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 \documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{graphicx} \usepackage{amsmath} \title{Homework 3} \begin{document} \maketitle \section*{Q1} Given the logical formula, we can follow the following process to convert it into strictly inverters, NOR, and NAND gates:   \begin{equation*}  \begin{aligned}  & \lnot((AB+C)D+E) \\  \Leftrightarrow & \lnot(\lnot\lnot(AB+C)D+E) & \text{(negation involutive)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(AB+C)+\lnot D)+E) & \text{(DeMorgan's Laws)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(\lnot\lnot AB+C)+\lnot D)+E) & \text{(negation involutive)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(\lnot(\lnot A + \lnot B)+C)+\lnot D)+E) & \text{(DeMorgan's Laws)} \\  \end{aligned} \end{equation*}   This corresponds to the following circuit:   \begin{figure}[h]  \centering  \includegraphics[width=0.7\linewidth]{Q1HW3.png} \end{figure}   \pagebreak \section*{Q2} Making Scott's adjustment (adding a top-level 'not' to the formula in the assignment) yields the following:   \begin{figure}[h]  \centering  \includegraphics[width=0.7\linewidth]{Q2.png} \end{figure}   \pagebreak \section*{Q3} The book gives the following equation for determing the ideal number of stages:   \begin{equation*}  \begin{aligned}  N &= \log_{\rho}F \\  0 &= p_\text{inv} + \rho(1-\ln\rho)  \end{aligned} \end{equation*}   Where $p_\text{inv}$ is the intrinsic delay of an inverter. For $p_\text{inv} = 5$, we have $\rho = 6.14$. We then compute $F$:   \begin{equation*}  \begin{aligned}  & F &= GBH \\  & G &= 1 \\  & B &= 1 \\  & H &= 1000 \\  \Rightarrow & F &= 1000  \end{aligned} \end{equation*}   The ideal number of stages is then:   \begin{equation*}  \log_\rho F = 3.8 \approx 4 \end{equation*}   For all inverters, then, we get the following:   \begin{equation*}  \begin{aligned}  \hat{f} &= \sqrt[4]{1000} \\  \text{sz}_4 &= 1000/\hat{f}^1 = 178 \\  \text{sz}_3 &= 1000/\hat{f}^2 = 31.6 \\  \text{sz}_2 &= 1000/\hat{f}^3 = 5.62 \\  \text{sz}_1 &= 1000/\hat{f}^4 = 1 \\  \end{aligned} \end{equation*}   \pagebreak \section*{Q4} First, to compute total effort $F$.   \begin{equation*}  \begin{aligned}  & F &= GBH \\  & G &= \left(\frac{4}{3}\right)\left(\frac{5}{3}\right)\left(\frac{5}{3}\right) \\  & B &= 3 \\  & H &= 1000 \\  \Rightarrow & F &= 11111  \end{aligned} \end{equation*}   Since we are using the same inverter as the one in Q3, we are once again using $p_\text{inv}=5$, and thus, have that $\rho=6.138$. From this, we can determine the ideal number of stages:   \begin{equation*}  \log_\rho F = 5.13 \approx 5 \end{equation*}   Since we currently have 3 stages, we should insert 2 inverters. I will insert these at the end of the path in question. From there, we once again compute stage effort $\hat{f}$, and work backwards to determine the optimal sizes for all of the stages.   \begin{equation*}  \begin{aligned}  \hat{f} &= \sqrt[5]{11111} \\  \text{sz}_\text{inv1} &= 1000/\hat{f}^1 = 155 \\  \text{sz}_\text{inv2} &= 1000/\hat{f}^2 = 24.1 \\  \text{sz}_\text{nand3} &= 1000/\hat{f}^3 * \left(\frac{5}{3}\right) = 6.23 \\  \text{sz}_\text{nor2} &= 1000/\hat{f}^4 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right) = 1.61 \\  \text{sz}_\text{nand2} &= 1000/\hat{f}^5 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)\left(\frac{4}{3}\right)3 = 1  \end{aligned} \end{equation*}   \end{document}