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\documentclass{article}
\usepackage{enumitem}
\usepackage{cancel}
\usepackage{amsmath}
\usepackage{listings}
\usepackage{soul}
\usepackage{geometry}
\geometry{portrait, margin=1in}
\title{Homework 1\\ EEET-331-01: Signals, Systems and Transforms}
\date{08/31/2021}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
\section{Exercise 1.} Put the following complex numbers into exponential form.
\begin{enumerate}[label=\alph*)]
\item
$3+4j$\\
$5e^{j0.93}$
\item
$\sqrt{5}+2j$\\
$3e^{j0.73}$
\item
$-\sqrt5+2j$\\
$3e^{j2.41}$
\item
$-4-3j$\\
$5e^{-j2.50}$
\item
$2-j$\\
$2.24e^{-j0.46}$
\item
$0$\\
$0e^{j0}=0$
\end{enumerate}
\section{Exercise 2.} Put the following complex numbers into polar form.
\begin{enumerate}[label=\alph*)]
\item
$3+4j$\\
$5\angle53.13^{\circ}$
\item
$-\sqrt5+2j$\\
$3\angle41.81^{\circ}$
\item
$4e^{j0.5}$\\
$4\angle28.65^{\circ}$
\item
$4e^{-j0.2}$\\
$4\angle11.46^{\circ}$
\item
$2$\\
$2\angle0^{\circ}$
\end{enumerate}
\section{Exercise 3.} Put the following complex numbers into rectangular form.
\begin{enumerate}[label=\alph*)]
\item
$4e^{-j0.5}$\\
$3.5103302+j1.9177022$
\item
$4e^{j0.2}$\\
$3.9202663+j0.79467732$
\item
$2\angle10^{\circ}$\\
$1.9696155+j0.34729636$
\end{enumerate}
\section{Exercise 4.} Put the following complex numbers into rectangular form. (Exponents can be reduced using the following equation.)
\begin{equation}
(r\angle\theta)^n=r^n\angle(\theta*n)
\end{equation}
\begin{enumerate}[label=\alph*)]
\item
$(3+4j)^{10}$\\
$-9653260+j1477156$
\item
$(5+2j)^4$\\
$41+j840$
\item
$(-1+1j)^{\frac{1}{3}}$\\
$0.794+j0.794$
\end{enumerate}
\section{Exercise 5.} Express as a sum of sine and cosine functions.\footnote{\label{1}
All of the following use a combination of the following values:\\
$4e^{-j3}=4cos3-4sin3$,
$4e^{j3}=4cos3+4sin3$,
$2e^{-j5}=2cos5-2sin5$,
$2e^{j5}=2cos5+2sin5$.\\
As such, all intermittent calculations are done using the following substitutions:\\
$a=cos3,\
b=sin3,\
c=cos5,\
d=sin5$}
\begin{enumerate}
\item
$-4e^{-j3}+2e^{j5}+2e^{-j5}+4e^{j3}$\\
\[-4(a-b)+2(c+d)+2(c-d)+4(a+b)\]
\[-4a+4b+2c+2d+2c-2d+4a+4b\]
\[\cancel{-4a}+4b+2c\cancel{+2d}+2c\cancel{-2d+4a}+4b\]
\[4b+2c+2c+4b\]
\[8b+4c\]
\[|\overline{\underline{4cos5+8jsin3}}|\]
\item
$4e^{-j3}+2e{j5}-2e^{-j5}-4e^{j3}$\\
\[4a-4b+2c+2d-(2c-2d)-(4a+4b)\]
\[\cancel{4a}-4b\cancel{+2c}+2d\cancel{-2c}+2d\cancel{-4a}-4b\]
\[-4b+2d+2d-4b\]
\[-8b+4d\]
\[|\overline{\underline{-8sin3+4sin5}}|\]
\item
$4e^{-j3}-2e^{j5}-2e{-j5}-4e^{j3}$\\
\[4a-4b-(2c+2d)-(2c-2d)-(4a+4b)\]
\[\cancel{4a}-4b-2c\cancel{-2d}-2c\cancel{+2d-4a}-4b\]
\[-4b-2c-2c-4b\]
\[|\overline{\underline{-8sin3-4cos5}}|\]
\end{enumerate}
\section{Exercise 6.} Evaluate the expression and leave the answer in polar form, given the following:
\begin{equation}
z_1=2+2j=2.828\angle45^{\circ}
\end{equation}
\begin{equation}
z_2=-1+3j=3.1622\overline7\angle108.43^{\circ}
\end{equation}
\begin{equation}
z_3=4.9240388+j0.86824089=5\angle10^{\circ}
\end{equation}
\begin{equation}
z_4=0.72471551+j1.8640782=2\angle68.75^{\circ}=2e^{j1.2}
\end{equation}
\begin{enumerate}[label=\alph*)]
\item
$\frac{z_1}{z_2}$\\
$0.8944\angle-63.43^{\circ}$
\item
$z_1+\frac{z_2}{z_3}$\\
$3.24\angle54.01^{\circ}$
\end{enumerate}
\section{Exercise 7.} Using the given values in question 6, and your research into MATLAB commands, calculate the following equation. Leave your answer in polar notation.
\begin{lstlisting}
conj(z1)abs(z4)+(real(z2)/imag(z3))
\end{lstlisting}
\paragraph .The command \verb|conj| transforms the input into its conjugate. The command \verb|abs| transforms its input into the magnitude of its input (also known as the absolute value). The command \verb|real| outputs the real component of the complex input, while the command \verb|imag| returns the imaginary portion of the complex input. Given this:
\[conj(z1)abs(z4)+\frac{real(z2)}{imag(z3)}\]
\[conj(2-2j)*abs(2\angle68.75^{\circ})+\frac{real(-1+3j)}{imag(4.9240388+j0.86824089)}\]
\[2+2j*2+\frac{-1}{0.86824089}\]
\[4+4j+\frac{-1}{0.86824089}\]
\[\overline{\underline{|4.91\angle54.54^{\circ}|}}\]
\end{document}

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N=1024;
t=linspace(0,6,N);
ya=zeros(size(t));
A=3;
ind=find(t>1&t<2);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>3&t<4);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>5&t<6);
ya(ind)=A*abs(sin(2*pi*t(ind)));
figure(110);
plot(t,ya);
axis([0 6 -1 A+1]); grid on;

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N=1024;
t=linspace(0,8,N);
ya=zeros(size(t));
A=3;
ind=find(t>1&t<2);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>3&t<4);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>5&t<6);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>7&t<8);
ya(ind)=A*abs(sin(2*pi*t(ind)));
figure(110);
plot(t,ya);
axis([0 8 -1 A+1]); grid on;

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N=1024;
T=2;
t=linspace(0,T,N);
ya=zeros(size(t));
ind=find(t>1&t<2);
ya(ind)=A*abs(sin(2*pi*t(ind)));
figure(111);
plot(t,ya);
axis([0 T -1 A+1]); grid on;

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N=1024;
t=linspace(0,3,N);
ya=zeros(size(t));
A=3;
ind=find(t>0&t<0.5);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>1&t<1.5);
ya(ind)=A*abs(sin(2*pi*t(ind)));
ind=find(t>2&t<2.5);
ya(ind)=A*abs(sin(2*pi*t(ind)));
figure(110);
plot(t,ya);
axis([0 3 -1 A+1]); grid on;

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N=1024;
t=linspace(0,1,N);
ya=zeros(size(t));
A=3;
ind=find(t>0&t<0.5);
ya(ind)=A*abs(sin(2*pi*t(ind)));
figure(110);
plot(t,ya);
axis([0 3 -1 A+1]); grid on;

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N=1024;
repeatTime=3;
%Each cycle takes 3t to complete
T=repeatTime*3;
t=linspace(0,T,N);
A=2;
yc=zeros(size(t));
for i=0:3:(T)
ind=find(t>(i)&t<(i+3));
yc(ind)=-A*triangularPulse(i,(i+1),(i+3),t(ind))+A;
end
figure(131);
plot(t,yc);
axis([0 T -1 A+1]); grid on;

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N=1024;
repeatTime=1;
%Each cycle takes 9t to complete
T=repeatTime*9;
t=linspace(0,T,N);
A=5;
yc=zeros(size(t));
for i=0:9:(T)
ind=find(t>(i+1)&t<(i+2)|t>(i+7)&t<(i+8));
yc(ind)=A;
end
figure(132);
plot(t,yc);
axis([0 T -1 A+1]); grid on;

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N=1024;
n=0:N-1;
Ts=613.59E-6;
t=n*Ts;
A=3;
Y=A*(sin(200*t));
cm=fft(y,N)/N;
make_stem(n,abs(cm),'Spectrum Amplitude','n','abs(cm)');

