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<nowiki>
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[[Category:bonus point project]]
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\documentclass[letterpaper,10pt]{report}
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[[Category:lecture notes]]
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\usepackage[utf8x]{inputenc}
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\usepackage{amsmath, amsthm, amssymb, mathabx}
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\usepackage{fullpage}
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\newenvironment{Observation}[2][Observation]{\begin{trivlist}
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=[[Lecture21ECE301S11|Lecture 21]]=
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\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries
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== Multiplication Property ==
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#2}]}{\end{trivlist}}
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<math> \mathcal{F}(x_1(t) x_2(t)) = \frac{1}{2\pi}\mathcal{F}(x_1(t))*\mathcal{F}(x_2(t))</math>
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\newenvironment{definition}[1][Definition]{\begin{trivlist}
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\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
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\newtheorem*{why}{Why}
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== Causal LTI system defined by cst coeff diff equations ==
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\newtheorem{property}{Property}
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<math> \sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t) =  \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t)</math>
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\newtheorem{fact}{Fact}
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−
\newtheorem*{corollary}{Corollary}
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% Title Page
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What is the frequency response of this system? Recall:
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\title{ECE 301}
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−
\author{Ethan Hall}
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 +
<math>
 +
\begin{align}
 +
\mathcal{H}(\omega) &= \mathcal{F}(h(t)) \\
 +
\mathcal{Y}(\omega) &= \mathcal{H}(\omega)\mathcal{X}(\omega)
 +
\end{align}
 +
</math>
  
−
\begin{document}
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Steps to solve:
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\maketitle
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−
\part*{Lecture 6}
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−
\section*{Memorkess/System with Memory}
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−
\begin{definition}
+
−
  A system is memoryless if, for any $t_0 \in \mathbb{R}$,
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−
  the output at $t_0$, $y(t_0)$ depends only on $x(t_0)$ at $t_0$ (not on $x(t)$ for $t > t_0$ or $t < t_0$ )
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−
\end{definition}
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−
Example:
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−
\begin{eqnarray*}
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−
  y(t) & = & x(t) + x(t - 1) \text{, memory} \\
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−
  y(t) & = & 2x(t) \text{, memoryless} \\
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−
  y(t) & = & (t-1)(x(t)) \text{, memoryless} \\
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−
  y(t) & = & \int_{-\infty}^{\infty}x(t')dt' \text{, memory}
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−
\end{eqnarray*}
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−
\begin{fact}
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−
  A memoryless system can be written as $ y(t) = f(t,x(t)) $ ( $Y[n] = f(n,x[n]) $ )
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−
\end{fact}
+
−
\section*{Causal/Non-Causal}
+
−
\begin{definition}
+
−
  A system is causal if, for any $t_0 \in \mathbb{R}$, the output of $y(t_0)$
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−
  depends only on the input $x(t_0)$ at or before $t_0$ ( $t \leq t_0 $ )
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−
\end{definition}
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−
Example:
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−
\begin{equation*}
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  y(t) = x(t^2)\text{, non-causal}
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−
\end{equation*}
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−
take $y(-1/2) == y(1/2)$
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−
\section*{Invertable Systems}
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−
\begin{definition}
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−
  A system is invertable if distinct input signals yield distinct output signals
+
−
\begin{eqnarray*}
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−
  X_1[n] \rightarrow &\fbox{\text{system}}& \rightarrow Y_1[n] \\
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−
  X_2[n] \rightarrow &\fbox{\text{system}}& \rightarrow Y_2[n] \\
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  \text{where } X_1[n] &\neq X_2[n]
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−
\end{eqnarray*}
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−
\end{definition}
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\begin{definition}
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<math>
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  System is invertable if there exists another system, called the ``inverse''
+
\begin{align}
−
  such that the cascade leaves the input unchanged
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\mathcal{F}\left(\sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t)\right) &= \mathcal{F}\left( \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) \right) \\
−
\begin{equation*}
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\sum_{k=0}^{N}a_k \mathcal{F}\left(\frac{d^k}{dt^k}y(t)\right) &= \sum_{k=0}^{M}b_k \mathcal{F}\left(\frac{d^k}{dt^k}x(t)\right) \\
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  x(t) \rightarrow \fbox{\text{system}} \rightarrow \fbox{\text{inverse}} \rightarrow x(t)
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& \text{recall: }\mathcal{F}\left(\frac{d^n}{dt^n}y(t)\right) = (j\omega)^n Y(\omega) \\
−
\end{equation*}
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\sum_{k=0}^{N}a_k \left( j \omega \right)^k \mathcal{Y}(\omega) &= \sum_{k=0}^{M}b_k \left( j \omega \right)^k \mathcal{X}(\omega) \\
−
\end{definition}
+
\mathcal{Y}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \mathcal{X}(\omega) \\
 +
\mathcal{H}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \\
 +
h(t) &= \mathcal{F}^{-1}\left(\mathcal{H}(\omega)\right)
 +
\end{align}
 +
</math>
  
−
Example:
+
==[[Table_DT_Fourier_Transforms|Def of DT F.T.]]==
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\begin{eqnarray*}
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Here are the practice problems that do this: [[Fourier_transform_3numinusn_DT_ECE301S11|Problem 1]], [[Fourier_transform_cosine_DT_ECE301S11|Problem 2]], [[Fourier_transform_window_DT_ECE301S11|Problem 3]]
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  y(t) & = & 2x(t) + 3 \text{ is invertable } \\
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−
  x(t) & = & \frac{y(t) - 3}{2}
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−
\end{eqnarray*}
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−
\begin{equation*}
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<math>
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  x(t) \rightarrow \fbox{\text{system}} \rightarrow \text{y(t) = 2x(t) + 3} \rightarrow \fbox{\text{inverse}} \rightarrow z(t)
+
\begin{align}
−
\end{equation*}
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x[n] &\xrightarrow{\mathcal{F}} \mathcal{X}(\omega) \text{ in book as }\mathcal{X}(e^{j\omega})& \\
−
\begin{equation*}
+
x[n] &\xleftarrow{\mathcal{F}^{-1}} \mathcal{X}(\omega) &\\
−
  z(t) = \frac{1}{2}y(t) + \frac{3}{2} \rightarrow \frac{1}{2}(2x(t) + 3) + \frac{3}{2} \Rightarrow x(t)
+
\mathcal{X}(\omega) & = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} &\\
−
\end{equation*}
+
x[n] &= \frac{1}{2\pi}\int_{0}^{2\pi} \mathcal{X}(\omega)e^{j \omega n} d\omega &
−
Example of non-invertable system $y(t) = (x(t))^2$
+
\end{align}
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\begin{eqnarray*}
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</math>
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  x_1(t) &= t \Rightarrow t^2 \\
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===Example===
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  x_2(t) &= -t \Rightarrow t^2
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Compute the FT of <math class="inline">x[n] = 2^{-n}u[n]</math>
−
\end{eqnarray*}
+
  
