Exercise 4 competed

5 files changed, 554 insertions, 10 deletions

diff --git a/common/settings.tex b/common/settings.tex
index 0e8e28c..5b5211c 100644
--- a/common/settings.tex
+++ b/common/settings.tex
@@ -292,6 +292,7 @@
 	toc-level = {subsection},
 	% solution/print = true % uncomment for tutors
 }
+\usepackage{tasks}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % PDF Metadata
diff --git a/exercise04/exercise04.tex b/exercise04/exercise04.tex
index 44dea36..ea80fda 100644
--- a/exercise04/exercise04.tex
+++ b/exercise04/exercise04.tex
@@ -15,7 +15,7 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{question}[subtitle={Sampling Periodic Signals}]
 	\begin{equation*}
-		u(t) = \SI{2}{V} \cos\left(2\pi \cdot \SI{2}{MHz} \cdot t + \SI{60}{\degree}\right)
+		u(t) = \SI{2}{V} \cdot \cos\left(2\pi \cdot \SI{2}{MHz} \cdot t + \SI{60}{\degree}\right)
 	\end{equation*}
 	The signal is sampled with a sampling period of $T_S = \SI{125}{\nano\second}$. The first sample taken is $u(t = 0)$.
 	
@@ -32,7 +32,7 @@
 		Hints:
 		\begin{equation*}
 			\begin{split}
-				x[n] = e^{-j a n} &= \underline{X}_{\frac{2\pi}{T_S}}\left(e^{-j T_S \omega}\right) = 2 \pi \cdot \delta \left(\omega + a\right) \\
+				x[n] = e^{-j a n} &= \underline{X}_{2\pi}\left(e^{-j \phi}\right) = 2 \pi \sum\limits_{l = -\infty}^{\infty} \delta \left(\phi + a - 2\pi l\right) \\
 				\cos\left(b\right) &= \frac{1}{2} \left(e^{j b} + e^{-j b}\right)
 			\end{split}
 		\end{equation*}
@@ -44,12 +44,325 @@
 		What is the longest possible sampling period? What must be considered at this sampling period?
 		
 		\task
-		Now, the sampling period is changed to $T_S = \SI{0.5}{\micro\second}$. There is no anti-aliasing filter. The reconstruction filter is an ideal low-pass filter with a cut-off frequency of \SI{50}{kHz}. Give the reconstructed output function in the time domain! Give an explanation in the frequency domain!
+		Now, the sampling period is changed to $T_S = \SI{0.5}{\micro\second}$. There is no anti-aliasing filter. The reconstruction filter is an ideal low-pass filter with a cut-off frequency of $\SI{2.5}{MHz}$. Give the reconstructed output function in the time domain! Give an explanation in the frequency domain!
 	\end{tasks}
 \end{question}
 
 \begin{solution}
 	\begin{tasks}
+		\task
+		\begin{figure}[H]
+			\centering
+			\begin{tikzpicture}
+				\begin{axis}[
+					height={0.25\textheight},
+					width=0.8\linewidth,
+					scale only axis,
+					xlabel={$t$ in \si{\micro\second}},
+					ylabel={$u(t)$},
+					%grid style={line width=.6pt, color=lightgray},
+					%grid=both,
+					grid=none,
+					legend pos=outer north east,
+					axis y line=middle,
+					axis x line=middle,
+					every axis x label/.style={
+						at={(ticklabel* cs:1.05)},
+						anchor=north,
+					},
+					every axis y label/.style={
+						at={(ticklabel* cs:1.05)},
+						anchor=east,
+					},
+					xmin=-0.1,
+					xmax=1.1,
+					ymin=-2.2,
+					ymax=2.2,
+					xtick={0,0.125,...,1},
+					%xticklabels={$- \omega_S$, $- \frac{\omega_S}{2}$, $0$, $\frac{\omega_S}{2}$, $\omega_S$},
+					%ytick={0},
+				]
+					\addplot[blue, dashed, smooth, domain=0:1, samples=50] plot(\x, {2*cos(deg(2*pi*2*\x)+60)});
+					%\addlegendentry{$u(t)$};
+					\pgfplotsinvokeforeach{0,0.125,...,1}{
+						\addplot[red] coordinates {(#1,0) (#1,{2*cos(deg(2*pi*2*#1)+60)})};