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% File: hw10.tex
% Created: 08:44:38 Wed, 03 Nov 2021 EDT
% Last Change: 08:44:38 Wed, 03 Nov 2021 EDT
%
\documentclass[letterpaper]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{cancel}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[shortlabels]{enumitem}
\usepackage{soul}
%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
\usepackage{minted}
\usepackage{geometry}
\usepackage{dirtytalk}
\usepackage{lplfitch}
\geometry{portrait, margin=1in}
%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
\date{11/03/2021}
\title{%
Homework 10\\
\large EEET--331--01:Signals, Systems, and Transforms}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
\section{Exercise 1}
\begin{enumerate}[(a)]
\item Complete the MATLAB code to create the waveform below.
\begin{figure}[h!]
\centering
\includegraphics{exercise1.png}
\end{figure}
\inputminted[linenos]{matlab}{exercise1a.m}
\newpage
\item Extend the waveform to 8 seconds. Attach the extended plot, and MATLAB code.
\begin{figure}[h!]
\centering
\includegraphics[width=\linewidth]{exercise1b.png}
\end{figure}
\inputminted[linenos]{matlab}{exercise1b.m}
\newpage
\item Complete the MATLAB code that draws one period ($0-T$)
\inputminted[linenos]{matlab}{exercise1c.m}
\end{enumerate}
\section{Exercise 2}
\begin{enumerate}[(a)]
\item Write MATLAB code that create the half wave rectified sine wave. Submit both the plot and the MATLAB code.
\inputminted[linenos]{matlab}{exercise2a.m}
\item Write MATLAB code that draws one period ($0-T$) of a half wave rectified sine wave. Submit both the plot and the MATLAB code.
\inputminted[linenos]{matlab}{exercise2b.m}
\end{enumerate}
\newpage
\section{Exercise 3}
The MATLAB code below created a single period of a triangular waveform.
Write MATLAB code to create the waveform within three cycles of the triangular signal.
{The code uses the \verb|triangularPulse| function.
The first 3 parameters are the time of the lower left, the peak, and lower right of the triangle.}
\inputminted[linenos]{matlab}{exercise3.m}
\section{Exercise 4}
Write MATLAB code that draws one period of the waveform. Submit a plot and MATLAB code.
\inputminted[linenos]{matlab}{exercise4.m}
\newpage
\section{Exercise 5}
MATLAB's \verb|fft| function is used in the code below to create an N-point DFT from a sine wave sampled every $T_s$. In the stem plot, what harmonic does the left point at $\frac{A}{2}$ represent?
\inputminted[linenos]{matlab}{exercise5.m}
\begin{equation}
\begin{split}
harmonic=\frac{\omega_0}{\frac{2\pi}{NT_s}}=\frac{200}{\frac{2\pi}{1024*(613.56E-6)}}\approx20
\end{split}
\end{equation}
\section{Exercise 7}
Remember the derivation of $v=\frac{2f}{f_s}$, the normalized of fractional frequency.
\begin{equation}
z=e^{sT_s}=e^{j\omega T_s}=e^{j2\pi f \frac{1}{f_s}}=e^{j\pi\frac{2f}{f_s}}=e^{j\pi v}
\end{equation}
\begin{enumerate}
\item If a notch filter is designed to eliminate all frequencies at $v=0.40$, and the sampling frequency is $44.1kHz$, what signal frequency will be removed?\\
$8.82kHz$.
\item What frequency will be removed if the sample frequency is increased to $45kHz$? Assume the same filter design.\\
$9kHz$.
\end{enumerate}
\newpage
\section{Exercise 6}
The positive spectrum of a square wave that is $1$ from $0-\frac{T}{2}$ and $-1$ from $\frac{T}{2}-T$ is shown below.
This odd function will have no even terms in its Fourier Series.
The angular frequency ($\omega_0=\frac{2\pi}{T}$) times $k$ is the frequency of each non-zero point.
\begin{figure}[h!]
\centering
\includegraphics{exercise6a.png}
\end{figure}
MATLAB's \verb|fft| function is used in the code below to create an N-point DFT from a square wave sampled every $T_s$. A stem plot is shown below.
\begin{figure}[h!]
\centering
\includegraphics{exercise6b.png}
\end{figure}
In the stem plot, the MATLAB index number of the first non-zero point is 21. The next point appears lower, and is 3 times the square wave's fundamental angular frequency ($200rad/sec$). Its index is 61. What is the index of the next point?\\
101
\newpage
\section{Exercise 8}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
Remember that $x(k)$ and $y(k)$ are always zero if $k<0$.
For example, $x(n-1),\ n=0\implies x(-1)=0$.
A unit step is $1$ for all $n\ge 0$.
Fill the table below from left to right, using $f(n)$ as an intermediate step.
\begin{equation}
y(n)-0.4y(n-1)=x(n)-0.4x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c}
& & & $f(n)$ & $+0.4y(n-1)$ & $=y(n)$\\
$n$ & $-0.4x(n-1)$ & $x(n)$ & $f(n)$ & &\\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ \\
\hline
$1$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$2$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$3$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$4$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$5$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
\end{tabular}
\end{table}
\section{Exercise 9}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
\begin{equation}
y(n)+0.2y(n-2)=x(n)-0.3x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c}
& & & $f(n)$ & $-0.2y(n-2)$ & $=y(n)$\\
$n$ & $-0.3x(n-1)$ & $+x(n)$ & $=f(n)$ & &\\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ \\
\hline
$1$ & $-0.3$ & $1$ & $0.7$ & $0$ & $0.7$ \\
\hline
$2$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$3$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$4$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$5$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
\end{tabular}
\end{table}
\newpage
\section{Exercise 10}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
\begin{equation}
y(n)-0.2y(n-1)+0.4y(n-2)=x(n)-0.2x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c|c}
& & & $f(n)$ & $+0.2y(n-1)$ & $-0.4y(n-2)$ & $=y(n)$\\
$n$ & $-0.2x(n-1)$ & $x(n)$ & $f(n)$ & & & \\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ & $1$\\
\hline
$1$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$2$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$3$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$4$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$5$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
\end{tabular}
\end{table}
\end{document}

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\begin{Verbatim}[commandchars=\\\{\}]
\PYG{n}{init}\PYG{p}{();}
\PYG{n}{N}\PYG{p}{=}\PYG{l+m+mi}{20}\PYG{p}{;}
\PYG{n}{n}\PYG{p}{=}\PYG{l+m+mi}{0}\PYG{p}{:}\PYG{n}{N}\PYG{o}{\PYGZhy{}}\PYG{l+m+mi}{1}\PYG{p}{;}
\PYG{n}{num}\PYG{p}{=[}\PYG{l+m+mi}{4} \PYG{o}{\PYGZhy{}}\PYG{l+m+mi}{4}\PYG{p}{];}
\PYG{n}{den}\PYG{p}{=[}\PYG{l+m+mi}{1} \PYG{l+m+mi}{1}\PYG{p}{];}
\PYG{n}{yn}\PYG{p}{=}\PYG{n}{dimpulse}\PYG{p}{(}\PYG{n}{num}\PYG{p}{,}\PYG{n}{den}\PYG{p}{,}\PYG{n}{N}\PYG{p}{);}
\PYG{n}{make\PYGZus{}stem}\PYG{p}{(}\PYG{n}{n}\PYG{p}{,}\PYG{n}{yn}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}Impulse Response\PYGZdq{}}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}n\PYGZdq{}}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}yn\PYGZdq{}}\PYG{p}{);}
\PYG{c}{\PYGZpc{}savefig(\PYGZsq{}exercise7c.png\PYGZsq{})}
\end{Verbatim}

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\begin{Verbatim}[commandchars=\\\{\}]
\PYG{n}{init}\PYG{p}{();}
\PYG{n}{N}\PYG{p}{=}\PYG{l+m+mi}{20}\PYG{p}{;}
\PYG{n}{n}\PYG{p}{=}\PYG{l+m+mi}{0}\PYG{p}{:}\PYG{n}{N}\PYG{o}{\PYGZhy{}}\PYG{l+m+mi}{1}\PYG{p}{;}
\PYG{n}{num}\PYG{p}{=[}\PYG{l+m+mf}{0.5} \PYG{l+m+mf}{0.1} \PYG{l+m+mi}{0}\PYG{p}{];}
\PYG{n}{den}\PYG{p}{=[}\PYG{l+m+mi}{2} \PYG{l+m+mf}{3.3} \PYG{l+m+mf}{2.8}\PYG{p}{];}
\PYG{n}{yn}\PYG{p}{=}\PYG{n}{dimpulse}\PYG{p}{(}\PYG{n}{num}\PYG{p}{,}\PYG{n}{den}\PYG{p}{,}\PYG{n}{N}\PYG{p}{);}
\PYG{n}{make\PYGZus{}stem}\PYG{p}{(}\PYG{n}{n}\PYG{p}{,}\PYG{n}{yn}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}Impulse Response\PYGZdq{}}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}n\PYGZdq{}}\PYG{p}{,}\PYG{l+s}{\PYGZdq{}yn\PYGZdq{}}\PYG{p}{);}
\end{Verbatim}

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\begin{Verbatim}[commandchars=\\\{\}]
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\makeatletter
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\def\PYGZti{\char`\~}
% for compatibility with earlier versions
\def\PYGZat{@}
\def\PYGZlb{[}
\def\PYGZrb{]}
\makeatother

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init();
N=20;
n=0:N-1;
num=[4 4];
den=[1 -1.7];
yn=dimpulse(num,den,N);
make_stem(n,yn,"Impulse Response","n","yn");
%savefig('exercise7a.png')

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init();
N=20;
n=0:N-1;
num=[4 -4];
den=[1 -0.7];
yn=dimpulse(num,den,N);
make_stem(n,yn,"Impulse Response","n","yn");
%savefig('exercise7b.png')

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init();
N=20;
n=0:N-1;
num=[4 -4];
den=[1 1];
yn=dimpulse(num,den,N);
make_stem(n,yn,"Impulse Response","n","yn");
%savefig('exercise7c.png')

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init();
N=20;
n=0:N-1;
num=[0.5 0.1 0];
den=[1 3.3 -2.8];
yn=dimpulse(num,den,N);
make_stem(n,yn,"Impulse Response","n","yn");
%savefig('exercise7d.png')

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init();
N=20;
n=0:N-1;
num=[0.5 0.1 0];
den=[2 3.3 2.8];
yn=dimpulse(num,den,N);
make_stem(n,yn,"Impulse Response","n","yn");