 +
<math>
 +
\begin{align}
 +
\mathcal{X}(\omega) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\
 +
&= \sum_{n=-\infty}^{\infty}2^{-n}u[n]e^{-j\omega n} \\
 +
&= \sum_{n=0}^{\infty}2^{-n}e^{-j\omega n} \\
 +
&= \sum_{n=0}^{\infty}\left(\frac{1}{2e^{-j\omega}}\right)^n \\
 +
&= \frac{1}{1-\frac{1}{2e^{j\omega}}}
 +
\end{align}
 +
</math>
  
−
\section*{Stability - (BIBO Stability)}
+
==[[Table_DT_Fourier_Transforms|Properties of DT FT]] ==
−
\begin{definition}
+
===Periodicity===
−
  A system is called BIBO-stable if a bounded input yields a bounded output
+
<math>
−
\end{definition}
+
\begin{align}
−
\begin{eqnarray*}
+
\mathcal{X}(\omega + 2\pi) &= \mathcal{X}\text{, for all } \omega \\
−
  \text{if there exists } &E \text{ suck that } &\|x(t)\| < E\text{, for all $t$} \\
+
\text{because} & \\
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  \text{then there exists } &M \text{ such that } &\|y(t)\| < M\text{, for all $t$}
+
\mathcal{X}(\omega + 2 \pi) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega + 2\pi)n} \\
−
\end{eqnarray*}
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&= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} e^{-j2\pi n} \\
 +
&= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\
 +
&= \mathcal{X}(\omega)
 +
\end{align}
 +
</math>
 +
===Linearity===
 +
<math> \text{for all }a,b,c \in \mathcal{C}\text{ and all } x_1[n],x_2[n]</math>
  
−
Example: $Y(t) = e^{x(t)}$ is \underline{stable} because if $\|x(t)\| < E \Rightarrow \|y(t)\| = \|e^{x(t)}\|$
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<math> \mathcal{F}(ax_1[n] + bx_2[n]) = a\mathcal{F}(n_1[n]) + b\mathcal{F}(n_2[n])</math>
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\section*{Time Invarrience}
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provided both FT's exist.
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\begin{definition}{1}
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−
If the cascade of
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−
\begin{eqnarray*}
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−
  x(t) \rightarrow \fbox{\text{system}} &\rightarrow \fbox{\text{time delay $t_0$}} \rightarrow y(t) \\
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  &\parallel \\
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  x(t) \rightarrow \fbox{\text{time delay $t_0$}} &\rightarrow \fbox{\text{system}} \rightarrow y(t)
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−
\end{eqnarray*}
+
−
\end{definition}
+
  
−
\begin{definition}{2}
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===The FT of DT periodic signals ===
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Time invarient means for any input sig $x(t)$ ( $x[n]$ )
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<math> x[n] \text{ periodic} \Rightarrow x[n] = \sum_{n=0}^{N-1}a_k e^{\omega_0 n k }</math>
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  and for any time $t_0$, the output for athe shifted input $x(t-t_0)$ is the shifted output $y(t-t_0)$
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−
\end{definition}
+
  
−
\part*{Lecture 7}
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<math> \mathcal{F}(x[n]) = \sum_{n=0}^{N-1}a_k \mathcal{F}\left(e^{\omega_0 n k }\right) </math>, by linearity
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\begin{definition}{3}
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−
\begin{eqnarray*}
+
−
  \text{A system is time invarient if} \\
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−
  x(t) &\rightarrow \fbox{\text{system}} &\rightarrow y(t) \\
+
−
  \text{then we also have} \\
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  x(t-t_0) &\rightarrow \fbox{\text{system}} &\rightarrow y(t-t_0) \text{ for any $t_0 \in \mathbb{C}$}
+
−
\end{eqnarray*}
+
−
\end{definition}
+
  
−
\begin{definition}{4}
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so all we need is the FT of <math class="inline"> e^{j k \omega_0 n} </math>
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  A system is time invarient if it comutes with a time delay
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−
\end{definition}
+
  
