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%\def\isac{${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}

%\def\sisac{{\footnotesize${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}}

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%\begin{document}

%\title{Interactive Course Material for Signal Processing based on Isabelle/\isac}

%\subtitle{Problemsolutions (Calculations)}

%\author{Walther Neuper, Jan Rocnik}

%\maketitle

%------------------------------------------------------------------------------

%FOURIER

\subsection{Fourier Transformation}

\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally fragment of the exam to \emph{Signaltransformationen VO} from 04.03.2011. Translated from German.}}

\textbf{(a)} Determine the fourier transform for the given rectangular impulse:

\begin{center}

$x(t)= \left\{

     \begin{array}{lr}

       1 & -1\leq t\geq1\\

       0 & else

     \end{array}

   \right.$

\end{center}

\textbf{(b)} Now consider the given delayed impulse, determine its fourie transformation and calculate phase and magnitude:

\begin{center}

$x(t)= \left\{

     \begin{array}{lr}

       1 & -1\leq t\leq1\\

       0 & else

     \end{array}

   \right.$

\end{center}

\subsubsection{Solution}

\textbf{(a)}

\onehalfspace{

\begin{tabbing}

000\=\kill

\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\

\`Insert Condition: $x(t) = 1\;$ for $\;\{-1\leq t\;\land\;t\leq 1\}\;$ and $\;x(t)=0\;$ otherwise\\

\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\

      \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\

\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\

       \` pbl: integration in $\cal C$\\

\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1\right)$\\

      \` $f\;t\;|_a^b = f\;b-f\;a$\\

\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1}$\\

\texttt{\footnotesize{06}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega} -  \frac{1}{-j\cdot\omega}\cdot e^{j\cdot\omega}$\\

\` Lift $\frac{1}{j\omega}$\\

\texttt{\footnotesize{07}} \> $\frac{1}{j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\

      \` trick~!\\

\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\

      \` table\\

\texttt{\footnotesize{09}} \> $2\cdot\frac{\sin\;\omega}{\omega}$

\end{tabbing}

}

\textbf{(b)}

\onehalfspace{

\begin{tabbing}

000\=\kill

\textsf{Subproblem 1}\\

\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\

\`Insert Condition: $x(t) = 1\;$ for $\;\{1\leq t\;\land\;t\leq 3\}\;$ and $\;x(t)=0\;$ otherwise\\

\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\

      \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\

\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{1}^3$\\

       \` pbl: integration in $\cal C$\\

\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{1}^3\right)$\\

      \` $f\;t\;|_a^b = f\;b-f\;a$\\

\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 3} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1}$\\

\`Lift $\frac{1}{-j\omega}$\\

\texttt{\footnotesize{06}} \> $\frac{1}{j\cdot\omega}\cdot(e^{-j\cdot\omega} - e^{-j\cdot\omega3})$\\

      \`Lift $e^{j\omega2}$ (trick)\\

\texttt{\footnotesize{07}} \> $\frac{1}{j\omega}\cdot e^{j\omega2}\cdot(e^{j\omega} - e^{-j\omega})$\\

\`Simplification (trick)\\

\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot e^{j\omega2}\cdot(\frac{e^{j\omega} - e^{-j\omega}}{j})$\\

      \` table\\

\texttt{\footnotesize{09}} \> $2\cdot e^{j\omega2}\cdot\frac{\sin\;\omega}{\omega}$\\

\textsf{Subproblem 2}\\

\`Definition: $X(j\omega)=|X(j\omega)|\cdot e^{arg(X(j\omega))}$\\

\`$|X(j\omega)|$ is called \emph{Magnitude}\\

\`$arg(X(j\omega))$ is called \emph{Phase}\\

\texttt{\footnotesize{10}} \> $|X(j\omega)|=\frac{2}{\omega}\cdot sin(\omega)$\\

\texttt{\footnotesize{11}} \> $arg(X(j\omega)=-2\omega$\\

\end{tabbing}

}

%------------------------------------------------------------------------------

%CONVOLUTION

\subsection{Convolution}

\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the SPSC Problem Class 2, Summer term 2008}}

Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response:

\begin{center}

$h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\

$h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$

\end{center}

The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$.

\subsubsection*{Solution}

\doublespace{

\begin{tabbing}

000\=\kill

\texttt{\footnotesize{01}} \> $h_c[n]=h_1[n]*h_2[n]$\\

\texttt{\footnotesize{02}} \> $h_c[n]=\left(\left(\frac{3}{5}\right)^n\cdot u[n]\right)*\left(\left(-\frac{2}{3}\right)^n\cdot u[n]\right)$\\

\`Definition: $a^n\cdot u[n]\,*\,b^n\cdot u[n]=\sum\limits_{k=-\infty}^{\infty}{a^k\cdot u[k]\cdot b^{n-k}\cdot u[n-k]}$\\

\texttt{\footnotesize{03}} \> $h_c[n]=\sum\limits_{k=-\infty}^{\infty}{\left(\frac{3}{5}\right)^k\cdot u[n]\,\cdot \,\left(-\frac{2}{3}\right)^{n-k}\cdot u[n-k]}$\\