+						\addplot[red, only marks, mark=o] coordinates {(#1,{2*cos(deg(2*pi*2*#1)+60)})};
+					}
+				\end{axis}
+			\end{tikzpicture}
+		\end{figure}
+	
+		\task
+		\begin{table}[H]
+			\centering
+			\begin{tabular}{|l|r|r|r|r|r|r|r|r|r|}
+				\hline
+				$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$
\\
+				\hline
+				$t$ in \si{\micro\second} & $0.0$ & $0.125$ & $0.25$ & $0.375$ & $0.5$ & $0.625$ & $0.75$ & $0.875$ & $1.0$
\\
+				\hline
+				\hline
+				$u[n]$ in \si{V} & $1.0$ & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ \\
+				\hline
+			\end{tabular}
+		\end{table}
+		
+		\task
+		$t = n T_S$ due to the sampling:
+		\begin{equation*}
+			\begin{split}
+				u[n] &= \SI{2}{V} \cdot \cos\left(2\pi \cdot \SI{2}{MHz} \cdot n T_S + \SI{60}{\degree}\right) \\
+				 &= \SI{1}{V} \cdot \left( e^{j \left(2\pi \cdot \SI{2}{MHz} \cdot n T_S + \SI{60}{\degree}\right)} + e^{-j \left(2\pi \cdot \SI{2}{MHz} \cdot n T_S + \SI{60}{\degree}\right)} \right) \\
+				 &= \SI{1}{V} \cdot \left( e^{j \SI{60}{\degree}} \cdot e^{j \cdot 2\pi \cdot \SI{2}{MHz} \cdot n T_S} + e^{-j \SI{60}{\degree}} \cdot e^{-j \cdot 2\pi \cdot \SI{2}{MHz} \cdot n T_S} \right)
+			\end{split}
+		\end{equation*}
+		
+		The DTFT is:
+		\begin{equation*}
+			\begin{split}
+				\underline{U}_{2\pi}\left(e^{-j \phi}\right) &= \SI{2}{V} \cdot \pi \sum\limits_{l = -\infty}^{\infty} \left( e^{j \SI{60}{\degree}} \cdot \delta\left(\phi - \left(2\pi \cdot \SI{2}{MHz} \cdot T_S\right) - 2\pi l\right)\right. \\ &\qquad + \left.e^{-j \SI{60}{\degree}} \cdot \delta\left(\phi + \left(2\pi \cdot \SI{2}{MHz} \cdot T_S\right) - 2\pi l\right) \right)
+			\end{split}
+		\end{equation*}
+		
+		\task
+		Yes, it is periodic with $T_0=\SI{500}{ns}$. This means:
+		\begin{equation*}
+			N = \frac{T_0}{T_S} = 4
+		\end{equation*}
+		
+		The DFT over $N=4$ is:
+		\begin{equation*}
+			\begin{split}
+				\underline{X}[k] &= \sum\limits_{n=0}^{N-1} \underline{x}[n] \cdot e^{-j 2\pi \frac{k}{N} n}
+			\end{split}
+		\end{equation*}
+		\begin{equation*}
+			\begin{split}
+				\phi[k] &= 2 \pi \frac{k}{N} \\
+				\omega[k] &= \frac{\phi[k]}{T_S} \\
+				f[k] &= \frac{\omega[k]}{2 \pi} \\
+			\end{split}
+		\end{equation*}
+		
+		\begin{table}[H]
+			\centering
+			\begin{tabular}{|l|r|r|r|r|}
+				\hline
+				$k$ & $0$ & $1$ & $2$ & $3$
\\
+				\hline
+				$k$ (alternate) & $0$ & $1$ & $-2$ & $-1$ \\
+				\hline
+				\hline
+				$\phi[k]$ & $0$ & $1.57 \approx \pi$ & $3.14 \approx 2 \pi \equiv -2\pi$ & $4.71 \approx 3 \pi \equiv -\pi$ \\
+				\hline
+				$\omega[k]$ & $\SI{0}{{\micro\second}^{-1}}$ & $\SI{12.57}{{\micro\second}^{-1}}$ & $\SI{25.13}{{\micro\second}^{-1}}$ & $\SI{37.7}{{\micro\second}^{-1}}$ \\
+				\hline
+				$f[k]$ & $\SI{0}{MHz}$ & $\SI{2}{MHz}$ & $\SI{4}{MHz} \equiv \SI{-4}{MHz}$ & $\SI{6}{MHz} \equiv \SI{-2}{MHz}$ \\
+				\hline
+				\hline
+				$\underline{U}[k]$ & $0$ & $(2+3.46j)$ & $0$ & $(2-3.46j)$ \\
+				\hline
+				$|\underline{U}[k]|$ & $0.0$ & $4.0$ & $0.0$ & $4.0$ \\
+				\hline
+				$\arg\left(\underline{U}[k]\right)$ & $\SI{3.14}{rad} \approx \SI{360}{\degree}$ & $\SI{1.05}{rad} \approx \SI{60}{\degree}$ & $\SI{3.14}{rad} \approx \SI{360}{\degree}$ & $\SI{-1.05}{rad} \approx \SI{-60}{\degree}$ \\
+				\hline
+			\end{tabular}
+		\end{table}
+	
+		\task
+		\begin{itemize}
+			\item Minimum possible sampling frequency is \SI{4}{MHz}. Anything below, will violate the Shannon-Nyquist theorem and cause aliasing.