−
Example 1: show time invarient $x(t) \rightarrow \fbox{\text{subject}} \rightarrow y(t) = 10x(t)$
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we want <math class="inline"> \mathcal{X}(\omega)</math> such that
−
\begin{eqnarray*}
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<math> \frac{1}{2\pi}\int_{0}^{2\pi}\mathcal{X}(\omega)e^{j\omega t} d\omega = e^{j\omega_0 n k}</math>
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  x(t) \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow &y(t) = x(t - t_0)&
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−
    \rightarrow \fbox{\text{system}} \rightarrow z_1(t) \\
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  & z_1(t) = 10y(t) &\\
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  & z_1(t) = 10 x(t - t_0)& \\
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  x(t) \rightarrow \fbox{\text{system}} \rightarrow &y(t) = 10 x(t) &
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−
    \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow z_2(t) \\
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  &z_2(t) = y(t - t_0) & \\
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  &z_2(t) = 10x(t - t_0)& \\
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  &z_1(t) \equiv z_2(t)&
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−
\end{eqnarray*}
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−
Example 2: show time invarient $t x(t)$
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try <math class="inline"> \mathcal{X}(\omega) = 2\pi \delta(\omega - k\omega_0)</math> and it works. So the real answer is
−
\begin{eqnarray*}
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−
  x(t) \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow &y(t) = x(t - t_0)&
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    \rightarrow \fbox{\text{system}} \rightarrow z_1(t) \\
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  & z_1(t) = t y(t) &\\
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  & z_1(t) = t x(t - t_0)& \\
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  x(t) \rightarrow \fbox{\text{system}} \rightarrow &y(t) = t x(t) &
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−
    \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow z_2(t) \\
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  &z_2(t) = y(t - t_0) & \\
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  &z_2(t) = (t - t_0)(x(t - t_0))& \\
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  &z_1(t) \neq  z_2(t)&
+
−
\end{eqnarray*}
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−
\section*{Linearity}
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<math> \mathcal{X}(\omega) = \sum_{m = -\infty}^{\infty} 2\pi \delta(\omega - \omega_0 k + 2\pi m)</math>
−
\begin{definition}{1}
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−
  A system is called ``linear'' if for any constant $a,b, \in \mathbb{C}$ and for any input signals $X_1(t), X_2(t)$
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    ( $X_1[n], X_2[n]$ ) with response $y_1(t), y_2(t)$ respectivly ( $Y_1[n], Y_2[n]$ ) the system's responce to any
+
−
    $a x_1(t) + b x_2(t)$ ( $a x_1[n] + b x_2[n] $ ) yields $a t_1(t) + b t_2(t)$ ( $a t_1[n] + b t_2[n] $ )
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−
\end{definition}
+
  
−
\begin{definition}{2}
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=[[Lecture22ECE301S11|Lecture 22]]=
−
  if
+
=== Time Shifting and Freq Shifting Property ===
−
\begin{eqnarray*}
+
<math>
−
  x_1(t) &\rightarrow \fbox{\text{system}} \rightarrow &y_1(t) \\
+
\begin{align}
−
  x_2(t) &\rightarrow \fbox{\text{system}} \rightarrow &y_2(t)  
+
\mathcal{F}(x[n-n_0]) &= e^{-j\omega n_0}\mathcal{F}(x[n]) \\
−
\end{eqnarray*}
+
\mathcal{F}\left( e^{j \omega_0 n}x[n] \right) &= \mathcal{X}(\omega - \omega_0)
 +
\end{align}
 +
</math>
  
−
\begin{equation*}
+
=== Conjugation and Conjugation Symmetry ===
−
\text{then } a x_1(t) + b x_2(t) \rightarrow \fbox{\text{system}} \rightarrow a t_1(t) + b t_2(t)
+
<math> \mathcal{F}\left( x^*[n] \right) = \mathcal{X}^*(-\omega)</math>
−
\end{equation*}
+
−
for any $a,b \in \mathbb{C}$, and any $x_1(t), x_2(t)$ then we say the system is linear
+
−
\end{definition}
+
  
−
\begin{definition}{3}
+
Important Corrilary
−
  A sytem is linear if both the follwoing yield the same output
+
−
\begin{eqnarray*}
+
−
\begin{matrix}
+
−
  x_1(t) & \rightarrow & \fbox{\text{system}} & \rightarrow & \otimes^a & \searrow \\
+
−
  x_2(t) & \rightarrow & \fbox{\text{system}} & \rightarrow & \otimes^b & \nearrow \\
+
−
\end{matrix}
+
−
\oplus \rightarrow y(t)
+
−
\end{eqnarray*}
+
  
−
\begin{eqnarray*}
+
if signal is real then <math class="inline"> \mathcal{X}(\omega) = \mathcal{X}^*(-\omega) </math> because <math class="inline"> x[n] </math> is real.
−
\begin{matrix}
+
−
  x_1(t) & \otimes^a & \searrow \\
+
−
  x_2(t) & \otimes^b & \nearrow \\
+
−
\end{matrix}
+
−
\oplus \rightarrow \fbox{\text{system}} \rightarrow y(t)
+
−
\end{eqnarray*}
+
  
−
\end{definition}
+
<math>
−
\setcounter{section}{0}
+
\begin{align}
−
\setcounter{chapter}{2}
+
x^*[n] &= x[n] \\
 +
\mathcal{X}^*[-\omega] &= \mathcal{X}(\omega)
 +
\end{align}
 +
</math>
  
−
\section{The convolution sum for LTI systems}
+
This mean that <math class="inline"> x[n] </math> real
−
Result: for LTI system the output $y[n] = x[n] \convolution h[n]$ where $h[n]$ is the
+
−
  systems responce to the input $\delta[n]$
+
−
\begin{Observation}{1}
+
−
Any DT signal can be written as a sum of shifted $\delta[n]$
+
−
\begin{equation}
+
−
  x[n] = \sum_{k=-\infty}^{\infty}x[n]\delta[n-k]
+
−
\end{equation}
+
  
−
\end{Observation}
+
=> Re <math class="inline"> \mathcal{X}(\omega) </math> is an odd function
  
−
%Lecture 8
+
=> Im <math class="inline"> \mathcal{X}(\omega) </math> is an odd function
−
\part*{Lecture 8}
+
  
−
Example: Write u[n] as a linear combanation of shifted $\delta$[n]
+
===Panseval's relation ===
−
\begin{equation*}
+
<math>
−
u[n] = \sum_{k=0}^{\infty}\delta[n-k]
+
\sum_{n=-\infty}^{\infty}| x[n] |^2 = \frac{1}{2\pi} \int_{0}^{2\pi} | x[\omega] |^2 d\omega
−
\end{equation*}
+
</math>
 +
===Convolution Property===
 +
<math>
 +
\begin{align}
 +
\mathcal{F}(x[n]*y[n]) &= \mathcal{F}(x[n])\mathcal{F}(y[n]) \\
 +
&= \mathcal{X}(\omega) \mathcal{X}(\omega)
 +
\end{align}
 +
</math>
  
−
\begin{Observation}{2}
+
so for any LTI system
−
The respoce ofa  DT linear system can be written as a sum $y[n] =
+
<math> x \rightarrow h[n] \rightarrow y[n] = x[n]*h[n] </math>
−
\sum_{k=-\infty}^{\infty}x[k]h_{k}[n]$; where $h_{k}[n]$ is the systems responce to $\delta[n-k]$
+
  