\`$u[n]= \left\{

     \begin{array}{lr}

       1 & for\ n>=0\\

       0 & else

     \end{array}

   \right.$\\

\`We can leave the unitstep through simplification.\\

\`So the lower limit is 0, the upper limit is n.\\

\texttt{\footnotesize{04}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n-k}}$\\

\`Expand\\

\texttt{\footnotesize{05}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n}\cdot \left(-\frac{2}{3}\right)^{-k}}$\\

\`Lift\\

\texttt{\footnotesize{06}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{\infty}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{-k}}$\\

\texttt{\footnotesize{07}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{3}{2}\right)^{k}}$\\

\texttt{\footnotesize{08}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(-\frac{9}{10}\right)^{k}}$\\

\`Geometric Series: $\sum\limits_{k=0}^{n}{q^k}=\frac{1-q^{n+1}}{1-q}$\\

\`Now we have to consider the limits again.\\

\`It is neccesarry to put the unitstep in again.\\

\texttt{\footnotesize{09}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n+1}}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\

\texttt{\footnotesize{10}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\

\texttt{\footnotesize{11}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{\left(\frac{19}{10}\right)}\cdot u[n]$\\

\texttt{\footnotesize{12}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot \left(1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)\right)\cdot\left(\frac{10}{19}\right)\cdot u[n]$\\

\`Lift $u[n]$\\

\texttt{\footnotesize{13}} \> $\left(\frac{10}{19}\cdot\left(-\frac{2}{3}\right)^n+\frac{9}{19}\cdot\left(\frac{3}{5}\right)^n\right)\cdot u[n]$\\

\end{tabbing}

}

%------------------------------------------------------------------------------

%Z-Transformation

\subsection{$\cal Z$-Transformation\label{sec:calc:ztrans}}

\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the signal processing problem class 5, summer term 2008.}}

Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion.

\begin{center}

$X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable

\end{center}

\subsubsection*{Solution}

\onehalfspace{

\begin{tabbing}

000\=\kill

\textsf{Main Problem}\\

\texttt{\footnotesize{01}} \> $\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}}$ \\

\`Divide through z, neccesary for z-transformation\\

\texttt{\footnotesize{02}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}$ \\

\`Start with partial fraction expansion\\

\texttt{\footnotesize{03}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}=\frac{A}{z-z_1}+\frac{B}{z-z_2}$ \\

\`Eliminate Fractions\\

\texttt{\footnotesize{04}} \> $3=A(z-z_2)+B(z-z_1)$ \\

\textsf{Subproblem 1}\\

\`Setup a linear equation system by inserting the zeros $z_1$ and $z_2$ for $z$\\

\texttt{\footnotesize{05}} \> $3=A(z_1-z_2)$ \& $3=B(z_2-z_1)$\\

\texttt{\footnotesize{06}} \> $\frac{3}{z_1-z_2}=A$ \& $\frac{3}{z_2-z_1}=B$\\

\textsf{Subproblem 2}\\

\`Determine $z_1$ and $z_2$\\

\texttt{\footnotesize{07}} \> $z_1=\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{8}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{8}}$\\

\texttt{\footnotesize{08}} \> $z_1=\frac{1}{8}+\sqrt{\frac{9}{64}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{9}{64}}$\\

\texttt{\footnotesize{09}} \> $z_1=\frac{1}{8}+\frac{3}{8}$ \& $z_2=\frac{1}{8}-\frac{3}{8}$\\

\texttt{\footnotesize{10}} \> $z_1=\frac{1}{2}$ \& $z_2=-\frac{1}{4}$\\

\textsf{Continiue with Subproblem 1}\\

\`Get the coeffizients $A$ and $B$\\

\texttt{\footnotesize{11}} \> $\frac{3}{\frac{1}{2}-(-\frac{1}{4})}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\

\texttt{\footnotesize{12}} \> $\frac{3}{\frac{1}{2}+\frac{1}{4}}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\

\texttt{\footnotesize{13}} \> $\frac{3}{\frac{3}{4}}=A$ \& $\frac{3}{-\frac{3}{4}}=B$\\

\texttt{\footnotesize{14}} \> $\frac{12}{3}=A$ \& $-\frac{12}{3}=B$\\

\texttt{\footnotesize{15}} \> $4=A$ \& $-4=B$\\

\textsf{Continiue with Main Problem}\\

\texttt{\footnotesize{16}} \> $\frac{A}{z-z_1}+\frac{B}{z-z_2}$\\

\texttt{\footnotesize{17}} \> $\frac{4}{z-\frac{1}{2}}+\frac{4}{z-\left(-\frac{1}{4}\right)}$ \\

\texttt{\footnotesize{18}} \> $\frac{4}{z-\frac{1}{2}}-\frac{4}{z+\frac{1}{4}}$ \\

\`Multiply with z, neccesary for z-transformation\\

\texttt{\footnotesize{19}} \> $\frac{4z}{z-\frac{1}{2}}-\frac{4z}{z+\frac{1}{4}}$ \\

\texttt{\footnotesize{20}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}$ \\

\`Transformation\\

\texttt{\footnotesize{21}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}\ \Ztransf\ 4\cdot\left(-\frac{1}{2}\right)^n\cdot u[n]+(-4)\cdot\left(\frac{1}{4}\right)^n\cdot u[n]$\\

\end{tabbing}

}

\theendnotes

%\end{document}