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jan@42246: %\def\isac{${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}
jan@42246: %\def\sisac{{\footnotesize${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}}
jan@42246: %
jan@42246: %\begin{document}
jan@42246: %\title{Interactive Course Material for Signal Processing based on Isabelle/\isac}
jan@42246: %\subtitle{Problemsolutions (Calculations)}
jan@42246: %\author{Walther Neuper, Jan Rocnik}
jan@42246: %\maketitle
jrocnik@42162: 
jan@42173: 
jan@42173: %------------------------------------------------------------------------------
jan@42173: %FOURIER
jan@42173: 
jan@42368: \subsection{Fourier Transformation}
jan@42368: \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally fragment of the exam to \emph{Signaltransformationen VO} from 04.03.2011. Translated from German.}}
jrocnik@42162: \textbf{(a)} Determine the fourier transform for the given rectangular impulse:
jrocnik@42162: 
jrocnik@42162: \begin{center}
jrocnik@42162: $x(t)= \left\{
jrocnik@42162:      \begin{array}{lr}
jrocnik@42162:        1 & -1\leq t\geq1\\
jrocnik@42162:        0 & else
jrocnik@42162:      \end{array}
jrocnik@42162:    \right.$
jrocnik@42162: \end{center}
jrocnik@42162: 
jan@42381: \textbf{\noindent (b)} Now consider the given delayed impulse, determine its fourie transformation and calculate phase and magnitude:
jrocnik@42162: 
jrocnik@42162: \begin{center}
jrocnik@42162: $x(t)= \left\{
jrocnik@42162:      \begin{array}{lr}
jan@42163:        1 & -1\leq t\leq1\\
jrocnik@42162:        0 & else
jrocnik@42162:      \end{array}
jrocnik@42162:    \right.$
jrocnik@42162: \end{center}
jrocnik@42162: 
jan@42368: \subsubsection{Solution}
jan@42381: \textbf{(a)} \textsf{Subproblem 1}
jan@42173: \onehalfspace{
jrocnik@42162: \begin{tabbing}
jrocnik@42162: 000\=\kill
jan@42173: \texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
jan@42173: \`Insert Condition: $x(t) = 1\;$ for $\;\{-1\leq t\;\land\;t\leq 1\}\;$ and $\;x(t)=0\;$ otherwise\\
jan@42173: \texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
jrocnik@42162:       \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
jan@42173: \texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\
jrocnik@42162:        \` pbl: integration in $\cal C$\\
jan@42173: \texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1\right)$\\
jrocnik@42162:       \` $f\;t\;|_a^b = f\;b-f\;a$\\
jan@42173: \texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1}$\\
jan@42173: \texttt{\footnotesize{06}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega} -  \frac{1}{-j\cdot\omega}\cdot e^{j\cdot\omega}$\\
jan@42173: \` Lift $\frac{1}{j\omega}$\\
jan@42173: \texttt{\footnotesize{07}} \> $\frac{1}{j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\
jrocnik@42162:       \` trick~!\\
jan@42173: \texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\
jrocnik@42162:       \` table\\
jan@42173: \texttt{\footnotesize{09}} \> $2\cdot\frac{\sin\;\omega}{\omega}$
jrocnik@42162: \end{tabbing}
jan@42173: }
jan@42173: 
jan@42381: \noindent\textbf{(b)} \textsf{Subproblem 1}
jan@42173: \onehalfspace{
jan@42173: \begin{tabbing}
jan@42173: 000\=\kill
jan@42173: \texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
jan@42173: \`Insert Condition: $x(t) = 1\;$ for $\;\{1\leq t\;\land\;t\leq 3\}\;$ and $\;x(t)=0\;$ otherwise\\
jan@42173: \texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
jan@42173:       \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
jan@42173: \texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{1}^3$\\
jan@42173:        \` pbl: integration in $\cal C$\\
jan@42173: \texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{1}^3\right)$\\
jan@42173:       \` $f\;t\;|_a^b = f\;b-f\;a$\\
jan@42173: \texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 3} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1}$\\
jan@42173: \`Lift $\frac{1}{-j\omega}$\\
jan@42173: \texttt{\footnotesize{06}} \> $\frac{1}{j\cdot\omega}\cdot(e^{-j\cdot\omega} - e^{-j\cdot\omega3})$\\
jan@42173:       \`Lift $e^{j\omega2}$ (trick)\\
jan@42173: \texttt{\footnotesize{07}} \> $\frac{1}{j\omega}\cdot e^{j\omega2}\cdot(e^{j\omega} - e^{-j\omega})$\\
jan@42173: \`Simplification (trick)\\
jan@42173: \texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot e^{j\omega2}\cdot(\frac{e^{j\omega} - e^{-j\omega}}{j})$\\
