finished exercise except for some parts of number 1

.direnv/flake-profile

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+flake-profile-1-link

.direnv/flake-profile-1-link

@@ -0,0 +1 @@
+/nix/store/zcsc63f1ncp6p1nf533q6m35imm42z78-latex-env

Assignment1_211.tex

@@ -105,7 +105,7 @@
 \begin{minipage}[b][4cm]{1.0\textwidth}
 \begin{center}
 \begin{bf}
-\begin{large} Digitale Signalverarbeitung WS 2025/26 -- 1.~Aufgabe\end{large} \\
+\begin{large} Digitale Signalverarbeitung SS 2025/26 -- 2.~Aufgabe\end{large} \\
 \vspace{0.3cm}
 \begin{Large} Analog Signals and Systems \end{Large} \\
 \vspace{0.3cm}
@@ -121,231 +121,115 @@ Quirin Ecker, k12310122 \\
 \noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em]
 %=======================================================================================
 
-
+\begin{aufgabe}
-
-\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
+		\item add back in
-\[
+		\item add back in
-\sin\left(2\pi f_i (t - 0.1)\right) = \sin\left(2\pi f_i t + \phi_i\right)
+		\item add back in
-\]
+		\item add back in
-
+		\item (see assignment2\_1.m)
-\[
+		\item table:
-2\pi f_i (t - 0.1) = 2\pi f_i t + \phi_i
+			\begin{table}[h]
-\]
+				\centering
-
+				\begin{tabular}{|c|c|c|}
-\[
+				\hline
-2\pi f_i (t - 0.1) - 2\pi f_i t = \phi_i
+				Signal & Energy & Power \\
-\]
+				\hline
-
+				$x_1[n]$ & 37  & --     \\
-\[
+				$x_2[n]$ & 48.0394  & --     \\
-2\pi f_i \big((t - 0.1) - t\big) = \phi_i
+				$x_3[n]$ & 1152  & 4.4825 \\
-\]
+				$x_4[n]$ & 128.9565  & --     \\
-
+				\hline
-\[
+				\end{tabular}
-\phi_i = 2\pi f_i (-0.1) = -0.1 \cdot 2\pi f_i
+				\caption{Energy and Power Values}
-\]
+				\label{tab:energy_power}
-
+			\end{table}
-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 $T [x_1(t)] + T [x_2(t)]$.
+		\item $|x[n]| + |h[n]| - 1 = 5 + 4 - 1 = 8$
-			You can also see in the error plot that the error is small.
+		\item
-			It is not exactyly zero becuase rounding errors.
+			\begin{gather*}
-		\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
+A=
-		\item The output sound of System B does differ when comparing $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
+\begin{bmatrix}
-			But is the same when using System A. This matches the differences in the graphs.
+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}

Assignment2_211/assignment2_1.m

@@ -0,0 +1,62 @@
+clear
+close all
+clc
+
+% x1[n], -5 <= n <= 10
+n1 = -5:10;
+x1 = -4*(n1 == -3) + 4*(n1 == 0) - (n1 == 3) + 2*(n1 == 7);
+
+% x2[n], -5 <= n <= 10
+n2 = -5:10;
+x2 = exp(-0.31*n2);
+
+% x3[n], 0 <= n <= 256
+n3 = 0:256;
+x3 = 3*sin(2*pi*(3.5/64)*n3);
+
+% x4[n], 0 <= n <= 256
+n4 = 0:256;
+x4 = -cos((9/64)*n4);
+
+
+% Visualization
+figure;
+subplot(121); hold on; grid on;
+stem(n1, x1);
+stem(n2, x2);
+xlabel('n');
+ylabel('x[n]');
+legend('x_1[n]', 'x_2[n]');
+
+subplot(122); hold on; grid on;
+plot(n3, x3);
+plot(n4, x4);
+xlabel('n');
+ylabel('x[n]');
+legend('x_3[n]', 'x_4[n]');
+
+% Custom power function
+function [ P ] = custom_power( x )
+    P = (1/length(x)) * sum(abs(x).^2);
+end
+
+% Calculate and display the power
+P3 = custom_power(x3);
+% Display the calculated powers
+disp(['Power of x_3[n]: ', num2str(P3)]);
+
+% Energy function
+function [ W ] = energy( x )
+    W = sum(abs(x).^2);
+end
+
+% Calculate and display the energy of each signal
+P1 = energy(x1);
+P2 = energy(x2);
+P3 = energy(x3);
+P4 = energy(x4);
+% Display the calculated powers
+disp(['Energy of x_1[n]: ', num2str(P1)]);
+disp(['Energy of x_2[n]: ', num2str(P2)]);
+disp(['Energy of x_3[n]: ', num2str(P3)]);
+disp(['Energy of x_4[n]: ', num2str(P4)]);

