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.~lock.Formula Sheet Controls 2020.docx#
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1stOrderUnreducedBlockDiagram.xcf
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Formula Sheet Controls 2020.docx
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PortfolioProjectRubric.docx
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diffEqReview/diffEqReview.tex
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diffEqReview/diffEqReview.tex
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% File: diffEqReview.tex
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% Created: 12:36:05 Fri, 27 Aug 2021 EDT
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% Last Change: 12:36:05 Fri, 27 Aug 2021 EDT
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%
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\documentclass[letterpaper]{article}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{cancel}
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\usepackage{amssymb}
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\usepackage{listings}
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\usepackage{enumitem}
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\date{08/27/2021}
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\title{%
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Differential Equations Overview\\
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\large EEET-427-01: Control Systems}
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\author{Blizzard MacDougall}
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\begin{document}
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\maketitle
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\pagenumbering{arabic}
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\section{General Steps}
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Given $F(x)=G(x)$
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\begin{enumerate}
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\item Solve for roots of function
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\item Using roots, fill in general solution of $F(x)$ (assuming $G(x)=0$). General solution options:
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\begin{itemize}
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\item Real roots ($c_1e^{r_1x}+c_2e^{r_2x}\dots$)
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\item repeated roots ($c_1e^{r_1x}+c_2xe^{r_1x}\dots$)
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\item imaginary roots ($e^{\alpha x}(c_1\cos(\beta x)+c_2\sin(\beta x))$)
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\end{itemize}
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\item Use initial conditions (if given) to find $c$ values.
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\item Generalize $G(x)$, take necessary derivatives, and plug in to $F(x)$, solve for coefficients
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\end{enumerate}
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\section{Example}
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\begin{equation}
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\begin{split}
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y''+5y'+6y=x^2\\
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y''+5y'+6y=0\\
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r^2+5r+6=0\\
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r=-2,\ -3\\
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y_g(x)=c_1e^{-2x}+c_2e^{-3x}\\
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G(x)=x^2\\
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y_p(x)=Ax^2+Bx+C\\
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y'(x)=2Ax+B\\
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y''(x)=2A\\
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y''+5y'+6y=x^2\\
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2A+5(2Ax+B)+6(Ax^2+Bx+C)=x^2\\
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2A+10Ax+5B+6Ax^2+6Bx+6C=x^2\\
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6Ax^2+(10A+6B)x+(2A+5B+6C)=x^2+0x+0\\
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6A=1\\
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10A+6B=0\\
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2A+5B+6C=0\\
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A=\frac16\\
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B=-\frac{10}{36}\\
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C=\frac{38}{216}\\
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y_p(x)=\frac16x^2-\frac{10}{36}x+\frac{38}{216}\\
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y(x)=\frac16x^2-\frac{10}{36}x+\frac{38}{216}+c_1e^{-2x}+c_2e^{-3x}
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\end{split}
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\end{equation}
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\end{document}
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\@writefile{toc}{\contentsline {section}{\numberline {1}Core Elements of Controls Systems}{1}{}\protected@file@percent }
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\title{%
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Final Review\\
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\large EEET--427--01: Controls Systems}
|
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\author{Blizzard MacDougall}
|
||||
\begin{document}
|
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\maketitle
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\pagenumbering{arabic}
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\section{Core Elements of Controls Systems}
|
||||
\begin{itemize}
|
||||
\item Making Models\\
|
||||
A visual or mathematical model that represents the control system. This shows how the system operates and what variables can be manipulated to tune the system.
|
||||
\item Trajectory Generation\\
|
||||
Using the model to predict how a system will react.
|
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This is all about finding the right input for the right output.
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Another way of wording it is putting bounds on the input to ensure that you don't attempt to overwork the system.
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\item Disturbance Decoupling\\
|
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Accounting for unwanted signals in a system such as drag or friction, by either feedback or feedforward.
|
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\item Feedforward\\
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||||
A signal using information what we already have from a model of the system to give the controller a more stable starting point.
|
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\item Feedback\\
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Using the output of a system as part of its model to obtain a more accurate response, introducing stability.
|
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\item Lead Compensation\\
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||||
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\begin{equation}
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\item Integral Compensation\\
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||||
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\item Lag Compensation\\
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\end{itemize}
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\section{First Order Systems}
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Homework 1\\
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\large EEET-427-01:Controls Systems}
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\author{Blizzard MacDougall}
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% File: homework1b.tex
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% Created: 19:03:39 Mon, 27 Sep 2021 EDT
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% Last Change: 19:03:39 Mon, 27 Sep 2021 EDT
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%
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\documentclass[letterpaper]{article}
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\date{09/27/2021}
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\title{%
|
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Homework 1b\\
|
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\large EEET-427-01:Controls Systems}
|
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\author{Blizzard MacDougall}
|
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\begin{document}
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\maketitle
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\pagenumbering{arabic}
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For each problem:\\
|
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\begin{enumerate}
|
||||
\item Find the differential equation that relates $V_{out}$ to $V_{in}$.
|
||||
\item Rewrite the differential equation in Laplace form.
|
||||
\item Make an unreduced block diagram.
|
||||
\item Reduce the block diagram into a single block.
|
||||
\item Find the transfer function of the system.
|
||||
\item Find the DC gain of the transfer function.
|
||||
\item Find the time constant of the circuit.
|
||||
\end{enumerate}
|
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\section{Problem 1}
|
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\begin{figure}[h!]
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\includegraphics{problem1.png}
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\end{figure}
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\begin{enumerate}[1)]
|
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\item \begin{equation}
|
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\begin{split}
|
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i_R=i_C\\
|
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\frac{v_R}{R}=C\frac{dv_C}{dt}\\
|
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v_R=v_i-v_o;\ v_C=v_o\\
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\frac{v_i-v_o}{R}=C\frac{dv_o}{dt}\\
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\frac1{RC}(v_i)=\frac{dv_o}{dt}+\frac1{RC}(v_o)
|
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\end{split}
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\end{equation}
|
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\item\begin{equation}
|
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\frac1{RC}{V_i}=sV_o+\frac1{RC}V_o
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\end{equation}
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\item.\\\begin{figure}[h!]
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\includegraphics{problem1UnreducedBlockDiagram.png}
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\caption{Problem 1 Unreduced Block Diagram}
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\end{figure}
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\item .\\
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\begin{figure}[h!]
