From de7027f4060b5047a69c84a5de58496f2a21793d Mon Sep 17 00:00:00 2001 From: Jan Benda Date: Fri, 12 Mar 2021 19:06:37 +0100 Subject: [PATCH 1/2] [spectral] note on how to compute the transfer function --- spectral/lecture/spectral.tex | 66 +++++++++++++++++++++++++++++++++++ 1 file changed, 66 insertions(+) diff --git a/spectral/lecture/spectral.tex b/spectral/lecture/spectral.tex index 32af20e..e00d405 100644 --- a/spectral/lecture/spectral.tex +++ b/spectral/lecture/spectral.tex @@ -28,6 +28,72 @@ 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} + \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. + +Computing the squared gain like this +\begin{equation} + |H(\omega)|^2 = \frac{R(\omega)R^*(\omega)}{S(\omega)S^*(\omega)} +\end{equation} +is not possible, it again requires to average over the noise +\begin{equation} + |H(\omega)|^2 = \frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} +\end{equation} +Subsequent averaging over stimuli leads to +\begin{equation} + |H(\omega)|^2 = \left\langle\frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} \right\rangle_s +\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} From 8b94ac1a7bab2e7fc7b1379aeb429efe9951e299 Mon Sep 17 00:00:00 2001 From: Jan Benda Date: Fri, 12 Mar 2021 22:58:55 +0100 Subject: [PATCH 2/2] [spectral] updated transfer function for noisy systems --- spectral/lecture/spectral.tex | 32 ++++++++++++-------------------- 1 file changed, 12 insertions(+), 20 deletions(-) diff --git a/spectral/lecture/spectral.tex b/spectral/lecture/spectral.tex index e00d405..eb745a5 100644 --- a/spectral/lecture/spectral.tex +++ b/spectral/lecture/spectral.tex @@ -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 of the same, frozen stimulus: \begin{equation} + \label{transfernoise} \langle R(\omega) \rangle_n = H(\omega) S(\omega) \end{equation} @@ -74,26 +75,17 @@ averages by $\langle \cdot \rangle$: The transfer function of a noisy system is estimated by dividing the cross spectrum by the power spectrum of the stimulus. -Computing the squared gain like this -\begin{equation} - |H(\omega)|^2 = \frac{R(\omega)R^*(\omega)}{S(\omega)S^*(\omega)} -\end{equation} -is not possible, it again requires to average over the noise -\begin{equation} - |H(\omega)|^2 = \frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} -\end{equation} -Subsequent averaging over stimuli leads to -\begin{equation} - |H(\omega)|^2 = \left\langle\frac{\langle R(\omega)R^*(\omega) \rangle_n}{S(\omega)S^*(\omega)} \right\rangle_s -\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. +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}