−
  \begin{why}
+
=[[Lecture23ECE301S11|Lecture 23]]=
−
    $x[n] = \sum_{k=-\infty}^{\infty} \underbrace{x[k]}_{\text{const}}\delta[n-k]\text{, by observation 1}$
+
===Multiplication Property===
−
  \end{why}
+
<math> x[n]y[n] \xrightarrow{\mathcal{F}} \frac{1}{2\pi} \mathcal{X}(\omega)*\mathcal{Y}(\omega)</math>
−
  by linearity $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$
+
−
\end{Observation}
+
  
−
\begin{Observation}{3}
+
=== Differentiation in frequency property ===
−
  The responce of an LTI system can be written as an even simpler function $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$
+
<math>
−
  where $h[n]$ is the responce to $\delta[n]$. (We call $h[n]$ the ``unit impulse responce'' of the system)
+
nx[n] \xrightarrow{\mathcal{F}} j\frac{d}{d\omega}\mathcal{X}(\omega)
−
  \begin{why}
+
</math>
−
  because if the system is time invarient, then $\delta[n-k] \rightarrow \fbox{system} \rightarrow h_{k}[n] = h[n-k] $
+
−
  \end{why}
+
−
\end{Observation}
+
  
−
Introduce '$\convolution$' the convolution between 2 DT function
+
====Example====
−
\begin{equation*}
+
Assume <math class="inline">|\alpha | < 1</math>
−
  Z_{1}[n] \convolution Z_{2}[n] = \sum_{k=-\infty}^{\infty} Z_{1}[k]Z_{2}[n-k]
+
# Compute the FT of <math class="inline">x_1[n] = \alpha^n u[n]</math>
−
\end{equation*}
+
# Use your andwer to compute the FT of <math class="inline">x_2[n] = (n+1)\alpha^n u[n]</math>
  
−
\begin{proof}
+
Answer:
−
\begin{eqnarray*}
+
1)
−
    y[n] & = & x[n] \convolution h[n] \\
+
−
    & = & \sum_{k=-\infty}^{\infty}x[k]h[n-k] \\
+
−
    & = & \sum_{k=-\infty}^{\infty}2^{k}u[k]u[n-k] \text{, but } u[n] =
+
−
    \begin{cases}
+
−
      1, & \text{$k \geq 0$}\\
+
−
      0, & \text{$k < 0$}
+
−
    \end{cases} \\
+
−
    & = & \sum_{k=-\infty}^{\infty}(2^{k})(1)(u[n-k]) \text{, but } u[n-k] =
+
−
    \begin{cases}
+
−
    1, & \text{$n-k \geq 0$} \\
+
−
    0, & \text{$n-k < 0$}
+
−
    \end{cases} \\
+
−
    & = &
+
−
    \begin{cases}
+
−
      \frac{1-2^{n+1}}{1-2}, & \text{$ n \geq 0 $}\\
+
−
      0, & \text{$ n < 0 $}
+
−
    \end{cases}
+
−
\end{eqnarray*}
+
−
\end{proof}
+
  
−
To know for the rest of your life:
+
<math>
−
\begin{eqnarray*}
+
\begin{align}
−
\sum_{k=0}^{n} \alpha^{k} = \begin{cases}
+
\mathcal{X}_1(\omega)  &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\
−
\frac{1 - \alpha^{n+1}}{1-\alpha},& \text{$\alpha \neq 1 $} \\
+
&= \sum_{n=-\infty}^{\infty} \alpha^nu[n]e^{-j\omega n} \\
−
n + 1,& \text{$ \alpha = 1$}
+
&= \sum_{n=0}^{\infty} \alpha^nu[n]e^{-j\omega n} \\
−
\end{cases}\\
+
&= \sum_{n=0}^{\infty} \left(\alpha u[n]e^{-j\omega } \right) ^n \\
−
\sum_{k=0}^{\infty} \alpha^{k} = \begin{cases}
+
&= \frac{1}{1-\alpha e^{-j\omega}}
−
\frac{1}{1-\alpha},& \text{$\|\alpha \| < 1 $} \\
+
\end{align}
−
diverdges,& \text{ $\|\alpha \| \geq 1 $}
+
</math>
−
\end{cases}
+
−
\end{eqnarray*}
+
  
−
\setcounter{section}{1}
+
2)
−
\setcounter{chapter}{2}
+
  
−
\section{CT LTI system and Confolution Intergral}
+
<math>
−
\begin{Observation}{1}
+
\begin{align}
−
Any CT system can be written as an intregral
+
\mathcal{X}_2(\omega) &= \mathcal{F}((x+1)\alpha^n u[n])\\
−
  \begin{equation*}
+
&= \mathcal{F}(n\alpha^n u[n]) + \mathcal{F}(\alpha^n u[n]) \\
−
    x(t) = \int_{-\infty}^{\infty} x(\tau)\delta(t - \tau)d\tau
+
&= j\frac{d}{d\omega}\mathcal{X}_1(\omega) + \mathcal{X}_1(\omega) \\
−
  \end{equation*}
+
&= j\frac{d}{d\omega}\left( \frac{1}{1-\alpha e^{-j\omega}} \right) + \frac{1}{1-\alpha e^{-j\omega}} \\
−
  \begin{why}
+
&= \frac{1}{\left( 1 - \alpha e^{-j\omega} \right)^2}
−
    \begin{equation*}
+
\end{align}
−
      x(\tau)\delta(t-\tau) = x(t)\delta(t-\tau)\text{, for any $t$}
+
</math>
−
    \end{equation*}
+
−
    \begin{eqnarray*}
+
−
      x(t) &=& \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau \\
+
−
      &=& \int_{-\infty}^{\infty}x(t)\delta(t-\tau)d\tau \\
+
−
      &=& x(t)\overbrace{\int_{-\infty}^{\infty}\delta(t-\tau)d\tau}^{1}
+
−
    \end{eqnarray*}
+
−
  \end{why}
+
−
\end{Observation}
+
  
