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\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com202}
\usepackage{pst-3dplot}
\begin{document}
\intertitle{autocorrelation \& friends}
\begin{frame} \frametitle{The quest for ``invariant'' signal features}
\begin{itemize}
\item deterministic signals are completely known; e.g. $x[n] = \sin((\pi/5)\,n)$
\item deterministic signals are not interesting!
\item signals that carry actual information are not exactly known in advance
\item usually, we know at least the signal's source (eg. voice, music, radio)
\item we can hope that each source produces signals with some common ``trait''
\item one good categorical descriptor is self-similarity
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Autocorrelation}
\[
r_x[k] = \langle x[n-k], x[n] \rangle = \sum_{n=-\infty}^{\infty} x^*[n-k]x[n]
\]
\vspace{2em}
\begin{itemize}
\item compare signal with a shifted copy of itself
\item measures signal's \textit{self-similarity} over time
\item well-defined for square-summable (energy) signals
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Properties of the autocorrelation}
\[
r_x[k] = \langle x[n-k], x[n] \rangle = \sum_{n=-\infty}^{\infty} x^*[n-k]x[n]
\]
\vspace{2em}
\begin{itemize}
\item if $y[n] = x[n-m]$, $r_y[k] = r_x[k]$ (shift-invariance)
\item $r_x[k] = r_x^*[-k]$ (Hermitian symmetry, symmetry for real signals)
\item $r_x[0] = \|x[n]\|^2$ (value in zero is energy)
\item $|r_x[0]| \ge |r_x[k]|$ (peak magnitude in zero)
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Quick proof of the last point}
\begin{itemize}
\item intuition: a signal is maximally similar to itself!
\item formally (assume $x[n] \in \mathbb{R}$ for simplicity):
\begin{align*}
(x[n-k] - x[n])^2 &= x^2[n] + x^2[n-k] - 2x[n-k]x[n] \\
x[n-k]x[n] &= [x^2[n] + x^2[n-k] - (x[n-k] - x[n])^2] / 2
\end{align*}
With this:
\begin{align*}
r_x[k] &= \sum_{n=-\infty}^{\infty} x[n-k]x[n] \\
&= \frac{1}{2} \sum_{n=-\infty}^{\infty} x^2[n] + \frac{1}{2} \sum_{n=-\infty}^{\infty} x^2[n-k] - \frac{1}{2} \sum_{n=-\infty}^{\infty} (x[n-k] - x[n])^2 \\
&= r_x[0] - \frac{1}{2} \sum_{n=-\infty}^{\infty} (x[n-k] - x[n])^2 \le r_x[0]
\end{align*}
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Example: delta sequence}
\begin{align*}
x[n] &= a \delta[n] \\ \\
r_x[k] &= a^2 \sum_{n=-\infty}^{\infty} \delta[n]\delta[n-k] = a^2\delta[-k] \\
&= a^2\delta[k]
\end{align*}
\end{frame}
\begin{frame} \frametitle{Example: exponential decay}
\begin{align*}
x[n] &= a^n u[n] \\ \\
r_x[k] &= \sum_{n=-\infty}^{\infty} x[n]x[n-k] = \sum_{n=0}^{\infty} a^n a^{n-k}u[n-k] = \sum_{n=\max\{0, k\}}^{\infty} a^{2n-k} \\
&= \begin{cases}
a^{-k} \sum_{n=0}^{\infty} a^{2n} & k \le 0 \\
a^{-k} \sum_{n=k}^{\infty} a^{2n} = a^{-k}\left(\sum_{n=0}^{\infty} a^{2n} - \sum_{n=0}^{k-1} a^{2n} \right) & k > 0
\end{cases} \\
&= \frac{a^{|k|}}{1-a^2}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Example: exponential decay}
\centering
\def\a{0.8 }
$x[n] = a^n u[n], \quad a = \a$
\begin{dspPlot}[height=2cm,xout=true]{-30,30}{-.5, 1.3}
\moocStyle
\dspSignal{x 0 ge {\a x exp} {0} ifelse}
\end{dspPlot}
$r_x[k] = a^{|k|}/(1-a^2)$
\begin{dspPlot}[height=2cm]{-30,30}{0, 4.3}
\moocStyle
\dspSignal{\a x abs exp 1 \a \a mul sub div}
\end{dspPlot}
\end{frame}
\begin{frame} \frametitle{Example: the autocorrelation is a robust descriptor}
\centering
\def\a{0.8 }
$y[n]$
\begin{dspPlot}[height=2cm,xout=true]{-30,30}{-.5, 1.3}
\moocStyle
\dspSignal[xmax=-1]{0}
\dspTapsAt{0}{0.81 0.648 0.5184 0.41472 0.331776 0.0944208
0.07553664 -0.11057069 -0.08845655 -0.07076524 -0.21051219 -0.16840975
-0.0986278 -0.07890224 -0.21702179 -0.31212744 -0.24970195 -0.16727156
-0.13381725 -0.0745638 -0.18431004 -0.28595803 -0.19952542 -0.15962034
-0.09845527 -0.19095732 -0.12352485 -0.07250298 -0.18266139 -0.11981221}
