path: root/exercise04/exercise04.tex

% SPDX-License-Identifier: CC-BY-SA-4.0
%
% Copyright (c) 2020 Philipp Le
%
% Except where otherwise noted, this work is licensed under a
% Creative Commons Attribution-ShareAlike 4.0 License.
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% Please find the full copy of the licence at:
% https://creativecommons.org/licenses/by-sa/4.0/legalcode

\phantomsection
\addcontentsline{toc}{section}{Exercise 4}
\section*{Exercise 4}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{question}[subtitle={Sampling Periodic Signals}]
	\begin{equation*}
		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 $u(t = 0)$.
	
	\begin{tasks}
		\task
		Plot the function from $t = 0$ to $t = \SI{1}{\micro\second}$!
		
		\task
		Calculate the samples $n = 0 \dots 8$!
		
		\task
		What is the DTFT of the signal?
		
		Hints:
		\begin{equation*}
			\begin{split}
				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*}
		
		\task
		Can the DFT directly applied to the signal? If yes, determine the smallest $N$ and give the values of all $\underline{U}[k]$!
		
		\task
		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 $\SI{2.5}{MHz}$. 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}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{question}[subtitle={Sampling Non-Periodic Signals}]
	The signal $x[n]$ is given in the time domain.
	\begin{table}[H]
		\centering
		\begin{tabular}{|l|r|r|r|r|r|r|r|r|}
			\hline
			$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
			\hline
			\hline
			$x[n]$ & 0.25 & 1 & -0.5 & 0.5 & -0.5 & -1 & -0.5 & -0.75 \\
			\hline
		\end{tabular}
	\end{table}
	
	\begin{tasks}
		\task
		The signal is windowed with $N = 4$ starting at $x[n = 0]$. A hamming window with $M = 2$ is applied. Calculate the values of $\underline{x}_W[n]$!
		
		Hamming window:
		\begin{equation*}
			w[n] = \begin{cases}0.54 - 0.46 \cos\left(\frac{2 \pi n}{M}\right), &\quad 0 \leq n \leq M,\\ 0, &\quad \text{otherwise}.\end{cases}
		\end{equation*}
		
		\task
		Calculate the discrete Fourier transform of the windowed signal!
		
		\task
		The signal has been sampled with $T_S = \SI{1}{ms}$. What frequency values do the $k$ represent?
	\end{tasks}
\end{question}

\begin{solution}
	\begin{tasks}
		\task
		At first the signal is truncated. Only the first $4$ samples are considered.
		
		The window function is:
		\begin{equation*}
			w[n] = \begin{cases}0.54 - 0.46 \cos\left(\frac{2 \pi n}{M}\right), &\quad 0 \leq n \leq M,\\ 0, &\quad \text{otherwise}.\end{cases}
		\end{equation*}
		
		The signal is then multiplied with the window:
		\begin{equation*}
			x_w[n] = x[n] \cdot w[n]
		\end{equation*}
		
		\begin{table}[H]
			\centering
			\begin{tabular}{|l|r|r|r|r|}
				\hline
				$n$ & 0 & 1 & 2 & 3 \\
				\hline
				\hline
				$w[n]$ & 0.08 & 1.0 & 0.08 & 0 \\
				\hline
				$x_w[n]$ & 0.02 & 1.0 & -0.04 & 0 \\
				\hline
			\end{tabular}
		\end{table}
		
		\task
		The signal is periodically repeated.
		
		The DFT is calculated over $N = 4$.
		\begin{equation*}
			\underline{X}_w[k] = \mathcal{F}_{\text{DFT}}\left\{\underline{x}[n]\right\} = \sum\limits_{n \in N} \underline{x}[n] \cdot e^{-j 2\pi \frac{k}{N} n}
		\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
				$\underline{X}_w[k]$ & $0.98$ & $(0.06-1j)$ & $-1.02$ & $(0.06+1j)$ \\
				\hline
				$|\underline{X}_w[k]|$ & $0.98$ & $1.00$ & $1.02$ & $1.00$ \\
				\hline
				$\arg\left(\underline{X}_w[k]\right)$ & $0$ & $-1.51 \approx -\pi$ & $3.14 \approx 2\pi$ & $1.51 \approx \pi$ \\
				\hline
			\end{tabular}
		\end{table}
	
		\task
		\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
				$\omega[k]$ & $\SI{0}{s^{-1}}$ & $\SI{1570.8}{s^{-1}}$ & $\SI{3141.6}{s^{-1}}$ & $\SI{4712.4}{s^{-1}}$ \\
				\hline
				\hline
				$f[k]$ & $\SI{0}{Hz}$ & $\SI{250}{Hz}$ & $\SI{500}{Hz} \equiv \SI{-500}{Hz}$ & $\SI{750}{Hz} \equiv \SI{-250}{Hz}$ \\
				\hline
			\end{tabular}
		\end{table}
	\end{tasks}
\end{solution}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}.
	
	\begin{tasks}
		\task
		Define a mapping from the value-continuous samples to the value-discrete samples!
		
		\task
		The value-discrete samples are now pulse-code modulated. How many bits are required?
		
		\task
		What is the resulting data (signal from Task 1b) )?
		
		\task
		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}

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%\begin{question}[subtitle={Decibel}]
%	\begin{tasks}
%	\end{tasks}
%\end{question}
%
%\begin{solution}
%	\begin{tasks}
%	\end{tasks}
%\end{solution}