1.1 --- a/doc-src/isac/jrocnik/present-1.tex Fri Jul 22 10:17:31 2011 +0200
1.2 +++ b/doc-src/isac/jrocnik/present-1.tex Fri Jul 22 10:20:24 2011 +0200
1.3 @@ -1,5 +1,3 @@
1.4 -
1.5 -% test
1.6
1.7 \documentclass{beamer}
1.8
1.9 @@ -128,6 +126,7 @@
1.10 \begin{frame}\frametitle{Fourier Transform 1: Specification}
1.11 {\footnotesize\it
1.12 Fourier Transform
1.13 +
1.14 \begin{tabbing}
1.15 1\=postcond \=: \= \= $\;\;\;\;$\=\kill
1.16 \>given \>:\> Time continiues, not periodic Signal \\
1.17 @@ -176,15 +175,24 @@
1.18
1.19 \begin{frame}\frametitle{Fourier Transform 2: Specification}
1.20 {\footnotesize\it
1.21 -Fourier Transform
1.22 +
1.23 +\textbf{(a)} Determine the fourier transform for the given rectangular impulse:
1.24 +
1.25 +\begin{center}
1.26 +$x(t)= \left\{
1.27 + \begin{array}{lr}
1.28 + 1 & -1\leq t\geq1\\
1.29 + 0 & else
1.30 + \end{array}
1.31 + \right.$
1.32 +\end{center}
1.33 +
1.34 \begin{tabbing}
1.35 1\=postcond \=: \= \= $\;\;\;\;$\=\kill
1.36 \>given \>:\> piecewise\_function \\
1.37 \> \> \> \>$(x (t::real), [(0,-\infty<t<1), (1,1\leq t\leq 3), (0, 3<t<\infty)])$\\
1.38 %?(iterativer) datentyp in Isabelle/HOL
1.39 \> \> \> translation $T=2$\\
1.40 -%WN these 2 inputs calculated to [(0,-\infty<t<-1), (1,-1\leq t\leq 1), (0, 1<t<\infty)]
1.41 -%WN translation helpful only, if solution available from other calculation - DROP T ???
1.42 \>precond \>:\> TODO\\
1.43 \>find \>:\> $X(j\cdot\omega)$\\
1.44 \>postcond \>:\> TODO\\
1.45 @@ -197,40 +205,38 @@
1.46 %% Transform expl 2 CALC %%
1.47 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.48
1.49 -\begin{frame}\frametitle{Fourier Transform 2: Calculation}
1.50 -\footnotesize{
1.51 -\begin{tabbing}
1.52 -000\=\kill
1.53 -%WN first
1.54 -01 \> ${\cal F}\;(x(t-2)) =$\\
1.55 - \`${\cal F}\;(x(t-T)) = e^{-j\cdot\omega\cdot T}\cdot X\;j\cdot\omega$\\
1.56 -02 \> $e^{-j\cdot\omega\cdot 2}\cdot X\;(j\cdot\omega)$\\
1.57 - \`definition $X\;(j\cdot\omega)$\\
1.58 -03 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-\infty}^\infty x\;t\;\cdot e^{-j\cdot\omega\cdot t} d t$\\
1.59 - \` $x\;t = 1\;{\it for}\;\{x.\;-1\leq t\;\land\;t\leq 1\}\;{\it and}\;x\;t=0\;{\it otherwise}$\\
1.60 -04 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-1}^1 1\cdot e^{-j\cdot\omega\cdot t} d t$\\
1.61 - \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
1.62 -05 \> $e^{-j\cdot\omega\cdot 2}\cdot \int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\
1.63 - %\` $\int e^{a\cdot t} = \frac{1}{a}\cdot e^{a\cdot t}$\\
1.64 - \` pbl: integration in $\cal C$\\
1.65 -06 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1)$\\
1.66 - \` $f\;t\;|_a^b = f\;b-f\;a$\\
1.67 -07 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1})$\\
1.68 -\vdots\` pbl: simplification+factorization in $\cal C$\\
1.69 -08 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{-j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\
1.70 - \` trick~!\\
1.71 -09 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\
1.72 - \` table\\
1.73 -10 \> $e^{-j\cdot\omega\cdot 2}\cdot 2\cdot\frac{\sin\;\omega}{\omega}$
1.74 -\end{tabbing}
1.75 -}
1.76 -\end{frame}
1.77 +%\begin{frame}\frametitle{Fourier Transform 2: Calculation}
1.78 +%\footnotesize{
1.79 +%\begin{tabbing}
1.80 +%000\=\kill
1.81 +%01 \> ${\cal F}\;(x(t-2)) =$\\
1.82 +% \`${\cal F}\;(x(t-T)) = e^{-j\cdot\omega\cdot T}\cdot X\;j\cdot\omega$\\
1.83 +%02 \> $e^{-j\cdot\omega\cdot 2}\cdot X\;(j\cdot\omega)$\\
