1 %\documentclass[a4paper]{scrartcl} |
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2 %\usepackage[top=2cm, bottom=2.5cm, left=3cm, right=2cm, footskip=1cm]{geometry} |
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3 %\usepackage[german]{babel} |
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4 %\usepackage[T1]{fontenc} |
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5 %\usepackage[latin1]{inputenc} |
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6 %\usepackage{endnotes} |
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7 %\usepackage{trfsigns} |
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8 %\usepackage{setspace} |
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9 % |
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10 %\setlength{\parindent}{0ex} |
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11 %\def\isac{${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$} |
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12 %\def\sisac{{\footnotesize${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}} |
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13 % |
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14 %\begin{document} |
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15 %\title{Interactive Course Material for Signal Processing based on Isabelle/\isac} |
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16 %\subtitle{Problemsolutions (Calculations)} |
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17 %\author{Walther Neuper, Jan Rocnik} |
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18 %\maketitle |
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19 |
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20 |
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21 %------------------------------------------------------------------------------ |
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22 %FOURIER |
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23 |
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24 \subsection{Fourier Transformation} |
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25 \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally fragment of the exam to \emph{Signaltransformationen VO} from 04.03.2011. Translated from German.}} |
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26 \textbf{(a)} Determine the fourier transform for the given rectangular impulse: |
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27 |
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28 \begin{center} |
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29 $x(t)= \left\{ |
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30 \begin{array}{lr} |
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31 1 & -1\leq t\geq1\\ |
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32 0 & else |
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33 \end{array} |
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34 \right.$ |
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35 \end{center} |
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36 |
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37 \textbf{\noindent (b)} Now consider the given delayed impulse, determine its fourie transformation and calculate phase and magnitude: |
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38 |
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39 \begin{center} |
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40 $x(t)= \left\{ |
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41 \begin{array}{lr} |
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42 1 & -1\leq t\leq1\\ |
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43 0 & else |
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44 \end{array} |
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45 \right.$ |
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46 \end{center} |
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47 |
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48 \subsubsection{Solution} |
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49 \textbf{(a)} \textsf{Subproblem 1} |
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50 \onehalfspace{ |
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51 \begin{tabbing} |
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52 000\=\kill |
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53 \texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\ |
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54 \`Insert Condition: $x(t) = 1\;$ for $\;\{-1\leq t\;\land\;t\leq 1\}\;$ and $\;x(t)=0\;$ otherwise\\ |
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55 \texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\ |
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56 \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\ |
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57 \texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\ |
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58 \` pbl: integration in $\cal C$\\ |
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59 \texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1\right)$\\ |
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60 \` $f\;t\;|_a^b = f\;b-f\;a$\\ |
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61 \texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1}$\\ |
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62 \texttt{\footnotesize{06}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega} - \frac{1}{-j\cdot\omega}\cdot e^{j\cdot\omega}$\\ |
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63 \` Lift $\frac{1}{j\omega}$\\ |
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64 \texttt{\footnotesize{07}} \> $\frac{1}{j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\ |
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65 \` trick~!\\ |
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66 \texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\ |
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67 \` table\\ |
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68 \texttt{\footnotesize{09}} \> $2\cdot\frac{\sin\;\omega}{\omega}$ |
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69 \end{tabbing} |
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70 } |
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71 |
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72 \noindent\textbf{(b)} \textsf{Subproblem 1} |
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73 \onehalfspace{ |
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74 \begin{tabbing} |
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75 000\=\kill |
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76 \texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\ |
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77 \`Insert Condition: $x(t) = 1\;$ for $\;\{1\leq t\;\land\;t\leq 3\}\;$ and $\;x(t)=0\;$ otherwise\\ |
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78 \texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\ |
