Difference between revisions of "2011 Spring ECE 301 Bouton Notes" - Rhea

Revision as of 18:15, 1 February 2011

Here are my lecture notes for ECE301 you can download both files from my dropbox account

Lecture.pdf

Lecture.tex

Notes 301 ( 5 - 10 ).pdf Lecture5.pdf

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-<nowiki>
+Here are my lecture notes for ECE301 you can download both files from my dropbox account
-\documentclass[letterpaper,10pt]{report}
-\usepackage[utf8x]{inputenc}
-\usepackage{amsmath, amsthm, amssymb, mathabx}
-\usepackage{fullpage}
-\newenvironment{Observation}[2][Observation]{\begin{trivlist}
+[http://dl.dropbox.com/u/16176877/Lecture.pdf Lecture.pdf]
-\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries
-#2}]}{\end{trivlist}}
-\newenvironment{definition}[1][Definition]{\begin{trivlist}
-\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
-\newtheorem*{why}{Why}
+[http://dl.dropbox.com/u/16176877/Lecture.tex Lecture.tex]
-\newtheorem{property}{Property}
-\newtheorem{fact}{Fact}
-\newtheorem*{corollary}{Corollary}
-% Title Page
+[http://dl.dropbox.com/u/16176877/Lecture Notes 301 ( 5 - 10 ).pdf Lecture5.pdf]
-\title{ECE 301}
-\author{Ethan Hall}
-\begin{document}
-\maketitle
-\part*{Lecture 6}
-\section*{Memorkess/System with Memory}
-\begin{definition}
-  A system is memoryless if, for any $t_0 \in \mathbb{R}$,
-  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 $t < t_0$ )
-\end{definition}
-Example:
-\begin{eqnarray*}
-  y(t) & = & x(t) + x(t - 1) \text{, memory} \\
-  y(t) & = & 2x(t) \text{, memoryless} \\
-  y(t) & = & (t-1)(x(t)) \text{, memoryless} \\
-  y(t) & = & \int_{-\infty}^{\infty}x(t')dt' \text{, memory}
-\end{eqnarray*}
-\begin{fact}
-  A memoryless system can be written as $ y(t) = f(t,x(t)) $ ( $Y[n] = f(n,x[n]) $ )
-\end{fact}
-\section*{Causal/Non-Causal}
-\begin{definition}
-  A system is causal if, for any $t_0 \in \mathbb{R}$, the output of $y(t_0)$
-  depends only on the input $x(t_0)$ at or before $t_0$ ( $t \leq t_0 $ )
-\end{definition}
-Example:
-\begin{equation*}
-  y(t) = x(t^2)\text{, non-causal}
-\end{equation*}
-take $y(-1/2) == y(1/2)$
-\section*{Invertable Systems}
-\begin{definition}
-  A system is invertable if distinct input signals yield distinct output signals
-\begin{eqnarray*}
-  X_1[n] \rightarrow &\fbox{\text{system}}& \rightarrow Y_1[n] \\
-  X_2[n] \rightarrow &\fbox{\text{system}}& \rightarrow Y_2[n] \\
-  \text{where } X_1[n] &\neq X_2[n]
-\end{eqnarray*}
-\end{definition}
-\begin{definition}
-  System is invertable if there exists another system, called the ``inverse''
-  such that the cascade leaves the input unchanged
-\begin{equation*}
-  x(t) \rightarrow \fbox{\text{system}} \rightarrow \fbox{\text{inverse}} \rightarrow x(t)
-\end{equation*}
-\end{definition}
-Example:
-\begin{eqnarray*}
-  y(t) & = & 2x(t) + 3 \text{ is invertable } \\
-  x(t) & = & \frac{y(t) - 3}{2}
-\end{eqnarray*}
-\begin{equation*}
-  x(t) \rightarrow \fbox{\text{system}} \rightarrow \text{y(t) = 2x(t) + 3} \rightarrow \fbox{\text{inverse}} \rightarrow z(t)
-\end{equation*}
-\begin{equation*}
-  z(t) = \frac{1}{2}y(t) + \frac{3}{2} \rightarrow \frac{1}{2}(2x(t) + 3) + \frac{3}{2} \Rightarrow x(t)
-\end{equation*}
-Example of non-invertable system $y(t) = (x(t))^2$
-\begin{eqnarray*}
-  x_1(t) &= t \Rightarrow t^2 \\
-  x_2(t) &= -t \Rightarrow t^2
-\end{eqnarray*}
-\section*{Stability - (BIBO Stability)}
-\begin{definition}
-  A system is called BIBO-stable if a bounded input yields a bounded output
-\end{definition}
-\begin{eqnarray*}
-  \text{if there exists } &E \text{ suck that } &\|x(t)\| < E\text{, for all $t$} \\
-  \text{then there exists } &M \text{ such that } &\|y(t)\| < M\text{, for all $t$}
-\end{eqnarray*}
-Example: $Y(t) = e^{x(t)}$ is \underline{stable} because if $\|x(t)\| < E \Rightarrow \|y(t)\| = \|e^{x(t)}\|$
-\section*{Time Invarrience}