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\includegraphics{problem1ReducedBlockDiagram.png}
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\caption{Problem 1 Reduced Block Diagram}
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\end{figure}
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\item\begin{equation}
|
||||
\frac{V_o}{V_i}=\frac{\frac1{RC}}{s+\frac1{RC}}
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\end{equation}
|
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\item DC gain = 1
|
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\item \begin{equation}
|
||||
\tau=RC
|
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\end{equation}
|
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\end{enumerate}
|
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\newpage
|
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\section{Problem 2}
|
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\begin{figure}[h!]
|
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\includegraphics{problem2.png}
|
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\end{figure}
|
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\begin{enumerate}[1)]
|
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\item \begin{equation}
|
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\begin{split}
|
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i_R=\frac{v_o}{R}=i_L\\
|
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v_L=L\frac{di_L}{dt}=v_i-v_o\\
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L\frac{d}{dt}(\frac{V_o}{R})=v_i-v_o\\
|
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v_i=v_o+\frac LRv_o\frac{d}{dt}\\
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\frac RLv_i=\frac RLv_o+\frac{dv_o}{dt}
|
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\end{split}
|
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\end{equation}
|
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\item\begin{equation}
|
||||
\frac RL{V_i}=sV_o+\frac R{L}V_o
|
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\end{equation}
|
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\item.\\\begin{figure}[h!]
|
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\includegraphics{problem2UnreducedBlockDiagram.png}
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\caption{Problem 2 Unreduced Block Diagram}
|
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\end{figure}
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\item \begin{equation}
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\begin{split}
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\end{equation}
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\newpage
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\includegraphics[height=5cm]{problem3.png}
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\newpage
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% File: homework2.tex
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% Created: 17:53:11 Sat, 02 Oct 2021 EDT
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% Last Change: 17:53:11 Sat, 02 Oct 2021 EDT
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%
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\documentclass[letterpaper]{article}
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||||
\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{cancel}
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\usepackage{amssymb}
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\usepackage{listings}
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\usepackage[shortlabels]{enumitem}
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\usepackage{lipsum}
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\usepackage{soul}
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\usepackage{geometry}
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\geometry{portrait, margin=1in}
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||||
\date{10/02/2021}
|
||||
\title{%
|
||||
Homework 2\\
|
||||
\large EEET-427-01: Controls Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
|
||||
\section{Question 1}
|
||||
Sketch the acceleration and position profiles associated with the given velocity.
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=5cm]{section1Accel.png}\\
|
||||
\includegraphics[height=5cm]{section1Vel.png}\\
|
||||
\includegraphics[height=5cm]{section1Pos.png}
|
||||
\end{figure}
|
||||
\newpage
|
||||
\section{Question 2}
|
||||
A force profile is applied that moves a sled from rest with a fixed acceleration of $1.0\frac{m}{s^2}$ to a plateau velocity of $2\frac ms$, and then back to rest with a fixed acceleration of $-0.5\frac{m}{s^2}$ such that the total distance traveled is $8m$.
|
||||
|
||||
Sketch the acceleration,velocity, and position profiles.
|
||||
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=5cm]{section2Accel.png}\\
|
||||
\includegraphics[height=5cm]{section2Vel.png}\\
|
||||
\includegraphics[height=5cm]{section2Pos.png}
|
||||
\end{figure}
|
||||
|
||||
\newpage
|
||||
\section{Question 3}
|
||||
Sketch the response of $\frac{dy}{dt}+by=ax$ given the inputs and initial conditions.
|
||||
|
||||
\begin{enumerate}[A.]
|
||||
\item $a=1.5$, $b=2$. Solve for $t=0.75$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Y=\frac{1.5}{s+2}X\\
|
||||
x(t)=u(t)\\
|
||||
X=\frac1s\\
|
||||
Y=\frac{1.5}{s(s+2)}\\
|
||||
\frac{1.5}{s(s+2)}=\frac As+\frac{B}{s+2}\\
|
||||
1.5=A(s+2)+Bs\\
|
||||
2A=1.5\\
|
||||
A=\frac34\\
|
||||
B=-A\\
|
||||
B=-\frac34\\
|
||||
y(t)=\frac34(u(t)-e^{-2t}u(t))\\
|
||||
y(0.75)=\frac34(1-e^{-2(\frac34)})\\
|
||||
y(0.75)\approx0.58
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=10cm]{section3a.png}
|
||||
\end{figure}
|
||||
|
||||
\item $a=4$, $b=3$; Solve for $t=0.25$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Y=\frac{4}{s+3}X\\
|
||||
x(t)=u(t)\\
|
||||
X=\frac1s\\
|
||||
Y=\frac{4}{s(s+3)}\\
|
||||
\frac{4}{s(s+3)}=\frac As+\frac{B}{s+3}\\
|
||||
4=A(s+3)+Bs\\
|
||||
3A=4\\
|
||||
A=\frac43\\
|
||||
B=-A\\
|
||||
B=-\frac43\\
|
||||
y(t)=\frac43(u(t)-e^{-3t}u(t))\\
|
||||
y(0.25)=\frac43(1-e^{-3(\frac14)})\\
|
||||
y(0.25)\approx0.7035
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=10cm]{section3b.png}
|
||||
\end{figure}
|
||||
|
||||
\item $a=3$, $b=5$; Solve for $t=0.25$, $t=0.75$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Y=\frac{3}{s+5}X\\
|
||||
x(t)=u(t)\\
|
||||
X=\frac1s\\
|
||||
Y=\frac{3}{s(s+5)}\\
|
||||
\frac{3}{s(s+5)}=\frac As+\frac{B}{s+5}\\
|
||||
3=A(s+5)+Bs\\
|
||||
5A=3\\
|
||||
A=\frac35\\
|
||||
B=-A\\
|
||||
B=-\frac35\\
|
||||
y(t)=\frac35(u(t)-e^{-5t}u(t))-\frac35(u(t-0.5)-e^{-5t}u(t-0.5))\\
|
||||
y(0.25)=\frac35(1-e^{-5(\frac14)})\\
|
||||
y(0.25)\approx0.7035\\
|
||||
y(0.75)=\frac35((1-1)-(e^{-5t}-e^{-5(t-0.5)}))\\
|
||||
y(0.75)=\frac35(e^{-5(t-0.5)}-e^{-5t})
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=9.75cm]{section3c.png}
|
||||
\end{figure}
|
||||
|
||||
\item $a=3$, $b=5$, Solve for $t=0.25$, $0.75$, given $y(0)=0$, solve $f(t)=0.5$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Y=\frac{3}{s+5}X\\
|
||||
x(t)=-2u(t)+3.5u(t-0.5)\\
|
||||
X_1=\frac{-2}s\\
|
||||
X_2=\frac{3.5}s\\
|
||||
Y_1=\frac{-6}{s(s+5)}\\
|
||||
-6=A_1(s+5)+B_1s\\
|
||||
5A=-6\\
|
||||
A_1=-\frac65\\
|
||||
B_1=-A_1\\
|
||||
B_1=\frac65\\
|
||||
Y_2=\frac{10.5}{s(s+5)}\\
|
||||
10.5=A_2(s+5)+B_2s\\
|
||||
5A_2=10.5\\
|
||||
A_2=\frac{21}{10}\\
|
||||
B_2=-A_2\\
|
||||
B_2=-\frac{21}{10}\\
|
||||
