[spectral] updated transfer function for noisy systems

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Jan Benda 2021-03-12 22:58:55 +01:00
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@ -46,6 +46,7 @@ predict the mean response $\langle R \rangle_n$, averaged over the
noise, i.e. averaged over responses evoked by several presentations noise, i.e. averaged over responses evoked by several presentations
of the same, frozen stimulus: of the same, frozen stimulus:
\begin{equation} \begin{equation}
\label{transfernoise}
\langle R(\omega) \rangle_n = H(\omega) S(\omega) \langle R(\omega) \rangle_n = H(\omega) S(\omega)
\end{equation} \end{equation}
@ -74,26 +75,17 @@ averages by $\langle \cdot \rangle$:
The transfer function of a noisy system is estimated by dividing the The transfer function of a noisy system is estimated by dividing the
cross spectrum by the power spectrum of the stimulus. cross spectrum by the power spectrum of the stimulus.
Computing the squared gain like this If we are interested in the gain of the transfer function, i.e. its magnitude, we get starting from Eq.~\eqref{transfernoise}
\begin{equation} \begin{eqnarray}
|H(\omega)|^2 = \frac{R(\omega)R^*(\omega)}{S(\omega)S^*(\omega)} |\langle R(\omega) \rangle_n| & = & |H(\omega)| |S(\omega)| \\
\end{equation} \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* & = & |H(\omega)|^2 S(\omega)S^*(\omega)\\
is not possible, it again requires to average over the noise \langle \langle R(\omega) \rangle_n \langle R(\omega) \rangle_n^* \rangle_s & = & |H(\omega)|^2 \langle S(\omega)S^*(\omega) \rangle_s \\
\begin{equation} |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 = \frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} |H(\omega)|^2 & \ne & \frac{\langle \langle R(\omega) R^*(\omega) \rangle_n \rangle_s}{\langle S(\omega)S^*(\omega) \rangle_s}
\end{equation} \end{eqnarray}
Subsequent averaging over stimuli leads to For noisy systems, dividing the power spectrum of the response by the
\begin{equation} power spectrum of the stimulus is not resulting in the squared gain of
|H(\omega)|^2 = \left\langle\frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} \right\rangle_s the transfer function. Only for noise-free systems does this work.
\end{equation}
which is \emph{not} just the power spectrum $\langle R R^* \rangle$ of
the response devided by the power spectrum $\langle S S^* \rangle$ of
the stimulus
\begin{equation}
|H(\omega)|^2 \ne \frac{\langle\langle R(\omega)R^*(\omega) \rangle_n\rangle_s}{\langle S(\omega)S^*(\omega)\rangle_s}
\end{equation}
The gain can not be computed by simply dividing the response spectrum
by the stimulus spectrum.
\section{Coherence function} \section{Coherence function}