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\course{Electrial Engineering}
\module{DigSig1}
\semester{Fall Semester 2021}

\authoremail{naoki.pross@ost.ch}
\author{\textsl{Naoki Pross} -- \texttt{\theauthoremail}}

\title{Digital Signal Processing Lecture Notes}
\date{\thesemester}

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\begin{document}

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\tableofcontents

\section*{License}
\doclicenseThis

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\section{Probability and stochastics}

\subsection{Random variables}

A \emph{random variable} (RV) is a function \(x : \Omega \to \mathbb{R}\). The \emph{distribution function} of a RV is a function \(F_x : \mathbb{R} \to [0,1]\) that is always monotonically increasing and given by
\[
  F_x(\alpha) = \Pr{x \leq \alpha}.
\]
The \emph{probability density function} (PDF) of a RV is
\[
  f_x(\alpha) = \frac{dF_x}{d\alpha}.
\]
The \emph{expectation} of a RV is
\[
  \E{x} = \int_\mathbb{R} \alpha f_x(\alpha) \,d\alpha,
\]
and in the case of a discrete RV
\[
  \E{x} = \sum_k \alpha_k \Pr{x = \alpha_k}.
\]
In general it holds that
\[
  \E{g(x)} = \int_\mathbb{R} g(\alpha) f_x(\alpha) \,d\alpha,
\]
for example
\begin{align*}
  \E{x^2} &= \int_\mathbb{R} \alpha^2 f_x(\alpha) \,d\alpha \\
  \E{|x|} &= \int_\mathbb{R} |\alpha| f_x(\alpha) \,d\alpha \\
    &= \int_0^\infty \alpha \left[ f_x(\alpha) + f_x(-\alpha) \right] \,d\alpha
\end{align*}
The \emph{variance} of a RV is
\[
  \Var{x} = \sigma^2 = \E{(x - \E{x})^2} = \E{x^2} - \E{x}^2,
\]
where \(\sigma\) is called the \emph{standard deviation}. The variance is sometimes also called the \emph{second moment} of a RV, the \emph{\(n\)-th moment} of a RV is \(\E{x^n}\).

\subsection{Jointly distributed RVs}

\section{Analog signals}

In this document we will use the notation \(\Omega = 2\pi f\) for physical analog frequencies (in radians / second), and \(\omega\) for digital frequencies (in radians / sample).

\subsection{Transformations}
Recall the three important operations for the analysis of analog signals.
\begin{flalign*}
  \textit{Fourier Transform} &&
  X(\Omega) &= \int_\mathbb{R} x(t) e^{j\Omega t} \,dt \\
  %
  \textit{Inverse Fourier Transform} &&
  x(t) &= \int_\mathbb{R} X(\Omega) e^{j\Omega t} \,\frac{d\Omega}{2\pi} \\
  %
  \textit{Laplace Transform} &&
  X(s) &= \int_\mathbb{R} x(t) e^{-st} \,dt
\end{flalign*}
The Laplace transform reduces to the Fourier transform under the substitution \(s = j\Omega\).

\subsection{Linear Systems}
Recall that superposition holds. Thus the system is characterized completely by the impulse response function \(h(t)\). The output in the time domain \(y(t)\) is given by the convolution product
\[
  y(t) = h(t) * x(t) = \int_\mathbb{R} h(t - t') x(t') \,dt',
\]
and in the frequency domain \(Y(\Omega) = H(\Omega) X(\Omega)\), where \(H(\Omega)\) is the Fourier transform of \(h(t)\).

% Analog signals:
% TODO: FT of eigenfunctions e^{j\Omega_k t\}

\section{Sampling and reconstruction}

To sample a signal \(x(t)\) it means to measure (take) the value at a periodic interval every \(T\) seconds. \(T\) is thus called the \emph{sample interval} and \(f_s =1/T\) is the \emph{sampling frequency}.

\subsection{Sampling theorem}

To represent a signal \(x(t)\) by its samples \(\hat{x}(nT)\) two conditions must be met:
\begin{enumerate}
  \item \(x(t)\) must be \emph{bandlimited}, i.e. there must be a frequency \(f_\text{max}\) after which the spectrum of \(x(t)\) is always zero.
  \item The sampling rate \(f_s\) must be chosen so that
    \[
      f_s \geq 2 f_\text{max}.
    \]
\end{enumerate}
In other words you need at least 2 samples / period to reconstruct a signal.  When \(f_s = 2 f_\text{max}\), the edge case, the sampling rate is called \emph{Nyquist rate}.  The interval \(\left[-f_s / 2, f_2 / 2\right]\), and its multiples are called \emph{Nyquist intervals}, as they are bounded by the Nyquist frequencies. It would be good to have an arbitrarily high sampling frequency but in reality there is upper limit given by processing time \(T_\text{proc}\). Thus \(2f_\text{max} \leq f_s \leq f_\text{proc}\).

\subsection{Discrete-Time Fourier Transform}

Mathematically speaking, to sample a signal is equivalent multiplying a function with the the so called \emph{impulse train distribution} (aka Dirac comb).
\[
  s(t) = \sum_{n = -\infty}^{\infty} \delta(t - nT),
\]
so we write \(\hat{x}(t) = s(t)\cdot x(t)\) to represent a sampled signal. Because of the special propertie of the Dirac delta, The spectrum of a sampled function \(\hat{x}\) is
\[
  \hat{X}(f) = \sum_{n = -\infty}^{\infty} x(nT) e^{-2\pi jfTn}.
\]
This can be thought as a numerical approximation of the real spectrum \(X(f)\) which gets better as \(T \to 0\), i.e.
\[
  X(f) = \lim_{T \to 0} T\hat{X}(f).
\]
If we have a finite number \(L\) of samples to work with, we will repeat them periodically and obtain what is known as the \emph{Discrete-Time Fourier Transform} (DTFT), i.e.
\[
  \hat{X}(f) \approx \hat{X}_L(f) = \sum_{n = 0}^{L -1} x(nT) e^{-2\pi jfTn}.
\]

\subsection{Spectrum replication and aliasing}

Notice that the impulse train is periodic, and has thus a Fourier series, whose coefficients all equal to \(1/T\) (\(= f_s\), the sampling rate). So the Fourier transform of a comb is also a comb. The consequence is that, because the Fourier of the product \(x(t)\cdot s(t)\) in the time domain becomes a convolution \(X(f) * S(f)\) where \(S(f)\) is an impulse train of Dirac deltas spaced \(1/T = f_s\) apart, what is called \emph{spectrum replication} happens, mathematically
\[
  \hat{X}(f)
    = \sum_{n = -\infty}^{\infty} x(nT) e^{2\pi jfTn}
    = \frac{1}{T}\sum_{m = -\infty}^\infty X(f - mf_s).
\]
In other words, the modulation of the property of the Fourier transform copies the baseband spectrum into integer multiples of the sampling frequency. This is why \(f_s \geq 2f_\text{max}\). The important result is that
\[
  X(f) = T \hat{X}(f), \quad 
  \text{for} \quad -\frac{f_2}{2} \leq f \leq \frac{f_s}{2},
\]
and if the sampling theorem is satisfied the exact original spectrum can be recovered with a low pass filter.


% Alias frequency \(f_a = f \pmod{f_s}\).
% Anti-aliasing: analog LP prefilter cutoff \@ \(f_s/2\)

\section{Quantization}


\end{document}
