DSP Formelsammlung
Författare
Jean Amadeus Elsner
Last Updated
6 years ago
Licens
Creative Commons CC BY 4.0
Sammanfattning
Für [IN2061] Einführung in die digitale Signalvereinbarung an der Technischen Universität München.
Für [IN2061] Einführung in die digitale Signalvereinbarung an der Technischen Universität München.
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\begin{document}
\twocolumn
\pagestyle{fancy}
\lhead{Einführung in die digitale Signalverarbeitung \\ Formelsammlung}
\rhead{\today \\ Elsner, Jean}
\section{Lineare zeitinvariante Systeme}
\subsection{Eigenschaften}
Eigenschaften LTI-Systeme
\begin{mdframed}[style=exercise]
\begin{enumerate}
\item Stabilität\\
$\abs{x(t)} < M < \infty \Rightarrow \abs{y(t)} < N < \infty$
\item Linearität\\
$W\qty{\sum_{k=1}^{N}a_n x_n(t)}=\sum_{n=1}^{N}W\qty{a_n x_n(t)}$
\item Zeitinvarianz\\
$W\qty{x(t-t_0)} =y(t-t_0)$
\item Kausalität\\
$t < 0 \Rightarrow x(t)=0 \land y(t)=0$
\end{enumerate}
\end{mdframed}
\subsection{Systemantwort}
Die Sprung-/Impulsantwort beschreibt Systemantwort vollständig
\begin{mdframed}[style=exercise]
\begin{align}
y(t) &= \int^{\infty}_{-\infty} a(t-\tau) x'(\tau) \dd{\tau}\\
a(t-\tau) &= W\qty{s(t-\tau)}\nonumber
\end{align}
\end{mdframed}
\begin{mdframed}[style=exercise]
\begin{align}
y(t) &= \int_{-\infty}^{\infty} h(t-\tau) x(\tau) \dd{\tau}\\
h(t-\tau) &= W\qty{\delta(t-\tau)}\nonumber
\end{align}
\end{mdframed}
\subsection{Abtasttheorem}
Durch die Abtastung wird das Spektrum von $f(t)$ unendlich oft um die Frequenzen $n\cdot \omega_a$ reproduziert.
\begin{mdframed}[style=exercise]
\begin{align}
F_A(\omega) &= \frac{1}{T_A} \sum_{n=-\infty}^{\infty} F(\omega-n\omega_A)\\
2\omega_g &\leq \omega_A\nonumber
\end{align}
\end{mdframed}
\section{Transformationen}
\subsection{Fourierreihe}
\begin{mdframed}[style=exercise]
\begin{align}
f(t) &= \sum_{n=0}^{\infty} \qty[a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t)]\\
a_n &= \frac{2}{T}\int_{-T/2}^{T/2}f(t)\cos(n\omega_0 t)\dd{t}\nonumber\\
b_n &= \frac{2}{T}\int_{-T/2}^{T/2}f(t)\sin(n\omega_0 t)\dd{t}\nonumber
\end{align}
\end{mdframed}
\pagebreak
\subsection{Fourierreihe, komplex}
\begin{mdframed}[style=exercise]
\begin{align}
f(t) &= \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_0 t}\\
c_n &= \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t}\dd{t}\nonumber
\end{align}
\end{mdframed}
\subsection{Fourierintegral}
\begin{mdframed}[style=exercise]
\begin{align}
f(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} \dd{\omega}\\
F(\omega) &= \int_{-\infty}^{\infty} f(t) e^{-j\omega t} \dd{t}
\end{align}
\end{mdframed}
\subsubsection{Eigenschaften}
\begin{mdframed}[style=exercise]
\begin{enumerate}