−
\begin{Observation}{2}
+
== LTI systems defined by linear, constant coef diff eq's ==
−
  If a system is linear, then its output can be written as an intergral
+
<math>
−
  \begin{equation*}
+
\sum_{k=0}^{N}a_k y[n-k] = \sum_{k=0}^{M}b_k y[n-k]
−
    y(t) = \int_{-\infty}^{\infty}x(\tau)h_{\tau}(t)d\tau \text{ where $h_{\tau}(t)$ is the systems responce to $\delta(t-\tau)$ }
+
</math>
−
  \end{equation*}
+
−
\end{Observation}
+
  
−
\begin{Observation}{3}
+
Example: <math class="inline"> y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n] = 2x[n] </math>
−
  \begin{equation*}
+
can look at this eq in freq domain.
−
    y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \text{where h(t) is the system's responce to $\delta(t)$ }
+
−
  \end{equation*}
+
−
  \begin{eqnarray*}
+
−
    \delta(t) & \rightarrow \fbox{system}  & \rightarrow h(t) \\
+
−
    \delta(t-\tau) & \rightarrow \fbox{system} & \rightarrow h(t-\tau)\text{, by time invarince}
+
−
  \end{eqnarray*}
+
−
\end{Observation}
+
  
−
Introduce '$\convolution$' the convolution between 2 CT signals
+
<math>
−
\begin{equation*}
+
\begin{align}
−
  X_{1}(t) \convolution X_{2}(t) = \int_{-\infty}^{\infty} X_{1}(\tau)X_{2}(t-\tau)d\tau
+
\mathcal{F}(y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n]) &= 2\mathcal{F}(x[n]) \\
−
\end{equation*}
+
\mathcal{Y} - \frac{3}{4}\mathcal{F}(y[n-1]) + \frac{1}{8}\mathcal{F}(y[n-8]) &= 2\mathcal{X}(\omega) \\
 +
\mathcal{Y}(\omega) - \frac{3}{4}e^{-j\omega}\mathcal{Y}(\omega) + \frac{1}{8}e^{-2j\omega}\mathcal{Y}(\omega) &= 2\mathcal{X}(\omega)
 +
\end{align}
 +
</math>
  
−
\part*{Lecture 9}
+
<math>
−
For LTI systems
+
\begin{align}
−
\begin{equation*}
+
\mathcal{Y}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}}\mathcal{X}(\omega) \\
−
y(t) = x(t) \convolution h(t) = h(t) \convolution x(t)
+
\mathcal{H}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}} \\
−
\end{equation*}
+
&= \frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}} \\
−
Example: the unit impuse resonse of an LTI system is $h(t) = u(t)$\newline
+
h[n] &= 4\left(\frac{1}{2}\right)^n u[n] - 2 \left( \frac{1}{4} \right)^n u[n]
−
Find the systems response to $x(t) = e^{-t}u(t)$
+
\end{align}
 +
</math>
  
−
\begin{eqnarray*}
+
Now find the systems responce to <math class = "inline" > x[n] = \left( \frac{1}{4} \right)^n u[n] </math>
−
  y(t) &=& x(t) \convolution h(t) \\
+
−
  &=& \int_{-\infty}^{\infty}x(\tau)h(t - \tau)d\tau \\
+
−
  &=& \int_{-\infty}^{\infty}e^{-\tau}u(\tau)u(t-\tau)d\tau\text{, but } u(\tau) =
+
−
    \begin{cases}
+
−
    0, &x < 0 \\
+
−
    1, &x \geq 1
+
−
    \end{cases}\\
+
−
  &=& \int_{0}^{\infty}e^{-\tau}u(t-\tau)d\tau\text{, but } u(t-\tau) =
+
−
\begin{cases}
+
−
    0, &t < \tau \\
+
−
    1, &t \geq \tau
+
−
\end{cases}\\
+
−
y(t)&=&
+
−
\begin{cases}
+
−
  /int_{0}^{t}e^{-\tau}d\tau, & t \geq 0 \\
+
−
  0, & t < 0
+
−
\end{cases}\\
+
−
&=& \dfrac{e^{-\tau}}{-1}\bigg|_{0}^{t}u(t) = (-e^{-\tau}-1)u(t)
+
−
\end{eqnarray*}
+
  
−
\setcounter{section}{2}
+
<math>
−
\setcounter{chapter}{2}
+
\begin{align}
 +
\mathcal{Y}(\omega) &= \mathcal{X}(\omega)\mathcal{Y}(\omega) \\
 +
&= \left(\frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}}\right)\frac{1}{1-\frac{1}{4}e^{-j\omega}} \\
 +
&= \frac{4}{\left(1-\frac{1}{2}e^{-j\omega}\right) \left(1-\frac{1}{4}e^{-j\omega}\right)} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\
 +
&= \frac{-4}{1-\frac{1}{2}e^{-j\omega}}  + \frac{8}{1-\frac{1}{4}e^{-j\omega}} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\
 +
y[n] &= -4\left(\frac{1}{4}\right)^n u[n]+8\left(\frac{1}{2}\right)^n u[n] -2(n+1)\left(\frac{1}{4}\right)^n u[n]
 +
\end{align}
 +
</math>
  
−
\section{Properties of an LTI System}
+
=[[Lecture24ECE301S11|Lecture 24]]=
−
\begin{eqnarray}
+
==Section 7.1 Sampling \ Representing a CT signal by Sampling==
−
y[n] & = & x[n] \convolution h[n] \nonumber \\
+
What is sampling? The process of measuring a CT signal is to take samples ( at the same interval for [[ECE301]] ) so you get a vector of values that fit the equation <math class ="inline"> x(nT) </math> where n is an int and the distance between the samples is T. This makes a DT signal <math class="inline"> x_d[n] = x(nT) </math>
−
y(t) & = & x(t) \convolution h(t)
+
−
\end{eqnarray}
+
  
