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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 WS 2025/26 -- 1.~Aufgabe\end{large} \\
\vspace{0.3cm}
\begin{Large} Analog Signals and Systems \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{align*}
c_1 &= -5 + 3j
= \sqrt{(-5)^2 + 3^2}\, e^{j \arctan\left(\frac{3}{-5}\right)}
= \sqrt{34}\, e^{-j\,0.54} \\
\end{align*}

\begin{align*}
c_2 &= \frac{\sqrt{2}}{2} e^{-j \frac{3\pi}{4}}
= \frac{\sqrt{2}}{2} \cos\left(-\frac{3\pi}{4}\right)
+ j \frac{\sqrt{2}}{2} \sin\left(-\frac{3\pi}{4}\right)
= -\frac{1}{2} - \frac{1}{2}j \\
\end{align*}

\begin{align*}
c_3 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}j
= \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}
\, e^{j \arctan\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)}
= e^{j \frac{\pi}{4}} \\
\end{align*}

\begin{align*}
c_4 &= c_1 + c_2
= (-5 + 3j) + \left(-\frac{1}{2} - \frac{1}{2}j\right)
= -\frac{11}{2} + \frac{5}{2}j \\
\end{align*}

\begin{align*}
c_5 &= c_1 \cdot c_2
= \left(\sqrt{34}\, e^{-j\,0.54}\right)\left(\frac{\sqrt{2}}{2} e^{-j \frac{3 \pi}{4}}\right)
\approx \sqrt{17} e^{-j\,2.8961} \\
\end{align*}

\begin{align*}
c_6 &= |c_3|^2
= \left(\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\right)^2
= 1 \\
\end{align*}

\begin{align*}
c_7 &= \arg(c_3) = \frac{\pi}{4} \\
\end{align*}

\begin{align*}
c_8 &= \frac{c_1}{c_2}
= \frac{-5 + 3j}{-\frac{1}{2} - \frac{1}{2}j}
\cdot \frac{-\frac{1}{2} + \frac{1}{2}j}{-\frac{1}{2} + \frac{1}{2}j} \\
&= \frac{(-5)(-\frac{1}{2}) + 3(-\frac{1}{2}) + j\left(3(-\frac{1}{2}) - (-5)(-\frac{1}{2})\right)}{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \\
&= 2 - 8j \\
\end{align*}

\begin{align*}
c_9 &= c_1 \cdot c_1^*
= \sqrt{34}\, e^{-j\,0.54} \cdot \sqrt{34}\, e^{j\,0.54}
= 34
\end{align*}
\end{aufgabe}

\begin{aufgabe}{}


\[
    x(t) = \hat{X} \cos(2\pi f_0 t) \xleftrightarrow{} \iff X(f) = \frac{\hat{X}}{2} \delta(f - f_0) + \frac{\hat{X}}{2} \delta(f + f_0) \\
\]

\[
	x(t) = \hat{X}( \frac{1}{2} \cdot e^{j2\pi f_0 t} + \frac{1}{2} \cdot e^{-j2\pi f_0 t} )
\]

\[
	X(f) = \hat{X}( \frac{1}{2} \cdot \delta (f-f_0) + \frac{1}{2} \cdot \delta (f+f_0) )
\]

\[
	X(f) = \frac{\hat{X}}{2} \cdot \delta (f-f_0) + \frac{\hat{X}}{2} \cdot \delta (f+f_0)
\]

\begin{center}
	\includegraphics[width=0.95\textwidth]{fig/aufgabe2.png}
\end{center}



\end{aufgabe}

\begin{aufgabe}{}

\begin{enumerate}
	\item
\[
\sin\left(2\pi f_i (t - 0.1)\right) = \sin\left(2\pi f_i t + \phi_i\right)
\]

\[
2\pi f_i (t - 0.1) = 2\pi f_i t + \phi_i
\]

\[
2\pi f_i (t - 0.1) - 2\pi f_i t = \phi_i
\]

\[
2\pi f_i \big((t - 0.1) - t\big) = \phi_i
\]

\[
\phi_i = 2\pi f_i (-0.1) = -0.1 \cdot 2\pi f_i
\]

The fourier transform shift theorem states $x(t - T) \;\longleftrightarrow\; X(f)\, e^{-j 2\pi f T}$, where the angle is $2\pi fT$ and $T = 0.1$ in this case. Combining everything, we get $-0.1 \cdot 2\pi f_i$ which is equivalent to the result we got.

\item Figures:
		\begin{center}
			\includegraphics[width=0.95\textwidth]{fig/aufgabe3_1.png}
			\includegraphics[width=0.95\textwidth]{fig/aufgabe3_2.png}
		\end{center}
\end{enumerate}

\end{aufgabe}


\begin{aufgabe}{}

\begin{itemize}
	\item $y(t) = (x(t))^2$

\[
y(t) = (x(t))^2
\]

\[
y_1(t) = (\alpha x_1(t))^2 + (\beta x_2(t))^2
\]

\[
= \alpha^2 (x_1(t))^2 + \beta^2 (x_2(t))^2
\]

\[
y_2(t) = \alpha (x_1(t))^2 + \beta (x_2(t))^2
\]

\[
y_1(t) \ne y_2(t)
\]

\[
\implies \text{Not linear}
\]


\[
    y_1(t) &= \left(x(t - \Delta)\right)^2 \\
\]

\[
    y_2(t) &= y_1(t - \Delta) = \left(x(t - \Delta)\right)^2 \\
\]

\[
    y_1 &= y_2 \\
\]

\[
\implies \text{Time invariant}
\]
	\item $y(t) = x(t) sin(\Omega_0 t)$

\[
	y(t) &= x(t) \sin(\Omega_0 t) \\
\]

\[
	y_1(t) &= \left( \alpha \, x(t) \sin(\Omega_0 t) \right) + \left( \beta \, x(t) \sin(\Omega_0 t) \right) \\
\]

\[
	y_t(t) &= \alpha \left( x(t) \sin(\Omega_0 t) \right) + \beta \left( x(t) \sin(\Omega_0 t) \right) \\
\]

\[
	y_1 &= y_2
\]

\[
	\implies \text{linear}
\]




\[
    y_1(t) = &\coloneqq x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
\]

\[
    y_2(t) &= y_1(t - \Delta) = x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
\]

\[
    y_1 &= y_2 \\
\]

\[
	\implies \text{Time invariant}
\]
\end{itemize}

\end{aufgabe}

\begin{aufgabe}

	\begin{enumerate}
		\item System A is linear because there is no difference between the plot $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
			You can also see in the error plot that the error is small.
			It is not exactyly zero becuase rounding errors.
		\item Input and output are not proportional in a non linear system and
			therefore the error change is greater than the one of the linear system
		\item The output sound of System B does differ when comparing $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
			But is the same when using System A. This matches the differences in the graphs.
	\end{enumerate}

\end{aufgabe}

\end{document}