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% File: hw11.tex
% Created: 09:56:59 Fri, 12 Nov 2021 EST
% Last Change: 09:56:59 Fri, 12 Nov 2021 EST
%
\documentclass[letterpaper]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{cancel}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[shortlabels]{enumitem}
\usepackage{soul}
%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
\usepackage{minted}
\usepackage{geometry}
\usepackage{dirtytalk}
\usepackage{lplfitch}
\usepackage{booktabs}
\geometry{portrait, margin=1in}
%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
\date{11/12/2021}
\title{%
Homework 11\\
\large EEET--331--01:Signals, Systems, and Transforms}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
%set section number to align with homework number
\setcounter{section}{9}
\section{Homework 10 Left-Overs}
%set subsection number to align with exercise number
\setcounter{subsection}{7}
\subsection{Exercise 8}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
Remember that $x(k)$ and $y(k)$ are always zero if $k<0$.
For example, $x(n-1),\ n=0\implies x(-1)=0$.
A unit step is $1$ for all $n\ge 0$.
Fill the table below from left to right, using $f(n)$ as an intermediate step.
\begin{equation}
y(n)-0.4y(n-1)=x(n)-0.4x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c}
& & & $f(n)$ & $+0.4y(n-1)$ & $=y(n)$\\
$n$ & $-0.4x(n-1)$ & $x(n)$ & $f(n)$ & &\\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ \\
\hline
$1$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$2$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$3$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$4$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$5$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
\end{tabular}
\end{table}
\subsection{Exercise 9}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
\begin{equation}
y(n)+0.2y(n-2)=x(n)-0.3x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c}
& & & $f(n)$ & $-0.2y(n-2)$ & $=y(n)$\\
$n$ & $-0.3x(n-1)$ & $+x(n)$ & $=f(n)$ & &\\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ \\
\hline
$1$ & $-0.3$ & $1$ & $0.7$ & $0$ & $0.7$ \\
\hline
$2$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$3$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$4$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
$5$ & $-0.3$ & $1$ & $0.7$ & $-0.2$ & $0.5$ \\
\hline
\end{tabular}
\end{table}
\newpage
\subsection{Exercise 10}
Use iteration to find the first five terms of the difference equation given $x(n)$ is a step function.
\begin{equation}
y(n)-0.2y(n-1)+0.4y(n-2)=x(n)-0.2x(n-1)
\end{equation}
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c|c}
& & & $f(n)$ & $+0.2y(n-1)$ & $-0.4y(n-2)$ & $=y(n)$\\
$n$ & $-0.2x(n-1)$ & $x(n)$ & $f(n)$ & & & \\
\hline
\hline
$0$ & $0$ & $1$ & $1.0$ & $0$ & $1$ & $1$\\
\hline
$1$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$2$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$3$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$4$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
$5$ & $-0.4$ & $1$ & $0.6$ & $0.4$ & $1$ \\
\hline
\end{tabular}
\end{table}
\section{Homework 11 Problems}
\subsection{Exercise 1}
\begin{enumerate}[(a)]
\item Find $H(z)=\frac{\overrightarrow Y}{\overrightarrow X}$ given the below equation :
\begin{equation}
y(n)=x(n)+0.4y(n-1)+0.4y(n-2)
\end{equation}
\begin{equation}
\begin{split}
y(n)-0.4(y(n-1)+y(n-2))=x(n)\\
\overrightarrow Y(1-0.4z^{-1}-0.4z^{-2})=\overrightarrow X(1)\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{1}{1-0.4z^{-1}-0.4z^{-2}}\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^2}{z^2-0.4z-0.4}
\end{split}
\end{equation}
\item Find the forced response for the difference equation found above, given the following input : $x(n)=10\cos(0.2167\pi n)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^2}{z^2-0.4z-0.4}\\
\overrightarrow Y=\frac{z^2}{z^2-0.4z-0.4}\overrightarrow X\\
\overrightarrow X=10\angle 0;\ z=e^{-0.2167j\pi}\\
\overrightarrow Y=11.31670654\angle -46.69444656^\circ\ (Found\ by\ calc)\\
y(n)=11.317\cos(0.2167\pi n-46.69^\circ)
\end{split}
\end{equation}
\end{enumerate}
\newpage
\subsection{Exercise 2}
\begin{enumerate}[(a)]
\item Find $H(z)=\frac{\overrightarrow Y}{\overrightarrow X}$ given the below equation :
\begin{equation}
y(n)=x(n)-0.3x(n-2)+0.2y(n-2)-0.3y(n-4)
\end{equation}
\begin{equation}
\begin{split}
y(n)=x(n)-0.3x(n-2)+0.2y(n-2)-0.3y(n-4)\\
y(n)-0.2y(n-2)+0.3y(n-4)=x(n)-0.3x(n-2)\\
\overrightarrow Y(1-0.2z^{-2}+0.3z^{-4})=\overrightarrow X(1-0.3z^{-2})\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{1-0.3z^{-2}}{1-0.2z^{-2}+0.3z^{-4}}\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^4-0.3z^2}{z^4-0.2z^2+0.3}
\end{split}
\end{equation}
\item Find the forced response for the difference equation found above, given the following input : $x(n)=12\cos(0.2222\pi n)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^4-0.3z^2}{z^4-0.2z^2+0.3}\\
\overrightarrow Y=\frac{z^4-0.3z^2}{z^4-0.2z^2+0.3}\overrightarrow X\\
\overrightarrow X=12\angle 0;\ z=e^{-0.2222j\pi}\\
\overrightarrow Y=17.27087332\angle-9.456934593^\circ\\
y(n)=17.27\cos(0.2222\pi n-9.46^\circ)
\end{split}
\end{equation}
\end{enumerate}
\subsection{Exercise 3}
\begin{enumerate}[(a)]
\item Use iteration to find the first 10 terms of the moving average difference equation given $x(n)$ is an impulse function.
Remember that an impulse function is $0$ for all values of $n$, except $n=0$.
$f(n)$ is skipped, since no old values of $y$ are included in the difference equation.
\begin{equation}
y(n)=\frac15(x(n)+x(n-1)+x(n-2)+x(n-3)+x(n-4))
\end{equation}
\begin{center}
\begin{tabular}{c||c|c|c|c|c||c}
$n$ & $x(n-4)$ & $x(n-3)$ & $x(n-2)$ & $x(n-1)$ & $x(n)$ & $y(n)$\\
\toprule
\bottomrule
$0$ & $0$ & $0$ & $0$ & $0$ & $1$ & $0.20$\\
\hline
$1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $0.20$\\
\hline
$2$ & $0$ & $0$ & $1$ & $0$ & $0$ & $0.20$\\
\hline
$3$ & $0$ & $1$ & $0$ & $0$ & $0$ & $0.20$\\
\hline
$4$ & $1$ & $0$ & $0$ & $0$ & $0$ & $0.20$\\
\hline
$5$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0.00$\\
\hline
$6$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0.00$\\
\hline
$7$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0.00$\\
\hline
$8$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0.00$\\
\hline
$9$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0.00$\\
\end{tabular}\\
\end{center}
\item Find $H(z)=\frac{\overrightarrow Y}{\overrightarrow X}$ for the moving average. Show your work.
\begin{equation}
\begin{split}
y(n)=\frac15(x(n)+x(n-1)+x(n-2)+x(n-3)+x(n-4))\\
\overrightarrow Y=\overrightarrow X\frac15(1+z^{-1}+z^{-2}+z^{-3}+z^{-4})\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{1}{5(1+z^{-1}+z^{-2}+z^{-3}+z^{-4})}\\
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^4}{5(z^4+z^{3}+z^{2}+z^{1}+1)}
\end{split}
\end{equation}
\item Find the forced response of $H(z)$ given the following input :$x(n)=14\cos(0.1278\pi n)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^4}{5(z^4+z^{3}+z^{2}+z^{1}+1)}\\
\overrightarrow Y=\frac{z^4}{5(z^4+z^{3}+z^{2}+z^{1}+1)}\overrightarrow X\\
\overrightarrow X=14\angle0;\ z=e^{-j0.1278\pi}\\
\overrightarrow Y=0.66199274246\angle46.008^\circ\\
y(n)=0.662\cos(0.1278\pi n+46^\circ)
\end{split}
\end{equation}
\end{enumerate}
\subsection{Exercise 4}
Find the difference equation for the following $H(z)$ :
\begin{equation}
H(z)=\frac{\overrightarrow Y}{\overrightarrow X}=\frac{z^4+4z^2+6z+6}{z^4-z}
\end{equation}
Show all work.
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{1+4z^{-2}+6z^{-3}+6z^{-4}}{1-z^{-3}}\\
\overrightarrow Y (1-z^{-3})=\overrightarrow X(1+4z^{-2}+6z^{-3}+6z^{-4})\\
y(n)-y(n-3)=x(n)+4x(n-2)+6x(n-3)+6x(n-4)
\end{split}
\end{equation}
\newpage
\subsection{Exercise 5}
\begin{enumerate}[(a)]
\item Complete the iteration table below:
\begin{table}[h!]
\centering
\begin{tabular}{c||c|c|c|c|c|c|c}
& & & & & $f(n)$ & $0.2 y(n-3)$ & y(n)\\
$n$ & $0.6x(n-4)$ & $0.2x(n-3)$ & $0.4x(n-2)$ & $x(n)$ & $=f(n)$ & \\
\toprule
\bottomrule
$0$ & $0.0$ & $0.0$ & $0.0$ & $1$ & $1.0$ & $0.000$ & $1.000$\\
\hline
$1$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.000$ & $0.000$\\
\hline
$2$ & $0.0$ & $0.0$ & $0.4$ & $0$ & $0.4$ & $0.000$ & $0.400$\\
\hline
$3$ & $0.0$ & $0.2$ & $0.0$ & $0$ & $0.2$ & $0.200$ & $0.400$\\
\hline
$4$ & $0.6$ & $0.0$ & $0.0$ & $0$ & $0.6$ & $0.000$ & $0.600$\\
\hline
$5$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.080$ & $0.080$\\
\hline
$6$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.080$ & $0.080$\\
\hline
$7$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.080$ & $0.080$\\
\hline
$8$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.120$ & $0.120$\\
\hline
$9$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.016$ & $0.016$\\
\hline
$10$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.024$ & $0.024$\\
\hline
$11$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.003$ & $0.003$\\
\hline
$12$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.003$ & $0.003$\\
\hline
$13$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.005$ & $0.005$\\
\hline
$14$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.001$ & $0.001$\\
\hline
$15$ & $0.0$ & $0.0$ & $0.0$ & $0$ & $0.0$ & $0.001$ & $0.001$
\end{tabular}
\end{table}
\end{enumerate}
\subsection{Exercise 6}
Confirm stability of the following difference equations by locating the poles of the transfer function ($H(z)$). Note all poles and zeros in each portion of the question.
\begin{enumerate}[(a)]
\item $y(n)=4x(n)+4x(n-1)+1.7y(n-1)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{4z+4}{z-1.7}\\
poles=+1.7;\ zeros=-1\\
unstable
\end{split}
\end{equation}
\item $y(n)=4x(n)-4x(n-1)+0.7y(n-1)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{4z-4}{z-0.7}\\
poles=+0.7;\ zeros=1\\
stable
\end{split}
\end{equation}
\item $y(n)=4x(n)-4x(n-1)-y(n-1)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{4z-4}{z-1}\\
poles=+1.0;\ zeros=1\\
unstable
\end{split}
\end{equation}
\item $y(n)=0.50x(n)+0.10x(n-1)-3.3y(n-1)+2.8y(n-2)$
\begin{equation}
\begin{split}
\frac{\overrightarrow Y}{\overrightarrow X}=\frac{0.50z^2-0.10z}{z^2+3.3z-2.8}\\
poles=0.7,-4;\ zeros=1\\
unstable
\end{split}
\end{equation}
\end{enumerate}
\subsection{Exercise 7}
Confirm stability of the following difference equations by plotting the impulse response in MATLAB. The equations are very similar (or even the same) as the previous problem.
\begin{enumerate}[(a)]
\item $y(n)=4x(n)+4x(n-1)+1.7y(n-1)$
\inputminted{matlab}{exercise7a.m}
\begin{figure}[h!]
\centering
\includegraphics[width=7cm]{exercise7a.png}
\end{figure}
\newpage
\item $y(n)=4x(n)-4x(n-1)+0.7y(n-1)$
\inputminted{matlab}{exercise7b.m}
\begin{figure}[h!]
\centering
\includegraphics[width=7cm]{exercise7b.png}
\end{figure}
\item $y(n)=4x(n)-4x(n-1)-y(n-1)$
\inputminted{matlab}{exercise7c.m}
\begin{figure}[h!]
\centering
\includegraphics[width=7cm]{exercise7c.png}
\end{figure}
\newpage
\item $y(n)=0.50x(n)+0.10x(n-1)-3.3y(n-1)+2.8y(n-2)$
\inputminted{matlab}{exercise7d.m}
\begin{figure}[h!]
\centering
\includegraphics[width=7cm]{exercise7d.png}
\end{figure}
\item Change one or more of the coefficients in the previous difference equation to make it stable. Submit the impulse response and the poles of your stable equation.
\inputminted{matlab}{exercise7e.m}
\begin{figure}[h!]
\centering
\includegraphics[width=7cm]{exercise7e.png}
\end{figure}
\end{enumerate}
\end{document}

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function init
clc; close all;
global fig_num
fig_num = 1;
end

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function make_stem(m,cm,graph_title,x_label,y_label)
global fig_num;
figure(fig_num);
fig_num = fig_num+1;
stem(m,abs(cm),'b');
grid on;
xlabel(x_label);
ylabel(y_label);
title(graph_title);
end