\end{dspPlot}
$r_y[k]$
\begin{dspPlot}[height=2cm]{-30,30}{0, 4.3}
\moocStyle
\dspSignal{\a x abs exp 1 \a \a mul sub div}
\end{dspPlot}
\end{frame}
\begin{frame} \frametitle{To understand better, move to the frequency domain (I)}
let's express the autocorrelation in a different way:
\begin{itemize}
\item autocorrelation: $\displaystyle r_x[k] = \sum_{n=-\infty}^{\infty} x[n]x^*[n-k]$
\item convolution: $\displaystyle (x \ast y)[k] = \sum_{n=-\infty}^{\infty} x[n]y[k-n]$
\item define $x^H[n] = x^*[-n]$
\end{itemize}
\[
r_x[k] = (x \ast x^H)[k]
\]
\end{frame}
\begin{frame} \frametitle{To understand better, move to the frequency domain (II)}
now let's compute the DTFT of the autocorrelation:
\begin{align*}
\DTFT{r_x[k]} &= \DTFT{(x \ast x^H)[k]} \\
&= \DTFT{x[n]} \cdot \DTFT{x^H[n]} \\
&= X(e^{j\omega})X^*(e^{j\omega}) \\
&= |X(e^{j\omega})|^2
\end{align*}
\end{frame}
\begin{frame} \frametitle{Energy spectral density}
\[
E_x(e^{j\omega}) = \DTFT{r_x[k]} = |X(e^{j\omega})|^2
\]
\vspace{1em}
\begin{itemize}
\item square magnitude of DTFT shows energy distribution (density) in frequency
\item DTFT of autocorrelation captures where the energy of the signal is
\item phase information is discarded: the shape of the signal in time does not matter!
\item autocorrelation is invariant to shifts and shape changes
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Example revisited: same magnitude, different phase}
\begin{center}
\begin{dspPlot}[height=2cm,xout=true]{-30,30}{-.5, 1.3}
\moocStyle
\dspSignal[xmax=-1]{0}
\dspTapsAt{0}{0.81 0.648 0.5184 0.41472 0.331776 0.0944208
0.07553664 -0.11057069 -0.08845655 -0.07076524 -0.21051219 -0.16840975
-0.0986278 -0.07890224 -0.21702179 -0.31212744 -0.24970195 -0.16727156
-0.13381725 -0.0745638 -0.18431004 -0.28595803 -0.19952542 -0.15962034
-0.09845527 -0.19095732 -0.12352485 -0.07250298 -0.18266139 -0.11981221}
\end{dspPlot}
\end{center}
\begin{columns}
\begin{column}{0.5\paperwidth}
\begin{align*}
& x[n] = a^n u[n] \\
& E_x(e^{j\omega}) = |X(e^{j\omega})|^2 = \left|\frac{1}{1-ae^{-j\omega}}\right|^2
\end{align*}
\end{column}
\begin{column}{0.5\paperwidth}
\begin{align*}
& y[n] = (h \ast x)[n] \\
& E_y(e^{j\omega}) = |Y(e^{j\omega})|^2 = |H(e^{j\omega})X(e^{j\omega})|^2 \\
& |H(e^{j\omega})| = 1 \qquad \mbox{\it (allpass filter)} \\
& E_y(e^{j\omega}) = E_x(e^{j\omega})
\end{align*}
\end{column}
\end{columns}
\end{frame}
%\begin{frame} \frametitle{We can work backwards as well}
% Compute the autocorrelation of $x[n] = (\omega_c/\pi)\sinc((\omega_c/\pi) n)$
%
% \begin{align*}
% E_x(\omega) &= |X(e^{j\omega})|^2 = [\rect(\omega/(2\omega_c))]^2 = \rect(\omega/(2\omega_c)) \\ \\
% r_x[k] &= \IDFT{\rect(\omega/(2\omega_c))} \\
% &= (\omega_c/\pi)\sinc((\omega_c/\pi) n
% \end{align*}
%\end{frame}
\begin{frame} \frametitle{What about power signals?}
\centering
power signals:
\begin{align*}
\lim_{N\rightarrow \infty}\sum_{n=-N}^{N}|x[n]|^2 = \infty \\
\lim_{N\rightarrow \infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2 < \infty
\end{align*}
\vspace{2em}
for power signals the autocorrelation needs a normalization term
\end{frame}
\begin{frame} \frametitle{Autocorrelation of power signals}
\centering
normalized, partial autocorrelation:
\[
r_{x,N}[k] = \langle x[n-k], x[n] \rangle_N = \frac{1}{2N+1} \sum_{n=-N}^{N} x^*[n-k]x[n]
\]
\vspace{2em}
the limit is well defined for power signals and we set
\[
r_x[k] = \lim_{N\rightarrow \infty} r_{x,N}[k]
\]
\end{frame}
\begin{frame} \frametitle{Power Spectral Density}
\centering
the Power Spectral Density (PSD) of a power signal $x[n]$ is \textit{defined} as
\[
P_x(e^{j\omega}) = \DTFT{r_x[k]}
\]
\vspace{1em}
\begin{itemize}
\item shows the power distribution in frequency for the signal
\item again, phase information is discarded and the shape of the signal in time does not matter!