−
\begin{property}
+
Problem: given <math class="inline"> x_d[n] </math> can you reconstruct <math class="inline"> x(t) </math>? In general no. However, we can approximate <math class="inline"> x(t) </math> by interpolation values of <math class="inline"> x(t) </math> between the samples.
−
  \begin{eqnarray*}
+
−
    x(t) & \rightarrow \fbox{h(t)} \rightarrow & y(t) \\
+
−
    \text{same as }h(t) & \rightarrow \fbox{x(t)} \rightarrow & y(t)
+
−
  \end{eqnarray*}
+
−
  \text{because $\convolution$ is communitive}
+
−
  \begin{eqnarray}
+
−
    X_1(t) \convolution X_2(t) & = & X_2(t) \convolution X_1(t)  \nonumber \\
+
−
    X_1[n] \convolution X_2[n] & = & X_2[n] \convolution X_1[n] 
+
−
  \end{eqnarray}
+
−
\end{property}
+
  
−
\begin{property}
+
Ex 1: Interpolation by using step function
−
Sence $ \convolution $ is distributive
+
−
\begin{equation*}
+
−
x(t)
+
−
\begin{matrix}
+
−
\nearrow & \fbox{\text{$h_1(t)$}} & \searrow \\
+
−
\searrow & \fbox{\text{$h_2(t)$}} & \nearrow \\
+
−
\end{matrix}
+
−
\bigoplus \rightarrow  y(t) = x(t) \rightarrow \fbox{\text{$h_1(t) + h_2(t)$}} \rightarrow y(t)
+
−
\end{equation*}
+
−
\end{property}
+
  
−
\begin{property}
+
<math>
−
Sence $\convolution$ is a linear operator
+
x(t) = \sum_{k =-\infty}^{\infty} x(kt)\left( u(t-kT) - u(t-(k+1)T) \right)
−
  \begin{equation}
+
</math>
−
    X_1(t) \convolution (X_2(t) + X_3(t)) = X_1(t) \convolution X_2(t) + X_1(t) \convolution X_3(t)
+
−
  \end{equation}
+
  
−
  \begin{equation*}
+
Ex 2: Interpolation by using piecewise linear function
−
  \begin{matrix}
+
−
      x_1(t) & \fbox{\text{h(t)}} & \searrow \\
+
−
      x_2(t) & \fbox{\text{h(t)}} & \nearrow \\
+
−
    \end{matrix}
+
−
    \bigoplus \rightarrow y(t) = X_1(t) \convolution h(t) + X_2(t) \convolution h(t)
+
−
  \end{equation*}
+
−
Same as
+
−
  \begin{equation*}
+
−
    X_1(t) + X_2(t) \rightarrow \fbox{\text{h(t)}} \rightarrow y(t) = (X_1(t) + X_2(t)) \convolution h(t)
+
−
  \end{equation*}
+
−
\end{property}
+
  
−
\begin{property}
+
Take take the current point and the next one and find a line that will connect the two
−
  \begin{eqnarray*}
+
−
    x(t) &\rightarrow &\fbox{\text{$h_1(t)$}} \rightarrow \fbox{\text{$h_2(t)$}} \rightarrow y(t) \\
+
−
    \text{Same as} & \\
+
−
    x(t) &\rightarrow &\fbox{\text{$h_1(t) \convolution h_2(t)$}} \rightarrow y(t) \\
+
−
    \text{Same as} & \\
+
−
    x(t) &\rightarrow &\fbox{\text{$h_2(t) \convolution h_1(t)$}} \rightarrow y(t) \\
+
−
    \text{Same as} & \\
+
−
    x(t) &\rightarrow &\fbox{\text{$h_2(t)$}} \rightarrow \fbox{\text{$h_1(t)$}} \rightarrow y(t)
+
−
  \end{eqnarray*}
+
−
\end{property}
+
  
−
\subsubsection{LTI systems w/ and w/o memory}
+
Conclusion: There are infanately many ways to reconstruct the signal from samples and get an approximation. All of these could have been the initial <math class="inline"> x(t) </math>. So it is impossible to reconstruct <math class="inline"> x(t) </math> from samples.
−
\begin{fact}
+
−
  If an LTI system is memoryless then its unit impusle response can be written as $h[n] = k\delta[n]$ ($h(t) = k\delta(t)$) for some
+
−
  $ k \in \mathbb{C}$.
+
−
\end{fact}
+
  
−
\begin{fact}
+
=== Sampling Theorem===
−
  If an LTI system is invertable then its inverse is also LTI.
+
# Let <math class="inline"> \omega_m </math> be a non-neg number
−
\begin{corollary}
+
# Let <math class="inline"> x(t) </math> be a signal such that <math class="inline"> \mathcal{X}(\omega) = 0 </math> when <math class="inline"> |\omega| > \omega_m </math> (signal is band limited to <math class="inline"> -\omega_m \leq \omega \leq \omega_m</math> )
−
  The Unit impulse responce $\hat{h}(t)$ of the inverse system satisfies $h(t) \convolution \hat{h}(t) = \delta(t)$
+
 
−
    ($h[n] \convolution \hat{h}[n] = \delta[n]$)
+
If <math class="inline"> T < \frac{1}{2}\left( \frac{2\pi}{\omega_m} \right) </math> then <math class="inline"> x(t) </math> can be uniquely recovered from its samples
−
\end{corollary}
+
 
−
because \begin{eqnarray*}
+
=[[Lecture25ECE301S11|Lecture 25]]=
−
  x(t) \rightarrow & \fbox{\text{$h(t)$}} \rightarrow \fbox{\text{$\hat{h}(t)$}} &\rightarrow x(t) \\
+
 
−
  x(t) \rightarrow & \fbox{\text{$h(t) \convolution \hat{h}(t) $}} &\rightarrow x(t)
+
 
−
        \end{eqnarray*}
+
----
−
\text{Example: time delay $y(t) = x(t-t_0)$ the inverse is $y(t) = x(t + t_0)$}
+
 
−
\begin{eqnarray*}
+
Here are my lecture notes from [[ECE301]] you can download both files from my dropbox account by Prof. Boutin
−
  h(t) &=& \delta(t-t_0)\\
+
 