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%% Routh-Hurwitz stability criterion
%
% The Routh-Hurwitz stability criterion is a necessary (and frequently
% sufficient) method to establish the stability of a single-input,
% single-output(SISO), linear time invariant (LTI) control system.
% More generally, given a polynomial, some calculations using only the
% coefficients of that polynomial can lead us to the conclusion that it
% is not stable.
% Instructions
% ------------
%
% in this program you must give your system coefficients and the
% Routh-Hurwitz table would be shown
%
% Farzad Sagharchi ,Iran
% 2007/11/12
%% Initialization
clear ; close all; clc
% Taking coefficients vector and organizing the first two rows
coeffVector = input('input vector of your system coefficients: \n i.e. [an an-1 an-2 ... a0] = ');
ceoffLength = length(coeffVector);
rhTableColumn = round(ceoffLength/2);
% Initialize Routh-Hurwitz table with empty zero array
rhTable = zeros(ceoffLength,rhTableColumn);
% Compute first row of the table
rhTable(1,:) = coeffVector(1,1:2:ceoffLength);
% Check if length of coefficients vector is even or odd
if (rem(ceoffLength,2) ~= 0)
% if odd, second row of table will be
rhTable(2,1:rhTableColumn - 1) = coeffVector(1,2:2:ceoffLength);
else
% if even, second row of table will be
rhTable(2,:) = coeffVector(1,2:2:ceoffLength);
end
%% Calculate Routh-Hurwitz table's rows
% Set epss as a small value
epss = 0.01;
% Calculate other elements of the table
for i = 3:ceoffLength
% special case: row of all zeros
if rhTable(i-1,:) == 0
order = (ceoffLength - i);
cnt1 = 0;
cnt2 = 1;
for j = 1:rhTableColumn - 1
rhTable(i-1,j) = (order - cnt1) * rhTable(i-2,cnt2);
cnt2 = cnt2 + 1;
cnt1 = cnt1 + 2;
end
end
for j = 1:rhTableColumn - 1
% first element of upper row
firstElemUpperRow = rhTable(i-1,1);
% compute each element of the table
rhTable(i,j) = ((rhTable(i-1,1) * rhTable(i-2,j+1)) - ....
(rhTable(i-2,1) * rhTable(i-1,j+1))) / firstElemUpperRow;
end
% special case: zero in the first column
if rhTable(i,1) == 0
rhTable(i,1) = epss;
end
end
%% Compute number of right hand side poles(unstable poles)
% Initialize unstable poles with zero
unstablePoles = 0;
% Check change in signs
for i = 1:ceoffLength - 1
if sign(rhTable(i,1)) * sign(rhTable(i+1,1)) == -1
unstablePoles = unstablePoles + 1;
end
end
% Print calculated data on screen
fprintf('\n Routh-Hurwitz Table:\n')
rhTable
% Print the stability result on screen
if unstablePoles == 0
fprintf('~~~~~> it is a stable system! <~~~~~\n')
else
fprintf('~~~~~> it is an unstable system! <~~~~~\n')
end
fprintf('\n Number of right hand side poles =%2.0f\n',unstablePoles)
reply = input('Do you want roots of system be shown? Y/N ', 's');
if reply == 'y' || reply == 'Y'
sysRoots = roots(coeffVector);
fprintf('\n Given polynomial coefficients roots :\n')
sysRoots
end

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\begin{Verbatim}[commandchars=\\\{\}]
\PYG{n}{test2}
\end{Verbatim}

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\begin{Verbatim}[commandchars=\\\{\}]
\PYG{n}{test}
\end{Verbatim}

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\makeatother

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[z,p,k]=butter(2,1,'s');
[num1,den1]=zp2tf(z,p,k);
w=1;
v=0.5;
c=w/(tan(v*pi/2));
[dnum1,dden1]=bilinear(num1,den1,C/2)
% dnum1 =
% 0.2929 0.5858 0.2929
% dden1 =
% 1.0000 -0.0000 0.1716

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% File: hw12.tex
% Created: 21:25:01 Wed, 17 Nov 2021 EST
% Last Change: 21:25:01 Wed, 17 Nov 2021 EST
%
\documentclass[letterpaper]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{cancel}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[shortlabels]{enumitem}
\usepackage{lipsum}
\usepackage{soul}
\usepackage{geometry}
\usepackage{minted}
\geometry{portrait, margin=1in}
\date{11/17/2021}
\title{%
Homework 12\\
\large EEET--331--01:Signals, Systems, and Transforms}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
\section{Exercise 1}
Convert a second order Chebyshev low-pass filter with $0.5dB$ of ripple to a digital filter using the bilinear transform. Set $\omega_{co}=1;\ v_{co}=0.30$.
\begin{equation}
\begin{split}
prototype=\frac{1.431\overline4}{p^2+1.4256p+1.5162}\\
\omega_{co}=C\tan\left(\frac{v\pi}{2}\right)\\
C=\frac{\omega_{co}}{\tan\left(\frac{v\pi}{2}\right)}\\
C=\frac{1}{\tan\left(\frac{0.3\pi}{2}\right)}\\
C\approx1.9626\\
P=\frac{\omega_{co}}{s}\\
s=C\left(\frac{z-1}{z+1}\right)\\
P=\frac{\omega_{co}}{C\left(\frac{z-1}{z+1}\right)}\\
P=\frac{\omega_{co}}{\frac{\omega_{co}}{\tan\left(\frac{v\pi}{2}\right)}\left(\frac{z-1}{z+1}\right)}\\
P=\frac{\tan\left(\frac{v\pi}{2}\right)(z+1)}{z-1}\\
P=\frac{1.9626(z+1)}{z-1}\\
H(z)=\frac{1.431\overline4 (z-1)^2}{(1.9626)^2(z+1)^2+1.4256*1.9296*(z+1)(z-1)+1.5162(z-1)^2}\\
H(z)=\frac{1.431\overline4 (z^2-2z+1)}{(1.9626)^2(z^2+2z+1)+1.4256*1.9296*(z^2-1)+1.5162(z^2-2z+1)}\\
H(z)=\frac{1.431\overline4 (z^2-2z+1)}{8.1659z^2+0.4464z-2.5701}\\
H(z)=\frac{1.431\overline4 (z-1)^2}{8.1659(z+0.6183)(z-0.5090)}
\end{split}
\end{equation}
\section{Exercise 2}
\begin{enumerate}[(a)]
\item Convert a second order Butterworth high-pass filter to a digital filter using the bilinear transform. Use the same $\omega$ and $v$ as in Exercise 1.
\begin{equation}
\begin{split}
prototype=\frac{1}{p^2+1.4142p+1}\\
p=\frac{s}{\omega_{co}}\\
s=C\left(\frac{z-1}{z+1}\right)\\
C=\frac{\omega_{co}}{\tan\left(\frac{v\pi}{2}\right)}\\
s=\frac{\omega_{co}}{\tan\left(\frac{v\pi}{2}\right)}\left(\frac{z-1}{z+1}\right)\\
p=\frac{1}{\tan\left(\frac{v\pi}{2}\right)}\left(\frac{z-1}{z+1}\right)\\
p=1.9626\left(\frac{z-1}{z+1}\right)\\
H(z)=\frac{(z+1)^2}{1.9626^2(z-1)^2+1.4142*1.9626*(z-1)(z+1)+(z+1)^2}\\
H(z)=\frac{(z+1)^2}{3.8518(z^z-2z+1)+2.7755(z^2-1)+z^2+2z+1}\\
H(z)=\frac{(z+1)^2}{7.6274z^2+5.7037z+2.0763}
\end{split}
\end{equation}
\item Confirm the transfer function using the MATLAB code below:
\inputminted{matlab}{exercise2.m}
\end{enumerate}
\newpage
\section{Exercise 3}
\begin{enumerate}[(a)]
\item Convert a first-order Butterworth low-pass filter to a bandpass digital filter using the bilinear transform. Set $\omega_{co}=1;\ v_l=0.40;\ v_h=0.80$.
\begin{equation}
\begin{split}
prototype=\frac{1}{p+1}\\
p=\frac{s^2+\omega_c^2}{Bs}\\
B=C\left(\tan\left(\frac{v_H\pi}{2}\right)+\tan\left(\frac{v_L\pi}{2}\right)\right)\\
C=\sqrt{\frac{1}{\tan\left(\frac{v_H\pi}{2}\right)+\tan\left(\frac{v_L\pi}{2}\right)}}\\
s=C\left(\frac{z-1}{z+1}\right)\\
C=0.5127;\
B=3.8042\\
s=0.5127\left(\frac{z-1}{z+1}\right)\\
p=\frac{\left(0.5127\left(\frac{z-1}{z+1}\right)\right)^2+1}{3.8042\left(0.5127\left(\frac{z-1}{z+1}\right)\right)}\\
H(z)=\frac{3.8042\left(0.5127\left(\frac{z-1}{z+1}\right)\right)}
{\left(0.5127\left(\frac{z-1}{z+1}\right)\right)^2+3.8042\left(0.5127\left(\frac{z-1}{z+1}\right)\right)}\\
H(z)=\frac{3.8042\left(0.5127\left(z-1\right)z+1\right)}
{0.5127\left(\left(z-1\right)\right)^2+3.8042\left(0.5127(z-1)*(z+1)\right)}\\
H(z)=\frac{(z-1)(z+1)}
{0.5127\left(\left(z-1\right)\right)^2+(z-1)*(z+1)}\\
H(z)=\frac{(z-1)(z+1)}{1.5127z^2-1.0254z+1.5127}
\end{split}
\end{equation}
\item Complete the MATLAB code that converts the given first-order BP Butterworth filter to a digital filter.
\begin{equation}
H_{bp}(s)=\frac{1.5723s}{s^2+1.5723s+1}
\end{equation}
\inputminted{matlab}{exercise3.m}
\end{enumerate}
\newpage
\section{Exercise 4}
\begin{enumerate}[(a)]
\item Convert a first-order Chebyshev $0.5dB$ low-pass filter to a stop-band digital filter using the bilinear transform. Set $\omega_{c}=1;\ v_l=0.10;\ v_h=0.30$.
\begin{equation}
\begin{split}
prototype=\frac{2.7028}{p+2.8628}\\
p=\frac{Bs}{s^2+\omega_c^2}\\
B=C\left(\tan\left(\frac{v_H\pi}{2}\right)+\tan\left(\frac{v_L\pi}{2}\right)\right)\\
C=\sqrt{\frac{1}{\tan\left(\frac{v_H\pi}{2}\right)+\tan\left(\frac{v_L\pi}{2}\right)}}\\
s=C\left(\frac{z-1}{z+1}\right)\\
C=1.2236;\
B=0.8173\\
s=1.2236\left(\frac{z-1}{z+1}\right)\\
p=\frac{0.8173(1.2236\left(\frac{z-1}{z+1}\right))}{\left(1.2236\left(\frac{z-1}{z+1}\right)\right)^2+1}\\
H(z)=\frac{2.7028}{\frac{0.8173(1.2236\left(\frac{z-1}{z+1}\right))}{\left(1.2236\left(\frac{z-1}{z+1}\right)\right)^2+1}+2.8628}\\
H(z)=\frac{2.7028(\left(1.2236\left(\frac{z-1}{z+1}\right)\right)^2+1)}
{{0.8173(1.2236\left(\frac{z-1}{z+1}\right))}+2.8628(\left(1.2236\left(\frac{z-1}{z+1}\right)\right)^2+1)}\\
H(z)=\frac{2.7028(\left(1.2236\left(z-1\right)\right)^2+(z+1)^2)}
{{0.8173(1.2236\left(z+1\right))}(z+1)+2.8628(\left(1.2236^2\left(z-1\right)^2\right)+(z+1)^2)}\\
H(z)=\frac{2.7028(\left(1.2236^2(z^2-2z+1)\right)+(z^2+2z+1))}
{{0.8173(1.2236(z^2-1))}+2.8628(\left(1.2236^2(z^2-2z+1)\right)+(z^2+2z+1))}\\
H(z)=\frac{4.0467(z^2-2z+1)+2.7028(z^2+2z+1)}
{{(z^2-1)}+4.2862(z^2-2z+1)+(z^2+2z+1)}\\
H(z)=\frac{6.7495z^2+1.3439z+6.7495}
{6.2862z^2-2.2862z+4.2862}\\
\end{split}
\end{equation}
\item Complete the MATLAB code that converts a first-order BP Butterworth filter to a digital filter.
\begin{equation}
H_{bp}(s)=\frac{s^2+1}{s^2+0.4317s+1}
\end{equation}
\inputminted{matlab}{exercise4.m}
\end{enumerate}
\end{document}

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\PYG{n}{iir2fir}\PYG{p}{=}\PYG{n}{dimpulse}\PYG{p}{([}\PYG{l+m+mi}{1} \PYG{l+m+mi}{0} \PYG{l+m+mi}{0}\PYG{p}{],[}\PYG{l+m+mi}{1} \PYG{l+m+mi}{0} \PYG{o}{\PYGZhy{}}\PYG{l+m+mf}{0.9}\PYG{p}{],}\PYG{l+m+mi}{8}\PYG{p}{)}
\end{Verbatim}

