\chapter{Spectral analysis}

\section{The Fourier Transform}
Complex numbers... magnitude and phase

Time domain --- Frequency domain, Fourier Space

\subsection{Fast Fourier transform}

\section{Power spectrum}
\[ S_{x,x} = |X(f)|^2 \]

Parceval theorem:
\[ \int_{-\infty}^{+\infty} x(t)^2 dt = \int_{-\infty}^{+\infty} |X(f)|^2 df \]

Autocorrelation:

Wiener-Kinchin theorem:
\[ {\cal F}\{Corr(x,x)\} = |X(f)|^2 \]

\section{Spectrogram}

\section{Cross spectrum}
\[ S_{x,y} = X(f)Y^*(f) \]
is complex valued (magnitude and phase)!

Correlation theorem:
\[ {\cal F}\{Corr(x,y)\} = X(f)Y^*(f) = S_{x,y} \]

\section{Transfer function}
The complex valued transfer function of a linear, noiseless system
relating stimulus $s(t)$ and response $r(t)$ is
\begin{equation}
  \label{transfer}
  H(\omega) = \frac{R(\omega)}{S(\omega)}
\end{equation}
where $S(\omega)$ and $R(\omega)$ are the Fourier transformed stimulus
and response, respectively. By means of the transfer function, the
response of the system to a stimulus can be predicted according to
\begin{equation}
  R(\omega) = H(\omega) S(\omega)
\end{equation}

Now, if the system is noisy, then the transfer function can only
predict the mean response $\langle R \rangle_n$, averaged over the
noise, i.e. averaged over responses evoked by several presentations
of the same, frozen stimulus:
\begin{equation}
  \label{transfernoise}
  \langle R(\omega) \rangle_n = H(\omega) S(\omega)
\end{equation}

Both sides of this equation can be multiplied by the complex conjugate
stimulus $S^*(\omega)$. Since the stimulus is always the same,
$S^*(\omega)$ can be pulled into the average over the noise and we get
\begin{equation}
  \langle R(\omega)S^*(\omega) \rangle_n = H(\omega) S(\omega)S^*(\omega)
\end{equation}
The right hand side can also be averaged over the noise, but it makes
no difference, because neither $S(\omega)$ nore $H(\omega)$ depend on
the noise. In addition, we can average both sides over different
realizations of the stimulus. We denote this average by $\langle \cdot
\rangle_s$. Because the transfer function does note depend on the
stimulus it can be pulled out of the stimulus average and we get
\begin{equation}
  \langle\langle R(\omega)S^*(\omega) \rangle_n\rangle_s = H(\omega) \langle \langle S(\omega)S^*(\omega) \rangle_n \rangle_s
\end{equation}

Finally, let's solve for the transfer function and denote both
averages by $\langle \cdot \rangle$:
\begin{equation}
  \label{transfercsd}
  H(\omega)  = \frac{\langle R(\omega)S^*(\omega) \rangle}{\langle S(\omega)S^*(\omega) \rangle}
\end{equation}
The transfer function of a noisy system is estimated by dividing the
cross spectrum by the power spectrum of the stimulus.

If we are interested in the gain of the transfer function, i.e. its magnitude, we get starting from Eq.~\eqref{transfernoise}
\begin{eqnarray}
  |\langle R(\omega) \rangle_n| & = & |H(\omega)| |S(\omega)| \\
  \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* & = & |H(\omega)|^2 S(\omega)S^*(\omega)\\
  \langle \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* \rangle_s & = & |H(\omega)|^2 \langle S(\omega)S^*(\omega) \rangle_s \\
  |H(\omega)|^2 & = & \frac{\langle \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* \rangle_s}{\langle S(\omega)S^*(\omega) \rangle_s} \\
  |H(\omega)|^2 & \ne & \frac{\langle \langle R(\omega) R^*(\omega) \rangle_n \rangle_s}{\langle S(\omega)S^*(\omega) \rangle_s}
\end{eqnarray}
For noisy systems, dividing the power spectrum of the response by the
power spectrum of the stimulus is not resulting in the squared gain of
the transfer function. Only for noise-free systems does this work.

\section{Coherence function}

\subsection{Forward and reverse filter}