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 \begin{document}

 \title{SPSC Specific Problems for ISAC}

 \subtitle{Solutions}

 \author{Walther Neuper, Jan Rocnik}

 \maketitle

 \section*{Fourier Transformation}

 \subsection*{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\geq1\\

        0 & else

      \end{array}

    \right.$

 \end{center}

 \subsection*{Solution}

 \begin{tabbing}

 000\=\kill

 01 \> ${\cal F}\;(x(t-2)) =$\\

       \`${\cal F}\;(x(t-T)) = e^{-j\cdot\omega\cdot T}\cdot X\;j\cdot\omega$\\

 02 \> $e^{-j\cdot\omega\cdot 2}\cdot X\;(j\cdot\omega)$\\

       \`definition $X\;(j\cdot\omega)$\\

 03 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-\infty}^\infty x\;t\;\cdot e^{-j\cdot\omega\cdot t} d t$\\

       \` $x\;t = 1\;{\it for}\;\{x.\;-1\leq t\;\land\;t\leq 1\}\;{\it and}\;x\;t=0\;{\it otherwise}$\\

 04 \> $e^{-j\cdot\omega\cdot 2}\cdot \int_{-1}^1 1\cdot e^{-j\cdot\omega\cdot t} d t$\\

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

 05 \> $e^{-j\cdot\omega\cdot 2}\cdot \int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\

       %\` $\int e^{a\cdot t} = \frac{1}{a}\cdot e^{a\cdot t}$\\

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

 06 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1)$\\

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

 07 \> $e^{-j\cdot\omega\cdot 2}\cdot (\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1})$\\

 \vdots\` pbl: simplification+factorization in $\cal C$\\

 08 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{-j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\

       \` trick~!\\

 09 \> $e^{-j\cdot\omega\cdot 2}\cdot \frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\

       \` table\\

 10 \> $e^{-j\cdot\omega\cdot 2}\cdot 2\cdot\frac{\sin\;\omega}{\omega}$

 \end{tabbing}

 \section*{Convolution}

 \subsection*{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 $h_c[n]$.

 \subsection*{Solution}

 \begin{tabbing}

 000\=\kill

 01 \> $h_c[n]=h_1[n]*h_2[n]$\\

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

 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.$\\

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

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

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

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

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

 \`Solving the sum... now u[n] again\\

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

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

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

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

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

 \end{tabbing}

 \section*{$\cal Z$-Transformation}

 \subsection*{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}

 \theendnotes

 \end{document}