y(t)=\frac65(e^{-5t}u(t)-u(t))+\frac{21} {10} (u(t-0.5)-e^{-5(t-0.5)}u(t-0.5))\\
|
||||
y(0.25)=\frac65(1-e^{-5(0.25)})\\
|
||||
y(0.25)=-0.8562\\
|
||||
y(0.75)=\frac65(e^{-5(0.75)}-1)+\frac{21}{10} (1-e^{-5(0.75-0.5)})\\
|
||||
y(0.75)=0.3266
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=6cm]{section3d.png}
|
||||
\end{figure}
|
||||
\newpage
|
||||
Solve $f(t)=0.5$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
y(t)=\frac65(e^{-5t}u(t)-u(t))+\frac{21} {10} (u(t-0.5)-e^{-5(t-0.5)}u(t-0.5))\\
|
||||
0.5=\frac65(e^{-5t}-1)\\
|
||||
\frac5{12}=e^{5t}-1\\
|
||||
\frac{17}{12}=e^{5t}\\
|
||||
5t=\ln(\frac{17}{12})\\
|
||||
t=\frac{\ln(\frac{17}{12})}{5}\\
|
||||
t\approx-0.06966\\
|
||||
This\ is\ impossible\\
|
||||
0.5=\frac65(e^{-5t}-1)+\frac{21} {10} (1-e^{-5(t-0.5)})\\
|
||||
0.5=(\frac{21}{10}-\frac{12}{10})+\frac65e^{-5t}-\frac{21}{10}e^{-5(t-0.5)}\\
|
||||
-0.4=\frac{12}{10}e^{-5t}-\frac{21}{10}e^{-5(t-0.5)}\\
|
||||
-0.4=\frac{12}{10e^{5t}}-\frac{21e^{-5(t-0.5)}}{10}\\
|
||||
-0.4=\frac{12}{10e^{5t}}-\frac{21e^{-5(t-0.5)}e^{5t}}{10e^{5t}}\\
|
||||
-0.4=\frac{12-21e^{-5(t-0.5)+5t}}{10e^{5t}}\\
|
||||
-4e^{5t}=12-21e^{2.5}\\
|
||||
e^{5t}=-3+5.25e^{2.5}\\
|
||||
5t=\ln(-3+5.25e^{2.5})\\
|
||||
t=\frac{\ln(-3+5.25e^{2.5})}{5}\\
|
||||
t\approx0.822
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\newpage
|
||||
|
||||
\item $a=3$, $b=5$, Solve for $t=0.25$, $0.75$, given $y(0)=STE ADY\ STATE$
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Copied\ From\ Part\ D:\\
|
||||
y(t)=\frac65(e^{-5t}u(t)-u(t))+\frac{21} {10} (u(t-0.5)-e^{-5(t-0.5)}u(t-0.5))\\
|
||||
y(t)=\frac65+\frac{21} {10} (u(t-0.5)-e^{-5(t-0.5)}u(t-0.5))\\
|
||||
y(0.25)=\frac65\\
|
||||
y(0.75)=\frac65+\frac{21}{10} (1-e^{-5(0.75-0.5)})\\
|
||||
y(0.75)=0.2983
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[height=10cm]{section3e.png}
|
||||
\end{figure}
|
||||
\end{enumerate}
|
||||
|
||||
\end{document}
|
||||
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\date{10/23/2021}
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\title{%
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Homework 3b\\
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\large EEET--427--01:Controls Systems}
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\author{Blizzard MacDougall}
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\begin{document}
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\maketitle
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\pagenumbering{arabic}
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\begin{enumerate}
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\item Find the root locations for a second order system that has a step response with $4.3\%$ overshoot and a time of peak overshoot of $0.44$ seconds.
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\begin{equation}
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Roots= 7.15\pm7.140j
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\end{equation}
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\item Find the damping ratio, the undamped natural frequency, and root locations for the system which has the following unit step response:
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\begin{figure}[h!]
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\centering
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\includegraphics[width=10cm]{exercise2.png}
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\end{figure}
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\begin{equation}
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\begin{split}
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\zeta=0.404\\
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\omega_n=9.811\\
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Roots=3.961\pm8.976j
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\end{split}
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\end{equation}
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\newpage
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\item Find the damping ratio, undamped natural frequency, and root locations for the system which has the following unit step response:
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\begin{figure}[h!]
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\centering
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\includegraphics[width=10cm]{exercise3.png}
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\end{figure}
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\begin{equation}
|
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\begin{split}
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\zeta=0.404\\
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\omega_n=3.346\\
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Roots=1.153\pm\pi j
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||||
\end{split}
|
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\end{equation}
|
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\item Find the damping ratio, undamped natural frequency, and root locations for the system which has the following unit step response:
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\begin{figure}[h!]
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\centering
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\includegraphics[width=10cm]{exercise4.png}
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\end{figure}
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\begin{equation}
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\begin{split}
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\zeta=0.404\\
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\omega_n=16.35\\
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Roots=6.601\pm14.96 j
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\end{split}
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\end{equation}
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\newpage
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\item Plot the root locations that you calculated for problems 1-4 in the s-plane. Label each pare of poles with its problem number.
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\begin{figure}[h!]
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\centering
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\includegraphics[width=10cm]{exercise5Answer.png}
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\end{figure}\\
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What is the same about all 3 of the step responses for these pole locations that lie on the same line?\\
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The $\zeta$ values for those 3 are the same.
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\item Find the damping ratio, undamped natural frequency, and root locations for the system which has an overshoot of $0.28\%$ and the time of peak response is $0.444$ seconds.
|
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\begin{equation}
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\begin{split}
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\zeta=0.882\\
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\omega_n=15.01\\
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Roots=13.24\pm7.076 j
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\end{split}
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\end{equation}
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\begin{figure}[h!]