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\makeatletter
\def\PYG@reset{\let\PYG@it=\relax \let\PYG@bf=\relax%
\let\PYG@ul=\relax \let\PYG@tc=\relax%
\let\PYG@bc=\relax \let\PYG@ff=\relax}
\def\PYG@tok#1{\csname PYG@tok@#1\endcsname}
\def\PYG@toks#1+{\ifx\relax#1\empty\else%
\PYG@tok{#1}\expandafter\PYG@toks\fi}
\def\PYG@do#1{\PYG@bc{\PYG@tc{\PYG@ul{%
\PYG@it{\PYG@bf{\PYG@ff{#1}}}}}}}
\def\PYG#1#2{\PYG@reset\PYG@toks#1+\relax+\PYG@do{#2}}
\expandafter\def\csname PYG@tok@w\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}}
\expandafter\def\csname PYG@tok@c\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYG@tok@cp\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.74,0.48,0.00}{##1}}}
\expandafter\def\csname PYG@tok@k\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@kp\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@kt\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}}
\expandafter\def\csname PYG@tok@o\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@ow\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
\expandafter\def\csname PYG@tok@nb\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@nf\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYG@tok@nc\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYG@tok@nn\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYG@tok@ne\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.82,0.25,0.23}{##1}}}
\expandafter\def\csname PYG@tok@nv\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@no\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.53,0.00,0.00}{##1}}}
\expandafter\def\csname PYG@tok@nl\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.63,0.63,0.00}{##1}}}
\expandafter\def\csname PYG@tok@ni\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.60,0.60,0.60}{##1}}}
\expandafter\def\csname PYG@tok@na\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.49,0.56,0.16}{##1}}}
\expandafter\def\csname PYG@tok@nt\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@nd\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
\expandafter\def\csname PYG@tok@s\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@sd\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@si\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
\expandafter\def\csname PYG@tok@se\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}}
\expandafter\def\csname PYG@tok@sr\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
\expandafter\def\csname PYG@tok@ss\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@sx\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@m\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@gh\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
\expandafter\def\csname PYG@tok@gu\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.50,0.00,0.50}{##1}}}
\expandafter\def\csname PYG@tok@gd\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.63,0.00,0.00}{##1}}}
\expandafter\def\csname PYG@tok@gi\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.63,0.00}{##1}}}
\expandafter\def\csname PYG@tok@gr\endcsname{\def\PYG@tc##1{\textcolor[rgb]{1.00,0.00,0.00}{##1}}}
\expandafter\def\csname PYG@tok@ge\endcsname{\let\PYG@it=\textit}
\expandafter\def\csname PYG@tok@gs\endcsname{\let\PYG@bf=\textbf}
\expandafter\def\csname PYG@tok@gp\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
\expandafter\def\csname PYG@tok@go\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.53,0.53,0.53}{##1}}}
\expandafter\def\csname PYG@tok@gt\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.27,0.87}{##1}}}
\expandafter\def\csname PYG@tok@err\endcsname{\def\PYG@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}}
\expandafter\def\csname PYG@tok@kc\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@kd\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@kn\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@kr\endcsname{\let\PYG@bf=\textbf\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@bp\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYG@tok@fm\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYG@tok@vc\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@vg\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@vi\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@vm\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYG@tok@sa\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@sb\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@sc\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@dl\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@s2\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@sh\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@s1\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYG@tok@mb\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@mf\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@mh\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@mi\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@il\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@mo\endcsname{\def\PYG@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYG@tok@ch\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYG@tok@cm\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYG@tok@cpf\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYG@tok@c1\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYG@tok@cs\endcsname{\let\PYG@it=\textit\def\PYG@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\def\PYGZbs{\char`\\}
\def\PYGZus{\char`\_}
\def\PYGZob{\char`\{}
\def\PYGZcb{\char`\}}
\def\PYGZca{\char`\^}
\def\PYGZam{\char`\&}
\def\PYGZlt{\char`\<}
\def\PYGZgt{\char`\>}
\def\PYGZsh{\char`\#}
\def\PYGZpc{\char`\%}
\def\PYGZdl{\char`\$}
\def\PYGZhy{\char`\-}
\def\PYGZsq{\char`\'}
\def\PYGZdq{\char`\"}
\def\PYGZti{\char`\~}
% for compatibility with earlier versions
\def\PYGZat{@}
\def\PYGZlb{[}
\def\PYGZrb{]}
\makeatother

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\makeatletter
\def\PYGdefault@reset{\let\PYGdefault@it=\relax \let\PYGdefault@bf=\relax%
\let\PYGdefault@ul=\relax \let\PYGdefault@tc=\relax%
\let\PYGdefault@bc=\relax \let\PYGdefault@ff=\relax}
\def\PYGdefault@tok#1{\csname PYGdefault@tok@#1\endcsname}
\def\PYGdefault@toks#1+{\ifx\relax#1\empty\else%
\PYGdefault@tok{#1}\expandafter\PYGdefault@toks\fi}
\def\PYGdefault@do#1{\PYGdefault@bc{\PYGdefault@tc{\PYGdefault@ul{%
\PYGdefault@it{\PYGdefault@bf{\PYGdefault@ff{#1}}}}}}}
\def\PYGdefault#1#2{\PYGdefault@reset\PYGdefault@toks#1+\relax+\PYGdefault@do{#2}}
\expandafter\def\csname PYGdefault@tok@w\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}}
\expandafter\def\csname PYGdefault@tok@c\endcsname{\let\PYGdefault@it=\textit\def\PYGdefault@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
\expandafter\def\csname PYGdefault@tok@cp\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.74,0.48,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@k\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@kp\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@kt\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}}
\expandafter\def\csname PYGdefault@tok@o\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@ow\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nb\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nf\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nc\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nn\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@ne\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.82,0.25,0.23}{##1}}}
\expandafter\def\csname PYGdefault@tok@nv\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@no\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.53,0.00,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nl\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.63,0.63,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@ni\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.60,0.60,0.60}{##1}}}
\expandafter\def\csname PYGdefault@tok@na\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.49,0.56,0.16}{##1}}}
\expandafter\def\csname PYGdefault@tok@nt\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@nd\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@s\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@sd\endcsname{\let\PYGdefault@it=\textit\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@si\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
\expandafter\def\csname PYGdefault@tok@se\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@sr\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}}
\expandafter\def\csname PYGdefault@tok@ss\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@sx\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@m\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@gh\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
\expandafter\def\csname PYGdefault@tok@gu\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.50,0.00,0.50}{##1}}}
\expandafter\def\csname PYGdefault@tok@gd\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.63,0.00,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@gi\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.63,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@gr\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{1.00,0.00,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@ge\endcsname{\let\PYGdefault@it=\textit}
\expandafter\def\csname PYGdefault@tok@gs\endcsname{\let\PYGdefault@bf=\textbf}
\expandafter\def\csname PYGdefault@tok@gp\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}}
\expandafter\def\csname PYGdefault@tok@go\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.53,0.53,0.53}{##1}}}
\expandafter\def\csname PYGdefault@tok@gt\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.27,0.87}{##1}}}
\expandafter\def\csname PYGdefault@tok@err\endcsname{\def\PYGdefault@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}}
\expandafter\def\csname PYGdefault@tok@kc\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@kd\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@kn\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@kr\endcsname{\let\PYGdefault@bf=\textbf\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@bp\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@fm\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}}
\expandafter\def\csname PYGdefault@tok@vc\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@vg\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@vi\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@vm\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}}
\expandafter\def\csname PYGdefault@tok@sa\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@sb\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@sc\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@dl\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@s2\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@sh\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@s1\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}}
\expandafter\def\csname PYGdefault@tok@mb\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@mf\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@mh\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@mi\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
\expandafter\def\csname PYGdefault@tok@il\endcsname{\def\PYGdefault@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}}
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num1=[0.0628 0];den1=[1 -2*0.9980 1];
y1=dimpulse(num1,den1);
figure(1);plot(y1);axis([0, 500, -1, 1]);title('Figure 1: Sine Wave');

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num1=[0.9980 1];den1=[1 -2*0.9980 1];
y1=dimpulse(num1,den1);
figure(1);plot(y1);axis([0, 500, -35, 35]);title('Figure 2: Sine Wave');
exportgraphics(gcf,'exercise1Bb.eps','ContentType','vector');

