\chapter{Fourier transformation convention}
\label{appendix:Fourier}
Fourier transformations come with a variety of normalizations. This appendix
lines out the definitions/conventions used in this text.

The Fourier transformation $\cal F$
\begin{equation}\label{eq:ap:FT}
  \tilde{a}(\omega) = {\cal F}[a(t)]  :=
  \int\limits_{-\infty}^{\infty}\!a(t)e^{-i\omega t}dt
\end{equation}
converts a time-domain function $a(t)$ into a frequency domain function
$\tilde{a}(\omega)$. The tilde above the function symbol is used to indicate a
function that depends on a frequency (or angular frequency $\omega$).

The inverse Fourier transformation  ${\cal F}^{-1}$is defined in the following way:
\begin{equation}
  a(t) = {\cal F}^{-1}[\tilde{a}(\omega)] :=
  \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!\tilde{a}(\omega)e^{i\omega t}d\omega
\end{equation}
With this combination, the forward and backward transformation reverse each
other:
\begin{equation}
  {\cal F}^{-1}\left[{\cal F}[a(t)]\right] = a(t)\quad \text{and}\quad
  {\cal F}\left[{\cal F}^{-1}[\tilde{a}(\omega)]\right] = \tilde{a}(\omega)
\end{equation}

A common convention used in signal processing is to replace the integration
over angular frequencies $\omega=2\pi f$ by integration over frequencies $f$.
Then, the Fourier transformation and its inverse look more symmetric:
\begin{eqnarray}
  \tilde{a}(f)={\cal F}[a(t)]&=&
  \int\limits_{-\infty}^{\infty}\!a(t)e^{-2\pi i f t}dt\\
  a(t)={\cal F}^{-1}[\tilde{a}(f)]&=&
  \int\limits_{-\infty}^{\infty}\!\tilde{a}(f)e^{2\pi i f t}df 
\end{eqnarray}
However, the values $\tilde{a}(\omega)$ and $\tilde{a}(f=\omega/2\pi)$ are the
same. 

\section{Fourier-transformations of real-valued signals}
Most signals are real-valued functions of time, following the identity
\begin{equation}
  f(t)=f^*(t)\,,
\end{equation}
where the asterisk $*$ denotes the complex conjugate. This leads to a simple
relation for the Fourier transformation:
\begin{equation}
  f(-\omega)=\int\limits_{-\infty}^{\infty}\!e^{+i\omega t}f(t)dt =
  \left[\int\limits_{-\infty}^{\infty}\!e^{-i\omega t}
    f^*(t)dt\right]^*=\tilde{f}^*(\omega)\,,
\end{equation}
This means that the Fourier transformation of real-valued functions have
always positive and negative frequency components of the same magnitude.

\section{Fourier-transformation of time-discrete signals}
\label{sec:FTdiscrete}
To obtain the Fourier transformation of a time-discrete signal, the
conventional definition (\ref{eq:ap:FT}) can be used, replacing the continuous
time-domain function by a weighted Dirac comb
\begin{equation}
  a(t)=\sum\limits_{k=-\infty}^{\infty}\!a_k\delta(t-k\Delta t)\,.
\end{equation}
Then, the integration can be carried out:
\begin{eqnarray}\label{eq:discreteFourierApp}
  \tilde{a}(\omega)&=&{\cal F}[a(t)]
                       =\int\limits_{-\infty}^{+\infty}\!a(t)e^{-i\omega
                       t}\,dt\nonumber \\
                   &=&\int\limits_{-\infty}^{+\infty}
                       \sum\limits_{k=-\infty}^{\infty}a_k
                       \delta(t-k \Delta t)\, e^{-i\omega t}\,dt\nonumber\\
                   &=&\sum\limits_{k=-\infty}^{\infty}a_k
                       \int\limits_{-\infty}^{+\infty}\! e^{-i\omega t}
                       \delta(t-k \Delta t)dt\,, \quad t'=t-k\Delta t\nonumber\\
                   &=&\sum\limits_{k=-\infty}^{\infty}a_k\,e^{-i\omega k\Delta t}
                       \int\limits_{-\infty}^{+\infty}e^{-i\omega t'}\delta(t')dt'\nonumber\\
                   &=&\sum\limits_{k=-\infty}^{\infty}a_k\,e^{-i\omega \Delta
                       t\cdot k}.
\end{eqnarray}
The Fourier transformation is periodic with a period of the sampling
frequency $\omega_S$:
\begin{equation}
  \tilde{a}(\omega)= \tilde{a}(\omega+k\cdot2\pi/\Delta t)= \tilde{a}(\omega+k\cdot\omega_S)\quad\text{for}\quad
  k\in\mathbb{Z}\,.
\end{equation}

