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\begin{document}


\title{Correlation function and time from power spectra}
\author{H-J. W\"unsche}
%\affiliation{% Humboldt-Universit\"at Berlin, Institut f\"ur Physik\\ Newtonstr.15  12489 Berlin, Germany \\ }
%
\date{\today}

\begin{abstract}
After having read Born-Wolff chapter 10.2, I came to conclusions very similar to Oleg's brief introduction. Here a summary.
\end{abstract}


\maketitle


\section{Preliminaries}
%----------------------

Be $P(t)$ a transient of optical intensities detected by the photodiode at the ESA.

Be $S(\nu)$ the power spectrum (spectral density) of $P(t)$, defined as \footnote{I follow chapter 10.2 of Born-Wolff.}
%
\begin{eqnarray}
S(\nu) \stackrel{def}{=} \lim_{T\rightarrow \infty} \overline{\frac{|F_T(\nu)|^2}{2T}} 
       \stackrel{(WK)}{=} \int_{-\infty}^\infty d\tau G(\tau )e^{i\omega \tau },
							~~~~ G(\tau )=\braket{P(t)P(t+\tau )} ,
							~~~~ \omega =2\pi \nu.
\end{eqnarray}
%
$F_T(\nu)$: Fourier transform of $P(t)$ truncated outside the interval $-T<t\le T$, 
(WK): Wiener-Khinchin-Theorem [formula (28) in Born-Wolff], 
$\overline{|\cdots|^2}$: ensemble average \footnote{this corresponds to averaging over many repetitions of the measurement, which can be done with the ESA. Thus, $S(\nu)$ corresponds to the limit of infinitely many repetitions. }, 
$\braket{f(t)}$: temporal average over $t$ \footnote{i.e. $\braket{f(t)}=\lim_{T\rightarrow\infty}\int_{-T}^{T}f(t+t')$dt'/2T}. 

Be $P(\nu_m)$ the spectrum measured with the ESA on an equidistant grid $\nu_m$ of frequencies. 

\underline{Assumption 1:} \hfill \mb{0.8}{
$P(\nu_m)$ is a reasonable approximation of the powerspectrum  of $P(t)$.
}

Nothing else makes sense.

\underline{Conclusion:}\hfill \mb{0.8}{
$P(\nu_m)$ must be quadratic in the power transient $P(t)$.
}

This conclusion is easily checked: look to a spectrum with at least one peak well above noise background. Attanuate the input signal by x dB. Then measured spectrum should go down by 2x dB. I'm sure that one of the usable scales (hopefully dBm) fulfills this requirement \footnote{I guess, the optical power $P$ is linearly transferred to a photo current I, which in turn is linearly related to the voltage $U$ at some internal resistivity (50$\Omega$?) analysed by the electronics of the ESA. The output with the choicable unit ''dBm'' is probably related to the electrical power consumed by the resistivity, i.e. $\sim U^2 \sim I^2 \sim P^2$, quadratic with the detected optical intensity. Let's look to the manual.  }. 

\underline{Assumption 2:} \hfill \mb{0.8}{
$P(\nu_m)$ represents the contribution of $S(\nu)$ to an grid interval $\nu_m-\nu_{m-1}$ around $\nu_m$, normalised to a certain resolution bandwith $\Delta \nu_\text{res}$ .
}

The corresponding formula is
%
\begin{eqnarray}
P(\nu_m) =
\frac{\Delta \nu_\text{res}}{\Delta \nu_\text{grid}} 
\int_{-\Delta \nu_\text{grid}/2}^{\Delta \nu_\text{grid}/2}
S(\nu_m+\nu')d\nu'.
~~~~~~ \Delta \nu_\text{grid} = \nu_m - \nu_{m-1}.
\end{eqnarray}
%
It seems to me the only reasonable choice by the following reasons: 
\begin{itemize}
\item
Sharp lines in between the frequency grid points should not be lost. Thus, I must integrate over all intervals between. 
\item
The magnitude of the spectrum should not depend on the choosen grid width. Thus, I must average over the grid intervals.
\item
The used unit is ''mW'', thus I must multiply the average with some standard frequency interval. I guess, it is the resolution band width, which can be choosen independently of the grid step.
\end{itemize}

\section{correlation function}
%-----------------------------

The (nonnormalised) correlation function is 
%
\begin{eqnarray}
C(\tau ) \stackrel{def}{=} G(\tau ) - \braket{P(t)}^2 
= \int_{-\infty}^{\infty}S(\nu)e^{-i\omega \tau }d\nu - \braket{P(t)}^2 .
\end{eqnarray}
%
Thus, it can be calculated from the power spectrum. Since the power spectrum is symmetric in $\nu$, this is a real quantity. Only how to calculate the subtracted second term from the power spectra, I don't know at present. However, except in ideal coherent cases it is just $G(\tau \rightarrow \infty)$, i.e. it can be taken from the large time limit of $G$. 

\end{document}