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% File: hw13.tex
% Created: 21:57:46 Thu, 02 Dec 2021 EST
% Last Change: 21:57:46 Thu, 02 Dec 2021 EST
%
\documentclass[letterpaper]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{cancel}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[shortlabels]{enumitem}
\usepackage{lipsum}
\usepackage{soul}
\usepackage{geometry}
\usepackage{minted}
\usepackage{tikz}
\usepackage{polynom}
\geometry{portrait, margin=1in}
\date{12/02/2021}
\title{%
Homework 13\\
\large EEET--331--01: Signals, Systems, and Transforms}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
\section{Exercise 1}
\begin{enumerate}[(a)]
\item Convert the following difference equations into transfer functions $H(z)$.
\begin{enumerate}[a.]
\item $y(n)=0.0628x(n-1)+2(0.9980)y(n-1)-y(n-2)$
\item $y(n)=x(n)-0.9980x(n-1)+2(0.9980)y(n-1)-y(n-2)$
\end{enumerate}
\item MATLAB Impulse Response
\begin{enumerate}[a.]
\item Study the MATLAB code that applies an impulse to the first difference equation (Exercise 1.a.a) and plot the results.\\
\inputminted[linenos]{matlab}{exercise1Ba.m}
\item Create a similar plot for the second difference equation.
\inputminted[linenos]{matlab}{exercise1Bb.m}
\begin{figure}[h!]
\centering
\includegraphics[width=5cm]{exercise1Bb.eps}
\end{figure}
\end{enumerate}
\end{enumerate}
\section{Exercise 2}
\begin{enumerate}[(a)]
\item Convert the difference equation below to a transfer function, $H(z)$.
\begin{equation}
y(n)=x(n)-0.90y(n-2)
\end{equation}
\begin{equation}
\begin{split}
H(z)=\frac{Y(z)}{X(z)}\\
H(z)=\frac{z^2}{z^2-0.9}
\end{split}
\end{equation}
\item Convert the transfer function to an eighth-order FIR difference equation using long-division.
\begin{equation}
1+0.9z^{-2}+0.81z^{-4}+0.729z^{-6}+0.6561z^{-8}
\end{equation}
\item Complete the MATLAB code to convert the transfer function to an eighth-order FIR difference equation.\\
\mintinline{matlab}{iir2fir=dimpulse([1 0 0],[1 0 -0.9],8)}
\end{enumerate}
\section{Exercise 3}
Use the definition for a LaPlace transform to show the trasform for the step function $6u(t)$ is $\frac{6}{s}$.
\begin{equation}
\begin{split}
F(s)=\int^{\infty}_{0}f(t)e^{-st}dt\\
f(t)=6u(t)\\
F(s)=\int^{\infty}_{0}(6u(t))e^{-st}dt\\
F(s)=\frac{6u(t)e^{-st}}{-s}|^{\infty}_{0}\\
F(s)=\frac{-1}{s}(6u(\infty)e^{-\infty}-6u(0)e^{0})\\
F(s)=\frac{-(0-6)}{s}\\
F(s)=\frac{6}{s}
\end{split}
\end{equation}
\section{Exercise 4}
Find the inverse LaPlace transform of the function below, then find $h(t); t=2s$.
\begin{equation}
H(s)=\frac{5s+2}{s(s+1)(s+2)}
\end{equation}
\begin{equation}
\begin{split}
5s+2=A(s+1)(s+2)+B(s^2+s)+C(s^2+2s)\\
A+B+C=0\\
3A+B+2C=5\\
2A=2\\
A=1\\
B+C=-1\\
B+2C=2\\
B=-1-C\\
-1-C+2C=2\\
C=3\\
B+C=-1\\
B=-4\\
H(s)=\frac{1}{s}+\frac{-4}{s+2}+\frac{3}{s+1}\\
H(t)=1-4(e^{-2t})+3(e^{-t})\\
H(2)=1-4e^{-4}+3e^{-2}\\
H(2)=1.3327
\end{split}
\end{equation}
\section{Exercise 5}
Find the inverse LaPlace transform of the below function, then find $h(t); t=2s$.
\begin{figure}[h!]
\centering
\includegraphics[width=12cm]{exercise5.jpg}
\end{figure}
\begin{equation}
\begin{split}
h(t)=0.5u(t)+3.54e^{-t}\cos(t-98.13)\\
h(2)\approx0.4488
\end{split}
\end{equation}
Clearly I'm not doing something right here, but I can't wrap my head around how this is "supposed" to be solved\dots
\section{Exercise 6}
Find the inverse LaPlace transform for the below function, then find $h(t);\ t=2s$.
\begin{equation}
\begin{split}
H(s)=\frac{6s-4}{s^2(s+1)}\\
6s-4=A(s(s+1))+B(s+1)+C(s^2)\\
6s-4=As^2+A+Bs+B+Cs^2\\
A+C=0\\
B=6\\
A+B=-4\\
A=-10\\
C=10\\
H(s)=\frac{-10}{s}+\frac{6}{s^2}+\frac{10}{s+1}\\
h(t)=-10+6t+10e^{-t}\\
h(2)=-10+12+10e^{-2}\\
h(2)=3.3534
\end{split}
\end{equation}
\section{Exercise 7}
A $3V$ power supply is turned on, powering a series RL circuit.
The circuit can be found using KVL:
\begin{equation}
3u(t)=Ri(t)+L\frac{di(t)}{dt}
\end{equation}
Given the following parameters, you can find current to be the following:
\begin{equation}
\begin{split}
R=8\Omega;\ L=1H;\ I_{0}=2A\\
\frac{3}{s}=8I+sI+2\\
I(s)=\frac{3+2s}{s(s+8)}
\end{split}
\end{equation}
Use the reverse LaPlace transform to find $i(t)$, and the current in the coil after 1 second.
Note that $i_L=i_R=i_{1s}$.
\begin{equation}
\begin{split}
3+2s=A(s+8)+B(s)\\
A+B=2\\
8A=3\\
A=\frac38\\
B=\frac{13}{8}\\
I(s)=\frac38(\frac{1}{s})+\frac{13}{8}(\frac{1}{s+8})\\
i(t)=\frac38+\frac{13}{8}(e^{-8t})\\
i(1)=\frac38+\frac{13}{8}(e^{-8})\\
i(1)=0.3755
\end{split}
\end{equation}
\end{document}

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\documentclass[letterpaper]{article}
\usepackage{enumitem}
\usepackage{cancel}
\usepackage{amsmath}
\usepackage{listings}
\usepackage{graphicx}
\title{Homework 2\\ EEET-331-01: Signals, Systems and Transformers}
\date{09/07/2021}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\section{Exercise 1.} Evaluate $f(x)$ at the given points.
\begin{enumerate}[label=\alph*)]
\item $f(x)=\frac{x^2-4}{x-2}@x=-2$ \[f(-2)=\frac{(-2)^2-4}{-2-2}\]
\[f(-2)=\frac{4-4}{-4}\] \[\overline{\underline{|f(-2)=0|}}\]
\item $f(x)=\frac{x^2-4}{x-2}@x=2$ \[f(2)=\frac{(2)^2-4}{2-2}\]
\[f(2)=\frac{4-4}{0}\]
\[lim_{x=2}\frac{\frac{d}{dx}(x^2-4)}{\frac{d}{dx}(x-2)}\]
\[lim_{x=2}\frac{2x}{1}\] \[\frac{2(2)}{1}=4\]
\[\overline{\underline{|f(2)=4|}}\]
\newpage \item $f(x)=\frac{1-\cos(3x)}{x}@x=0$\\
\[lim_{x=0}\frac{1-\cos(3x)}{x}\]
\[lim_{x=0}\frac{\frac{d}{dx}(1-\cos(3x))}{\frac{d}{dx}(x)}\]
\[lim_{x=0}\frac{3\sin(3x)}{1}\] \[f(0)=3(\sin(3(0)))\]
\[\overline{\underline{|f(0)=0|}}\]
\item $f(x)=\frac{sin(3x)}{6x}@x=0$ \[lim_{x=0}\frac{\sin(3x)}{6x}\]
\[lim_{x=0}\frac{\frac{d}{dx}(\sin3x)}{\frac{d}{dx}(6x)}\]
\[lim_{x=0}\frac{3\cos(3x)}{6}\] \[lim_{x=0}\frac12\cos(3x)\]
\[f(0)=\frac12\cos(3(0))\] \[\overline{\underline{|f(0)=\frac12|}}\]
\newpage
\item $f(x)=\frac{x^3-9x^2-12x+20}{x^2-4x+3}@=1$
\[lim_{x=1}=\frac{x^3-9x^2-12x+20}{x^2-4x+3}\]
\[lim_{x=1}=\frac{1^3-9(1)^2-12(1)+20}{1^2-4(1)+3}=\frac00\]
\[lim_{x=1}=\frac{\frac{d}{dx}(x^3-9x^2-12x+20)}{\frac{d}{dx}(x^2-4x+3)}\]
\[lim_{x=1}=\frac{3x^2-18x-12}{2x-4}\]
\[f(1)=\frac{3(1)^2-18(1)-12}{2(1)-4}\] \[f(1)=\frac{3-18-12}{2-4}\]
\[f(1)=\frac{-27}{-2}\]
\[\overline{\underline{|f(1)=\frac{27}{2}=13.5|}}\] \end{enumerate}
\section{Exercise 2.} Determine the value of $x$ in the following lines of
MATLAB code.
\begin{enumerate}[label=\alph*)]
\item
\verb|x1=linspace(0,9,5)|
\begin{lstlisting}
x1=[0 2.25 4.5 6.75 9]
\end{lstlisting}
\item
\verb|x2=0:5:15|
\begin{lstlisting}
x2=[0 5 10 15]
\end{lstlisting}
\item
\verb|x3=[3 4 5]+[1 2 4]|
\begin{lstlisting}
x3=[4 6 9]
\end{lstlisting}
\item
\verb|x4=[3 4 5].+[1 2 4]|
\begin{lstlisting}
x4=[4 6 9]
\end{lstlisting}
\item
\verb|x5=[3 4 5].*[1 2 4]|
\begin{lstlisting}
x5=[3 8 20]
\end{lstlisting}
\end{enumerate}
\section{Exercise 3.}
\begin{enumerate}[label=\alph*)]
\item
Using \verb|linspace|, create a vector starting at 2, and ending at 6, that has 5 points.
\begin{lstlisting}
vector1=linspace(2,6,5)
\end{lstlisting}
\item
Using the colon notation in exercise 2, question b, create a vector starting at 2, and ending at 6, that has 5 points.
\begin{lstlisting}
vector2=2:6
\end{lstlisting}
\item
Is the dot required in exercise 2, question d?\\
The dot is not required. The results in both instances are the same, in theory. When you attempt to run the line in exercise 2, question d, the line will fail, and throw an error.
\item
Is the dot required in 2e?\\
The dot is not necessarily required. However, the equation with the dot outputs a different result than the equation without the dot.
\end{enumerate}
\section{Exercise 4.} Plug in the Euler Lite vectors into the differential equations. Take the derivatives and leave as a function of $s$.
\begin{enumerate}[label=\alph*)]
\item
$v(t)=L\frac{di(t)}{dt}$ where $i(t)=\overrightarrow Ie^{st}$, $L=3H$, and $v(t)=\overrightarrow Ve^{st}$\\
\[\overrightarrow Ve^{st}=3s\overrightarrow Ie^{st}\]
\[\overline{\underline{|3s=\overrightarrow V/\overrightarrow I|}}\]
\item
$v(t)=3\frac{d^2i(t)}{dt^2}+8\frac{di(t)}{dt}+2i(t)$ where $i(t)=\overrightarrow Ie^{st}$ and $v(t)=\overrightarrow Ve^{st}$
\[\overrightarrow V=3s^2\overrightarrow I+8s\overrightarrow I+2\overrightarrow I\]
\[\overline{\underline{|\overrightarrow V/\overrightarrow I=3s^2+8s+2|}}\]
\item
$\frac{d^3v(t)}{dt^3}+2\frac{dv(t)}{dt}+v(t)=7\frac{d^2i(t)}{dt^2}+\frac{di(t)}{dt}+9i(t)$ where $i(t)=\overrightarrow Ie^{st}$ and $v(t)=\overrightarrow Ve^{st}$\\
\[s^3\overrightarrow V+2s^2\overrightarrow V+\overrightarrow V=7s^2\overrightarrow I + s\overrightarrow I + 9\overrightarrow I\]
\[\overline{\underline{|\frac{\overrightarrow V}{\overrightarrow I}=\frac{7s^2+s+9}{s^3+2s^2+1}|}}\]
\end{enumerate}
\section{Exercise 5.}Determine the Euler Phasor, the Euler Lite Phasor, and $s$.
\begin{enumerate}[label=\alph*)]
\item
$i(t)=3e^{-3t}$\\
Euler Phasor: $\frac32\angle0$\\
Euler Lite Phasor: $3\angle0$\\
$s=-3$
\item
$i(t)=2$\\
Euler Phasor: $1\angle0$\\
Euler Lite Phasor: $2\angle0$\\
$s=0$
\item
$i(t)=5\cos(2t+10)$\\
Euler Phasor: $\frac52\angle10^{\circ}$\\
Euler Lite Phasor: $5\angle10^{\circ}$\\
$s=j2$
\item
$i(t)=5e^{-2t}\cos(2t+10)$\\
Euler Phasor: $\frac52\angle10^{\circ}$\\
Euler Lite Phasor: $5\angle10^{\circ}$\\
$s=-2+j2$
\end{enumerate}
\section{Exercise 6.}Convert the differential equations to transfer functions.
\begin{enumerate}[label=\alph*)]
\item
$\frac{d^2i}{dt^2}+3\frac{di}{dt}+4i=4\frac{d^2v}{dt^2}+2\frac{dv}{dt}+8v$ where $v(t)=\overrightarrow Ve^{st}$ and $i(t)=\overrightarrow Ie^{st}$\\
\[\overrightarrow I(s^2+3s+4)=\overrightarrow V(4s^2+2s+8)\]
\[\overline{\underline{|\frac{\overrightarrow V}{\overrightarrow I}=\frac{s^2+3s+4}{4s^2+2s+8}|}}\]
\item
$\frac{d^3i}{dt^3}-6\frac{d^2i}{dt^2}+4i=4v$ given $v(t)=\overrightarrow Ve^{st}$ and $i(t)=\overrightarrow Ie^{st}$\\
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{4}{s^3-6s+4}\]
\end{enumerate}
\newpage
\section{Exercise 7.}For the functions in Exercise 5, problems b and d, use the transfer function in Exercise 6, problem a to find $v(t)$.
\begin{enumerate}[label=\alph*)]
\item
$\overrightarrow I=2\angle0$, $s=0$\\
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{s^2+3s+4}{4s^2+2s+8}\]
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{(0)^2+3(0)+4}{4(0)^2+2(0)+8}\]
\[\overrightarrow V= \frac12*\overrightarrow I\]
\[\overrightarrow V=\frac12*(2\angle0)\]
\[\overline{\underline{|v(t)=1|}}\]
\item
$\overrightarrow I=5\angle10^{\circ}$, $s=-2+j2$\\
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{s^2+3s+4}{4s^2+2s+8}\]
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{(-2+j2)^2+3(-2+j2)+4}{4(-2+j2)^2+2(-2+j2)+8}\]
\[\frac{\overrightarrow V}{\overrightarrow I}=\frac{-2-j2}{4-j28}=0.1\angle-53.13^{\circ}\]
\[\overrightarrow V=(0.1\angle-53.13^{\circ})*\overrightarrow I=0.5\angle-43.13^{\circ}\]
\[\overline{\underline{|v(t)=0.5e^{-2t}\cos(2t-43.13^{\circ})|}}\]
\end{enumerate}
\newpage
\section{Exercise 8.}Write the differential equations for the circuits in the figure below.
\begin{figure}[h!]
\includegraphics{exercise8Circuits.png}
%\begin{center}
%\begin{circuitikz}[american]
%\draw (0,0)
%to [I,i=$I_s$];
%\end{circuitikz}
%\end{center}
\end{figure}
\newpage
\begin{enumerate}[label=\alph*)]
\item
Use Kirchhoff's Current Law to write an equation for circuit 1 in terms of $i_s$ and $i_c$.
\[i_s=i_c+i_r\]
\[i_s=i_c+\frac{v_r}{R_r}\]
Since the capacitor and the resistor are in parallel, $V_r=V_c$:
\[i_s=i_c+\frac{V_c}{R_r}\]
Since current is the derivative of voltage with respect to time:
\[i_s=i_c+\frac{\int i_cdt}{R_r}\]
Since $R_r=5\Omega$:
\[\overline{\underline{|i_s=i_c+\frac15\int i_cdt|}}\]
\item
Use Kirchhoff's Voltage Law to write an equation for circuit 2 in terms of $v_s$ and $i$.
\[v_s=v_l+v_r\]
Since $v_L=L\frac{di}{dt}$ and $v_R=Ri$:
\[v_s=L\frac{di}{dt}+Ri\]
Since $R=3\Omega$ and $L=1H$:
\[\overline{\underline{|v_s=1H\frac{di}{dt}+3\Omega i|}}\]
\end{enumerate}
\newpage
\section{Exercise 9.}The differential equations for the RLC circuit in the image below are given in the following equations. When a $5V$ step is applied for $V_{IN}$, various output waveforms with different $R_1$ values are shown in Figures 3 through 5. Which resistor value creates the largest peak output voltage? What is that peak output voltage?\footnote{\label{1} Figures 2-5 have been omitted to reduce file size.}
The resistor value with the highest output voltage is $R=0.05\Omega$. The voltage peak for this resistor value is $\approx 6V$.
\end{document}