−
  \hat{h}(t) &=& \delta(t + t_0)\\
+
There are bound to be a few errors in the document, if you find them please let me know and I'll fix them ASAP.
−
  h(t) \convolution \hat{h}(t) &=& \delta(t - t_0) \convolution \delta(t + t_0) \\
+
 
−
  &=& \int_{-\infty}^{\infty}\delta(\tau - t_0)\delta(t-\tau-t_0)d\tau \\
+
Lecture.pdf contains all lectures after lecture 5.
−
  &=& \delta(t) \text{ by siffting property}
+
 
−
\end{eqnarray*}
+
[http://dl.dropbox.com/u/16176877/Lecture.pdf Lecture.pdf]
−
\end{fact}
+
 
−
\end{document}         
+
[http://dl.dropbox.com/u/16176877/Lecture.tex Lecture.tex]
−
</nowiki>
+
 
 +
[http://dl.dropbox.com/u/16176877/Lecture_Notes_301_(5_-_10).pdf Lecture5.pdf]
 +
 
 +
[http://dl.dropbox.com/u/16176877/Lecture_Notes_301_(1_-_4).pdf Lectures 1 - 4]

Latest revision as of 08:22, 11 April 2013


Lecture 21

Multiplication Property

$ \mathcal{F}(x_1(t) x_2(t)) = \frac{1}{2\pi}\mathcal{F}(x_1(t))*\mathcal{F}(x_2(t)) $

Causal LTI system defined by cst coeff diff equations

$ \sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t) = \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) $

What is the frequency response of this system? Recall:

$ \begin{align} \mathcal{H}(\omega) &= \mathcal{F}(h(t)) \\ \mathcal{Y}(\omega) &= \mathcal{H}(\omega)\mathcal{X}(\omega) \end{align} $

Steps to solve:

$ \begin{align} \mathcal{F}\left(\sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t)\right) &= \mathcal{F}\left( \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) \right) \\ \sum_{k=0}^{N}a_k \mathcal{F}\left(\frac{d^k}{dt^k}y(t)\right) &= \sum_{k=0}^{M}b_k \mathcal{F}\left(\frac{d^k}{dt^k}x(t)\right) \\ & \text{recall: }\mathcal{F}\left(\frac{d^n}{dt^n}y(t)\right) = (j\omega)^n Y(\omega) \\ \sum_{k=0}^{N}a_k \left( j \omega \right)^k \mathcal{Y}(\omega) &= \sum_{k=0}^{M}b_k \left( j \omega \right)^k \mathcal{X}(\omega) \\ \mathcal{Y}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \mathcal{X}(\omega) \\ \mathcal{H}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \\ h(t) &= \mathcal{F}^{-1}\left(\mathcal{H}(\omega)\right) \end{align} $

Def of DT F.T.

Here are the practice problems that do this: Problem 1, Problem 2, Problem 3

$ \begin{align} x[n] &\xrightarrow{\mathcal{F}} \mathcal{X}(\omega) \text{ in book as }\mathcal{X}(e^{j\omega})& \\ x[n] &\xleftarrow{\mathcal{F}^{-1}} \mathcal{X}(\omega) &\\ \mathcal{X}(\omega) & = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} &\\ x[n] &= \frac{1}{2\pi}\int_{0}^{2\pi} \mathcal{X}(\omega)e^{j \omega n} d\omega & \end{align} $

Example

Compute the FT of $ x[n] = 2^{-n}u[n] $

$ \begin{align} \mathcal{X}(\omega) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty}2^{-n}u[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty}2^{-n}e^{-j\omega n} \\ &= \sum_{n=0}^{\infty}\left(\frac{1}{2e^{-j\omega}}\right)^n \\ &= \frac{1}{1-\frac{1}{2e^{j\omega}}} \end{align} $

Properties of DT FT

Periodicity

$ \begin{align} \mathcal{X}(\omega + 2\pi) &= \mathcal{X}\text{, for all } \omega \\ \text{because} & \\ \mathcal{X}(\omega + 2 \pi) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega + 2\pi)n} \\ &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} e^{-j2\pi n} \\ &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \mathcal{X}(\omega) \end{align} $

Linearity

$ \text{for all }a,b,c \in \mathcal{C}\text{ and all } x_1[n],x_2[n] $

$ \mathcal{F}(ax_1[n] + bx_2[n]) = a\mathcal{F}(n_1[n]) + b\mathcal{F}(n_2[n]) $ provided both FT's exist.

The FT of DT periodic signals

$ x[n] \text{ periodic} \Rightarrow x[n] = \sum_{n=0}^{N-1}a_k e^{\omega_0 n k } $

$ \mathcal{F}(x[n]) = \sum_{n=0}^{N-1}a_k \mathcal{F}\left(e^{\omega_0 n k }\right) $, by linearity

so all we need is the FT of $ e^{j k \omega_0 n} $

we want $ \mathcal{X}(\omega) $ such that $ \frac{1}{2\pi}\int_{0}^{2\pi}\mathcal{X}(\omega)e^{j\omega t} d\omega = e^{j\omega_0 n k} $

try $ \mathcal{X}(\omega) = 2\pi \delta(\omega - k\omega_0) $ and it works. So the real answer is

$ \mathcal{X}(\omega) = \sum_{m = -\infty}^{\infty} 2\pi \delta(\omega - \omega_0 k + 2\pi m) $

Lecture 22

Time Shifting and Freq Shifting Property

$ \begin{align} \mathcal{F}(x[n-n_0]) &= e^{-j\omega n_0}\mathcal{F}(x[n]) \\ \mathcal{F}\left( e^{j \omega_0 n}x[n] \right) &= \mathcal{X}(\omega - \omega_0) \end{align} $

Conjugation and Conjugation Symmetry

$ \mathcal{F}\left( x^*[n] \right) = \mathcal{X}^*(-\omega) $

Important Corrilary

if signal is real then $ \mathcal{X}(\omega) = \mathcal{X}^*(-\omega) $ because $ x[n] $ is real.