\end{itemize}
\vspace{3em}
\footnotesize
[Note that the PSD $P_x(e^{j\omega})$ is \textit{not} the square magnitude of $X(e^{j\omega})$ because power signals don't have a ``normal'' DTFT. The generalized DTFT of a power signal contains Dirac deltas and it makes no sense to square a Dirac delta!]
\end{frame}
\begin{frame} \frametitle{Example: constant signal}
\begin{align*}
x[n] &= a \\ \\
r_x[k] &= \lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum_{n=-N}^{N} |a|^2\\
&= |a|^2 \\ \\
P_x(e^{j\omega}) &= |a|^2 \tilde{\delta}(\omega)
\end{align*}
\end{frame}
\begin{frame} \frametitle{Example: unit step}
\begin{align*}
x[n] &= a\, u[n] \\ \\
r_x[k] &= \lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum_{n=0}^{N} |a|^2 u[n-k] \\
&= |a|^2 \lim_{N\rightarrow \infty} \frac{N+1 - \max\{0,k\}}{2N+1} \\
&= \frac{|a|^2}{2} \\ \\
P_x(e^{j\omega}) &= (|a|^2/2) \tilde{\delta}(\omega)
\end{align*}
\end{frame}
\begin{frame} \frametitle{Example: complex exponential}
\begin{align*}
x[n] &= a\,e^{j\omega_0 n} \\ \\
r_x[k] &= \lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum_{n=-N}^{N} |a|^2 e^{-j\omega_0 (n-k)}e^{j\omega_0 n} \\
&= |a|^2 e^{j\omega_0 k} \lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum_{n=-N}^{N} 1\\
&= |a|^2 e^{j\omega_0 k} \\ \\
P_x(e^{j\omega}) &= |a|^2 \tilde{\delta}(\omega - \omega_0)
\end{align*}
\end{frame}
\begin{frame} \frametitle{So which is it? Energy or Power?}
In practice, it doesn't matter:
\begin{itemize}
\item real-world signals have a finite amount of samples
\item we can only compute an estimate of the autocorrelation
\item estimates are always normalized
\end{itemize}
\end{frame}
\begin{frame} \frametitle{The autocorrelation in practice}
\begin{itemize}
\item assume we know $x[n]$ only for $n=0, 1, \ldots, N-1$
\item the empirical \textit{sample autocorrelation} is defined as:
\[
\hat{r}_x[k] = \frac{1}{N}\sum_{n=0}^{N-1-|k|}x^*[n+|k|]x[n], \qquad -N < k < N
\]
\vspace{1em}
\item number of terms in the sum for $\hat{r}_x[k]$ is $N-|k|$
\item as $|k|$ grows towards $N$ the estimates become less accurate
\item rule of thumb: $N > 4k_{\max}$, with $k_{\max}$ the maximum needed lag
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Autocorrelation of a filtered signal (I)}
\centering
\begin{align*}
y[n] &= (x \ast h)[n] \\ \\
r_y[k] &= \sum_{n = -\infty}^{\infty} y[n]y^*[n - k] \\
&= \sum_{n = -\infty}^{\infty} \left[ \sum_{m = -\infty}^{\infty} h[m] x[n - m] \, \sum_{i = -\infty}^{\infty} h^*[i] x^*[(n-k) - i] \right]\\
&= \sum_{m = -\infty}^{\infty} \sum_{i = -\infty}^{\infty} h[m] h^*[i] \sum_{n = -\infty}^{\infty} x[n - m]x^*[n - (k+i)] \qquad (\text{lag} = k+i-m) \\
&= \sum_{m = -\infty}^{\infty} \sum_{i = -\infty}^{\infty} h[m] h^*[i] \sum_{n = -\infty}^{\infty} r_x[k+i-m]
\end{align*}
\end{frame}
\begin{frame} \frametitle{Autocorrelation of a filtered signal (II)}
\centering
\begin{align*}
r_y[k] &= \sum_{m = -\infty}^{\infty} \sum_{i = -\infty}^{\infty} h[m] h^*[i] r_x[k - (m-i)] \\
&= \sum_{m = -\infty}^{\infty} \sum_{d = -\infty}^{\infty} h[m] h^*[m-d] r_x[k - d] \qquad \leftarrow d=m-i\\
&= \sum_{d = -\infty}^{\infty} r_x[k - d] \sum_{m = -\infty}^{\infty} h[m] h^*[m-d] \\
&= \sum_{d = -\infty}^{\infty} r_x[k - d] r_h[d] \\
&= (r_x \ast r_h)[k]
\end{align*}
\end{frame}
\begin{frame} \frametitle{Autocorrelation of a filtered signal (III)}