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\centering
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\includegraphics[width=10cm]{exercise6Answer.png}
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\end{figure}\\
|
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Which pole location has the greatest damping ratio?\\
|
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Question 5 has the greatest damping ratio.
|
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\newpage
|
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\item For a system with root locations of $s=-2.3\pm j11.5$, find the percent overshoot and time to peak response.
|
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\begin{equation}
|
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\begin{split}
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T_p=0.273s\\
|
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Overshoot\ \%=50.3\%
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\end{split}
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\end{equation}
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\item Find the roots of the denominator of the third order polynomial given below:
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\frac{\theta}{\theta_{ref}}=\frac{aK_ps+aK_i}{s^3+bs^2+aK_ps+aK_i}\\
|
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a=1800;\ b=6;\ K_p=0.08;\ K_i=0.3\\
|
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Roots=-1.0139\pm 11.6154j;\ -3.9722+0j
|
||||
\end{split}
|
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\end{equation}
|
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Then, find the approximate percent overshoot and time to peak response.\\
|
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\begin{equation}
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\begin{split}
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Overshoot\ \%=75.13\%\\
|
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T_p=0.27
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\end{split}
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\end{equation}
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\end{enumerate}
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\end{document}
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\date{11/16/2021}
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\title{%
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Homework \# 4\\
|
||||
\large EEET--427--01:Controls Systems}
|
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\author{Blizzard MacDougall}
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\begin{document}
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\maketitle
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\pagenumbering{arabic}
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\begin{equation}
|
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M_r=\frac{1}{2\zeta\sqrt{1-\zeta^2}}
|
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\end{equation}
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\begin{equation}
|
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\frac{BW}{\omega_n}=\sqrt{\left(1-2\zeta^2\right)+\sqrt{4\zeta^4-4\zeta^2+2}}
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\end{equation}
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\begin{enumerate}
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\item Plot a fully labeled $M_r$ as the damping ratio varies from $0.1-0.7$.
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\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{question1Answer.eps}
|
||||
\end{figure}
|
||||
\item Plot and fully label $\frac{BW}{\omega_n}$ as the damping ratio varies from $0-1.2$.
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{question2Answer.eps}
|
||||
\end{figure}
|
||||
\newpage
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=8cm]{question3.png}
|
||||
\caption{Open Loop Gain Plot}
|
||||
\end{figure}
|
||||
\item What is the phase margin in Figure 1?\\
|
||||
$\approx10^\circ$.
|
||||
\item What is the approximate damping ratio, given Figure 1?\\
|
||||
$\approx0.1$.
|
||||
\item What is the approximate steady state error in Figure 1, assuming $H=1.0$?\\
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
Steady\ State=E\\
|
||||
E=R-CH\\
|
||||
E=R-EGH\\
|
||||
E+EGH=R\\
|
||||
E(1+GH)=R\\
|
||||
\frac ER =\frac{1}{1+GH}\\
|
||||
GH=55dB=10^{\frac{55dB}{20}}\approx562.3\\
|
||||
\frac ER = \frac{1}{1+562.3}\\
|
||||
\frac ER \approx 0.001775
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item What is the closed loop bandwidth of this system in Figure 1, assuming $H=1.0$?\\
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
BW_{CL}\approxGH=1\\
|
||||
\omega=1
|
||||
\end{split} \end{equation} \item If you reduce the loop gain by $20dB$, what will the phase margin be? What will the damping ratio be? What will the magnitude peaking ratio be in the closed loop frequency response?
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
PM\approx5^\circ\\
|
||||
\zeta=0.05\\
|
||||
\frac{E}{R}=\frac{1}{1+GH}\\
|
||||
GH=35dB=10^{\frac{35}{20}}\approx 56.23\\
|
||||
\frac{E}{R}=\frac{1}{1+56.23}\approx 0.01747
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item Draw a control system block diagram for simple inverting amplifier op-amp circuit (shown below).
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=5cm]{question8.png}
|
||||
\end{figure}
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
V=\frac{V_{in}R_f+V_{out}R_i}{R_f}\\
|
||||
V_{out}=-A_v\left(\frac{V_{in}R_f+V_{out}R_i}{R_f}\right)\\
|
||||
V_{out}=\frac{\frac{-A_vR_f}{R_f+R_i}}{1+\frac{-A_vR_f}{R_f+R_i}\times\frac{R_i}{R_f}}\\
|
||||
H=\frac{R_i}{R_f}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{question8BlockDiagram.png}
|
||||
\end{figure}
|
||||
\newpage
|
||||
\item Sketch a bode plot of the transfer function of the inverting practical integrator amplifier circuit shown, assuming an ideal op-amp. Use $R_f=10M\Omega;\ C_f=2\mu F;\ R_{in}=10k\Omega$.
|
||||
After you have the sketch, plot the bode plot in MATLAB with the frequency ranging from $10^{-3}-10^{3}$.
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=5cm]{question9.png}
|
||||
\end{figure}
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
TF=\frac{R_f}{R_{in}(sCR_f+1)}\\
|
||||
s<1;\ \frac{R_f}{R_{in}};\ Mag=60dB;\ phase=0\\
|
||||
1<s;\ \frac{R_f}{R_{in}(sCR_f)};\ Mag=34dB-20\log(\omega);\ phase=-90^\circ
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{question9Graphs.png}
|
||||
\end{figure}
|
||||
\newpage
|
||||
\item At $1rad/sec$, how much phase does this circuit have, and what is the gain?
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=5cm]{question10.png}
|
||||
\end{figure}
|
||||
\begin{equation}
|
||||
Gain=37.5dB;\ Phase-87.5^\circ
|
||||
\end{equation}
|
||||
\item For the practical Op-Amp differentiator shown, assuming an ideal op-amp, sketch the bode plot over the frequency range $10^{-3}-10^{3}$. Use $R_f=100k\Omega;\ C_a=20\mu F;\ R_a=1k\Omega$. After you sketch the plot by hand, use matlab to confirm your results.