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79 \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\ |
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80 \texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{1}^3$\\ |
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81 \` pbl: integration in $\cal C$\\ |
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82 \texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{1}^3\right)$\\ |
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83 \` $f\;t\;|_a^b = f\;b-f\;a$\\ |
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84 \texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 3} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1}$\\ |
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85 \`Lift $\frac{1}{-j\omega}$\\ |
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86 \texttt{\footnotesize{06}} \> $\frac{1}{j\cdot\omega}\cdot(e^{-j\cdot\omega} - e^{-j\cdot\omega3})$\\ |
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87 \`Lift $e^{j\omega2}$ (trick)\\ |
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88 \texttt{\footnotesize{07}} \> $\frac{1}{j\omega}\cdot e^{j\omega2}\cdot(e^{j\omega} - e^{-j\omega})$\\ |
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89 \`Simplification (trick)\\ |
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90 \texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot e^{j\omega2}\cdot(\frac{e^{j\omega} - e^{-j\omega}}{j})$\\ |
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91 \` table\\ |
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92 \texttt{\footnotesize{09}} \> $2\cdot e^{j\omega2}\cdot\frac{\sin\;\omega}{\omega}$\\ |
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93 \noindent\textbf{(b)} \textsf{Subproblem 2}\\ |
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94 \`Definition: $X(j\omega)=|X(j\omega)|\cdot e^{arg(X(j\omega))}$\\ |
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95 \`$|X(j\omega)|$ is called \emph{Magnitude}\\ |
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96 \`$arg(X(j\omega))$ is called \emph{Phase}\\ |
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97 \texttt{\footnotesize{10}} \> $|X(j\omega)|=\frac{2}{\omega}\cdot sin(\omega)$\\ |
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98 \texttt{\footnotesize{11}} \> $arg(X(j\omega)=-2\omega$\\ |
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99 \end{tabbing} |
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100 } |
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101 %------------------------------------------------------------------------------ |
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102 %CONVOLUTION |
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103 |
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104 \subsection{Convolution} |
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105 \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the SPSC Problem Class 2, Summer term 2008}} |
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106 Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response: |
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107 |
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108 \begin{center} |
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109 $h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\ |
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110 $h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$ |
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111 \end{center} |
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112 |
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113 The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$. |
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114 \subsubsection*{Solution} |
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115 |
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116 \doublespace{ |
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117 \begin{tabbing} |
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118 000\=\kill |
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119 \texttt{\footnotesize{01}} \> $h_c[n]=h_1[n]*h_2[n]$\\ |
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120 \texttt{\footnotesize{02}} \> $h_c[n]=\left(\left(\frac{3}{5}\right)^n\cdot u[n]\right)*\left(\left(-\frac{2}{3}\right)^n\cdot u[n]\right)$\\ |
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121 \`Definition: $a^n\cdot u[n]\,*\,b^n\cdot u[n]=\sum\limits_{k=-\infty}^{\infty}{a^k\cdot u[k]\cdot b^{n-k}\cdot u[n-k]}$\\ |
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122 \texttt{\footnotesize{03}} \> $h_c[n]=\sum\limits_{k=-\infty}^{\infty}{\left(\frac{3}{5}\right)^k\cdot u[n]\,\cdot \,\left(-\frac{2}{3}\right)^{n-k}\cdot u[n-k]}$\\ |
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123 \`$u[n]= \left\{ |
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124 \begin{array}{lr} |
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125 1 & for\ n>=0\\ |
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126 0 & else |
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127 \end{array} |
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128 \right.$\\ |
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129 \`We can leave the unitstep through simplification.\\ |
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130 \`So the lower limit is 0, the upper limit is n.\\ |
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131 \texttt{\footnotesize{04}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n-k}}$\\ |
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132 \`Expand\\ |
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133 \texttt{\footnotesize{05}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n}\cdot \left(-\frac{2}{3}\right)^{-k}}$\\ |
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134 \`Lift\\ |
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135 \texttt{\footnotesize{06}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{\infty}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{-k}}$\\ |
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136 \texttt{\footnotesize{07}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{3}{2}\right)^{k}}$\\ |
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137 \texttt{\footnotesize{08}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(-\frac{9}{10}\right)^{k}}$\\ |
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138 \`Geometric Series: $\sum\limits_{k=0}^{n}{q^k}=\frac{1-q^{n+1}}{1-q}$\\ |
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139 \`Now we have to consider the limits again.\\ |
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140 \`It is neccesarry to put the unitstep in again.\\ |
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141 \texttt{\footnotesize{09}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n+1}}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\ |