+			\item The longest possible sampling period is therefore \SI{250}{ns}.
+			\item At $T_S = \SI{250}{ns}$, the sampling phase must be considered, too.
+			\begin{itemize}
+				\item It must be shifted by $\SI{+60}{\degree}$, so that the maxima of $u(t)$ are sampled.
+				\item This retains the amplitude of \SI{2}{V}.
+				\item A timing error will effectively reduce the amplitude of the sampled signal in relation to the original signal.
+				\item At a timing error of $\Delta T_S = \SI{125}{ns}$, all samples will be zero, because the Dirac comb used for sampling is orthogonal to the original signal.
+			\end{itemize}
+		\end{itemize}
+	
+		\task
+		The sampling period of \SI{500}{ns} violates the Shannon-Nyquist theorem and causes aliasing.
+		
+		\begin{figure}[H]
+			\subfloat[Original signal $\underline{U}\left(j\omega\right)$] {
+				\centering
+				\begin{tikzpicture}
+					\begin{axis}[
+						height={0.15\textheight},
+						width=0.25\linewidth,
+						scale only axis,
+						xlabel={$\omega$},
+						ylabel={$|\underline{U}\left(j\omega\right)|$},
+						%grid style={line width=.6pt, color=lightgray},
+						%grid=both,
+						grid=none,
+						legend pos=north east,
+						axis y line=middle,
+						axis x line=middle,
+						every axis x label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=north,
+						},
+						every axis y label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=east,
+						},
+						xmin=-1.5,
+						xmax=1.5,
+						ymin=0,
+						ymax=1.2,
+						xtick={-1, -0.5, 0, 0.5, 1},
+						xticklabels={$- \omega_S$, $- \frac{\omega_S}{2}$, $0$, $\frac{\omega_S}{2}$, $\omega_S$},
+						ytick={0},
+						%ytick={0, 1},
+						%yticklabels={0, $\SI{}$}
+					]
+						\draw[-latex, red, very thick] (axis cs:1,0) -- (axis cs:1,0.8); %node[left,align=right,anchor=north,rotate=90,yshift=-5mm]{$\arg\left(\underline{U}\left(j\omega_S\right)\right) = \SI{60}{\degree}$};
+						\draw[-latex, green, very thick] (axis cs:-1,0) -- (axis cs:-1,0.8); %node[left,align=right,anchor=north,rotate=90,yshift=-5mm]{$\arg\left(\underline{U}\left(-j\omega_S\right)\right) = \SI{-60}{\degree}$};
+					\end{axis}
+				\end{tikzpicture}
+			}
+			\hfill
+			\subfloat[Spectrum of the Dirac comb $\frac{2 \pi}{T} \Sha_{\frac{2 \pi}{T}}(\omega)$] {
+				\centering
+				\begin{tikzpicture}
+					\begin{axis}[
+						height={0.15\textheight},
+						width=0.27\linewidth,
+						scale only axis,
+						xlabel={$t$},
+						ylabel={$|\frac{2 \pi}{T} \Sha_{\frac{2 \pi}{T}}(\omega)|$},
+						%grid style={line width=.6pt, color=lightgray},
+						%grid=both,
+						grid=none,
+						legend pos=north east,
+						axis y line=middle,
+						axis x line=middle,
+						every axis x label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=north,