jan@42173:       \` table\\
jan@42246: \texttt{\footnotesize{09}} \> $2\cdot e^{j\omega2}\cdot\frac{\sin\;\omega}{\omega}$\\
jan@42381: \noindent\textbf{(b)} \textsf{Subproblem 2}\\
jan@42173: \`Definition: $X(j\omega)=|X(j\omega)|\cdot e^{arg(X(j\omega))}$\\
jan@42173: \`$|X(j\omega)|$ is called \emph{Magnitude}\\
jan@42173: \`$arg(X(j\omega))$ is called \emph{Phase}\\
jan@42173: \texttt{\footnotesize{10}} \> $|X(j\omega)|=\frac{2}{\omega}\cdot sin(\omega)$\\
jan@42173: \texttt{\footnotesize{11}} \> $arg(X(j\omega)=-2\omega$\\
jan@42173: \end{tabbing}
jan@42173: }
jan@42173: %------------------------------------------------------------------------------
jan@42173: %CONVOLUTION
jrocnik@42162: 
jan@42368: \subsection{Convolution}
jan@42368: \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the SPSC Problem Class 2, Summer term 2008}}
jrocnik@42162: Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response:
jrocnik@42162: 
jrocnik@42162: \begin{center}
jrocnik@42162: $h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\
jrocnik@42162: $h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$
jrocnik@42162: \end{center}
jrocnik@42162: 
jrocnik@42162: The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$.
jan@42368: \subsubsection*{Solution}
jrocnik@42162: 
jan@42173: \doublespace{
jrocnik@42162: \begin{tabbing}
jrocnik@42162: 000\=\kill
jan@42173: \texttt{\footnotesize{01}} \> $h_c[n]=h_1[n]*h_2[n]$\\
jan@42173: \texttt{\footnotesize{02}} \> $h_c[n]=\left(\left(\frac{3}{5}\right)^n\cdot u[n]\right)*\left(\left(-\frac{2}{3}\right)^n\cdot u[n]\right)$\\
jrocnik@42162: \`Definition: $a^n\cdot u[n]\,*\,b^n\cdot u[n]=\sum\limits_{k=-\infty}^{\infty}{a^k\cdot u[k]\cdot b^{n-k}\cdot u[n-k]}$\\
jan@42173: \texttt{\footnotesize{03}} \> $h_c[n]=\sum\limits_{k=-\infty}^{\infty}{\left(\frac{3}{5}\right)^k\cdot u[n]\,\cdot \,\left(-\frac{2}{3}\right)^{n-k}\cdot u[n-k]}$\\
jrocnik@42162: \`$u[n]= \left\{
jrocnik@42162:      \begin{array}{lr}
jrocnik@42162:        1 & for\ n>=0\\
jrocnik@42162:        0 & else
jrocnik@42162:      \end{array}
jrocnik@42162:    \right.$\\
jan@42173: \`We can leave the unitstep through simplification.\\
jan@42173: \`So the lower limit is 0, the upper limit is n.\\
jan@42173: \texttt{\footnotesize{04}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n-k}}$\\
jan@42173: \`Expand\\
jan@42173: \texttt{\footnotesize{05}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n}\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
jan@42173: \`Lift\\
jan@42173: \texttt{\footnotesize{06}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{\infty}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
jan@42173: \texttt{\footnotesize{07}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{3}{2}\right)^{k}}$\\
jan@42173: \texttt{\footnotesize{08}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(-\frac{9}{10}\right)^{k}}$\\
jan@42173: \`Geometric Series: $\sum\limits_{k=0}^{n}{q^k}=\frac{1-q^{n+1}}{1-q}$\\
jan@42173: \`Now we have to consider the limits again.\\
jan@42173: \`It is neccesarry to put the unitstep in again.\\
jan@42173: \texttt{\footnotesize{09}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n+1}}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
jan@42173: \texttt{\footnotesize{10}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
jan@42173: \texttt{\footnotesize{11}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{\left(\frac{19}{10}\right)}\cdot u[n]$\\
jan@42173: \texttt{\footnotesize{12}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot \left(1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)\right)\cdot\left(\frac{10}{19}\right)\cdot u[n]$\\
jan@42173: \`Lift $u[n]$\\
jan@42173: \texttt{\footnotesize{13}} \> $\left(\frac{10}{19}\cdot\left(-\frac{2}{3}\right)^n+\frac{9}{19}\cdot\left(\frac{3}{5}\right)^n\right)\cdot u[n]$\\
jrocnik@42162: \end{tabbing}
jan@42173: }
jan@42173: 
jan@42173: %------------------------------------------------------------------------------
jan@42173: %Z-Transformation
jrocnik@42162: 
jan@42381: \subsection{Z-Transformation\label{sec:calc:ztrans}}
jan@42368: \subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the signal processing problem class 5, summer term 2008.}}
jrocnik@42162: Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion.