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142 \texttt{\footnotesize{10}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\ |
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143 \texttt{\footnotesize{11}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{\left(\frac{19}{10}\right)}\cdot u[n]$\\ |
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144 \texttt{\footnotesize{12}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot \left(1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)\right)\cdot\left(\frac{10}{19}\right)\cdot u[n]$\\ |
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145 \`Lift $u[n]$\\ |
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146 \texttt{\footnotesize{13}} \> $\left(\frac{10}{19}\cdot\left(-\frac{2}{3}\right)^n+\frac{9}{19}\cdot\left(\frac{3}{5}\right)^n\right)\cdot u[n]$\\ |
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147 \end{tabbing} |
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148 } |
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149 |
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150 %------------------------------------------------------------------------------ |
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151 %Z-Transformation |
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152 |
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153 \subsection{Z-Transformation\label{sec:calc:ztrans}} |
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154 \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the signal processing problem class 5, summer term 2008.}} |
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155 Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion. |
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156 |
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157 \begin{center} |
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158 $X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable |
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159 \end{center} |
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160 |
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161 \subsubsection*{Solution} |
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162 \onehalfspace{ |
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163 \begin{tabbing} |
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164 000\=\kill |
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165 \textsf{Main Problem}\\ |
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166 \texttt{\footnotesize{01}} \> $\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}}$ \\ |
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167 \`Divide through z, neccesary for z-transformation\\ |
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168 \texttt{\footnotesize{02}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}$ \\ |
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169 \`Start with partial fraction expansion\\ |
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170 \texttt{\footnotesize{03}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}=\frac{A}{z-z_1}+\frac{B}{z-z_2}$ \\ |
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171 \`Eliminate Fractions\\ |
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172 \texttt{\footnotesize{04}} \> $3=A(z-z_2)+B(z-z_1)$ \\ |
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173 \textsf{Subproblem 1}\\ |
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174 \`Setup a linear equation system by inserting the zeros $z_1$ and $z_2$ for $z$\\ |
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175 \texttt{\footnotesize{05}} \> $3=A(z_1-z_2)$ \& $3=B(z_2-z_1)$\\ |
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176 \texttt{\footnotesize{06}} \> $\frac{3}{z_1-z_2}=A$ \& $\frac{3}{z_2-z_1}=B$\\ |
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177 \textsf{Subproblem 2}\\ |
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178 \`Determine $z_1$ and $z_2$\\ |
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179 \texttt{\footnotesize{07}} \> $z_1=\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{8}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{8}}$\\ |
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180 \texttt{\footnotesize{08}} \> $z_1=\frac{1}{8}+\sqrt{\frac{9}{64}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{9}{64}}$\\ |
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181 \texttt{\footnotesize{09}} \> $z_1=\frac{1}{8}+\frac{3}{8}$ \& $z_2=\frac{1}{8}-\frac{3}{8}$\\ |
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182 \texttt{\footnotesize{10}} \> $z_1=\frac{1}{2}$ \& $z_2=-\frac{1}{4}$\\ |
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183 \textsf{Continiue with Subproblem 1}\\ |
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184 \`Get the coeffizients $A$ and $B$\\ |
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185 \texttt{\footnotesize{11}} \> $\frac{3}{\frac{1}{2}-(-\frac{1}{4})}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\ |
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186 \texttt{\footnotesize{12}} \> $\frac{3}{\frac{1}{2}+\frac{1}{4}}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\ |
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187 \texttt{\footnotesize{13}} \> $\frac{3}{\frac{3}{4}}=A$ \& $\frac{3}{-\frac{3}{4}}=B$\\ |
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188 \texttt{\footnotesize{14}} \> $\frac{12}{3}=A$ \& $-\frac{12}{3}=B$\\ |
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189 \texttt{\footnotesize{15}} \> $4=A$ \& $-4=B$\\ |
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190 \textsf{Continiue with Main Problem}\\ |
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191 \texttt{\footnotesize{16}} \> $\frac{A}{z-z_1}+\frac{B}{z-z_2}$\\ |
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192 \texttt{\footnotesize{17}} \> $\frac{4}{z-\frac{1}{2}}+\frac{4}{z-\left(-\frac{1}{4}\right)}$ \\ |
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193 \texttt{\footnotesize{18}} \> $\frac{4}{z-\frac{1}{2}}-\frac{4}{z+\frac{1}{4}}$ \\ |
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194 \`Multiply with z, neccesary for z-transformation\\ |
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195 \texttt{\footnotesize{19}} \> $\frac{4z}{z-\frac{1}{2}}-\frac{4z}{z+\frac{1}{4}}$ \\ |
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196 \texttt{\footnotesize{20}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}$ \\ |
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197 \`Transformation\\ |
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198 \texttt{\footnotesize{21}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}\ \Ztransf\ 4\cdot\left(-\frac{1}{2}\right)^n\cdot u[n]+(-4)\cdot\left(\frac{1}{4}\right)^n\cdot u[n]$\\ |
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199 \end{tabbing} |
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200 } |
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201 \theendnotes |
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202 %\end{document} |
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