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\begin{document}
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%===  This is the header section ============================================================
\thispagestyle{empty}
\noindent
\begin{minipage}[b][4cm]{1.0\textwidth}
\begin{center}
\begin{bf}
\begin{large} Digitale Signalverarbeitung SS 2025/26 -- 2.~Aufgabe\end{large} \\
\vspace{0.3cm}
\begin{Large} Discrete Time Signals, Convolution, DTFT \end{Large} \\
\vspace{0.3cm}
\end{bf}
\begin{large}
Gruppennummer 211 \\
Manuel Illmayer,  k12308149 \\
Quirin Ecker, k12310122 \\
\end{large}
\end{center}
\end{minipage}

\noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em]
%=======================================================================================

\begin{aufgabe}{}

	\begin{enumerate}
		\item figure: \\
			\includegraphics[width=0.95\textwidth]{fig/aufgabe1_a.png}

		\item
			\begin{gather*}
				\Omega_3 = 2 \pi \frac{3.5}{64} \\
				\Omega_4 = \frac{9}{64}
			\end{gather*}
		\item
			\begin{gather*}
				\frac{\Omega_3}{2 \pi} = \frac{3.5}{64} = \frac{7}{128} \\
				N_{0,3} = 128 \\
				\frac{\Omega_4}{2 \pi} = \frac{9}{64 \cdot 2 \pi} = \frac{9}{128 \pi} \\
				\implies \text{Irational because of $\pi$} \\
				\implies \text{Not periodic} \\
			\end{gather*}
		\item (see assignment2\_1.m)
		\item (see assignment2\_1.m)
		\item table:
			\begin{table}[h]
				\centering
				\begin{tabular}{|c|c|c|}
				\hline
				Signal & Energy & Power \\
				\hline
				$x_1[n]$ & 37  & --     \\
				$x_2[n]$ & 48.0394  & --     \\
				$x_3[n]$ & 1152  & 4.4825 \\
				$x_4[n]$ & 128.9565  & --     \\
				\hline
				\end{tabular}
				\caption{Energy and Power Values}
				\label{tab:energy_power}
			\end{table}
	\end{enumerate}
\end{aufgabe}

\begin{aufgabe}{}

	\begin{enumerate}
		\item $|x[n]| + |h[n]| - 1 = 5 + 4 - 1 = 8$
		\item
			\begin{gather*}
A=
\begin{bmatrix}
3 & 0 & 0 & 0\\
-1 & 3 & 0 & 0\\
2 & -1 & 3 & 0\\
0 & 2 & -1 & 3\\
1 & 0 & 2 & -1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
\qquad
x=
\begin{bmatrix}
2\\
3\\
4\\
1
\end{bmatrix} \\
Ax=
\begin{bmatrix}
6\\
7\\
13\\
5\\
9\\
5\\
4\\
1
\end{bmatrix}
			\end{gather*}
	\end{enumerate}
\end{aufgabe}

\begin{aufgabe}{}

	\begin{enumerate}
		\item $L_y = |x[n]| + |h[n]| - 1 = 50 + 3 - 1 = 52$
		\item (see assignment2\_3.m)
		\item see d)
		\item figure: \\
			\includegraphics[width=0.95\textwidth]{fig/aufgabe3_cd.png}
	\end{enumerate}
\end{aufgabe}

\begin{aufgabe}{}

	\begin{enumerate}
		\item (see assignment2\_4.m)
		\item Phase Response is very small in the realm of $10^{-15}$. The signal contians multiple bigger spikes.  \\
			\includegraphics[width=0.95\textwidth]{fig/aufgabe4_b.png}
		\item The magnitued seems to be repeating and the phase response seems to be mirrored accross the x and y axis.
	\end{enumerate}
\end{aufgabe}


\begin{aufgabe}{}

	\begin{enumerate}
		\item (siehe assignment2\_5.m)
		\item (siehe assignment2\_5.m)
		\item figure: \\
			\includegraphics[width=0.95\textwidth]{fig/aufgabe5_c.png}
		\item original energy: 41343 \\ reverb energy: 2069.0506 \\ rir energy: $9.9997 \cdot 10^{3}$ \\
			explanation: The rir energy is less than one which would explain why the outcoming signal is smaller than the original, because convulution scales proportionally.
	\end{enumerate}
\end{aufgabe}

\end{document}