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[spectral] note on how to compute the transfer function

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Jan Benda 2021-03-12 19:06:37 +01:00
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spectral/lecture/spectral.tex
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@ -28,6 +28,72 @@ Correlation theorem:
\[ {\cal F}\{Corr(x,y)\} = X(f)Y^*(f) = S_{x,y} \] \[ {\cal F}\{Corr(x,y)\} = X(f)Y^*(f) = S_{x,y} \]
\section{Transfer function} \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} \section{Coherence function}