\begin{itemize}
\item if $y[n] = (x \ast h)[n]$ then $r_y[k] = (r_x \ast r_h)[k]$
\item autocorrelation of the output is autocorrelation of the input filtered by the autocorrelation of the filter's impulse response:
\[
r_h[k] = \langle h[n], h[n-k] \rangle = (h \ast h^H)[k]
\]
\item same results hold for power signals by normalizing and taking limits
\end{itemize}
\end{frame}
\begin{frame} \frametitle{In the frequency domain}
\centering
\begin{align*}
y[n] &= (x \ast h)[n] \\ \\
P_y(e^{j\omega}) &= \DTFT{(h \ast h^H \ast r_x)[k]} \\
&= H(e^{j\omega}) \cdot H^*(e^{j\omega}) \cdot P_x(e^{j\omega}) \\
&= |H(e^{j\omega})|^2 P_x(e^{j\omega})
\end{align*}
\vspace{1em}
\begin{itemize}
\item filters act on a signal's PSD ``as intended'' (lowpass, highpass, etc)
\item again, phase information is discarded since $|H(e^{j\omega})|^2$ is real-valued
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Example revisited: same magnitude, different phase}
\centering
\def\a{0.8 }
$x[n] = a^n\,u[n], \quad y[n] = (h\ast x)[n], \quad h[n] \text{ allpass}$
\begin{dspPlot}[height=2cm,xout=true]{-30,30}{-.5, 1.3}
\moocStyle
\dspSignal[xmax=-1]{0}
\dspTapsAt{0}{0.81 0.648 0.5184 0.41472 0.331776 0.0944208
0.07553664 -0.11057069 -0.08845655 -0.07076524 -0.21051219 -0.16840975
-0.0986278 -0.07890224 -0.21702179 -0.31212744 -0.24970195 -0.16727156
-0.13381725 -0.0745638 -0.18431004 -0.28595803 -0.19952542 -0.15962034
-0.09845527 -0.19095732 -0.12352485 -0.07250298 -0.18266139 -0.11981221}
\end{dspPlot}
$P_y(e^{j\omega}) = |1 - ae^{j\omega}|^{-2}$
\begin{dspPlot}[xtype=freq,height=2cm,yticks=4]{-1,1}{0, 6}
\moocStyle
\dspFunc{x 180 mul dup sin \a mul dup mul exch cos \a mul 1 exch sub dup mul add sqrt 1 exch div}
\end{dspPlot}
\end{frame}
\begin{frame} \frametitle{Intuition}
\begin{itemize}
\item signals same energy distribution in frequency can look very different in time
\item spectral energy distribution is a more robust characterization of a signal
\item autocorrelation captures this robust feature
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Cross-correlation}
\centering
\begin{itemize}
\item autocorrelation measures self-similarity over time shifts
\item cross-correlation measures similarity between two different signals over relative shifts
\[
r_{xy}[k] = \langle y[n-k], x[n] \rangle = \sum_{n=-\infty}^{\infty} x[n]y^*[n-k]
\]
\item orthogonality: when two signals have nothing in common, $r_{xy}[k] = 0$
\item for power signals, just normalize and take the limit
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Delay estimation via cross-correlation}
\begin{itemize}
\item assume $y[n] = x[n - D]$, with $x[n]$ known and $D$ unknown
\item we want to find $D$
\item the cross-correlation is
\[
r_{xy}[k] = \langle y[n-k], x[n] \rangle = \langle x[n-D-k], x[n] \rangle = r_x[k+D]
\]
\item we know $|r_x[0] \ge |r_x[m]|$ for all $m \neq 0$
\item we can find $D$ by looking at the peak of $r_{xy}[k]$
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Detection in noise via cross-correlation}
\centering
$x[n] = a^n\,u[n], \quad y[n] = x[n-D] + \eta[n], \qquad D=100, \quad \eta[n] = \text{random noise}$
\includegraphics[height=3cm]{buried.eps}
$r_y[k]$
\includegraphics[height=3cm]{ccdet.eps}
\end{frame}
\end{document}
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