93 lines
3.6 KiB
TeX
93 lines
3.6 KiB
TeX
\chapter{Spectral analysis}
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\section{The Fourier Transform}
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Complex numbers... magnitude and phase
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Time domain --- Frequency domain, Fourier Space
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\subsection{Fast Fourier transform}
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\section{Power spectrum}
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\[ S_{x,x} = |X(f)|^2 \]
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Parceval theorem:
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\[ \int_{-\infty}^{+\infty} x(t)^2 dt = \int_{-\infty}^{+\infty} |X(f)|^2 df \]
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Autocorrelation:
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Wiener-Kinchin theorem:
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\[ {\cal F}\{Corr(x,x)\} = |X(f)|^2 \]
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\section{Spectrogram}
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\section{Cross spectrum}
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\[ S_{x,y} = X(f)Y^*(f) \]
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is complex valued (magnitude and phase)!
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Correlation theorem:
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\[ {\cal F}\{Corr(x,y)\} = X(f)Y^*(f) = S_{x,y} \]
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\section{Transfer function}
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The complex valued transfer function of a linear, noiseless system
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relating stimulus $s(t)$ and response $r(t)$ is
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\begin{equation}
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\label{transfer}
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H(\omega) = \frac{R(\omega)}{S(\omega)}
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\end{equation}
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where $S(\omega)$ and $R(\omega)$ are the Fourier transformed stimulus
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and response, respectively. By means of the transfer function, the
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response of the system to a stimulus can be predicted according to
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\begin{equation}
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R(\omega) = H(\omega) S(\omega)
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\end{equation}
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Now, if the system is noisy, then the transfer function can only
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predict the mean response $\langle R \rangle_n$, averaged over the
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noise, i.e. averaged over responses evoked by several presentations
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of the same, frozen stimulus:
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\begin{equation}
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\label{transfernoise}
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\langle R(\omega) \rangle_n = H(\omega) S(\omega)
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\end{equation}
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Both sides of this equation can be multiplied by the complex conjugate
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stimulus $S^*(\omega)$. Since the stimulus is always the same,
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$S^*(\omega)$ can be pulled into the average over the noise and we get
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\begin{equation}
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\langle R(\omega)S^*(\omega) \rangle_n = H(\omega) S(\omega)S^*(\omega)
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\end{equation}
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The right hand side can also be averaged over the noise, but it makes
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no difference, because neither $S(\omega)$ nore $H(\omega)$ depend on
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the noise. In addition, we can average both sides over different
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realizations of the stimulus. We denote this average by $\langle \cdot
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\rangle_s$. Because the transfer function does note depend on the
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stimulus it can be pulled out of the stimulus average and we get
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\begin{equation}
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\langle\langle R(\omega)S^*(\omega) \rangle_n\rangle_s = H(\omega) \langle \langle S(\omega)S^*(\omega) \rangle_n \rangle_s
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\end{equation}
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Finally, let's solve for the transfer function and denote both
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averages by $\langle \cdot \rangle$:
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\begin{equation}
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\label{transfercsd}
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H(\omega) = \frac{\langle R(\omega)S^*(\omega) \rangle}{\langle S(\omega)S^*(\omega) \rangle}
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\end{equation}
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The transfer function of a noisy system is estimated by dividing the
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cross spectrum by the power spectrum of the stimulus.
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If we are interested in the gain of the transfer function, i.e. its magnitude, we get starting from Eq.~\eqref{transfernoise}
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\begin{eqnarray}
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|\langle R(\omega) \rangle_n| & = & |H(\omega)| |S(\omega)| \\
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\langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* & = & |H(\omega)|^2 S(\omega)S^*(\omega)\\
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\langle \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* \rangle_s & = & |H(\omega)|^2 \langle S(\omega)S^*(\omega) \rangle_s \\
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|H(\omega)|^2 & = & \frac{\langle \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* \rangle_s}{\langle S(\omega)S^*(\omega) \rangle_s} \\
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|H(\omega)|^2 & \ne & \frac{\langle \langle R(\omega) R^*(\omega) \rangle_n \rangle_s}{\langle S(\omega)S^*(\omega) \rangle_s}
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\end{eqnarray}
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For noisy systems, dividing the power spectrum of the response by the
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power spectrum of the stimulus is not resulting in the squared gain of
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the transfer function. Only for noise-free systems does this work.
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\section{Coherence function}
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\subsection{Forward and reverse filter}
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