+						},
+						every axis y label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=east,
+						},
+						xmin=-2.5,
+						xmax=2.5,
+						ymin=0,
+						ymax=1.2,
+						xtick={-2, ..., 2},
+						xticklabels={$-2 \omega_S$, $- \omega_S$, $0$, $\omega_S$, $2 \omega_S$},
+						ytick={0},
+					]
+						\pgfplotsinvokeforeach{-3, -2, ..., 3}{
+							\draw[-latex, blue, very thick] (axis cs:#1,0) -- (axis cs:#1,1);
+						}
+					\end{axis}
+				\end{tikzpicture}
+			}
+			\hfill
+			\subfloat[Sampled signal $\underline{U}_S\left(j\omega\right)$, showing aliasing] {
+				\centering
+				\begin{tikzpicture}
+					\begin{axis}[
+						height={0.15\textheight},
+						width=0.27\linewidth,
+						scale only axis,
+						xlabel={$\omega$},
+						ylabel={$|\underline{U}_S\left(j\omega\right)|$},
+						%grid style={line width=.6pt, color=lightgray},
+						%grid=both,
+						grid=none,
+						legend pos=north east,
+						axis y line=middle,
+						axis x line=middle,
+						every axis x label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=north,
+						},
+						every axis y label/.style={
+							at={(ticklabel* cs:1.05)},
+							anchor=east,
+						},
+						xmin=-2.5,
+						xmax=2.5,
+						ymin=0,
+						ymax=1.2,
+						xtick={-2, -1, -0.5, 0, 0.5, 1, 2},
+						xticklabels={$-2 \omega_S$, $- \omega_S$, $- \frac{\omega_S}{2}$, $0$, $\frac{\omega_S}{2}$, $\omega_S$, $2 \omega_S$},
+						ytick={0},
+					]
+						\pgfplotsinvokeforeach{-3, -2, ..., 3}{
+							\draw[-latex, blue, dashed, very thick] (axis cs:#1,0) -- (axis cs:#1,1);
+							\draw[-latex, green, very thick] (axis cs:{#1-0.01},0) -- (axis cs:{#1-0.01},0.8);
+							\draw[-latex, red, very thick] (axis cs:{#1+0.01},0) -- (axis cs:{#1+0.01},0.8);
+						}
+					\end{axis}
+				\end{tikzpicture}
+			}
+		\end{figure}
+	
+		\begin{itemize}
+			\item The spectral components of $\SI{+2}{MHz}$ (with $e^{j \SI{60}{\degree}}$) and $\SI{-2}{MHz}$ (with $e^{j \SI{-60}{\degree}}$) superimpose at \SI{0}{Hz} (DC).
+			\item The resulting component at \SI{0}{Hz} is the addition if two complex numbers (see Part c) ):
+			\begin{equation*}
+				\underline{U}_S\left(j 0\right) = \SI{2}{V} \cdot \pi \underbrace{\left(e^{j \SI{60}{\degree}} + e^{j \SI{-60}{\degree}}\right)}_{= 1}  \delta(\omega) = \SI{2}{V} \cdot \pi \cdot  \delta(\omega)
+			\end{equation*}
+			\item Applying the reconstruction filter gives the spectrum of the reconstructed signal:
+			\begin{equation*}
+				\underline{U}_R\left(j \omega\right) = \SI{2}{V} \cdot \pi \cdot \delta(\omega)
+			\end{equation*}
+			\item The inverse DTFT is:
+			\begin{equation*}
+				u_R(t) = \SI{1.0}{V}
+			\end{equation*}
+			\item The reconstructed signal is a DC signal of \SI{1.0}{V}.
