[spectral] updated transfer function for noisy systems
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@ -46,6 +46,7 @@ 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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@ -74,26 +75,17 @@ averages by $\langle \cdot \rangle$:
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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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Computing the squared gain like this
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\begin{equation}
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|H(\omega)|^2 = \frac{R(\omega)R^*(\omega)}{S(\omega)S^*(\omega)}
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\end{equation}
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is not possible, it again requires to average over the noise
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\begin{equation}
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|H(\omega)|^2 = \frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)}
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\end{equation}
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Subsequent averaging over stimuli leads to
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\begin{equation}
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|H(\omega)|^2 = \left\langle\frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} \right\rangle_s
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\end{equation}
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which is \emph{not} just the power spectrum $\langle R R^* \rangle$ of
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the response devided by the power spectrum $\langle S S^* \rangle$ of
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the stimulus
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\begin{equation}
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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{equation}
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The gain can not be computed by simply dividing the response spectrum
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by the stimulus spectrum.
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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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