\documentclass[letterpaper,10pt]{report} \usepackage[utf8x]{inputenc} \usepackage{amsmath, amsthm, amssymb, mathabx} \usepackage{fullpage} \newenvironment{Observation}[2][Observation]{\begin{trivlist} \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} \newtheorem{property}{Property} \newtheorem{fact}{Fact} \newtheorem*{corollary}{Corollary} % Title Page \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} 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 $} \end{cases} \end{eqnarray*} \end{proof} To know for the rest of your life: \begin{eqnarray*} \sum_{k=0}^{n} \alpha^{k} = \begin{cases} \frac{1 - \alpha^{n+1}}{1-\alpha},& \text{$\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} 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} \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}