+		\end{itemize}
+	
+		\begin{figure}[H]
+			\centering
+			\begin{tikzpicture}
+				\begin{axis}[
+					height={0.15\textheight},
+					width=0.8\linewidth,
+					scale only axis,
+					xlabel={$t$ in \si{\micro\second}},
+					ylabel={$u(t)$},
+					%grid style={line width=.6pt, color=lightgray},
+					%grid=both,
+					grid=none,
+					legend pos=outer north east,
+					axis y line=middle,
+					axis x line=middle,
+					every axis x label/.style={
+						at={(ticklabel* cs:1.05)},
+						anchor=north,
+					},
+					every axis y label/.style={
+						at={(ticklabel* cs:1.05)},
+						anchor=east,
+					},
+					xmin=-0.1,
+					xmax=1.1,
+					ymin=-2.2,
+					ymax=2.2,
+					xtick={0,0.125,...,1},
+					%xticklabels={$- \omega_S$, $- \frac{\omega_S}{2}$, $0$, $\frac{\omega_S}{2}$, $\omega_S$},
+					%ytick={0},
+				]
+					\addplot[blue, dashed, smooth, domain=0:1, samples=50] plot(\x, {2*cos(deg(2*pi*2*\x)+60)});
+					\addlegendentry{$u(t)$};
+					\addplot[olive] coordinates {(0,1) (0.5,1) (1,1)};
+					\addlegendentry{$u_R(t)$};
+					\pgfplotsinvokeforeach{0,0.5,1}{
+						\addplot[red] coordinates {(#1,0) (#1,{2*cos(deg(2*pi*2*#1)+60)})};
+						\addplot[red, only marks, mark=o] coordinates {(#1,{2*cos(deg(2*pi*2*#1)+60)})};
+					}
+				\end{axis}
+			\end{tikzpicture}
+		\end{figure}
 	\end{tasks}
 \end{solution}
 
@@ -131,9 +444,9 @@
 				$k$ (alternate) & 0 & 1 & -2 & -1 \\
 				\hline
 				\hline
-				$\underline{X}_w[k]$ & $0.98$ & $(0.06-1j)$ & $-1.02$ & $(0.06+1j)$
 \\
+				$\underline{X}_w[k]$ & $0.98$ & $(0.06-1j)$ & $-1.02$ & $(0.06+1j)$ \\
 				\hline
-				$|\underline{X}_w[k]|$ & $0.98$ & $1.00$ & $1.02$ & $1.00$
 \\
+				$|\underline{X}_w[k]|$ & $0.98$ & $1.00$ & $1.02$ & $1.00$ \\
 				\hline
 				$\arg\left(\underline{X}_w[k]\right)$ & $0$ & $-1.51 \approx -\pi$ & $3.14 \approx 2\pi$ & $1.51 \approx \pi$ \\
 				\hline
@@ -160,7 +473,7 @@
 				\hline
 				$\phi[k]$ & $0$ & $1.57 \approx \pi$ & $3.14 \approx 2 \pi \equiv -2\pi$ & $4.71 \approx 3 \pi \equiv -\pi$ \\
 				\hline
-				$\omega[k]$ & $\SI{0}{s^{-1}}$ & $\SI{1570.8}{s^{-1}}$ & $\SI{3141.6}{s^{-1}}$ & $\SI{4712.4}{s^{-1}}$
\\
+				$\omega[k]$ & $\SI{0}{s^{-1}}$ & $\SI{1570.8}{s^{-1}}$ & $\SI{3141.6}{s^{-1}}$ & $\SI{4712.4}{s^{-1}}$ \\
 				\hline
 				\hline
 				$f[k]$ & $\SI{0}{Hz}$ & $\SI{250}{Hz}$ & $\SI{500}{Hz} \equiv \SI{-500}{Hz}$ & $\SI{750}{Hz} \equiv \SI{-250}{Hz}$ \\
@@ -172,7 +485,7 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{question}[subtitle={Quantization}]
-	The signal of task 1b) is now quantized. The quantizer has $8$ discrete values. These values are equally distributed between \SI{-2}{V} and \SI{2}{V}. Prior to sampling, the original time-continuous signal passed through an ideal low-pass filter with a cut-off frequency of \SI{4}{MHz}.
+	The signal of Task 1b) is now quantized. The quantizer has $8$ discrete values. These values are equally distributed between \SI{-2}{V} and \SI{2}{V}. Prior to sampling, the original time-continuous signal passed through an ideal low-pass filter with a cut-off frequency of \SI{4}{MHz}.
 	
 	\begin{tasks}
 		\task
@@ -182,15 +495,130 @@
 		The value-discrete samples are now pulse-code modulated. How many bits are required?