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=5cm]{question11.png}
|
||||
\end{figure}
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
TF=\frac{sCR_f}{sCR_{in}+1}\\
|
||||
s<1;\ sCR_f;\ Mag=26dB;\ Phase=90\\
|
||||
1<s;\ \frac{R_f}{R_{in}};\ Mag=60dB-20\log(\omega);\ Phase=0
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=5cm]{question11Graphs.png}
|
||||
\end{figure}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
5
homework/homework4/hw4/init.m
Normal file
5
homework/homework4/hw4/init.m
Normal file
|
|
@ -0,0 +1,5 @@
|
|||
function init
|
||||
clc; close all;
|
||||
global fig_num
|
||||
fig_num = 1;
|
||||
end
|
||||
12
homework/homework4/hw4/make_bode.m
Normal file
12
homework/homework4/hw4/make_bode.m
Normal file
|
|
@ -0,0 +1,12 @@
|
|||
function make_bode (mag_plot, phase_plot, H, log_xmin, log_xmax, ymin, ymax)
|
||||
global fig_num;
|
||||
figure(fig_num);
|
||||
fig_num = fig_num + 1;
|
||||
|
||||
global options;
|
||||
options.Ylim = [ymin ymax];
|
||||
options.MagVisible = mag_plot;
|
||||
options.PhaseVisible = phase_plot;
|
||||
w=logspace(log_xmin,log_xmax,1000);
|
||||
bode(H,options,w); grid on;
|
||||
end
|
||||
14
homework/homework4/hw4/make_plot.m
Normal file
14
homework/homework4/hw4/make_plot.m
Normal file
|
|
@ -0,0 +1,14 @@
|
|||
function make_plot(x_data,y_data,graph_title,x_label,y_label,x2_data,y2_data,y2_marker)
|
||||
global fig_num;
|
||||
figure(fig_num);
|
||||
fig_num=fig_num+1;
|
||||
|
||||
plot(x_data,y_data);
|
||||
grid on;
|
||||
xlabel(x_label); ylabel(y_label);
|
||||
title(graph_title);
|
||||
if nargin==8
|
||||
hold on;
|
||||
plot(x2_data,y2_data,y2_marker);
|
||||
end
|
||||
end
|
||||
6
homework/homework4/hw4/question1.m
Normal file
6
homework/homework4/hw4/question1.m
Normal file
|
|
@ -0,0 +1,6 @@
|
|||
init();
|
||||
zeta=0.1:0.7/2048:0.7;
|
||||
zetaSqrt=sqrt(1-zeta.^2);
|
||||
mr=1./(2.*zeta.*zetaSqrt);
|
||||
make_plot(zeta,mr,"Mr Plot","zeta","Mr");
|
||||
exportgraphics(gcf,'question1Answer.eps','ContentType','vector')
|
||||
BIN
homework/homework4/hw4/question10.png
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homework/homework4/hw4/question10.png
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homework/homework4/hw4/question11.png
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homework/homework4/hw4/question11.png
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homework/homework4/hw4/question11Graphs.png
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homework/homework4/hw4/question11Graphs.png
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homework/homework4/hw4/question1Answer-eps-converted-to.pdf
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homework/homework4/hw4/question1Answer-eps-converted-to.pdf
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homework/homework4/hw4/question1Answer.eps
Normal file
2954
homework/homework4/hw4/question1Answer.eps
Normal file
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6
homework/homework4/hw4/question2.m
Normal file
6
homework/homework4/hw4/question2.m
Normal file
|
|
@ -0,0 +1,6 @@
|
|||
init();
|
||||
zeta=0:1.2/2048:1.2;
|
||||
zetaSqrt=sqrt(4.*(zeta.^4)-4.*(zeta.^2)+2);
|
||||
BW=sqrt(1-2.*zeta.^2+zetaSqrt);
|
||||
make_plot(zeta,BW,"BW/wn Plot","zeta","BW/wn");
|
||||
exportgraphics(gcf,'question2Answer.eps','ContentType','vector')
|
||||
BIN
homework/homework4/hw4/question2Answer-eps-converted-to.pdf
Normal file
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homework/homework4/hw4/question2Answer-eps-converted-to.pdf
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Binary file not shown.
3118
homework/homework4/hw4/question2Answer.eps
Normal file
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homework/homework4/hw4/question2Answer.eps
Normal file
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homework/homework4/hw4/question3.png
Normal file
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homework/homework4/hw4/question3.png
Normal file
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|
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BIN
homework/homework4/hw4/question8.png
Normal file
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homework/homework4/hw4/question8.png
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|
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BIN
homework/homework4/hw4/question8BlockDiagram.png
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homework/homework4/hw4/question8BlockDiagram.png
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|
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10
homework/homework4/hw4/question9.m
Normal file
10
homework/homework4/hw4/question9.m
Normal file
|
|
@ -0,0 +1,10 @@
|
|||
init();
|
||||
Rf=1E7;
|
||||
Cf=2E-6;
|
||||
Rin=1E4;
|
||||
global s;
|
||||
TF=(Rf/Rin)*(1./(s*Cf*Rf+1));
|
||||
make_bode('on','off',TF,-3,3,-120,45);
|
||||
exportgraphics(gcf,'question9Magnitude.eps','ContentType','vector');
|
||||
make_bode('off','on',TF,-3,3,-180,180);
|
||||
exportgraphics(gcf,'question9Phase.eps','ContentType','vector');
|
||||
BIN
homework/homework4/hw4/question9.png
Normal file
BIN
homework/homework4/hw4/question9.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 30 KiB |
BIN
homework/homework4/hw4/question9Graphs.png
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homework/homework4/hw4/question9Graphs.png
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|
After Width: | Height: | Size: 9.9 KiB |
BIN
homework/homework5/Bode Lead Homework HANDOUT.docx
Normal file
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homework/homework5/Bode Lead Homework HANDOUT.docx
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Binary file not shown.
5
homework/homework5/hw5/hw5.aux
Normal file
5
homework/homework5/hw5/hw5.aux
Normal file
|
|
@ -0,0 +1,5 @@
|
|||
\relax
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {1}Section 1}{1}{}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {2}Section 2}{1}{}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {3}Section 3}{1}{}\protected@file@percent }
|
||||
\gdef \@abspage@last{2}
|
||||
338
homework/homework5/hw5/hw5.log
Normal file
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homework/homework5/hw5/hw5.log
Normal file
|
|
@ -0,0 +1,338 @@
|
|||
This is pdfTeX, Version 3.14159265-2.6-1.40.21 (TeX Live 2020/Debian) (preloaded format=pdflatex 2021.9.12) 17 NOV 2021 21:13
|
||||
entering extended mode
|
||||
\write18 enabled.
|
||||
file:line:error style messages enabled.
|
||||
%&-line parsing enabled.
|
||||
**hw5.tex
|
||||
(./hw5.tex
|
||||
LaTeX2e <2020-10-01> patch level 4
|
||||
L3 programming layer <2021-01-09> xparse <2020-03-03>
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/base/article.cls
|
||||
Document Class: article 2020/04/10 v1.4m Standard LaTeX document class
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/base/size10.clo
|
||||
File: size10.clo 2020/04/10 v1.4m Standard LaTeX file (size option)
|
||||
)
|
||||
\c@part=\count177
|
||||
\c@section=\count178
|
||||
\c@subsection=\count179
|
||||
\c@subsubsection=\count180
|
||||
\c@paragraph=\count181
|
||||
\c@subparagraph=\count182
|
||||
\c@figure=\count183
|
||||
\c@table=\count184
|
||||
\abovecaptionskip=\skip47
|
||||
\belowcaptionskip=\skip48
|
||||
\bibindent=\dimen138
|
||||
)
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsmath.sty
|
||||
Package: amsmath 2020/09/23 v2.17i AMS math features
|
||||
\@mathmargin=\skip49
|
||||
|
||||
For additional information on amsmath, use the `?' option.