$ \begin{align} x^*[n] &= x[n] \\ \mathcal{X}^*[-\omega] &= \mathcal{X}(\omega) \end{align} $

This mean that $ x[n] $ real

=> Re $ \mathcal{X}(\omega) $ is an odd function

=> Im $ \mathcal{X}(\omega) $ is an odd function

Panseval's relation

$ \sum_{n=-\infty}^{\infty}| x[n] |^2 = \frac{1}{2\pi} \int_{0}^{2\pi} | x[\omega] |^2 d\omega $

Convolution Property

$ \begin{align} \mathcal{F}(x[n]*y[n]) &= \mathcal{F}(x[n])\mathcal{F}(y[n]) \\ &= \mathcal{X}(\omega) \mathcal{X}(\omega) \end{align} $

so for any LTI system $ x \rightarrow h[n] \rightarrow y[n] = x[n]*h[n] $

Lecture 23

Multiplication Property

$ x[n]y[n] \xrightarrow{\mathcal{F}} \frac{1}{2\pi} \mathcal{X}(\omega)*\mathcal{Y}(\omega) $

Differentiation in frequency property

$ nx[n] \xrightarrow{\mathcal{F}} j\frac{d}{d\omega}\mathcal{X}(\omega) $

Example

Assume $ |\alpha | < 1 $

  1. Compute the FT of $ x_1[n] = \alpha^n u[n] $
  2. Use your andwer to compute the FT of $ x_2[n] = (n+1)\alpha^n u[n] $

Answer: 1)

$ \begin{align} \mathcal{X}_1(\omega) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty} \alpha^nu[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty} \alpha^nu[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty} \left(\alpha u[n]e^{-j\omega } \right) ^n \\ &= \frac{1}{1-\alpha e^{-j\omega}} \end{align} $

2)

$ \begin{align} \mathcal{X}_2(\omega) &= \mathcal{F}((x+1)\alpha^n u[n])\\ &= \mathcal{F}(n\alpha^n u[n]) + \mathcal{F}(\alpha^n u[n]) \\ &= j\frac{d}{d\omega}\mathcal{X}_1(\omega) + \mathcal{X}_1(\omega) \\ &= j\frac{d}{d\omega}\left( \frac{1}{1-\alpha e^{-j\omega}} \right) + \frac{1}{1-\alpha e^{-j\omega}} \\ &= \frac{1}{\left( 1 - \alpha e^{-j\omega} \right)^2} \end{align} $

LTI systems defined by linear, constant coef diff eq's

$ \sum_{k=0}^{N}a_k y[n-k] = \sum_{k=0}^{M}b_k y[n-k] $

Example: $ y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n] = 2x[n] $ can look at this eq in freq domain.

$ \begin{align} \mathcal{F}(y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n]) &= 2\mathcal{F}(x[n]) \\ \mathcal{Y} - \frac{3}{4}\mathcal{F}(y[n-1]) + \frac{1}{8}\mathcal{F}(y[n-8]) &= 2\mathcal{X}(\omega) \\ \mathcal{Y}(\omega) - \frac{3}{4}e^{-j\omega}\mathcal{Y}(\omega) + \frac{1}{8}e^{-2j\omega}\mathcal{Y}(\omega) &= 2\mathcal{X}(\omega) \end{align} $

$ \begin{align} \mathcal{Y}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}}\mathcal{X}(\omega) \\ \mathcal{H}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}} \\ &= \frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}} \\ h[n] &= 4\left(\frac{1}{2}\right)^n u[n] - 2 \left( \frac{1}{4} \right)^n u[n] \end{align} $

Now find the systems responce to $ x[n] = \left( \frac{1}{4} \right)^n u[n] $

$ \begin{align} \mathcal{Y}(\omega) &= \mathcal{X}(\omega)\mathcal{Y}(\omega) \\ &= \left(\frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}}\right)\frac{1}{1-\frac{1}{4}e^{-j\omega}} \\ &= \frac{4}{\left(1-\frac{1}{2}e^{-j\omega}\right) \left(1-\frac{1}{4}e^{-j\omega}\right)} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\ &= \frac{-4}{1-\frac{1}{2}e^{-j\omega}} + \frac{8}{1-\frac{1}{4}e^{-j\omega}} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\ y[n] &= -4\left(\frac{1}{4}\right)^n u[n]+8\left(\frac{1}{2}\right)^n u[n] -2(n+1)\left(\frac{1}{4}\right)^n u[n] \end{align} $

Lecture 24

Section 7.1 Sampling \ Representing a CT signal by Sampling

What is sampling? The process of measuring a CT signal is to take samples ( at the same interval for ECE301 ) so you get a vector of values that fit the equation $ x(nT) $ where n is an int and the distance between the samples is T. This makes a DT signal $ x_d[n] = x(nT) $

Problem: given $ x_d[n] $ can you reconstruct $ x(t) $? In general no. However, we can approximate $ x(t) $ by interpolation values of $ x(t) $ between the samples.

Ex 1: Interpolation by using step function

$ x(t) = \sum_{k =-\infty}^{\infty} x(kt)\left( u(t-kT) - u(t-(k+1)T) \right) $

Ex 2: Interpolation by using piecewise linear function

Take take the current point and the next one and find a line that will connect the two

Conclusion: There are infanately many ways to reconstruct the signal from samples and get an approximation. All of these could have been the initial $ x(t) $. So it is impossible to reconstruct $ x(t) $ from samples.

Sampling Theorem

  1. Let $ \omega_m $ be a non-neg number
  2. Let $ x(t) $ be a signal such that $ \mathcal{X}(\omega) = 0 $ when $ |\omega| > \omega_m $ (signal is band limited to $ -\omega_m \leq \omega \leq \omega_m $ )

If $ T < \frac{1}{2}\left( \frac{2\pi}{\omega_m} \right) $ then $ x(t) $ can be uniquely recovered from its samples

Lecture 25

Here are my lecture notes from ECE301 you can download both files from my dropbox account by Prof. Boutin

There are bound to be a few errors in the document, if you find them please let me know and I'll fix them ASAP.

Lecture.pdf contains all lectures after lecture 5.

Lecture.pdf

Lecture.tex

Lecture5.pdf

Lectures 1 - 4

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