 		
 		\task
-		Determine the quantization error for each value-discrete sample! How much is the signal-to-noise ratio?
+		What is the resulting data (signal from Task 1b) )?
 		
 		\task
-		3 bits are a very poor resolution. How many bits are appropriate for the quantizer to obtain the best signal-to-noise ratio? Effects of the window filter are neglected. Assume that the signal has passed through a processing chain with a total gain of \SI{25}{dB} and noise figure of \SI{12}{dB} prior to quantization. The input of the quantizer has an impedance of \SI{50}{\ohm}. % 14 bits
+		Determine the quantization error for each value-discrete sample! How much is the signal-to-quantization-noise ratio?
+		
+		\task
+		3 bits are a very poor resolution. How many bits are appropriate for the quantizer to obtain the best signal-to-noise ratio? Effects of the window filter are neglected. Assume that the signal has passed through a processing chain with a total gain of \SI{24}{dB}, noise figure of \SI{12}{dB} and bandwidth of \SI{4}{MHz} prior to quantization. The input of the quantizer has an impedance of \SI{50}{\ohm}. % 14 bits
 	\end{tasks}
 \end{question}
 
 \begin{solution}
 	\begin{tasks}
+		\task
+		\begin{itemize}
+			\item $\hat{U}_L = \SI{-2}{V}$
+			\item $\hat{U}_H = \SI{2}{V}$
+			\item $K = 8$
+		\end{itemize}
+		\begin{equation*}
+			\Delta \hat{U} = \frac{\hat{U}_H - \hat{U}_L}{K - 1} = \SI{0.57}{V}
+		\end{equation*}
+		
+		The boundaries for the rounding quantizer are distributed $\pm \frac{1}{2} \Delta \hat{U}$ around each mean.
+		\begin{table}[H]
+			\begin{tabular}{|r|r|r|r|}
+				\hline
+				From & To & Maps to & PCM data \\
+				\hline
+				\hline
+				$-\infty$ & $\SI{-1.71}{V}$ & $\SI{-2.0}{V}$ & 0
\\
+				$\SI{-1.71}{V}$ & $\SI{-1.14}{V}$ & $\SI{-1.43}{V}$ & 1
\\
+				$\SI{-1.14}{V}$ & $\SI{-0.57}{V}$ & $\SI{-0.86}{V}$ & 2
\\
+				$\SI{-0.57}{V}$ & $\SI{-0.0}{V}$ & $\SI{-0.29}{V}$ & 3
\\
+				$\SI{0.0}{V}$ & $\SI{0.57}{V}$ & $\SI{0.29}{V}$ & 4
\\
+				$\SI{0.57}{V}$ & $\SI{1.14}{V}$ & $\SI{0.86}{V}$ & 5
\\
+				$\SI{1.14}{V}$ & $\SI{1.71}{V}$ & $\SI{1.43}{V}$ & 6
\\
+				$\SI{1.71}{V}$ & $\infty$ & $\SI{2.0}{V}$ & 7 \\
+				\hline
+			\end{tabular}
+		\end{table}
+	
+		\task
+		\begin{equation*}
+			B \geq \log_2 \left(K\right) = 3
+		\end{equation*}
+		Minimum 3 bits.