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amstext.sty
|
||||
Package: amstext 2000/06/29 v2.01 AMS text
|
||||
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsgen.sty
|
||||
File: amsgen.sty 1999/11/30 v2.0 generic functions
|
||||
\@emptytoks=\toks15
|
||||
\ex@=\dimen139
|
||||
))
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsbsy.sty
|
||||
Package: amsbsy 1999/11/29 v1.2d Bold Symbols
|
||||
\pmbraise@=\dimen140
|
||||
)
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsopn.sty
|
||||
Package: amsopn 2016/03/08 v2.02 operator names
|
||||
)
|
||||
\inf@bad=\count185
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
\DOTSCASE@=\count189
|
||||
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|
||||
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|
||||
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|
||||
\Mathstrutbox@=\box47
|
||||
\strutbox@=\box48
|
||||
\big@size=\dimen141
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
\c@parentequation=\count192
|
||||
\dspbrk@lvl=\count193
|
||||
\tag@help=\toks16
|
||||
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|
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|
||||
\maxfields@=\count196
|
||||
\andhelp@=\toks17
|
||||
\eqnshift@=\dimen142
|
||||
\alignsep@=\dimen143
|
||||
\tagshift@=\dimen144
|
||||
\tagwidth@=\dimen145
|
||||
\totwidth@=\dimen146
|
||||
\lineht@=\dimen147
|
||||
\@envbody=\toks18
|
||||
\multlinegap=\skip50
|
||||
\multlinetaggap=\skip51
|
||||
\mathdisplay@stack=\toks19
|
||||
LaTeX Info: Redefining \[ on input line 2923.
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)
|
||||
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|
||||
Package: graphicx 2020/09/09 v1.2b Enhanced LaTeX Graphics (DPC,SPQR)
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|
||||
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|
||||
Package: keyval 2014/10/28 v1.15 key=value parser (DPC)
|
||||
\KV@toks@=\toks20
|
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)
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/graphics/graphics.sty
|
||||
Package: graphics 2020/08/30 v1.4c Standard LaTeX Graphics (DPC,SPQR)
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|
||||
Package: trig 2016/01/03 v1.10 sin cos tan (DPC)
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||||
)
|
||||
(/usr/share/texlive/texmf-dist/tex/latex/graphics-cfg/graphics.cfg
|
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File: graphics.cfg 2016/06/04 v1.11 sample graphics configuration
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)
|
||||
Package graphics Info: Driver file: pdftex.def on input line 105.
|
||||
|
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(/usr/share/texlive/texmf-dist/tex/latex/graphics-def/pdftex.def
|
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File: pdftex.def 2020/10/05 v1.2a Graphics/color driver for pdftex
|
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))
|
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\Gin@req@height=\dimen148
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)
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Package: amssymb 2013/01/14 v3.01 AMS font symbols
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Package: amsfonts 2013/01/14 v3.01 Basic AMSFonts support
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|
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|
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|
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|
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|
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|
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|
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|
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|
||||
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|
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|
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|
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\lst@framebox=\box50
|
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)
|
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))
|
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Package: listings 2020/03/24 1.8d (Carsten Heinz)
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|
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|
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|
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\documentclass[letterpaper]{article}
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\date{11/17/2021}
|
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\title{%
|
||||
Homework \#5\\
|
||||
\large EEET--427--01:Controls Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
\section{Section 1}
|
||||
\begin{enumerate}
|
||||
\item Create a bode plot showing the magnitude in dB and phase of the below lead compensator. Clearly mark on the plot the peak phase in degrees created by the compensator and at what frequency the peak phase is created (in rad/sec).
|
||||
\begin{equation}
|
||||
G_c(s)=10\frac{s+50}{s+500}
|
||||
\end{equation}
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{question1Graph.eps}
|
||||
\end{figure}
|
||||
\item Using the design formula for $\varnothing_{max}$ in terms of $\alpha$, what is the theoretical max value of phase that the compensator adds? You will need to figure out what $\alpha$ is and plug it into the formula.
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\alpha = 0.1\\
|
||||
\varnothing_{max} = \sin^{-1}\left(\frac{1-\alpha}{1+\alpha}\right)=\sin^{-1}\left(\frac{9}{11}\right)\\
|
||||
\varnothing_{max} \approx 55^\circ
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item At what frequency is the phase max?\\
|
||||
$\approx 150rad/sec$
|
||||
\newpage
|
||||
\item Using the formula of the lead compensator, calculate (using a calculator) what the magnitude of the compensator gain is at $\omega=5rad/sec$.
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
T=\frac{1}{\omega\sqrt{\alpha}}\approx0.632\\
|
||||
G_{c}(s)=\frac{0.632s+1}{0.0632s+1}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item \dots at $\omega=5000rad/sec$.
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
T\approx6.32_E-4;\ G_{c}(s)=\frac{(6.32_E-4) s+1}{(6.32_E-5)s+1}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\end{enumerate}
|
||||
\section{Section 2}
|
||||
\begin{enumerate}[resume]
|
||||
\item Design a phase lead compensator to produce a max of $30^\circ$ of phase shift at a frequency of $\omega=10rad/sec$. Include $5^\circ$ of margin in your design. What is the value of $\alpha$?
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\varnothing_{max}=35^\circ\\
|
||||
\alpha=\frac{1-\sin(\varnothing_{max})}{1+\sin(\varnothing_{max})}\\
|
||||
\alpha\approx0.271
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item What is the value of $T$ for this lead compensator?
