finished exercise except for some parts of number 1
This commit is contained in:
@@ -105,7 +105,7 @@
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\begin{minipage}[b][4cm]{1.0\textwidth}
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\begin{center}
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\begin{bf}
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\begin{large} Digitale Signalverarbeitung WS 2025/26 -- 1.~Aufgabe\end{large} \\
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\begin{large} Digitale Signalverarbeitung SS 2025/26 -- 2.~Aufgabe\end{large} \\
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\vspace{0.3cm}
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\begin{Large} Analog Signals and Systems \end{Large} \\
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\vspace{0.3cm}
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@@ -121,231 +121,115 @@ Quirin Ecker, k12310122 \\
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\noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em]
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%=======================================================================================
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\begin{aufgabe}
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\begin{aufgabe}{}
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\begin{align*}
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c_1 &= -5 + 3j
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= \sqrt{(-5)^2 + 3^2}\, e^{j \arctan\left(\frac{3}{-5}\right)}
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= \sqrt{34}\, e^{-j\,0.54} \\
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\end{align*}
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\begin{align*}
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c_2 &= \frac{\sqrt{2}}{2} e^{-j \frac{3\pi}{4}}
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= \frac{\sqrt{2}}{2} \cos\left(-\frac{3\pi}{4}\right)
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+ j \frac{\sqrt{2}}{2} \sin\left(-\frac{3\pi}{4}\right)
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= -\frac{1}{2} - \frac{1}{2}j \\
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\end{align*}
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\begin{align*}
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c_3 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}j
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= \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}
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\, e^{j \arctan\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)}
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= e^{j \frac{\pi}{4}} \\
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\end{align*}
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\begin{align*}
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c_4 &= c_1 + c_2
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= (-5 + 3j) + \left(-\frac{1}{2} - \frac{1}{2}j\right)
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= -\frac{11}{2} + \frac{5}{2}j \\
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\end{align*}
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\begin{align*}
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c_5 &= c_1 \cdot c_2
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= \left(\sqrt{34}\, e^{-j\,0.54}\right)\left(\frac{\sqrt{2}}{2} e^{-j \frac{3 \pi}{4}}\right)
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\approx \sqrt{17} e^{-j\,2.8961} \\
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\end{align*}
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\begin{align*}
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c_6 &= |c_3|^2
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= \left(\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\right)^2
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= 1 \\
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\end{align*}
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\begin{align*}
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c_7 &= \arg(c_3) = \frac{\pi}{4} \\
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\end{align*}
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\begin{align*}
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c_8 &= \frac{c_1}{c_2}
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= \frac{-5 + 3j}{-\frac{1}{2} - \frac{1}{2}j}
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\cdot \frac{-\frac{1}{2} + \frac{1}{2}j}{-\frac{1}{2} + \frac{1}{2}j} \\
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&= \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} \\
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&= 2 - 8j \\
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\end{align*}
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\begin{align*}
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c_9 &= c_1 \cdot c_1^*
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= \sqrt{34}\, e^{-j\,0.54} \cdot \sqrt{34}\, e^{j\,0.54}
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= 34
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\end{align*}
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\end{aufgabe}
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\begin{aufgabe}{}
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\[
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x(t) = \hat{X} \cos(2\pi f_0 t) \xleftrightarrow{} \iff X(f) = \frac{\hat{X}}{2} \delta(f - f_0) + \frac{\hat{X}}{2} \delta(f + f_0) \\
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\]
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\[
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x(t) = \hat{X}( \frac{1}{2} \cdot e^{j2\pi f_0 t} + \frac{1}{2} \cdot e^{-j2\pi f_0 t} )
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\]
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\[
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X(f) = \hat{X}( \frac{1}{2} \cdot \delta (f-f_0) + \frac{1}{2} \cdot \delta (f+f_0) )
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\]
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\[
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X(f) = \frac{\hat{X}}{2} \cdot \delta (f-f_0) + \frac{\hat{X}}{2} \cdot \delta (f+f_0)
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\]
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\begin{center}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe2.png}
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\end{center}
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\end{aufgabe}
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\begin{aufgabe}{}
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\begin{enumerate}
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\item
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\[
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\sin\left(2\pi f_i (t - 0.1)\right) = \sin\left(2\pi f_i t + \phi_i\right)
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\]
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\[
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2\pi f_i (t - 0.1) = 2\pi f_i t + \phi_i
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\]
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\[
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2\pi f_i (t - 0.1) - 2\pi f_i t = \phi_i
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\]
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\[
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2\pi f_i \big((t - 0.1) - t\big) = \phi_i
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\]
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\[
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\phi_i = 2\pi f_i (-0.1) = -0.1 \cdot 2\pi f_i
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\]
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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.
