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| -rw-r--r-- | chapter04/content_ch04.tex | 200 |
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diff --git a/chapter04/content_ch04.tex b/chapter04/content_ch04.tex index 6e678d8..c40e76c 100644 --- a/chapter04/content_ch04.tex +++ b/chapter04/content_ch04.tex @@ -13,12 +13,12 @@ height={0.25\textheight}, width=0.6\linewidth, scale only axis, - xlabel={$t$ or $n$, respectively}, + xlabel={$t$}, ylabel={$x$}, %grid style={line width=.6pt, color=lightgray}, %grid=both, grid=none, - legend pos=north east, + legend pos=north west, axis y line=middle, axis x line=middle, every axis x label/.style={ @@ -34,9 +34,14 @@ %ymin=0, %ymax=3, %xtick={0, 1, ..., 6}, - %ytick={0, 0.5, ..., 2.5} + %ytick={0, 0.5, ..., 2.5}, + xmin=0, + xmax=6.5, + xtick={0, 1, ..., 6}, + xticklabels={$0$, $T_S$, $2 T_S$, $3 T_S$, $4 T_S$, $5 T_S$, $6 T_S$}, ] \addplot[smooth, blue, dashed] coordinates {(0, 1.1) (1, 1.8) (2, 2.1) (3, 1.0) (4, 0.8) (5, 1.7) (6, 2.4)}; + \addlegendentry{$\underline{x}{t}$}; \addplot[red, thick] coordinates {(0, 0) (0, 1.1)}; \addplot[red, thick] coordinates {(1, 0) (1, 1.8)}; \addplot[red, thick] coordinates {(2, 0) (2, 2.1)}; @@ -45,6 +50,7 @@ \addplot[red, thick] coordinates {(5, 0) (5, 1.7)}; \addplot[red, thick] coordinates {(6, 0) (6, 2.4)}; \addplot[only marks, red, thick, mark=o] coordinates {(0, 1.1) (1, 1.8) (2, 2.1) (3, 1.0) (4, 0.8) (5, 1.7) (6, 2.4)}; + \addlegendentry{$\underline{x}_S{t}$}; \end{axis} \end{tikzpicture} \caption{Sampling of a time-continuous signal} @@ -72,78 +78,109 @@ Sampling parameters: \begin{equation} f_S = \frac{1}{T_S} \end{equation} + \item The \index{sampling angular frequency} \textbf{sampling angular frequency} $\omega_S$. + \begin{equation} + \omega_S = \frac{2 \pi}{T_S} + \end{equation} \end{itemize} Ideal sampling: \begin{itemize} - \item The samples are truly equidistant. The sampling period $T_S$ is constant and is \underline{not} subject to fluctuations. - \item The sample is the value of the original signal $\underline{x}(t)$ at \underline{exactly} the time instance where has been taken. + \item The samples are ideally equidistant. The sampling period $T_S$ is constant and is \underline{not} subject to fluctuations. + \item The sample has the value of the original signal $\underline{x}(t)$ at \underline{exactly} the time instance where has been taken. \end{itemize} Some corollaries can be deducted from these two points: \begin{itemize} - \item The sampled signal at the discrete time $n$ is the value of the original signal at time $t = n T_S$: $\underline{x}[n] = \underline{x}\left(n T_S\right)$ - \item The sampled signal consists of a chain of indefinitely narrow pulses. + \item The sampled signal $\underline{x}_S(n T_S)$ at the discrete time $n$ is the value of the original signal at time $t = n T_S$. + \begin{equation} + \underline{x}[n] = \underline{x}\left(n T_S\right) + \end{equation} + \item The sampled signal $\underline{x}_S(t)$ consists of a chain of equidistant, indefinitely narrow pulses. \begin{itemize} \item The pulses are equidistant with $T_S$. \item The pulses have the value of $\underline{x}\left(n T_S\right)$ as their amplitudes. + \item The value of the sampled signal is zero in between the pulses. + \begin{equation} + \underline{x}_S(t) = \begin{cases} + \underline{x}\left(n T_S\right) & \quad \forall \; t = n T_S, n \in \mathbb{Z}, \\ + 0 & \quad \forall \; n T_S < t < \left(n+1\right) T_S, n \in \mathbb{Z}. + \end{cases} + \end{equation} \end{itemize} \end{itemize} -\begin{proof}{} - We know already indefinitely narrow pulses. They are Dirac delta functions $\delta\left(t - n T_S\right)$. - - Taking out \underline{exactly one} sample out of $\underline{x}(t)$ is a convolution of $\underline{x}(t)$ with $\delta(t)$. - \begin{equation} - \begin{split} - \underline{x}[n] &= \underline{x}(t) * \delta(t) \\ - &= \int\limits_{-\infty}^{\infty} \underline{x}(t) \cdot \delta\left(n T_S - t\right) \, \mathrm{d} t \\ - & \text{$\delta(t)$ is symmetric} \\ - &= \int\limits_{-\infty}^{\infty} \underline{x}(t) \cdot \delta\left(t - n T_S\right) \, \mathrm{d} t \\ - &= \underline{x}\left(n T_S\right) - \end{split} - \label{eq:ch04:one_sample} - \end{equation} - - \begin{figure}[H] - \centering - \begin{tikzpicture} - \begin{axis}[ - height={0.25\textheight}, - width=0.6\linewidth, - scale only axis, - xlabel={$t$ or $n$, respectively}, - ylabel={$x$}, - %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=0, - %xmax=7, - %ymin=0, - %ymax=3, - %xtick={0, 1, ..., 6}, - %ytick={0, 0.5, ..., 2.5} - ] - \addplot[smooth, blue, dashed] coordinates {(0, 1.1) (1, 1.8) (2, 2.1) (3, 1.0) (4, 0.8) (5, 1.7) (6, 2.4)}; - \addplot[red, thick] coordinates {(2, 0) (2, 2.1)}; - \addplot[only marks, red, thick, mark=o] coordinates {(2, 2.1)}; - \end{axis} - \end{tikzpicture} - \caption{Taking out exactly one sample out of $\underline{x}(t)$} - \end{figure} -\end{proof} +We know already indefinitely narrow pulses. They are Dirac delta functions $\delta\left(t - n T_S\right)$. + +Taking out \underline{exactly one} sample out of $\underline{x}(t)$ is a multiplication of $\underline{x}(t)$ with $\delta\left(t - n T_S\right)$. +\begin{equation} + \underline{x}_{S,n}(t) = \underline{x}(t) \delta\left(t - n T_S\right) + \label{eq:ch4:one_sample_1} +\end{equation} +The Dirac delta function is zero expect at $t = n T_S$. So, \eqref{eq:ch4:one_sample_1} can be further reduced. +\begin{equation} + \underline{x}_{S,n}(t) = \underline{x}(n T_S) \delta\left(t - n T_S\right) + \label{eq:ch4:one_sample_2} +\end{equation} + +\begin{figure}[H] + \centering + \begin{tikzpicture} + \begin{axis}[ + height={0.25\textheight}, + width=0.6\linewidth, + scale only axis, + xlabel={$t$}, + ylabel={$x$}, + %grid style={line width=.6pt, color=lightgray}, + %grid=both, + grid=none, + legend pos=north west, + 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, + %xmax=7, + %ymin=0, + %ymax=3, + %xtick={0, 1, ..., 6}, + %ytick={0, 0.5, ..., 2.5}, + xmin=0, + xmax=6.5, + xtick={0, 1, ..., 6}, + xticklabels={$0$, $T_S$, $2 T_S$, $3 T_S$, $4 T_S$, $5 T_S$, $6 T_S$}, + ] + \addplot[smooth, blue, dashed] coordinates {(0, 1.1) (1, 1.8) (2, 2.1) (3, 1.0) (4, 0.8) (5, 1.7) (6, 2.4)}; + \addlegendentry{$\underline{x}{t}$}; + \addplot[red, thick] coordinates {(2, 0) (2, 2.1)}; + \addplot[only marks, red, thick, mark=o] coordinates {(2, 2.1)}; + \addlegendentry{$\underline{x}_{S,n}{t}$}; + \end{axis} + \end{tikzpicture} + \caption[Taking out exactly one sample out of $\underline{x}(t)$]{Taking out exactly one sample out of $\underline{x}(t)$ -- in this example $n = 2$.} +\end{figure} + +To obtain the sampled signal, the sampling process $\underline{x}_{S,n}(t)$ \eqref{eq:ch4:one_sample_1} needs to be repeated for each $n \in \mathbb{Z}$. All individual sample processes $\underline{x}_{S,n}(t)$ are then superimposed to form the complete sampled signal $\underline{x}_S(t)$. +\begin{equation} + \begin{split} + \underline{x}_S(t) &= \sum\limits_{n = -\infty}^{\infty} \underline{x}_{S,n}(t) \\ + &= \sum\limits_{n = -\infty}^{\infty} \underline{x}\left(t\right) \delta\left(t - n T_S\right) \\ + &= \underline{x}\left(t\right) \cdot \underbrace{\sum\limits_{n = -\infty}^{\infty} \delta\left(t - n T_S\right)}_{= \Sha_{T_S}(t)} + \end{split} +\end{equation} + +The sum of Dirac delta functions +\begin{itemize} + \item forms a series of equidistant pulses repeating at a period of $T_S$, + \item is called \index{Dirac comb} \textbf{Dirac comb} $\Sha_{T_S}(t)$ or \index{impulse train} \textbf{impulse train}. +\end{itemize} -These Dirac pulses are repeated with a period of $T_S$ and form a \index{Dirac comb} \textbf{Dirac comb} $\Sha_{T_S}(t)$ -- also called \index{impulse train} \textbf{impulse train}. \begin{equation} \Sha_{T_S}(t) = \sum\limits_{n = -\infty}^{\infty} \delta\left(t - n T_S\right) \end{equation} @@ -191,10 +228,23 @@ These Dirac pulses are repeated with a period of $T_S$ and form a \index{Dirac c A \index{sampler} \textbf{sampler} is a system which \begin{itemize} \item applies the Dirac comb $\Sha_{T_S}(t)$ - \item to a time-continuous signal $\underline{x}(t)$ and - \item output a series of equidistant pulses $\underline{x}_S(t)$. + \item to a time-continuous signal $\underline{x}(t)$ (multiplication) and + \item outputs a series of equidistant pulses $\underline{x}_S(t)$. \end{itemize} -The chain of pulses can then be reinterpreted as a time-discrete signal $\underline{x}[n]$. +\begin{equation} + \begin{split} + \underline{x}_S(t) &= \underline{x}(t) \cdot \Sha_{T_S}(t) \\ + &= \sum\limits_{n = -\infty}^{\infty} \underline{x}\left(t\right) \delta\left(t - n T_S\right) \\ + &= \sum\limits_{n = -\infty}^{\infty} \underline{x}\left(n T_S\right) \delta\left(t - n T_S\right) + \end{split} + \label{eq:ch04:ideal_sampling} +\end{equation} + +The sampled signal $\underline{x}_S(t)$ is a chain of pulses (red signal in Figure \ref{fig:ch04:sampling_of_signal}). The chain of pulses can then be reinterpreted as a time-discrete signal $\underline{x}[n]$. The value of $\underline{x}[n]$ is: +\begin{equation} + \underline{x}[n] = \underline{x}_S\left(n T_S\right) = \underline{x}\left(n T_S\right) \qquad \forall \; n \in \mathbb{Z} + \label{eq:ch04:sample_value} +\end{equation} \begin{figure}[H] \centering @@ -215,13 +265,6 @@ The chain of pulses can then be reinterpreted as a time-discrete signal $\underl \caption{An abstract view on sampling} \end{figure} -The ideal sampler multiplies the time-continuous signal $\underline{x}(t)$ with the Dirac comb $\Sha_{T_S}(t)$ in order to obtain the sampled signal $\underline{x}_S(t)$. -\begin{equation} - \underline{x}_S(t) = \underline{x}(t) \cdot \Sha_{T_S}(t) = \sum\limits_{n = -\infty}^{\infty} \underline{x}\left(n T_S\right) \delta\left(t - n T_S\right) - \label{eq:ch04:ideal_sampling} -\end{equation} -In Figure \ref{fig:ch04:sampling_of_signal}, the chain of pulses is red. - \begin{fact} The act of sampling is irreversible. \end{fact} @@ -230,18 +273,25 @@ There is a way to obtain the sampled signal: \begin{equation*} \underline{x}_S(t) = \mathrm{Sampling} \left(\underline{x}(t)\right) \end{equation*} -But there is no way back to reconstruct the original signal. $\mathrm{Sampling}^{-1} \left(\underline{x}_S(t)\right)$ does not exist. +But there is generally no way back to reconstruct the original signal. $\mathrm{Sampling}^{-1} \left(\underline{x}_S(t)\right)$ does not exist. +\begin{equation*} + \underline{x}(t) \neq \underbrace{\mathrm{Sampling}^{-1}}_{\text{Does not exist}} \left(\underline{x}_S(t)\right) +\end{equation*} + +Sampling is always lossy in general. + +\subsection{Sampling Theorem and Aliasing} -Sampling is always lossy. \subsection{Discrete-Time Fourier Transform} % TODO -Using \eqref{eq:ch04:ideal_sampling} and \eqref{eq:ch04:one_sample}, a expression depending on the time-discrete signal $\underline{x}[n]$ can be formulated: +Using \eqref{eq:ch04:ideal_sampling} and \eqref{eq:ch04:sample_value}, a expression depending on the time-discrete signal $\underline{x}[n]$ can be formulated: \begin{equation} \underline{x}_S(t) = \sum\limits_{n = -\infty}^{\infty} \underline{x}[n] \cdot \delta(t - n T_S) \end{equation} +The Fourier transform of the sampled signal $\underline{x}_S(t)$ is: \begin{equation} \begin{split} \underline{X}_S \left(j \omega\right) &= \mathcal{F} \left\{\underline{x}_S(t)\right\} \\ @@ -257,8 +307,6 @@ Redefining $\phi = T_S \omega$: \underline{X}_S \left(j \omega\right) = \underline{X} \left(e^{j \phi}\right) = \sum\limits_{n = -\infty}^{\infty} \underline{x}[n] \cdot e^{-j \phi n} \end{equation} -\subsection{Sampling Theorem and Aliasing} - \subsection{Discrete Fourier Transform} \section{Analogies Of Time-Continuous and Time-Discrete Signals and Systems} |