+		
+		\task
+		\begin{table}[H]
+			\centering
+			\begin{tabular}{|l|r|r|r|r|r|}
+				\hline
+				$n$ & $0$ & $1$ & $2$ & $3$ & $4$
\\
+				\hline
+				\hline
+				$u[n]$ in \si{V} & $1.0$ & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ \\
+				\hline
+				Quantized values in \si{V} & $0.86$ & $-2.0$ & $-0.86$ & $2.0$ & $0.86$ \\
+				\hline
+				PCM data & $\left(101\right)_2$ & $\left(000\right)_2$ & $\left(010\right)_2$ & $\left(111\right)_2$ & $\left(101\right)_2$ \\
+				\hline
+				\hline
+				$n$ & $5$ & $6$ & $7$ & $8$ &
\\
+				\hline
+				\hline
+				$u[n]$ in \si{V} & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ & \\
+				\hline
+				Quantized values in \si{V} & $-2.0$ & $-0.86$ & $2.0$ & $0.86$ & \\
+				\hline
+				PCM data & $\left(000\right)_2$ & $\left(010\right)_2$ & $\left(111\right)_2$ & $\left(101\right)_2$ & \\
+				\hline
+			\end{tabular}
+		\end{table}
+	
+		\task
+		\begin{table}[H]
+			\centering
+			\begin{tabular}{|l|r|r|r|r|r|}
+				\hline
+				$n$ & $0$ & $1$ & $2$ & $3$ & $4$
\\
+				\hline
+				\hline
+				$u[n]$ in \si{V} & $1.0$ & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ \\
+				\hline
+				Quantized values in \si{V} & $0.86$ & $-2.0$ & $-0.86$ & $2.0$ & $0.86$ \\
+				\hline
+				Quantization error in \si{V} & $0.14$ & $0.27$ & $0.14$ & $0.27$ & $0.14$ \\
+				\hline
+				\hline
+				$n$ & $5$ & $6$ & $7$ & $8$ &
\\
+				\hline
+				\hline
+				$u[n]$ in \si{V} & $-1.73$ & $-1.0$ & $1.73$ & $1.0$ & \\
+				\hline
+				Quantized values in \si{V} & $-2.0$ & $-0.86$ & $2.0$ & $0.86$ & \\
+				\hline
+				Quantization error in \si{V} & $0.27$ & $0.14$ & $0.27$ & $0.14$ & \\
+				\hline
+			\end{tabular}
+		\end{table}
+	
+		The signal-to-quantization-noise ratio for the sine wave is:
+		\begin{equation*}
+			\mathrm{SQNR} = \SI{1.761}{dB} + B \cdot \SI{6.02}{dB} = \SI{19.82}{dB}
+		\end{equation*}
+		
+		\task
+		\begin{itemize}
+			\item The RMS value of the signal is $U_{\mathrm{RMS}} = \frac{\SI{2}{V}}{\sqrt{2}} = \SI{1.41}{V}$
+			\item The signal power is $P_S = \frac{U_{\mathrm{RMS}}^2}{R} = \SI{39.8}{mW} \equiv L_{P,S} = \SI{16}{dBm}$
+			\item The thermal noise floor is $\SI{-174}{dBm/Hz}$
+			\item At $\SI{4}{MHz} \equiv \SI{66}{dBHz}$, the thermal noise power is $\SI{-174}{dBm/Hz} + \SI{66}{dBHz} = \SI{-108}{dBm}$
+			\item The gain and noise factor is applied to the thermal noise, resulting in a noise power of $L_{P,N} = \SI{-108}{dBm} + \SI{24}{dB} + \SI{12}{dB} = \SI{-70}{dBm}$
+			\item \textit{Note that the gain is not applied to the signal power, because it is already known/given at the quantizer input.}
+			\item The signal-to-noise ratio is $L_{\mathrm{SNR}} = L_{P,S} - L_{P,N} = \SI{16}{dBm} - \SI{-71}{dBm} = \SI{86}{dB}$
+			\item The signal-to-quantization-noise ratio should not be grater than the signal-to-noise ratio, because otherwise the thermal noise would dominate the quantization noise.
+			\begin{equation*}
+				L_{\mathrm{SQNR}} \stackrel{!}{=} L_{\mathrm{SNR}}
+			\end{equation*}
+			\item The number of bits must be at least
+			\begin{equation*}
+				B \geq \frac{L_{\mathrm{SQNR}} - \SI{1.761}{dB}}{\SI{6.02}{dB}} = 14
+			\end{equation*}
+		\end{itemize}
 	\end{tasks}
 \end{solution}
 
diff --git a/exercise04/helpers/ex_4_1.py b/exercise04/helpers/ex_4_1.py
new file mode 100644
index 0000000..2208737
--- /dev/null
+++ b/exercise04/helpers/ex_4_1.py
@@ -0,0 +1,63 @@
+#!/usr/bin/env python3
+
+# SPDX-License-Identifier: BSD-3-Clause
+#
+# Copyright 2020 Philipp Le
+#
+# Redistribution and use in source and binary forms, with or without
+# modification, are permitted provided that the following conditions
+# are met:
+#
+# 1. Redistributions of source code must retain the above copyright
+#    notice, this list of conditions and the following disclaimer.