-\begin{definition}{1}
- If the cascade of
-\begin{eqnarray*}
-  x(t) \rightarrow \fbox{\text{system}} &\rightarrow \fbox{\text{time delay $t_0$}} \rightarrow y(t) \\
-  &\parallel \\
-  x(t) \rightarrow \fbox{\text{time delay $t_0$}} &\rightarrow \fbox{\text{system}} \rightarrow y(t)
-\end{eqnarray*}
-\end{definition}
-\begin{definition}{2}
- Time invarient means for any input sig $x(t)$ ( $x[n]$ )
-  and for any time $t_0$, the output for athe shifted input $x(t-t_0)$ is the shifted output $y(t-t_0)$
-\end{definition}
-\part*{Lecture 7}
-\begin{definition}{3}
-\begin{eqnarray*}
-  \text{A system is time invarient if} \\
-  x(t) &\rightarrow \fbox{\text{system}} &\rightarrow y(t) \\
-  \text{then we also have} \\
-  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}
-  A system is time invarient if it comutes with a time delay
-\end{definition}
-Example 1: show time invarient $x(t) \rightarrow \fbox{\text{subject}} \rightarrow y(t) = 10x(t)$
-\begin{eqnarray*}
-  x(t) \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow &y(t) = x(t - t_0)&
-    \rightarrow \fbox{\text{system}} \rightarrow z_1(t) \\
-  & z_1(t) = 10y(t) &\\
-  & z_1(t) = 10 x(t - t_0)& \\
-  x(t) \rightarrow \fbox{\text{system}} \rightarrow &y(t) = 10 x(t) &
-    \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow z_2(t) \\
-  &z_2(t) = y(t - t_0) & \\
-  &z_2(t) = 10x(t - t_0)& \\
-  &z_1(t) \equiv z_2(t)&
-\end{eqnarray*}
-Example 2: show time invarient $t x(t)$
-\begin{eqnarray*}
-  x(t) \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow &y(t) = x(t - t_0)&
-    \rightarrow \fbox{\text{system}} \rightarrow z_1(t) \\
-  & z_1(t) = t y(t) &\\
-  & z_1(t) = t x(t - t_0)& \\
-  x(t) \rightarrow \fbox{\text{system}} \rightarrow &y(t) = t x(t) &
-    \rightarrow \fbox{\text{time delay $t_0$}} \rightarrow z_2(t) \\
-  &z_2(t) = y(t - t_0) & \\
-  &z_2(t) = (t - t_0)(x(t - t_0))& \\
-  &z_1(t) \neq  z_2(t)&
-\end{eqnarray*}
-\section*{Linearity}
-\begin{definition}{1}
-  A system is called ``linear'' if for any constant $a,b, \in \mathbb{C}$ and for any input signals $X_1(t), X_2(t)$
-    ( $X_1[n], X_2[n]$ ) 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] $ )
-\end{definition}
-\begin{definition}{2}
-  if
-\begin{eqnarray*}
-  x_1(t) &\rightarrow \fbox{\text{system}} \rightarrow &y_1(t) \\
-  x_2(t) &\rightarrow \fbox{\text{system}} \rightarrow &y_2(t)
-\end{eqnarray*}
-\begin{equation*}
- \text{then } a x_1(t) + b x_2(t) \rightarrow \fbox{\text{system}} \rightarrow a t_1(t) + b t_2(t)
-\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}
-  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*}
-\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}
-\setcounter{section}{0}
-\setcounter{chapter}{2}
-\section{The convolution sum for LTI systems}
-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}
-%Lecture 8
-\part*{Lecture 8}
-Example: Write u[n] as a linear combanation of shifted $\delta$[n]
-\begin{equation*}
-u[n] = \sum_{k=0}^{\infty}\delta[n-k]
-\end{equation*}
-\begin{Observation}{2}
-The respoce ofa  DT linear system can be written as a sum $y[n] =
-\sum_{k=-\infty}^{\infty}x[k]h_{k}[n]$; where $h_{k}[n]$ is the systems responce to $\delta[n-k]$
-  \begin{why}
-    $x[n] = \sum_{k=-\infty}^{\infty} \underbrace{x[k]}_{\text{const}}\delta[n-k]\text{, by observation 1}$
-  \end{why}
-  by linearity $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$
-\end{Observation}
-\begin{Observation}{3}
-  The responce of an LTI system can be written as an even simpler function $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$
-  where $h[n]$ is the responce to $\delta[n]$. (We call $h[n]$ the ``unit impulse responce'' of the system)
-  \begin{why}
-   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
-\begin{equation*}
-  Z_{1}[n] \convolution Z_{2}[n] = \sum_{k=-\infty}^{\infty} Z_{1}[k]Z_{2}[n-k]