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
T=\frac{1}{\omega\sqrt{\alpha}}\\
|
||||
T\approx0.192
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item What is the transfer function of the compensator? Specifically, where are the poles and zeroes? Create a bode plot (both gain and phase) for the compensator.
|
||||
\begin{equation}
|
||||
G_c(s)=\frac{0.192s+1}{0.052s+1};\ pole=-19.23;\ zero=-5.21
|
||||
\end{equation}
|
||||
\item What is the gain of the compensator (in $dB$) at $\omega=0.1rad/sec$?
|
||||
$\approx -60dB$
|
||||
\item \dots $\omega=10rad/sec$?
|
||||
$\approx 15dB$
|
||||
\item \dots $\omega=1000rad/sec$?
|
||||
$\approx 21dB$
|
||||
\end{enumerate}
|
||||
\newpage
|
||||
\section{Section 3}
|
||||
\begin{enumerate}[resume]
|
||||
\item A control system has an open loop gain transfer function of:
|
||||
\begin{equation}
|
||||
G_{ol}(s)=\frac{9}{s(s+1.5)}
|
||||
\end{equation}
|
||||
The open loop gain bode plot is shown below:
|
||||
\begin{figure}[h!]
|
||||
\centering
|
||||
\includegraphics[width=7cm]{section3BodePlot.png}
|
||||
\end{figure}\\
|
||||
Design a phase lead compensator to achieve a closed loop step response with a $10\%$ overshoot. Note that the percent overshoot is related to the damping ratio through the formula given in the design overview section above. What does the damping ratio have to be?
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\zeta=\frac{-\ln\left(\frac{OS}{100}\right)}{\sqrt{\pi^2+\left(\ln\left(\frac{OS}{100}\right)\right)^2}}\\
|
||||
\zeta\approx0.591
|
||||
\end{split}
|
||||
\end{equation}
|
||||
\item What does the phase margin have to be to achieve this damping ratio?\\
|
||||
$\approx 59^\circ$
|
||||
\item How much phase in degrees must the lead compensator contribute to achieve this damping ratio? Be sure to include an additional $5^\circ$ of margin in your answer.\\
|
||||
$\approx 36^\circ$
|
||||
\item What is the value of $\alpha$?\\
|
||||
$\approx 0.2596$
|
||||
\item What is the value of $T$?
|
||||
\item What are the poles and zeros of the final lead compensator transfer function?
|
||||
\item What is the gain of the compensator (in $dB$) at $\omega=0.1rad/sec$?
|
||||
\item \dots $\omega=3rad/sec$?
|
||||
\item \dots $\omega=1000rad/sec$?
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\end{document}
|
||||
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<head>
|
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</head>
|
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<body>
|
||||
<p>for example:</p>
|
||||
<p><a href="https://www.mathworks.com/videos/control-systems-in-practice-part-3-what-is-feedforward-control-1535712796292.html">https://www.mathworks.com/videos/control-systems-in-practice-part-3-what-is-feedforward-control-1535712796292.html</a></p>
|
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</body>
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</html>
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54
week1/24-08-2021/24-08-2021.tex
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|
|||
% File: 23-08-2021.tex
|
||||
% Created: 14:00:00 Tue, 24 Aug 2021 EDT
|
||||
% Last Change: 15:00:00 Tue, 24 Aug 2021 EDT
|
||||
%
|
||||
\documentclass[letterpaper]{article}
|
||||
|
||||
\usepackage{graphicx}
|
||||
\usepackage{cancel}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{listings}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{inputenc}
|
||||
\usepackage{amsmath}
|
||||
|
||||
\date{08/23/2021}
|
||||
\title{%
|
||||
Day One Notes\\
|
||||
\large EEET-427-01: Control Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
Textbook not required. Helpful, but not required. This class is more theoretical than CPET.\\
|
||||
In lecture, we'll be using problem solving studio approach.\\
|
||||
\\
|
||||
In lab, we'll be building control systems experiments. We'll be cutting foam core, mounting motors, sensors, wiring power, sensors, actuators, microcontrollers, etc. There will be real-world control code thats actively in real-world use. \\
|
||||
\\
|
||||
Syllabus is on MyCourses. Changes/updates will be announced.\\
|
||||
Big terms for this course:
|
||||
\begin{itemize}
|
||||
\item control
|
||||
\item feedback
|
||||
\item compensation
|
||||
\item system
|
||||
\item analysis
|
||||
\item Laplace
|
||||
\end{itemize}
|
||||
[import syllabus rough schedule]\\
|
||||
\\
|
||||
Primary example of a control system: balancing a ball on a beam\\
|
||||
\\
|
||||
Quizzes will be after lecture. The questions on the daily quizzes will be basically the same as the exams (midterm and final). Quizzes are 20\% of the course, 15\% is the final, anf 15\% is the midterm. There will be a portfolio as a final project; this will include a write-up. \\
|
||||
\\
|
||||
\section{Control System Introduction}
|
||||
HMI - Human Mechanical Interface\\
|
||||
Controller - the thing that tells the actuators what to do, based on sensor input\\
|
||||
Actuator - the item that creates change in the system\\
|
||||
Plant - the object that the actuator modifies\\
|
||||
Trajectory - what you ultimately want to happen with the plant/process\\
|
||||
Sensor - monitors the plant, sends information to controller\\
|
||||
\\
|
||||
This class will use a good amount of block diagrams. As such, each of the lines and boxes should be labeled.
|
||||
|
||||
\end{document}
|
||||
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|
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% Created: 14:03:03 Thu, 26 Aug 2021 EDT
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%
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||||
\documentclass[letterpaper]{article}
|
||||
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|
||||
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|
||||
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||||
\usepackage{listings}
|
||||
\usepackage{enumitem}
|
||||
|
||||
\date{08/26/2021}
|
||||
\title{%
|
||||
Day Two Notes\\
|
||||
\large EEET-427-01:Control Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
\section{Controls Intro, and Why}
|
||||
Control Systems Engineering is the analysis of a system to determine how it responds to inputs, mathematical models that support the analysis of the systems, and the design of a controller that changes inputs to a system to achieve a desired output.\\
|
||||
Controls engineers get to see the full scope of a project, and interact with all parts of a system. They also save systems, by taking existing systems and making them better, more efficient, or expand its usefulness at minimal cost.\\
|
||||
Controls engineers are also involved from the beginning of a project to ensure that systems can perform safely at the required level. \\
|
||||
Using the controls engineer mindset brings a unique perspective to life.\\
|
||||
\\
|
||||
Tools of controls engineers:\\
|
||||
\begin{itemize}
|
||||
\item Models (physics, electrical, chemical, etc)
|
||||
\item Diff.Eq.