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\documentclass[letterpaper]{article}
\usepackage{enumitem}
\usepackage{cancel}
\usepackage{amsmath}
\usepackage{listings}
\usepackage{graphicx}
\title{%
Homework 3\\
\large EEET-331-01: Signals, Systems and Transformers}
\date{09/07/2021}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\section{Exercise 1.}Find the forced response ($i(t)$), and draw its pole-zero plot for each of the following transfer functions.
\begin{enumerate}[label=\alph*)]
\item
\begin{equation}
v(t)=10e^{-2t}\cos(2t+5)\\
\end{equation}
\begin{equation}
\frac{d^2i}{dt^2}+3\frac{di}{dt}+2i=2\frac{dv}{dt}+4v
\end{equation}
\[\overrightarrow V=10\angle5^\circ,\ s=-2+j2\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2s+4}{s^2+3s+2}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2(-2+j2)+4}{(-2+j2)^2+3(-2+j2)+2}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{j4}{-4-j2}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=-0.4-j0.8\]
\[\overrightarrow I=\overrightarrow V(-0.4-j0.8)\]
\[\overrightarrow I=(10\angle5^\circ)(-0.4-j0.8)\]
\[\overrightarrow I=8.944271910\angle-111.5650512^\circ\]
\[\overline{\underline{|i(t)=8.944271910e^{-2t}\cos(2t-111.5650512^\circ)|}}\]
\newpage
\item
\begin{equation}
v(t)=10\cos(4t)
\end{equation}
\begin{equation}
\frac{d^3i}{dt^3}+10\frac{d^2i}{dt^2}+2\frac{di}{dt}+10i=2\frac{d^2v}{dt^2}+4v
\end{equation}
\[\overrightarrow V=10\angle0,\ s=j4\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2s^2+4}{s^3+10s^2+2s+10}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2(j4)^2+4}{(j4)^3+10(j4)^2+2(j4)+10}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{-28}{-150-j56}\]
\[\overrightarrow I=\overrightarrow V(0.1638321111-j0.06116398814)\]
\[\overrightarrow I=(10\angle0)(0.1748770827\angle-20.47227952^\circ)\]
\[\overrightarrow I=1.748770827\angle-20.47227952^\circ\]
\[\overline{\underline{|i(t)=1.748770827\cos(4t-20.47227952^\circ)|}}\]
\item
\begin{equation}
v(t)=10e^{-3t}
\end{equation}
\begin{equation}
\frac{\overrightarrow I}{\overrightarrow V}=2(\frac{s-1}{s+2})-(\frac{s-3}{s+5})
\end{equation}
\[\overrightarrow V=10\angle0,\ s=-3\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2(s-1)}{s+2}-\frac{s-3}{s+5}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{2(-3-1)}{-3+2}-\frac{-3-3}{-3+5}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=\frac{-8}{-1}-\frac{-6}{2}\]
\[\frac{\overrightarrow I}{\overrightarrow V}=8+3=11\]
\[\overrightarrow I=11\overrightarrow V\]
\[\overrightarrow I=110\angle0\]
\[\overline{\underline{|i(t)=110e^{-3t}|}}\]
\end{enumerate}
\section{Exercise 2.}Find the natural response of the transfer functions in Exercise 1. In other words, given $v(t)=0$, solve $i(t)$.
\begin{enumerate}[label=\alph*)]
\item
\begin{equation}
\begin{split}
\frac{d^2i}{dt^2}+3\frac{di}{dt}+2i=2\frac{dv}{dt}+4v\\
\frac{d^2i}{dt^2}+3\frac{di}{dt}+2i=0\\
s^2+3s+2=0\\
s=-2,\ -1
\end{split}
\end{equation}
\item
\begin{equation}
\begin{split}
\frac{d^3i}{dt^3}+10\frac{d^2i}{dt^2}+2\frac{di}{dt}+10i=2\frac{d^2v}{dt^2}+4v\\
\frac{d^3i}{dt^3}+10\frac{d^2i}{dt^2}+2\frac{di}{dt}+10i=0\\
s^3+10s^2+2s+10=0\\
s=-9.90001020094,\\ (-0.049994899531-1.0037930460594i),\\ (-0.049994899531011+1.0037930460594i)
\end{split}
\end{equation}
(Final answer found using the TI-nspire CX CAS \verb|cPolyRoots| function.)
\item
\begin{equation}
\begin{split}
\frac{\overrightarrow I}{\overrightarrow V}=2(\frac{s-1}{s+2})-(\frac{s-3}{s+5})\\
\frac{\overrightarrow I}{\overrightarrow V}=\frac{2(s-1)(s+5)-(s-3)(s+2)}{(s+2)(s+5)}\\
\overrightarrow I((s+2)(s+5))=\overrightarrow V(2(s-1)(s+5)-(s-3)(s+2))\\
\overrightarrow I(s^2+7s+10)=0\\
s^2+7s+10=0\\
s=-5,\ -2
\end{split}
\end{equation}
\end{enumerate}
\newpage
\section{Exercise 3.}Use MATLAB's \verb|root| function to verify the poles and zeroes of the transfer functions in question 1. The \verb|root| function will be done once for each numerator and each denominator. Place a screenshot of the transcript window below.
\begin{enumerate}[label=\alph*)]
\item
\begin{lstlisting}[language=matlab]
p1=[2 4];
p2=[1 3 2];
eqzero=roots(p1)
eqroot=roots(p2)
eqzero=-2
eqroot=-2
-1
\end{lstlisting}
\item
\begin{lstlisting}[language=matlab]
p1=[2 0 4];
p2=[2 10 2 10];
eqzero=roots(p1)
eqroot=roots(p2)
eqzero= 0.0000 + 1.4142i ,
0.0000 - 1.4142i
eqroot=-9.9000 +0.0000i ,
-0.0500 + 1.0038i ,
-0.0500 - 1.0038i
\end{lstlisting}
\item
\begin{lstlisting}[language=matlab]
p1=[1 9 -4];
p2=[1 7 10];
eqzero=roots(p1)
eqroot=roots(p2)
eqzero=-9.4244 ,
0.4244
eqroot=-5 ,
-2
\end{lstlisting}
\end{enumerate}
\newpage
\section{Exercise 4.}What MATLAB command is used to evaluate $H=s^2+3s+10$ when $s=3$? Submit the MATLAB code and a screenshot of the transcript window.\\
\begin{lstlisting}[language=matlab]
s=3;
H=s^(2)+3*s+10
ans=
28
\end{lstlisting}
\section{Exercise 5.}What will MATLAB produce for the following operations, given \verb|x=[3 6 9]| and \verb|y=[5 3 0]|? If the command generates an error, describe how to fix the command, and what the intended results were.
\begin{enumerate}[label=\alph*)]
\item
\verb|x.+y|\\
This equation fails. To run it properly, it needs to be run without the period.
\verb|[8 9 9]|
\item
\verb|x*y|\\
\verb|[15 18 0]|
\item
\verb|x+y|\\
This answer is the same as the answer for part a.
\item
\verb|x./y|\\
\verb|[0.6000 2.0000 Inf]|
\end{enumerate}
\end{document}