Assignment2_211/assignment2_3.m

@@ -0,0 +1,39 @@
+clear
+close all
+clc
+
+
+% Convolution function
+function [ Y ] = convolution( h, x )
+    Ny = length(x)+length(h)-1
+    Y = zeros(1, Ny);
+    for n = 1:Ny
+        for i = 1:length(h)
+            if (n-i+1 >= 1) && (n-i+1 <= length(x))
+                Y(n) = Y(n) + h(i) * x(n-i+1)
+            end
+        end
+    end
+end
+
+n = 0:49
+hn = [0.25, 0.5, 0.25]
+xn = cos(((2*pi)/20)*n)
+
+% Perform convolution of xn and hn
+Y = convolution(hn, xn);
+Y2 = conv(xn, hn)
+
+% Plot the results
+plot(Y);
+xlabel('Sample Index');
+ylabel('Amplitude');
+title('Convolution Result');
+grid on;
+
+% Plot Y2 on same graph
+hold on;
+plot(Y2, '--');
+legend('Custom Convolution', 'MATLAB conv Function');
+hold off;
+

Assignment2_211/assignment2_4.m

@@ -0,0 +1,29 @@
+function X = dtft ( x , n , w )
+    % DTFT Computes Discrete-time Fourier transform
+    % @param x : finite duration sequence (vector) over n
+    % @param n : sample indices vector
+    % @param w : frequency vector
+    % @return X : DTFT of x[n] at frequencies w
+    % this function does not use for loops
+    X = sum(x(:) .* exp(-1j * n(:) * w), 1);
+end
+
+n = -10:10;
+x = (0.8) .^ abs(n) .* ((n >= -10) & (n < 11));
+w = linspace(-pi, pi, 1000);
+
+X = dtft(x, n, w);
+
+magnitude = abs(X);
+phase = angle(X);
+figure;
+subplot(2, 1, 1);
+plot(w, magnitude);
+title('Magnitude Response');
+xlabel('Frequency (rad/sample)');
+ylabel('|X(w)|');
+subplot(2, 1, 2);
+plot(w, phase);
+title('Phase Response');
+xlabel('Frequency (rad/sample)');
+ylabel('Phase(rad)');

Assignment2_211/assignment2_5.m

@@ -0,0 +1,42 @@
+% Read and play original audio
+[audio, fs] = audioread('./audio.wav');
+%sound(audio, fs);
+
+% Read and play room impulse response (RIR)
+[rir, fs_rir] = audioread('./rir.wav');
+%sound(rir, fs_rir);
+
+% Apply convolution (reverb effect)
+audio_reverb = conv(audio, rir);
+
+%sound(audio_reverb, fs);
+
+% Plot the waveforms
+figure;
+
+% Original audio
+subplot(2,1,1);
+plot(audio);
+title('Original Audio');
+xlabel('Sample Number');
+ylabel('Amplitude');
+
+% Reverberated audio
+subplot(2,1,2);
+plot(audio_reverb);
+title('Audio with Reverb');
+xlabel('Sample Number');
+ylabel('Amplitude');
+
+% Calculate signal energy
+function [ W ] = energy( x )
+    W = sum(abs(x).^2);
+end
+
+energy_rir = energy(rir)
+energy_original = energy(audio)
+energy_reverb = energy(audio_reverb)
+
+% Display energy values
+fprintf('Original Audio Energy: %.4f\n', energy_original);
+fprintf('Reverberated Audio Energy: %.4f\n', energy_reverb);

flake.lock

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