\item Linearität\\
$a f_1(t) + b f_2(t) \laplace a F_1(\omega) + b F_2(\omega)$
\item Zeitverschiebung\\
$f(t-t_0) \laplace F(\omega)e^{-j\omega t_0}$
\item Frequenzverschiebung\\
$f(t) e^{\pm j\omega_0 t}\laplace F(\omega\mp \omega_0)$
\item Faltung\\
$f_1(t)*f_2(t)\laplace F_1(\omega)\cdot F_2(\omega)$\\
$f_1(\omega)\cdot f_2(\omega)\laplace \frac{1}{2\pi}F_1(t)*F_2(t)$
\end{enumerate}
\end{mdframed}
\subsection{DFT}
\begin{mdframed}[style=exercise]
\begin{align}
x_n &= \frac{1}{N} \sum_{k=0}^{N-1} X_k\cdot e^{i 2 \pi k n / N}\\
X_k &= \sum_{n=0}^{N-1} x_n\cdot e^{-i 2 \pi k n / N}
\end{align}
\end{mdframed}
\subsubsection{FFT}
\begin{center}
\includegraphics[width=.35\textwidth]{butterfly}
\captionof{figure}{FFT}
\end{center}
\pagebreak
\subsection{Hilbert Transformation}
\begin{mdframed}[style=exercise]
\begin{align}
x_{\mathrm{ht}}(t) &= x_{\mathrm{r}}(t) * h(t)\\
H(\omega) &= -j \, \text{sgn}(\omega)
\end{align}
\end{mdframed}
\subsection{z Transformation}
\begin{mdframed}[style=exercise]
\begin{align}
X(z) &= \sum_{n=-\infty}^{\infty} x(n) z^{-n}\\
x(n) &= \frac{1}{2\pi j} \oint_c X(Z) z^{n-1}\dd{z}
\end{align}
\end{mdframed}
\subsubsection{Übertragungsfunktion}
\begin{mdframed}[style=exercise]
\begin{align}
H(Z) &=\frac{Y(Z)}{X(z)}=\frac{\sum_{k=0}^q b_k z^{-k}}{\sum_{k=0}^p a_k z^{-k}}=k\frac{\prod_{k=1}^q (1-z_k z^{-1})}{\prod_{k=1}^p (1-p_k z^{-1})}
\end{align}
\end{mdframed}
\subsubsection{Verschiebung im Zeitbereich}
\begin{mdframed}[style=exercise]
\begin{align}
Y(z) &= \sum_{n=0}^{\infty} \qty[x(n-m)] z^{-n} = z^{-m} X(z)\\
Y(z) &= \sum_{n=0}^{\infty}\qty[x(n+m)]z^{-n} = z^{m}\qty[x(t)-\sum_{n=0}^{m-1} x(n) z^{-n}]
\end{align}
\end{mdframed}
\section{Filter}
\subsection{FIR}
\begin{mdframed}[style=exercise]
\begin{align}
y[n] &= \sum_{k=0}^{q} b_k x(n-k)
\end{align}
\end{mdframed}
\begin{center}
\includegraphics[width=.4\textwidth]{fir}
\end{center}
\subsection{IIR}
\begin{mdframed}[style=exercise]
\begin{align}
y[n] &= \sum_{k=0}^{q} b_k x(n-k) - \sum_{k=1}^{p} a_k y(n-k)
\end{align}
\end{mdframed}
\begin{center}
\includegraphics[width=.35\textwidth]{df1}
\captionof{figure}{Direkt Form 1}
\end{center}
\begin{center}
\includegraphics[width=.35\textwidth]{df2}
\captionof{figure}{Direkt Form 2}
\end{center}
\section{Entropie}
\subsection{Informationsgehalt}
\begin{mdframed}[style=exercise]
\begin{align}
I_i &= \log_2 \frac{1}{p_i}
\end{align}
\end{mdframed}
\subsection{Entropie}
Mittlerer Informaionsgehalt
\begin{mdframed}[style=exercise]
\begin{align}
H(s) &= E\qty{I} = -\sum_{i=0}^{n-1} p_i \log_2 p_i\\
0 &\leq H(s) \leq \log_2 n
\end{align}
\end{mdframed}
Für $H(s)=\log_2 n$ Gleichverteilung und völlige Ungewissheit.
\end{document}