1.84 +% \`definition $X\;(j\cdot\omega)$\\
1.85 +%03 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-\infty}^\infty x\;t\;\cdot e^{-j\cdot\omega\cdot t} d t$\\
1.86 +% \` $x\;t = 1\;{\it for}\;\{x.\;-1\leq t\;\land\;t\leq 1\}\;{\it and}\;x\;t=0\;{\it otherwise}$\\
1.87 +%04 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-1}^1 1\cdot e^{-j\cdot\omega\cdot t} d t$\\
1.88 +% \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
1.89 +%05 \> $e^{-j\cdot\omega\cdot 2}\cdot \int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\
1.90 +% %\` $\int e^{a\cdot t} = \frac{1}{a}\cdot e^{a\cdot t}$\\
1.91 +% \` pbl: integration in $\cal C$\\
1.92 +%06 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1)$\\
1.93 +% \` $f\;t\;|_a^b = f\;b-f\;a$\\
1.94 +%07 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1})$\\
1.95 +%\vdots\` pbl: simplification+factorization in $\cal C$\\
1.96 +%08 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{-j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\
1.97 +% \` trick~!\\
1.98 +%09 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\
1.99 +% \` table\\
1.100 +%10 \> $e^{-j\cdot\omega\cdot 2}\cdot 2\cdot\frac{\sin\;\omega}{\omega}$
1.101 +%\end{tabbing}
1.102 +%}
1.103 +%\end{frame}
1.104
1.105 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.106 %% Transform expl 2 REQ %%
1.107 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.108
1.109 -%WN ...
1.110 \begin{frame}\frametitle{Fourier Transform 2: Development effort}
1.111 {\small
1.112 \begin{center}
1.113 @@ -275,7 +281,16 @@
1.114
1.115 \begin{frame}\frametitle{Convolution: Specification}
1.116 {\footnotesize\it
1.117 -Convolution
1.118 +
1.119 +Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response:
1.120 +
1.121 +\begin{center}
1.122 +$h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\
1.123 +$h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$
1.124 +\end{center}
1.125 +
1.126 +The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$.
1.127 +
1.128 \begin{tabbing}
1.129 1\=postcond \=: \= \= $\;\;\;\;$\=\kill
1.130 \>given \>:\> Signals h1[n], h2[n] \\
1.131 @@ -293,9 +308,9 @@
1.132 %% DTS CALC %%
1.133 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.134
1.135 -\begin{frame}\frametitle{Convolution: Calculation}
1.136 -TODO
1.137 -\end{frame}
1.138 +%\begin{frame}\frametitle{Convolution: Calculation}
1.139 +%TODO
1.140 +%\end{frame}
1.141
1.142 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.143 %% DTS REQ %%
1.144 @@ -349,7 +364,13 @@
1.145
1.146 \begin{frame}\frametitle{(Inverse) ${\cal Z}$-Transformation: Specification}
1.147 {\footnotesize\it
1.148 -Convolution
1.149 +
1.150 +Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion.
1.151 +
1.152 +\begin{center}
1.153 +$X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable
1.154 +\end{center}
1.155 +
1.156 \begin{tabbing}
1.157 1\=postcond \=: \= \= $\;\;\;\;$\=\kill
1.158 \>given \>:\> Expression of z \\
1.159 @@ -367,9 +388,9 @@
1.160 %% Z expl CALC %%
1.161 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.162
1.163 -\begin{frame}\frametitle{(Inverse) ${\cal Z}$-Transformation: Calculation}
1.164 -TODO
1.165 -\end{frame}
1.166 +%\begin{frame}\frametitle{(Inverse) ${\cal Z}$-Transformation: Calculation}
1.167 +%TODO
1.168 +%\end{frame}
1.169
1.170 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1.171 %% Z expl REQ %%