|
||||
\item Laplace transforms
|
||||
\item Block diagrams
|
||||
\item Bode plots
|
||||
\item Measurements of system response
|
||||
\item Complex number math (roots of polynomials)
|
||||
\item Software for digital control
|
||||
\end{itemize}
|
||||
\newpage
|
||||
\section{Language of Controls}
|
||||
\begin{itemize}
|
||||
\item process: the thing being controlled
|
||||
\item actuator: manipulates an input of the process (in the lab kit, its the motor)
|
||||
\item controller: designed system that controls the actuator
|
||||
\item sensor: measures the system
|
||||
\item summing junction: compares desired output to the output measured by the sensor, passes the error to the controller
|
||||
\end{itemize}
|
||||
The controller has to account for sense-errors and other disturbances. (Disturbances include change in power supply voltage, environmental temperature, sensor noise, etc)\\
|
||||
A process can have multiple output signals, and can have multiple control loops. These loops can be independent or dependent.\\
|
||||
Open-loop control is where the output isn't sensed, its just assumed that everything is working properly. Closed-loop control includes a sensor that checks the output.\\
|
||||
\\
|
||||
To start in the process of designing a system, you have to write some differential equations that describe the system you're modeling.\\
|
||||
\section{Example Problem 1}
|
||||
Skydiver of mass 100kg, $g=9.8m/s$, $v_i=0m/s$, drag coefficient $D_L$, $F_D=vD_L$\\
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
mg+D_Lv(t)=m\frac{dv}{dt}\\
|
||||
980N-D_Lv(t)=m\frac{dv}{dt}\\
|
||||
Terminal\ velocity=\frac{980N}{D_L}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
Given a terminal velocity of $v_1=50m/s$\\
|
||||
\begin{equation}
|
||||
D_L=19.6\frac{N\times sec}m
|
||||
\end{equation}
|
||||
|
||||
\end{document}
|
||||
33
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|
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|
|||
% File: 26-10-2021.tex
|
||||
% Created: 14:07:20 Tue, 26 Oct 2021 EDT
|
||||
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|
||||
%
|
||||
\documentclass[letterpaper]{article}
|
||||
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|
||||
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|
||||
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|
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|
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|
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|
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|
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|
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\usepackage{geometry}
|
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|
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\usepackage{lplfitch}
|
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|
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\geometry{portrait, margin=1in}
|
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|
||||
%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
|
||||
|
||||
\date{10/26/2021}
|
||||
\title{%
|
||||
Notes\\
|
||||
\large EEET--427--01:Controls Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
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|
||||
|
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\end{document}
|
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96
week11/02-11-2021/02-11-2021.tex
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|
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|
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% File: 11-02-2021.tex
|
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% Created: 14:10:43 Tue, 02 Nov 2021 EDT
|
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|
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%
|
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|
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|
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|
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|
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|
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\geometry{portrait, margin=1in}
|
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|
||||
%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
|
||||
|
||||
\date{11/02/2021}
|
||||
\title{%
|
||||
\say{Fun With Control Systems}\\
|
||||
\large EEET--427--01:Controls Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
You now have a ball on a beam. What are the forces on the ball?
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\sum F=ma\\
|
||||
mg\sin\theta=m\ddot{x}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
Now, whats handy about what we're working with, is it can follow the Small Angle Theorem (which roughly says $\sin\theta\approx\theta$). Meaning\dots
|
||||
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
mg\theta=m\ddot{x}\\
|
||||
g\theta=\ddot{x}\\
|
||||
g\theta(s)=s^2X(s)\\
|
||||
\frac{X(s)}{\theta(s)}=\frac{g}{s^2}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
But this results in a\dots nasty RH Table.
|
||||
So, lets redo the physics math here to include drag.
|
||||
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\sum F=ma\\
|
||||
mg\sin\theta-bv=m\ddot{x}\\
|
||||
mg\theta-b\dot{x}=m\ddot{x}\\
|
||||
mg\theta(s)-bsX(s)=ms^2X(s)\\
|
||||
mg\theta(s)=ms^2X(s)+bsX(s)\\
|
||||
mg\theta(s)=(ms^2+bs)X(s)\\
|
||||
\frac{X(s)}{\theta(s)}=\frac{mg}{ms^2+bs}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
|
||||
But this also results in a system that is unstable.
|
||||
So lets add a simple controller, make it stable.
|
||||
In a block diagram, this is adding feedback, and a $K_p$ gain.
|
||||
This results in\dots another unstable system.
|
||||
|
||||
\begin{equation}
|
||||
\frac{X(s)}{X_{ref}(s)}=\frac{K_p g}{s^2+K_p g}
|
||||
\end{equation}
|
||||
|
||||
What if we added control to the drag system stated before?
|
||||
(And yes, this as just as tedious for you as it is for me.)
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\frac{G}{1+GH}\\
|
||||
G=\frac{K_p mg}{ms^2+bs}\\
|
||||
\frac{X(s)}{X_{ref}(s)}=\frac{K_p mg}{ms^2+bs+K_p mg}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
|
||||
This turns out to be stable. (I'm not writing out a RH table here, sorry)
|
||||
But how stable is it?
|
||||
This involves a shift\dots and I'm sorry that it does
|
||||
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
m{(z-\tau)}^2+b(z-\tau)+K_p mg\\
|
||||
mz^2-2m\tau z+m\tau^2+bz-b\tau+K_p mg\\
|
||||
mz^2+(b-2m\tau)z+(K_p mg-b\tau+m\tau^2)\\
|
||||
\tau=1;\
|
||||
mz^2+(b-2m)z+(K_p mg-b+m)
|
||||
\end{split}
|
||||
\end{equation}
|
||||
|
||||
\end{document}
|
||||
36
week11/04-11-2021/04-11-2021.tex
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|
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% File: 11-04-2021.tex
|
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% Created: 14:12:25 Thu, 04 Nov 2021 EDT
|
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|
||||
%
|
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|
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|
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|
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|
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|
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|
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|
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%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
|
||||
|
||||
\date{11/04/2021}
|
||||
\title{%
|
||||
Bode Analysis\\
|
||||
\large EEET--427--01:Controls Systems}
|
||||
\author{Blizzard MacDougall}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\pagenumbering{arabic}
|
||||
Routh Hurwitz isn't the most helpful thing in the world, so we'll use Bode as well when analyzing systems.
|
||||
Rather than doing the whole nine yards, we can just analyze GH.
|
||||
So, the only thing we really need to ensure is that $GH\ne-1$.
|
||||
|
||||
|
||||
\end{document}
|
||||
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|
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\relax
|
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|
||||
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|
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This is pdfTeX, Version 3.141592653-2.6-1.40.22 (TeX Live 2022/dev/Debian) (preloaded format=pdflatex 2021.12.7) 9 DEC 2021 19:08
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|
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\write18 enabled.
|
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**02-09-2021.tex
|
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[]
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[]
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[]
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[1
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msy7.pfb>
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PDF statistics:
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36 compressed objects within 1 object stream
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BIN
week2/02-09-2021/02-09-2021.pdf
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week2/02-09-2021/02-09-2021.pdf
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week2/02-09-2021/02-09-2021.synctex.gz
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week2/02-09-2021/02-09-2021.synctex.gz
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