The inverse Fourier transformation for time-discrete signals makes use of the
periodicity in the Fourier domain:
\begin{eqnarray}
  a(t) &=& {\cal F}^{-1}[\tilde{a}(\omega)] =
           \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!\tilde{a}(\omega)
           e^{i\omega t}\,d\omega \nonumber\\
       &=&\frac{1}{2\pi}\sum\limits_{k=-\infty}^{\infty}\int\limits_{-\omega_S/2}^{\omega_S/2}\!\tilde{a}(\omega)e^{i(\omega+k\omega_S)t}d\omega
           \nonumber\\
       &=&\frac{1}{2\pi}\int\limits_{-\omega_S/2}^{\omega_S/2}\!\!\!\tilde{a}(\omega)
           \left[\sum\limits_{k=-\infty}^{\infty}e^{ik\omega_St}\right] e^{i\omega t}d\omega\nonumber\\
       &=&\frac{1}{2\pi}
           \int\limits_{-\omega_S/2}^{\omega_S/2}\!\!\!\tilde{a}(\omega) e^{i\omega
           t}\left[\sum\limits_{k=-\infty}^{\infty}\delta(t-k\Delta t))\right]d\omega\nonumber\\
       &=&\sum\limits_{k=-\infty}^{\infty}\delta(t-k\Delta t)) 
           \underbrace{\frac{1}{2\pi}
           \int\limits_{-\omega_S/2}^{\omega_S/2}\!\!\!\tilde{a}(\omega)e^{i\omega k
           \Delta t}d\omega}_{=:a_k}
\end{eqnarray}
The last line makes use of a representation of the sampled function, and can
be read as a definition of the time-discrete inverse Fourier transform:
\begin{equation}
  a_k = \frac{1}{2\pi}
  \int\limits_{-\omega_S/2}^{\omega_S/2}\!\!\!\tilde{a}(\omega)e^{i\omega k 
    \Delta t}d\omega\quad\text{with}\quad k\in\mathbb{Z}\,.
\end{equation}
\section{Parseval's theorem, power and total energy of signals}
For square-integrable functions $f(t), g(t)$, i.e., functions that do not have
an infinite extent, Parseval's theorem states 
\begin{equation}
  \int\limits_{-\infty}^{\infty}\!\left|f(t)\cdot g^*(t)\right|dt =
  \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!
    \tilde{f}(\omega)\cdot\tilde{g}^*(\omega)d\omega\,,
\end{equation}
This allows to define a total energy $E$ of a signal,
\begin{equation}
  E=\int\limits_{-\infty}^{\infty}\!\left|f(t)^2\right|dt=  \frac{1}{2\pi}
  \int\limits_{-\infty}^{\infty}\!\left|\tilde{f}^2(\omega)\right|d\omega\,.
\end{equation}
The  integrand in the second integral can be interpreted as a
{\em spectral power density}.

For time-discrete signals, Parseval's theorem reads
\begin{equation}
  \Delta t\sum\limits_{k=-\infty}^{\infty}\!f_kg^*_k =
  \frac{1}{2\pi}\int\limits_{-\omega_S/2}^{\omega_S/2}\!
  \tilde{f}(\omega)\cdot\tilde{g}^*(\omega)d\omega\,.
\end{equation}
One can again define a total signal energy,
\begin{equation}
  E=\Delta t\sum\limits_{k=-\infty}^{\infty}\!\left|f_k^2\right| =
  \frac{1}{2\pi}\int\limits_{-\omega_S/2}^{\omega_S/2}\!
  \left|\tilde{f}^2(\omega)\right|d\omega\,.
\end{equation}
The quantity $E$ in this expression has the ugly property that for continuous
signals, it is not finite. A connection to concepts of power and spectral
power density becomes then painful.