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\item Figures:
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\begin{center}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe3_1.png}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe3_2.png}
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\end{center}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}{}
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\begin{itemize}
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\item $y(t) = (x(t))^2$
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\[
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y(t) = (x(t))^2
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\]
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\[
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y_1(t) = (\alpha x_1(t))^2 + (\beta x_2(t))^2
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\]
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\[
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= \alpha^2 (x_1(t))^2 + \beta^2 (x_2(t))^2
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\]
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\[
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y_2(t) = \alpha (x_1(t))^2 + \beta (x_2(t))^2
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\]
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\[
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y_1(t) \ne y_2(t)
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\]
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\[
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\implies \text{Not linear}
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\]
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\[
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y_1(t) &= \left(x(t - \Delta)\right)^2 \\
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\]
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\[
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y_2(t) &= y_1(t - \Delta) = \left(x(t - \Delta)\right)^2 \\
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\]
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\[
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y_1 &= y_2 \\
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\]
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\[
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\implies \text{Time invariant}
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\]
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\item $y(t) = x(t) sin(\Omega_0 t)$
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\[
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y(t) &= x(t) \sin(\Omega_0 t) \\
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\]
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\[
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y_1(t) &= \left( \alpha \, x(t) \sin(\Omega_0 t) \right) + \left( \beta \, x(t) \sin(\Omega_0 t) \right) \\
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\]
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\[
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y_t(t) &= \alpha \left( x(t) \sin(\Omega_0 t) \right) + \beta \left( x(t) \sin(\Omega_0 t) \right) \\
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\]
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\[
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y_1 &= y_2
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\]
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\[
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\implies \text{linear}
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\]
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\[
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y_1(t) = &\coloneqq x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
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\]
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\[
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y_2(t) &= y_1(t - \Delta) = x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
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\]
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\[
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y_1 &= y_2 \\
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\]
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\[
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\implies \text{Time invariant}
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\]
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\end{itemize}
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\begin{enumerate}
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\item add back in
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\item add back in
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\item add back in
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\item add back in
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\item (see assignment2\_1.m)
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\item table:
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\begin{table}[h]
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\centering
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\begin{tabular}{|c|c|c|}
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\hline
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Signal & Energy & Power \\
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\hline
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$x_1[n]$ & 37 & -- \\
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$x_2[n]$ & 48.0394 & -- \\
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$x_3[n]$ & 1152 & 4.4825 \\
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$x_4[n]$ & 128.9565 & -- \\
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\hline
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\end{tabular}
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\caption{Energy and Power Values}
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\label{tab:energy_power}
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\end{table}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}
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\item System A is linear because there is no difference between the plot $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
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You can also see in the error plot that the error is small.
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It is not exactyly zero becuase rounding errors.
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\item Input and output are not proportional in a non linear system and
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therefore the error change is greater than the one of the linear system
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\item The output sound of System B does differ when comparing $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
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But is the same when using System A. This matches the differences in the graphs.
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\item $|x[n]| + |h[n]| - 1 = 5 + 4 - 1 = 8$
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\item
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\begin{gather*}
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A=
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\begin{bmatrix}
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3 & 0 & 0 & 0\\
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-1 & 3 & 0 & 0\\
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2 & -1 & 3 & 0\\
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0 & 2 & -1 & 3\\
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1 & 0 & 2 & -1\\
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0 & 1 & 0 & 2\\
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0 & 0 & 1 & 0\\
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0 & 0 & 0 & 1
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\end{bmatrix},
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\qquad
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x=
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\begin{bmatrix}
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2\\
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3\\
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4\\
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1
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\end{bmatrix}
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Ax=
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\begin{bmatrix}
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6\\
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7\\
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13\\
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5\\
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9\\
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5\\
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4\\
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1
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\end{bmatrix}
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\end{gather*}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}
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\item $L_y = |x[n]| + |h[n]| - 1 = 50 + 3 - 1 = 52$
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\item (see assignment2\_3.m)
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\item see d)
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\item figure: \\
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\includegraphics[width=0.95\textwidth]{fig/aufgabe3_cd.png}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}
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\item (see assignment2\_4.m)
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\item Phase Response is very small in the realm of $10^{-15}$. The signal contians multiple bigger spikes. \\
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\includegraphics[width=0.95\textwidth]{fig/aufgabe4_b.png}
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\item The magnitued seems to be repeating and the phase response seems to be mirrored accross the x and y axis.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}
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\item (siehe assignment2\_5.m)
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\item (siehe assignment2\_5.m)
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\item figure: \\
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\includegraphics[width=0.95\textwidth]{fig/aufgabe5_c.png}
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\item original energy: 41343 \\ reverb energy: 2069.0506 \\ rir energy: $9.9997 \cdot 10^{3}$ \\
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explanation: The rir energy is less than one which would explain why the outcoming signal is smaller than the original, because convulution scales proportionally.
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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