-\end{equation*}
-\begin{proof}
-\begin{eqnarray*}
-    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}
-, & \text{$k \geq 0$}\\
-, & \text{$k < 0$}
-    \end{cases} \\
-    & = & \sum_{k=-\infty}^{\infty}(2^{k})(1)(u[n-k]) \text{, but } u[n-k] =
-    \begin{cases}
-, & \text{$n-k \geq 0$} \\
-, & \text{$n-k < 0$}
-    \end{cases} \\
-    & = &
-    \begin{cases}
-      \frac{1-2^{n+1}}{1-2}, & \text{$ n \geq 0 $}\\
-, & \text{$ n < 0 $}
-    \end{cases}
-\end{eqnarray*}
-\end{proof}
-To know for the rest of your life:
-\begin{eqnarray*}
-\sum_{k=0}^{n} \alpha^{k} = \begin{cases}
-\frac{1 - \alpha^{n+1}}{1-\alpha},& \text{$\alpha \neq 1 $} \\
-n + 1,& \text{$ \alpha = 1$}
-\end{cases}\\
-\sum_{k=0}^{\infty} \alpha^{k} = \begin{cases}
-\frac{1}{1-\alpha},& \text{$\|\alpha \| < 1 $} \\
-diverdges,& \text{ $\|\alpha \| \geq 1 $}
-\end{cases}
-\end{eqnarray*}
-\setcounter{section}{1}
-\setcounter{chapter}{2}
-\section{CT LTI system and Confolution Intergral}
-\begin{Observation}{1}
- Any CT system can be written as an intregral
-  \begin{equation*}
-    x(t) = \int_{-\infty}^{\infty} x(\tau)\delta(t - \tau)d\tau
-  \end{equation*}
-  \begin{why}
-    \begin{equation*}
-      x(\tau)\delta(t-\tau) = x(t)\delta(t-\tau)\text{, for any $t$}
-    \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}
-  If a system is linear, then its output can be written as an intergral
-  \begin{equation*}
-    y(t) = \int_{-\infty}^{\infty}x(\tau)h_{\tau}(t)d\tau \text{ where $h_{\tau}(t)$ is the systems responce to $\delta(t-\tau)$ }
-  \end{equation*}
-\end{Observation}
-\begin{Observation}{3}
-  \begin{equation*}
-    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
-\begin{equation*}
-  X_{1}(t) \convolution X_{2}(t) = \int_{-\infty}^{\infty} X_{1}(\tau)X_{2}(t-\tau)d\tau
-\end{equation*}
-\part*{Lecture 9}
-For LTI systems
-\begin{equation*}
- y(t) = x(t) \convolution h(t) = h(t) \convolution x(t)
-\end{equation*}
-Example: the unit impuse resonse of an LTI system is $h(t) = u(t)$\newline
-Find the systems response to $x(t) = e^{-t}u(t)$
-\begin{eqnarray*}
-  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}
-, &x < 0 \\
-, &x \geq 1
-    \end{cases}\\
-  &=& \int_{0}^{\infty}e^{-\tau}u(t-\tau)d\tau\text{, but } u(t-\tau) =
-\begin{cases}
-, &t < \tau \\
-, &t \geq \tau
-\end{cases}\\
-y(t)&=&
-\begin{cases}
-  /int_{0}^{t}e^{-\tau}d\tau, & t \geq 0 \\
-, & t < 0
-\end{cases}\\
-&=& \dfrac{e^{-\tau}}{-1}\bigg|_{0}^{t}u(t) = (-e^{-\tau}-1)u(t)
-\end{eqnarray*}
-\setcounter{section}{2}
-\setcounter{chapter}{2}
-\section{Properties of an LTI System}
-\begin{eqnarray}
- y[n] & = & x[n] \convolution h[n] \nonumber \\
- y(t) & = & x(t) \convolution h(t)
-\end{eqnarray}
-\begin{property}
-  \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}
-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}
-Sence $\convolution$ is a linear operator
-  \begin{equation}
-    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*}
-   \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}
-  \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}
-\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}
-  If an LTI system is invertable then its inverse is also LTI.
-\begin{corollary}
-  The Unit impulse responce $\hat{h}(t)$ of the inverse system satisfies $h(t) \convolution \hat{h}(t) = \delta(t)$
-    ($h[n] \convolution \hat{h}[n] = \delta[n]$)
-\end{corollary}
-because \begin{eqnarray*}
-	  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*}
-  h(t) &=& \delta(t-t_0)\\
-  \hat{h}(t) &=& \delta(t + t_0)\\
-  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 \\
-  &=& \delta(t) \text{ by siffting property}
-\end{eqnarray*}
-\end{fact}
-\end{document}
-</nowiki>

Difference between revisions of "2011 Spring ECE 301 Bouton Notes" - Rhea

Revision as of 18:15, 1 February 2011

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