + B \times C \times (\bn{A} + A \times D) =$ + +\paragraph{then apply the inverse law:} +$\bn{B} \times \bn{C \times D} \times 1 + B \times C \times (\bn{A} + A \times D) =$ + +\paragraph{then apply the identity law:} +$\bn{B} \times \bn{C \times D} + B \times C \times (\bn{A} + A \times D) =$ + +\paragraph{then apply De Morgan's law:} +$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{A \times D}} =$ + +\paragraph{then apply it again:} +$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times (\bn{A} + \bn{D})} =$ + +\paragraph{then apply the distributive law:} +$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{A} + A \times \bn{D}} =$ + +\paragraph{then apply the inverse law:} +$\bn{B} \times \bn{C \times D} + B \times C \times \bn{0 + A \times \bn{D}} =$ + +\paragraph{then apply the identity law:} +$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{D}} =$ + +\paragraph{then, finally, apply De Morgan's law, we find the result:} +$\bn{B} \times \bn{C \times D} + B \times C \times (\bn{A} + D)$ + +\subsection{Sub-question 2} +Using the minterm normal form given in section \ref{sec:11}, we can deduct the truth table of the function. Using that, we can find the following Karnaugh map and the following prime implicants: + +\begin{figure}[h] +\centering{ + \begin{karnaugh-map}[4][4][1][AB][CD] + \maxterms{1,3,5,7,11,12,14} + \minterms{0,2,4,6,8,9,10,13,15} + \implicant{1}{7} + \implicantedge{12}{12}{14}{14} + \implicantedge{3}{3}{11}{11} + \end{karnaugh-map} +} +\end{figure} + +Using those prime implicants we find the boolean function $(\bn{B} + C) \times (\bn{A} + \bn{B} ++ D) \times (B + \bn{C} + \bn{D})$. + +\section{Question 2} +\subsection{Sub-question 1} +The truth table of the function is: + +\begin{figure}[h] +\centering{ + \begin{tabular}{cccc|c} + \textbf{$X_3$} & \textbf{$X_2$} & \textbf{$X_1$} & \textbf{$X_0$} & \textbf{$Y$} \\ \hline + 0 & 0 & 0 & 0 & 0 \\ + 0 & 0 & 0 & 1 & 1 \\ + 0 & 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 1 & 0 \\ + 0 & 1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 1 & 0 \\ + 0 & 1 & 1 & 0 & 1 \\ + 0 & 1 & 1 & 1 & 1 \\ + 1 & 0 & 0 & 0 & 1 \\ + 1 & 0 & 0 & 1 & 1 \\ + 1 & 0 & 1 & 0 & 1 \\ + 1 & 0 & 1 & 1 & 1 \\ + 1 & 1 & 0 & 0 & 1 \\ + 1 & 1 & 0 & 1 & 1 \\ + 1 & 1 & 1 & 0 & 1 \\ + 1 & 1 & 1 & 1 & 1 \\ + \end{tabular} +} +\end{figure} + +\subsection{Sub-question 2} +The Conjunctive Normal Form, or the maxterm expansion of the function, is: + +\begin{figure}[h] +$(X_3 + X_2 + X_1 + X_0) \times +(X_3 + X_2 + \bn{X_1} + X_0) \times (X_3 + X_2 + \bn{X_1} + \bn{X_0}) \times +(X_3 + \bn{X_2} + X_1 + X_0) \times (X_3 + \bn{X_2} + X_1 + \bn{X_0})$ +\end{figure} + +The Disjunctive Normal Form, or the minterm expansion of the function, is: + +\begin{figure}[h] +$(\bn{X_3} \times \bn{X_2} \times \bn{X_1} \times X_0) + (\bn{X_3} \times X_2 \times X_1 \times \bn{X_0}) + +(\bn{X_3} \times X_2 \times X_1 \times X_0) + (X_3 \times \bn{X_2} \times \bn{X_1} \times \bn{X_0}) + +(X_3 \times \bn{X_2} \times \bn{X_1} \times X_0) + (X_3 \times \bn{X_2} \times X_1 \times \bn{X_0}) + +(X_3 \times \bn{X_2} \times X_1 \times X_0) + (X_3 \times X_2 \times \bn{X_1} \times \bn{X_0}) + +(X_3 \times X_2 \times \bn{X_1} \times X_0) + (X_3 \times X_2 \times X_1 \times \bn{X_0}) + +(X_3 \times X_2 \times X_1 \times X_0)$ +\end{figure} + +\subsection{Sub-question 3} +The maxterm expansion of the function above is clearly the best approach between the two, + since it contains fewer terms. + +\subsection{Sub-question 4} +TODO + +\subsection{Sub-question 5} +TODO + +\section{Question 3} +The bomb will not explode if either the first, the second or the fourth cable from the left +are cut. For the first and the second cable this happens because the NOR inside the circuit will +get at least a 1 as input, and therefore it will produce a 0 as output, pulling the final AND +output to 0. When the fourth cable is cut, the NOT will give always 0 as output and therefore +the final AND output will be always 0. +\end{document} \ No newline at end of file