jrocnik@42162: 
jrocnik@42162: \begin{center}
jrocnik@42162: $X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable
jrocnik@42162: \end{center}
jrocnik@42162: 
jan@42368: \subsubsection*{Solution}
jan@42173: \onehalfspace{
jan@42173: \begin{tabbing}
jan@42173: 000\=\kill
jan@42173: \textsf{Main Problem}\\
jan@42173: \texttt{\footnotesize{01}} \> $\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}}$ \\
jan@42173: \`Divide through z, neccesary for z-transformation\\
jan@42173: \texttt{\footnotesize{02}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}$ \\
jan@42173: \`Start with partial fraction expansion\\
jan@42173: \texttt{\footnotesize{03}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}=\frac{A}{z-z_1}+\frac{B}{z-z_2}$ \\
jan@42173: \`Eliminate Fractions\\
jan@42173: \texttt{\footnotesize{04}} \> $3=A(z-z_2)+B(z-z_1)$ \\
jan@42173: \textsf{Subproblem 1}\\
jan@42173: \`Setup a linear equation system by inserting the zeros $z_1$ and $z_2$ for $z$\\
jan@42173: \texttt{\footnotesize{05}} \> $3=A(z_1-z_2)$ \& $3=B(z_2-z_1)$\\
jan@42173: \texttt{\footnotesize{06}} \> $\frac{3}{z_1-z_2}=A$ \& $\frac{3}{z_2-z_1}=B$\\
jan@42173: \textsf{Subproblem 2}\\
jan@42173: \`Determine $z_1$ and $z_2$\\
jan@42173: \texttt{\footnotesize{07}} \> $z_1=\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{8}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{8}}$\\
jan@42173: \texttt{\footnotesize{08}} \> $z_1=\frac{1}{8}+\sqrt{\frac{9}{64}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{9}{64}}$\\
jan@42173: \texttt{\footnotesize{09}} \> $z_1=\frac{1}{8}+\frac{3}{8}$ \& $z_2=\frac{1}{8}-\frac{3}{8}$\\
jan@42173: \texttt{\footnotesize{10}} \> $z_1=\frac{1}{2}$ \& $z_2=-\frac{1}{4}$\\
jan@42173: \textsf{Continiue with Subproblem 1}\\
jan@42173: \`Get the coeffizients $A$ and $B$\\
jan@42173: \texttt{\footnotesize{11}} \> $\frac{3}{\frac{1}{2}-(-\frac{1}{4})}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
jan@42173: \texttt{\footnotesize{12}} \> $\frac{3}{\frac{1}{2}+\frac{1}{4}}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
jan@42173: \texttt{\footnotesize{13}} \> $\frac{3}{\frac{3}{4}}=A$ \& $\frac{3}{-\frac{3}{4}}=B$\\
jan@42173: \texttt{\footnotesize{14}} \> $\frac{12}{3}=A$ \& $-\frac{12}{3}=B$\\
jan@42173: \texttt{\footnotesize{15}} \> $4=A$ \& $-4=B$\\
jan@42173: \textsf{Continiue with Main Problem}\\
jan@42173: \texttt{\footnotesize{16}} \> $\frac{A}{z-z_1}+\frac{B}{z-z_2}$\\
jan@42173: \texttt{\footnotesize{17}} \> $\frac{4}{z-\frac{1}{2}}+\frac{4}{z-\left(-\frac{1}{4}\right)}$ \\
jan@42173: \texttt{\footnotesize{18}} \> $\frac{4}{z-\frac{1}{2}}-\frac{4}{z+\frac{1}{4}}$ \\
jan@42173: \`Multiply with z, neccesary for z-transformation\\
jan@42173: \texttt{\footnotesize{19}} \> $\frac{4z}{z-\frac{1}{2}}-\frac{4z}{z+\frac{1}{4}}$ \\
jan@42173: \texttt{\footnotesize{20}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}$ \\
jan@42173: \`Transformation\\
jan@42173: \texttt{\footnotesize{21}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}\ \Ztransf\ 4\cdot\left(-\frac{1}{2}\right)^n\cdot u[n]+(-4)\cdot\left(\frac{1}{4}\right)^n\cdot u[n]$\\
jan@42173: \end{tabbing}
jan@42173: }
jrocnik@42162: \theendnotes
jan@42246: %\end{document}



