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Swept-sine calibration information

0

The swept sine calibration files are 3-column ASCII files, of the form:

$f_0$ $\qquad$ $r_0$ $\qquad$ $i_0$
$f_1$ $\qquad$ $r_1$ $\qquad$ $i_1$
$f_2$ $\qquad$ $r_2$ $\qquad$ $i_2$
$\cdots$
$f_m$ $\qquad$ $r_m$ $\qquad$ $i_m$
where the $f_j$ are frequencies, in Hz, and $r_j$ and $i_j$ are dimensionless ratios of voltages. There are typically $m=801$ lines in these files. The data from these files (as well as one additional line of the form
0.0 0.0 0.0
showing vanishing response at DC) have been included in the frames. Each line gives the ratio of the IFO output voltage to a calibration coil driving voltage, at a different frequency. The $r_j$ are the ``real part" of the response, i.e. the ratio of the IFO output in phase with the coil driving voltage, to the coil driving voltage. The $i_j$ are the ``imaginary part" of the response, $90$ degrees out of phase with the coil driving voltage. The sign of the phase (or equivalently, the sign of the imaginary part of the response) is determined by the following convention. Suppose that the driving voltage (in volts) is
\begin{displaymath}
V_{\rm coil} = 10 \cos( \omega t) = 10 \Re {\rm e}^{i \omega t}
\end{displaymath} (4.7.14)

where $\omega= 2 \pi \times 60 \> {\rm radians/sec}$ is the angular frequency of a 60 Hz signal. Suppose the response of the interferometer output to this is (again, in volts)
$\displaystyle V_{\rm IFO}$ $\textstyle =$ $\displaystyle 6.93 \; \cos(\omega t) + 4\; \sin(\omega t)\cr$ (4.7.15)

This is shown in Figure [*]. An electrical engineer would describe this situation by saying that the phase of the response $V_{\rm IFO}$ is lagging the phase of the driving signal $V_{\rm coil}$ by $30^\circ$. The corresponding line in the swept sine calibration file would read:
$\cdots$
$60.000$ $\qquad$ $0.6930$ $\qquad$ $-0.40000$
$\cdots$
Hence, in this example, the real part is positive and the imaginary part is negative. We will denote this entry in the swept sine calibration file by $S(60) = 0.8 \; {\rm e}^{ -i\pi/6} = 0.693 - 0.400
i$. Because the interferometer output is real, there is also a value implied at negative frequencies which is the complex conjugate of the positive frequency value: $S(-60) = S^*(60) = 0.8 \; {\rm e}^{
i\pi/6} = 0.693 + 0.400 i$.

Because the interferometer has no DC response, it is convenient for us to add one additional point at frequency $f=0$ into the output data arrays, with both the real and imaginary parts of the response set to zero. Hence the output arrays contain one element more than the number of lines in the input files. Note that both of these arrays are arranged in order of increasing frequency; after adding our one additional point they typically contain 802 points at frequencies from 0 Hz to 5001 Hz.

For the data runs of interest in this section (from November 1994) typically a swept sine calibration curve was taken immediately before each data tape was generated.

Figure: This shows a driving voltage $V_{\rm coil}$ (solid curve) and the response voltage $V_{\rm IFO}$ (dotted curve) as functions of time (in sec). Both are 60 Hz sinusoids; the relative amplitude and phase of the in-phase and out-of-phase components of $V_{\rm IFO}$ are contained in the swept-sine calibration files.
\begin{figure}\begin{center}
\epsfig{file=Figures/figure8.ps,height=5cm,bbllx=60pt,bblly=250pt,
bburx=550pt,bbury=530pt}\end{center}\end{figure}

We will shortly address the following question. How does one use the dimensionless data in the swept-sine calibration curve to reconstruct the differential motion $\Delta l(t)$ (in meters) of the interferometer arms? Here we address the closely related question: given $V_{\rm IFO}$, how do we reconstruct $V_{\rm coil}$? We choose the sign convention for the Fourier transform which agrees with that of Numerical Recipes: equation (12.1.6) of [1]. The Fourier transform of a function of time $V(t)$ is

\begin{displaymath}
{\tilde V}(f) = \int {\rm e}^{2 \pi i f t} V(t) dt.
\end{displaymath} (4.7.16)

The inverse Fourier transform is
\begin{displaymath}
V(t)= \int {\rm e}^{-2 \pi i f t} {\tilde V}(f) df.
\end{displaymath} (4.7.17)

With these conventions, the signals ([*]) and ([*]) shown in in Figure [*] have Fourier components:
$\displaystyle {\tilde V}_{\rm coil}(60) = 5 \quad$ $\textstyle {\rm and}$ $\displaystyle \quad {\tilde V}_{\rm coil}(-60) = 5,$ (4.7.18)
$\displaystyle {\tilde V}_{\rm IFO}(60) = 4{\rm e}^{i \pi/6} \quad$ $\textstyle {\rm and}$ $\displaystyle \quad {\tilde V}_{\rm IFO}(-60) = 4 {\rm e}^{-i \pi/6}.$ (4.7.19)

At frequency $f_0=60$ Hz the swept sine file contains
\begin{displaymath}
S(60) = 0.8 \; {\rm e}^{-i \pi/6} \Rightarrow S(-60) = S^*(60) =
0.8 \; {\rm e}^{i \pi/6}.
\end{displaymath} (4.7.20)

since $S(-f) = S^*(f)$.

With these choices for our conventions, one can see immediately from our example (and generalize to all frequencies) that

\begin{displaymath}
{\tilde V}_{\rm coil}(f) = {{\tilde V}_{\rm IFO} \over S^*(f)}.
\end{displaymath} (4.7.21)

In other words, with the Numerical Recipes [1] conventions for forward and reverse Fourier Transforms, the (FFT of the) calibration-coil voltage is the (FFT of the) IFO-output voltage divided by the complex conjugate of the swept sine response.
Author: Bruce Allen, ballen@dirac.phys.uwm.edu
Comments: The swept-sine calibration curves are usually quite smooth but sometimes they contain a ``glitch" in the vicinity of 1 kHz; this may be due to drift of the unity-gain servo point.


next up previous contents
Next: Function: GRcalibrate() Up: GRASP Routines: Reading/using FRAME Previous: Example: animateF program   Contents
Bruce Allen 2000-11-19