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diff --git a/chapter02/content_ch02.tex b/chapter02/content_ch02.tex index 9dc7df2..41ddc15 100644 --- a/chapter02/content_ch02.tex +++ b/chapter02/content_ch02.tex @@ -588,14 +588,24 @@ The inner integral is the \textbf{continuous Fourier transform}, also called onl The \index{continuous Fourier transform} \textbf{continuous Fourier transform} of the function $\underline{x}(t)$ is: \begin{equation} \underline{X}(j \omega) = \mathcal{F} \left\{\underline{x}(t)\right\} = \int\limits_{t = -\infty}^{\infty} \underline{x}(t) \cdot e^{-j \omega t} \, \mathrm{d} t + \label{eq:ch02:def_fourier_transform} \end{equation} The \index{inverse Fourier transform} \index{inverse continuous Fourier transform} \textbf{inverse (continuous) Fourier transform} is: \begin{equation} \underline{x}(t) = \mathcal{F}^{-1} \left\{\underline{X}(j \omega)\right\} = \frac{1}{2 \pi} \int\limits_{\omega = -\infty}^{\infty} \underline{X}(j \omega) \cdot e^{+j \omega t} \, \mathrm{d} \omega + \label{eq:ch02:def_inv_fourier_transform} \end{equation} \end{definition} +The Fourier transform $\mathcal{F} \left\{\underline{x}(t)\right\}$ and its inverse $\mathcal{F}^{-1} \left\{\underline{X}(j \omega)\right\}$ both yield functions which depend on $t$ or $\omega$, respectively. This relation is sometimes emphasized by appending $(t)$ or $\left(j \omega\right)$. +\begin{subequations} + \begin{align} + \mathcal{F} \left\{\underline{x}(t)\right\} &= \mathcal{F} \left\{\underline{x}(t)\right\} \left(j \omega\right) \\ + \mathcal{F}^{-1} \left\{\underline{X}(j \omega)\right\} &= \mathcal{F}^{-1} \left\{\underline{X}(j \omega)\right\} (t) + \end{align} +\end{subequations} + \subsection{Amplitude and Phase Spectra} The value-continuous complex frequency variable $j \omega$ in the continuous Fourier transforms replaced the value-discrete index $n$ of the Fourier series. Due to their similarity, the constraints for all signals and the \index{spectrum!symmetry rules} \textbf{symmetry rules} for real-valued signals apply analogously. @@ -645,7 +655,7 @@ where $\mathrm{sinc}(t)$ is the \emph{normalized} sinc function. \end{equation*} \end{attention} -The resulting spectra of $\underline{X}\left(j \omega\right)$ can now be drawn. The rectangular function is special. The imaginary part $\Im\left\{\underline{X}\left(j \omega\right)\right\} = 0$ is zero. Thus, the phase can only be $0$ or $\pm \pi$. However, this is a special property of the sinc function, but not of every function. +The resulting spectra of $\underline{X}\left(j \omega\right)$ can now be drawn. The rectangular function is special. The imaginary part $\Im\left\{\underline{X}\left(j \omega\right)\right\} = 0$ is zero. Thus, the phase can only be $0$ or $\pm \pi$. However, this is a special property of all functions which are real-valued and even in the time domain like the sinc function. \begin{figure}[H] \centering @@ -841,21 +851,163 @@ Using the Dirac measure, the Fourier transform can be calculated: \end{equation} The Fourier transform of the Dirac delta function is the frequency-independent constant $1$. -\subsection{Operations 1: Linearity} +\subsection{Basic Properties} + +All properties of the Fourier transform can be proven using the definition of the Fourier transform \eqref{eq:ch02:def_fourier_transform}. + +\subsubsection{Linearity} + +%\begin{equation} + +%\end{equation} + +\begin{definition}{Linearity of the Fourier transform} + \begin{equation} + \mathcal{F}\left\{\underline{a} \cdot \underline{f}(t) + \underline{b} \cdot \underline{g}(t)\right\} = \underline{a} \cdot \mathcal{F}\left\{\underline{f}(t)\right\} + \underline{b} \cdot \mathcal{F}\left\{\underline{g}(t)\right\} + \label{eq:ch02:op_lin} + \end{equation} + where + \begin{itemize} + \item $\underline{a} \in \mathbb{C}$ and $\underline{b} \in \mathbb{C}$ are complex numbers and + \item $\underline{f}(t)$ and $\underline{g}(t)$ are Fourier-transformable functions. + \end{itemize} +\end{definition} + +\subsubsection{Differentiation and Integration} + +\begin{definition}{Differentiation of the Fourier transform} + \begin{equation} + \mathcal{F}\left\{\frac{\mathrm{d}^n}{\mathrm{d} t^n} \underline{f}(t)\right\} = \left(j \omega\right)^n \underbrace{\underline{F} \left(j \omega\right)}_{= \mathcal{F}\left\{\underline{f}(t)\right\}} + \label{eq:ch02:op_diff} + \end{equation} +\end{definition} + +\begin{definition}{Integration of the Fourier transform} + \begin{equation} + \mathcal{F}\left\{\int\limits_{t'= -\infty}^{t} \underline{f}(t') \, \mathrm{d} t' \right\} = \frac{1}{j \omega} \underbrace{\underline{F} \left(j \omega\right)}_{= \mathcal{F}\left\{\underline{f}(t)\right\}} + \label{eq:ch02:op_int} + \end{equation} +\end{definition} + +\begin{excursus}{Network analysis of reactive electrical circuits} + Linear, reactive electrical networks are analysed using the Fourier transform. + + For example, voltage and current have following relation in time domain at a capacity: + \begin{equation} + u(t) = \frac{1}{C} \int i(t) \, \mathrm{d} t + \end{equation} + The expression in complex-valued phasors (frequency domain) is: + \begin{equation} + \underline{U} = \underline{Z}_C \cdot \underline{I} + \end{equation} + Using the Fourier transform, the impedance $\underline{Z}_C$ can be determined to: + \begin{equation} + \underline{Z}_C = \frac{1}{j \omega C} + \end{equation} + + The calculation is analogous for inductances. The volatge-current relation in the time domain is: + \begin{equation} + u(t) = L \cdot \frac{\mathrm{d}}{\mathrm{d} t} i(t) + \end{equation} + The complex-valued impedance $\underline{Z}_L$ (frequency domain) is: + \begin{equation} + \underline{Z}_L = j \omega L + \end{equation} +\end{excursus} + +\subsubsection{Multiplication} + +\begin{definition}{Convolution theorem} + A multiplication in the time-domain becomes a convolution in the frequency domain. + \begin{equation} + \mathcal{F}\left\{ \underline{f}(t) \cdot \underline{g}(t) \right\} = \frac{1}{2 \pi} \mathcal{F}\left\{\underline{f}(t)\right\} * \mathcal{F}\left\{\underline{g}(t)\right\} + \label{eq:ch02:op_mult} + \end{equation} +\end{definition} -\subsection{Operations 2: Differentiation and Integration} +\begin{excursus}{Convolution} + The convolution is defined to: + \begin{equation} + f(t) * g(t) = \left(f * g\right) (t) = \int_{\tau = -\infty}^{\infty} f(\tau) g(t - \tau) \, \mathrm{d} \tau + \label{eq:ch02:def_convolution} + \end{equation} +\end{excursus} -\subsection{Operations 3: Multiplication} +\subsubsection{Time Shift} -\subsection{Operations 4: Time Shift} +%Let +%\begin{equation} +% h(t) = \underline{f}(t - t_0) +%\end{equation} + +\begin{definition}{Translation} + \begin{equation} + \mathcal{F}\left\{\underline{f}(t - t_0)\right\} = e^{-j t_0 \omega} \cdot \underbrace{\underline{F} \left(j \omega\right)}_{= \mathcal{F}\left\{\underline{f}(t)\right\}} + \label{eq:ch02:op_time_shift} + \end{equation} + where + \begin{itemize} + \item $t_0 \in \mathbb{R}$ is a real number and + \item $\underline{f}(t)$ is a Fourier-transformable function. + \end{itemize} +\end{definition} \subsection{Duality} +\begin{definition}{Duality} + Suppose $\underline{g}(t)$ has a Fourier transform $\underline{G}\left(j \omega\right)$, i.e., $\underline{g} \TransformHoriz \underline{G}$. The Fourier transform of $\underline{G}(t)$ is: + \begin{equation} + \mathcal{F}\left\{\underline{G}(t)\right\} = \underline{g} \left(- j \omega\right) + \label{eq:ch02:op_duality} + \end{equation} +\end{definition} + +An example for the duality is, the frequency shift. We already know the Fourier transform of the time shift \eqref{eq:ch02:op_time_shift}. +\begin{equation} + \mathcal{F}\left\{e^{j \omega_0 t} \underline{f}(t)\right\} = \underbrace{\underline{F} \left(j (\omega - \omega_0)\right)}_{= \mathcal{F}\left\{\underline{f}(t)\right\} \left( j (\omega - \omega_0) \right)} + \label{eq:ch02:op_freq_shift} +\end{equation} + +Another example is the convolution in time-domain. Due to the duality, it becomes a multiplication the frequency domain. +\begin{equation} + \mathcal{F}\left\{ \underline{f}(t) * \underline{f}(t) \right\} = \mathcal{F}\left\{\underline{f}(t)\right\} \cdot \mathcal{F}\left\{\underline{g}(t)\right\} + \label{eq:ch02:op_conv} +\end{equation} + +\begin{figure}[H] + \centering + \begin{tikzpicture} + \node[align=center, minimum width=2.5cm, minimum height=1.5cm] (TD1) {$\underline{f}(t) \cdot \underline{g}(t)$}; + \node[align=center, minimum width=2.5cm, minimum height=1.5cm, right=3.5cm of TD1] (TD2) {$\underline{f}(t) * \underline{f}(t)$}; + \node[align=center, minimum width=2.5cm, minimum height=1.5cm, below=2cm of TD1] (FD1) {$\frac{1}{2 \pi} \left(\underline{F}\left(j \omega\right) * \underline{G}\left(j \omega\right)\right)$}; + \node[align=center, minimum width=2.5cm, minimum height=1.5cm, below=2cm of TD2] (FD2) {$\underline{F}\left(j \omega\right) \cdot \underline{G}\left(j \omega\right)$}; + + \node[align=right, anchor=east, left=3cm of TD1] (LabelTD) {\textbf{Time domain}}; + \node[align=right, anchor=east, below=2cm of LabelTD] (LabelFD) {\textbf{Frequency domain}}; + \node[align=right, above=1cm of TD1] (Func1) {\textbf{Function 1}}; + \node[align=right, above=1cm of TD2] (Func2) {\textbf{Function 2}}; + + %\draw (TD1) node[midway, align=right, rotate=-90]{$\TransformHoriz$} (FD1); + %\draw (TD2) node[midway, align=right, rotate=-90]{$\TransformHoriz$} (FD2); + \draw[o-*, thick] (TD1.south) -- (FD1.north); + \draw[o-*, thick] (TD2.south) -- (FD2.north); + + \draw[thick] (TD1.south east) -- (FD2.north west); + \draw[thick] (TD2.south west) -- (FD1.north east); + \end{tikzpicture} + \caption{Duality} +\end{figure} + +The duality also affects the units of the time variable $t$ and the frequency variable $\omega$. The units must be inverse. If $t$ is in seconds, $\omega$ must be \si{1/s}. + + \section{\acs{LTI} Systems} -\subsection{Transfer Function and Impulse Response} +\subsection{Transfer Function} + +\subsection{Impulse Response} -\subsection{Convolution} +% Convolution \subsection{Poles and Zeroes} |