This can be addressed by introducing a time-continuous window function
$\window_T(t)$, or its time-discrete equivalent $\window_{T,k}$ for a large
time window $T$
\begin{equation}
  \window_T(t)=\begin{cases}1\,,&-T/2<2<T/2\\0&\text{otherwise}\end{cases}\,,
  \qquad
  \window_{T,k}=\begin{cases}1\,,&-T/2\Delta t<2<T/2\Delta t\\0&\text{otherwise}\end{cases}\,.
\end{equation}
Its Fourier tranform in time-continuous and time-discrete versions are
\begin{equation}
  {\cal F}\left[\window_T(t)\right] =
  \int\limits_{-\infty}^{\infty}\!\window_T(t)e^{-i\omega t}dt
  =\int\limits_{-T/2}^{T/2}\! e^{-i\omega t}dt
  =T\cdot \sinc\left(\frac{\omega T}{2}\right)
\end{equation}
and
\begin{eqnarray}
  {\cal F}\left[\window_{T,k}\right]&=& \sum\limits_{k=-\infty}^{\infty}\!
                                        \window_{T,k}e^{-i\omega\Delta t k}=
                                        \sum\limits_{k=-T/(2\Delta
                                        t)}^{T/(2\Delta t)}
                                        \!e^{-i\omega\Delta t k}
                                        \nonumber\\
                                    &=&e^{i\omega T/2}
                                        \sum\limits_{k=0}^{T/\Delta t}
                                        \!e^{-i\omega\Delta t k} \nonumber\\
                                    &=&e^{i\omega T/2}\frac{1-e^{-i\omega
                                        T}}{1-e^{-i\omega\Delta t}}
                                        \nonumber\\
                                    &=&e^{i\omega \Delta t /2}
                                        \frac{\sin\omega T/2}
                                        {\sin\omega\Delta t /2}
\end{eqnarray}

\chapter{Dirac comb properties}
\label{appendix:DiracComb}
The definition of a Dirac comb of a certain separation $\Delta t$ in time,
\begin{equation}
  \comb_{\Delta t}(t):=\sum\limits_{k=-\infty}^{\infty}\delta(t-k\Delta t)\,,
\end{equation}
deserves some clarification on the notation. The definition above includes a
timing scale $\Delta t$ as an index of the function. Without an index, it is
supposed to be one, or dimensionless:
\begin{equation}
  \comb(x):=\sum\limits_{k=-\infty}^{\infty}\delta(x-k)\,.
\end{equation}
The relationship between the indexed and non-indexed comb function are
determined by the scaling property of the Dirac function:
\begin{equation}
  \comb_{\Delta t}(t)=\frac{1}{\Delta t}\comb\left(\frac{t}{\Delta t}\right)
\end{equation}

The Dirac comb can be represented as a Fourier series,
\begin{equation}
  \comb_{\Delta t}(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-k\Delta t)=
  \frac{1}{\Delta t}\sum\limits_{n=-\infty}^{+\infty}e^{i\frac{2\pi}{\Delta
      t}nt}\,,
\end{equation}
Result where I want to get:
\begin{equation}
  \tilde{F}(\omega)={\cal F}\left[\comb_{\Delta t}(t)\right](\omega)=
  \frac{2\pi}{\Delta t}
  \sum\limits_{k=-\infty}^{\infty}\delta(\omega-k\frac{2\pi}{\Delta t})=
  \omega_S\sum\limits_{k=-\infty}^{\infty}\delta(\omega-k\omega_S)
\end{equation}

\chapter{Convolution of two functions}
\label{appendix:convolution}
A very useful operation is the convolution (or convolution product) of two
functions. For two functions $a(t)$ and $b(t)$, the convolution function is defined as
\begin{equation}
  c(t) = a(t)\ast b(t) := \int\limits_{-\infty}^{\infty}\!a(t')b(t-t')dt' =
  \int\limits_{-\infty}^{\infty}\!a(t-t')b(t')dt' \,.
\end{equation}
The last identity indicates that the convolution is a commutative operation.

The {\em convolution theorem} makes a statement about the Fourier
transformation of the convolution of two functions:
\begin{equation}\label{eq:convolution1}
  {\cal F}[a(t)\ast b(t)] = \tilde{a}(\omega)\cdot\tilde{b}(\omega)
\end{equation}
This means that the Fourier transformation of the convolution of two functions
is the direct product of the Fourier transformations of each function.

This can be shown by explicit calculation and making use of a property of the
Kronecker delta:
\begin{equation}
  \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!e^{i\omega t}dt= \delta(\omega)
\end{equation}

Using the convention of Fourier transformations, the convolution of two
functions in the frequency domain leads to
\begin{equation}
  {\cal F}^{-1}[\tilde{a}(\omega)\ast \tilde{b}(\omega)] = 2\pi\,a(t)\cdot b(t)
\end{equation}
or equivalently
\begin{equation}
  a(t)\cdot b(t) =  {\cal F}^{-1}\left[\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!\tilde{a}(\omega')\tilde{b}(\omega-\omega')d\omega'\right]
\end{equation}
When using the convention of integration over frequencies rather angular
frequencies, both relations become more symmetric again:
\begin{eqnarray}
  {\cal F}[a(t)\ast b(t)] &=& \tilde{a}(f)\cdot\tilde{b}(f)\\
  {\cal F}^{-1}[\tilde{a}(f)\ast \tilde{b}(f)] &=& a(t)\cdot b(t)
\end{eqnarray}


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