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% File: hw4.tex
% Created: 17:42:01 Tue, 14 Sep 2021 EDT
% Last Change: 17:42:01 Tue, 14 Sep 2021 EDT
%
\documentclass[letterpaper]{article}
\usepackage{amsmath}
\usepackage{minted}
\usepackage{graphicx}
\usepackage{cancel}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[shortlabels]{enumitem}
\usepackage{lipsum}
\usepackage{soul}
\usepackage{geometry}
\geometry{portrait, margin=1in}
\date{09/14/2021}
\title{%
Homework 4\\
\large EEET-331-01:Systems, Signals, and Transforms}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
\section{Exercise 1}
Find the following transfer functions for Figure 1, given the following:
\begin{equation}
R_1=2\Omega;\ R_2=5\Omega;\ L_1=3H
\end{equation}
\begin{figure}[h!]
\centering
\includegraphics[height=5cm,keepaspectratio]{figure1.png}
\caption{Circuit}
\end{figure}
\begin{enumerate}[(a)]
\item Find $Z_{in}$ using algebra.
\begin{equation}
\begin{split}
Z_{in}=R_1+(\frac 1{\frac1R +\frac1{3s}})\\
Z_P=\frac1{\frac1R+\frac1{3s}}\\
Z_P=\frac1{\frac{3s+R}{3sR}}\\
Z_P=\frac{3Rs}{3s+R}\\
Z_{in}=R_1+Z_P\\
Z_{in}=R_1+\frac{LR_2s}{Ls+R_2}\\
Z_{in}=\frac{R_1(Ls+R_2)+LR_2s}{Ls+R_2}\\
Z_{in}=\frac{s(LR_2+LR_1)+R_2R_1}{Ls+R_2}
\end{split}
\end{equation}
\begin{equation}
\begin{split}
Z_{in}=\frac{s(LR+LR_1)+RR_1}{Ls+R}\\
Z_{in}=\frac{s(3(5)+3(2))+5(2)}{3s+5}\\
Z_{in}=\frac{s(5+2)+\frac{5(2)}3}{s+\frac53}\\
Z_{in}=\frac{7s+\frac{10}{3}}{s+\frac53}\\
Z_{in}=\frac{7(s+\frac{10}{21})}{s+\frac53}
\end{split}
\end{equation}
\item Find $Z_{in}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=2;
R2=5;
ZL=3*s;
ZP= (R2*ZL)/(R2+ZL);
Zin = minreal(R1+ZP)
\end{minted}
\item Find $\frac{V_{out}}{V_{in}}$ by hand using algebra.
\begin{equation}
\begin{split}
\frac{V_{out}}{V_{in}}=\frac{R_1}{Z_{in}}\\
\frac{V_{out}}{V_{in}}=\frac{R_1}{\frac{s(LR_2+LR_1)+R_2R_1}{Ls+R_2}}\\
\frac{V_{out}}{V_{in}}=\frac{R_1(Ls+R_2)}{s(LR_2+LR_1)+R_2R_1}\\
\frac{V_{out}}{V_{in}}=\frac{2(3s+5)}{7s+\frac{10}3}\\
\frac{V_{out}}{V_{in}}=\frac{10(s+\frac53)}{7(s+\frac{10}{21})}
\end{split}
\end{equation}
\item Find $\frac{V_{out}}{V_{in}}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=3;
R2=5;
ZL=3*s;
ZP= (R2*ZL)/(R2+ZL);
Zin = R1+ZP;
VGain=minreal(R1/Zin)
\end{minted}
\end{enumerate}
\newpage
\section{Exercise 2}
Find The following transfer functions given the circuit and the following values:
\begin{equation}
C_1=2F;\ R_2=5\Omega;\ L_1=3H
\end{equation}
\begin{figure}[h!]
\includegraphics{figure2.png}
\caption{Circuit}
\end{figure}
\begin{enumerate}[(a)]
\item Find $Z_{in}$ by hand using algebra.
\begin{equation}
\begin{split}
Z_{in}=\frac {1}{Cs}+Z_p\\
Z_p=\frac{RLs}{R+Ls}\\
Z_{in}=\frac{R+Ls+RLs^2}{CLs^2+CRs}\\
Z_{in}=\frac{15(s^2+\frac 15s+\frac 13)}{6s(s+\frac{5}{3})}
\end{split}
\end{equation}
\item Find $Z_{in}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R=5;
C=1/(2*s);
ZL=3*s;
Zp=(R*ZL)(R+ZL);
Zin=minreal(C+Zp)
\end{minted}
\item Find $\frac{V_{out}}{V_{in}}$ by hand using algebra.
\begin{equation}
\begin{split}
\frac{V_{out}}{V_{in}}=\frac{\frac {1}{Cs}}{Z_{in}}\\
\frac{V_{out}}{V_{in}}=\frac{1}{Cs}*\frac{1}{Z_{in}}\\
\frac{V_{out}}{V_{in}}=\frac{1}{Cs}*\frac{CLs^2+CRs}{R+Ls+RLs^2}\\
\frac{V_{out}}{V_{in}}=\frac{(Ls+R)}{RLs^2+Ls+R}\\
\frac{V_{out}}{V_{in}}=\frac{3(s+\frac53)}{15(s^2+\frac{1}{5}s+\frac{1}{3})}
\end{split}
\end{equation}
\newpage
\item Find $\frac{V_{out}}{V_{in}}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R=5;
ZC=1/(2*s);
ZL=3*s;
Zp=(R*ZL)(R+ZL);
Zin=ZC+Zp;
VGain=minreal(ZC/Zin)
\end{minted}
\end{enumerate}
\section{Exercise 3}
Find The following transfer functions given the circuit and the following values:
\begin{equation}
R_1=1\Omega;\ C=4F;\ R_2=4\Omega;\ L=2H
\end{equation}
\begin{center}
\begin{figure}[h!]
\includegraphics{figure3.png}
\caption{Circuit}
\end{figure}
\end{center}
\begin{enumerate}[(a)]
\item Find $Z_{in}$ by hand using algebra.
\begin{equation}
\begin{split}
Z_{in}=Z_1+Z_2\\
Z_1=\frac{R_1Ls}{R_1+Ls};\ Z_2=R_2+\frac 1{Cs}\\
Z_2=\frac{R_2Cs+1}{Cs}\\
Z_{in}=\frac{(R_1LsCs)+(R_2Cs+1)(R_1+Ls)}{Cs(R_1+Ls)}\\
Z_{in}=\frac{R_1LCs^2+R_1R_2Cs+LR_2Cs^2+Ls+R_1}{Cs(R_1+Ls)}\\
Z_{in}=\frac{40s^2+18s+1}{8s(s+\frac12)}=\frac{40(s+0.3851)(s+0.0649)}{8s(s+\frac12)}
\end{split}
\end{equation}
\item Find $Z_{in}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=1;
R2=4;
ZC=1/(4*s);
ZL=2*s;
Zp=(R1*ZL)(R1+ZL);
Zin=minreal(R2+Zp+ZC)
\end{minted}
\item Find $\frac{V_{out}}{V_{in}}$ by hand using algebra.
\begin{equation}
\begin{split}
\frac{V_o}{V_i}=\frac{Z_1}{Z_{in}}\\
\frac{V_o}{V_i}=\frac{R_1Ls(8s(s+\frac12))}{(R_1+Ls)(40s^2+18s+1)}\\
\frac{V_o}{V_i}=\frac{2s(8s(s+\frac12))}{(1+2s)(40s^2+18s+1)}\\
\frac{V_o}{V_i}=\frac{16s^2(s+\frac12)}{80s^3+76s^2+20s+1}\\
\frac{V_o}{V_i}=\frac{s^2(s+\frac12)}{5(s+0.5)(s+0.3851)(s+0.0649)}
\end{split}
\end{equation}
\item Find $\frac{V_{out}}{V_{in}}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=1;
R2=4;
ZC=1/(4*s);
ZL=2*s;
Zp=(R1*ZL)(R1+ZL);
Zin=R2+Zp+ZC;
VGain=minreal(Zp/Zin)
\end{minted}
\end{enumerate}
\newpage
\section{Exercise 4}
Find The following transfer functions given the circuit and the following values:
\begin{equation}
R_1=10\Omega;\ R_2=4\Omega;\ R_3=8\Omega;\ C=3F;\ L=2H
\end{equation}
\begin{figure}[h!]
\includegraphics{figure4.png}
\caption{Circuit}
\end{figure}
\begin{enumerate}[(a)]
\item Find $Z_{in}$ by hand using algebra.
\begin{equation}
\begin{split}
Z_{in}=Z_1+Z_2\\
Z_1=\frac{R_1R_2}{R_1+R_2}\\
Z_1=\frac{40}{14}\\
Z_2=\frac{\frac 1{Cs}(R_3+Ls)}{\frac1{Cs}+(R_3+Ls)}\\
Z_2=\frac{R_3+Ls}{Cs}*\frac{Cs}{1+R_3Cs+LCs^2}\\
Z_2=\frac{2(s+4)}{6s^2+24s+1}\\
Z_{in}=\frac{40(6s^2+24s+1)+2(14)(s+4)}{14(6s^2+24s+1)}\\
Z_{in}=\frac{240s^2+988s+152}{84s^2+336s+14}\\
Z_{in}=\frac{20(s^2+\frac{499}{120}s+\frac{19}{30})}{7(s^2+4s+\frac16)}
\end{split}
\end{equation}
\item Find $Z_{in}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=10;
R2=4;
R3=8;
ZC=1/(3*s);
ZL=2*s;
ZP1=(R1*R2)/(R1+R2);
ZP2=(ZC*(R3+ZL))/(ZC+R3+ZL);
Zin=minreal(ZP1+ZP2)
\end{minted}
\item Find $\frac{V_{out}}{V_{in}}$ by hand using algebra.
\begin{equation}
\begin{split}
\frac{V_1}{V_i}=\frac{Z_1}{Z_{in}}\\
\frac{V_o}{V_1}=\frac{R_3}{R_3+sL}\\
\frac{V_o}{V_i}=\frac{Z_1R_3}{Z_{in}(R_3+sL)}\\
\frac{V_o}{V_i}=\frac{160}{7}*\frac{84s^2+336s+14}{(240s^2+988s+152)(2s+8)}\\
\frac{V_o}{V_i}=\frac{160(7(s^2+4s+\frac{1}{6}))}{7(480s^3+3896s^2+8208s+1216)}\\
\frac{V_o}{V_i}=\frac{160(s^2+4s+\frac{1}{6})}{480(s^3+\frac{487}{60}s^2+\frac{171}{10}s+\frac{38}{15})}
\end{split}
\end{equation}
\item Find $\frac{V_{out}}{V_{in}}$ using MATLAB.
\begin{minted}[linenos]{matlab}
s=tf('s');
R1=10;
R2=4;
R3=8;
ZC=1/(3*s);
ZL=2*s;
ZP1=(R1*R2)/(R1+R2);
ZP2=(ZC*(R3+ZL))/(ZC+R3+ZL);
Zin=ZP1+ZP2;
V1Gain=ZP1/Zin;
V0Gain=R3/(R3+ZL);
VTotalGain=minreal(V1Gain*V0Gain)
\end{minted}
\end{enumerate}
\newpage
\section{Exercise 5}
Given the block diagram below, and $H=\frac13$,
\begin{figure}[h!]
\includegraphics{figure5.png}
\caption{Block Diagram}
\end{figure}
\begin{enumerate}[(a)]
\item What is $\frac{\overrightarrow C}{\overrightarrow R}$ if $G=500$?
\begin{equation}
\begin{split}
\frac CR=\frac{G}{1+GH}\\
\frac CR=\frac{500}{1+\frac{500}{3}}\\
\frac CR=2.982\\
\frac CR\approx 3
\end{split}
\end{equation}
\item What is $\frac{\overrightarrow C}{\overrightarrow R}$ if $G=50,000$?
\begin{equation}
\begin{split}
\frac CR=\frac{G}{1+GH}\\
\frac CR=\frac{50000}{1+\frac{50000}{3}}\\
\frac CR=2.99982\\
\frac CR\approx3
\end{split}
\end{equation}
\item What is $\frac{\overrightarrow C}{\overrightarrow R}$ if $G=\infty$?
\begin{equation}
\frac CR=3
\end{equation}
\item What value of $G$ will make $\frac{\overrightarrow C}{\overrightarrow R}$ 98\% of the value when $G=\infty$?
\begin{equation}
\begin{split}
\frac CR=0.98*3=2.94\\
G=2.94(1+\frac G3)\\
G(1-\frac{2.94}{3})=2.94\\
G=147
\end{split}
\end{equation}
\end{enumerate}
\end{document}

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