+#
+# 2. Redistributions in binary form must reproduce the above copyright
+#    notice, this list of conditions and the following disclaimer in the
+#    documentation and/or other materials provided with the distribution.
+#
+# 3. Neither the name of the copyright holder nor the names of its
+#    contributors may be used to endorse or promote products derived from
+#    this software without specific prior written permission.
+#
+# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
+# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
+# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
+# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
+# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
+# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
+# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
+# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF
+# THE POSSIBILITY OF SUCH DAMAGE.
+
+
+import numpy
+
+def latex_table(ar):
+    return " & ".join(["$"+str(it)+"$" for it in ar])
+
+print("$n$ & "+latex_table([int(num) for num in numpy.linspace(0,8,9)]))
+
+t=numpy.linspace(0,1e-6,9)
+print("$t$ in \\si{\\micro\\second} & "+latex_table(t*1e6))
+
+x=2*numpy.cos(2*numpy.pi*2e6*t+(numpy.pi/3))
+print("$u[n]$ in \\si{V} & "+latex_table(numpy.round(x*100)/100))
+
+print("")
+
+phi=2*numpy.pi*numpy.array([0,1,2,3])/4
+omega=phi/125e-9
+f=omega/(2*numpy.pi)
+print("$\\phi[k]$ & "+latex_table(numpy.round(phi*100)/100))
+print("$\\omega[k]$ & "+latex_table(numpy.round(omega*100)/100))
+print("$f[k]$ & "+latex_table(f))
+
+
+Xft = numpy.fft.fft(x[0:4])
+Xft_abs = numpy.abs(Xft)
+Xft_phase = numpy.angle(Xft)
+print("$\\underline{U}[k]$ & "+latex_table(numpy.round(Xft*100)/100))
+print("$|\\underline{U}[k]|$ & "+latex_table(numpy.round(Xft_abs*100)/100))
+print("$\\arg\\left(\\underline{U}[k]\\right)$ & "+latex_table(numpy.round(Xft_phase*100)/100))
diff --git a/exercise04/helpers/ex_4_3.py b/exercise04/helpers/ex_4_3.py
new file mode 100644
index 0000000..1079ce9
--- /dev/null
+++ b/exercise04/helpers/ex_4_3.py
@@ -0,0 +1,52 @@
+#!/usr/bin/env python3
+
+# SPDX-License-Identifier: BSD-3-Clause
+#
+# Copyright 2020 Philipp Le
+#
+# Redistribution and use in source and binary forms, with or without
+# modification, are permitted provided that the following conditions
+# are met:
+#
+# 1. Redistributions of source code must retain the above copyright
+#    notice, this list of conditions and the following disclaimer.
+#
+# 2. Redistributions in binary form must reproduce the above copyright
+#    notice, this list of conditions and the following disclaimer in the
+#    documentation and/or other materials provided with the distribution.
+#
+# 3. Neither the name of the copyright holder nor the names of its
+#    contributors may be used to endorse or promote products derived from
+#    this software without specific prior written permission.
+#
+# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
+# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
+# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
+# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
+# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
+# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
+# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
+# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF
+# THE POSSIBILITY OF SUCH DAMAGE.
+
+
+import numpy
+
+def latex_table(ar):
+    return " & ".join(["$"+str(it)+"$" for it in ar])
+
+U_L = -2
+U_H = 2
+K = 8
+
+dU = (U_H - U_L)/(K-1)
+
+for step in range(K):
+    mean = numpy.round((U_L + (step * dU))*100)/100
+    lower = numpy.round((mean - (dU/2))*100)/100
+    upper = numpy.round((mean + (dU/2))*100)/100
+    print("$\\SI{"+str(lower)+"}{V}$ & $\\SI{"+str(upper)+"}{V}$ & $\\SI{"+str(mean)+"}{V}$")
+
+
diff --git a/main/exercise04.tex b/main/exercise04.tex
index 1002d6e..788ee0a 100644
--- a/main/exercise04.tex
+++ b/main/exercise04.tex
@@ -14,7 +14,7 @@
 \def\thekindofdocument{Exercise}
 \def\thesubtitle{Chapter 4: Sampling and Time-Discrete Signals and Systems}
 \def\therevision{1}
-\def\therevisiondate{2020-05-26}
+\def\therevisiondate{2020-05-30}
 
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