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R&D (FilterCavity)
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YuhangZhao - 19:19, Tuesday 08 May 2018 (760)Get code to link to this report
Comparison of old and new servo

The result of previous comparison is wired. So we decide to change back the old servo and measured error signal again. At the same time, we also did calibration again.

Note here, the calibration method for green likes entry 750. The only difference is that we consider frequency much higher than unity gain frequency. Then we got result like attached picture 1 and 2.

The calibration for infrared got from attached picture 3 and 4. For the problem of saturation, we changed the demodulation phase.

Finally we got result as attached picture 5. We can make them overlap by multiplying a factor of 11 like picture 6.

As a result, we find the new servo make noise level lower than before. So now I put back the new servo, we can get the green transmission as 1.5V and very stable like picture 6.

Images attached to this report
760_20180508121118_greenca.png 760_20180508121133_greencafinal.png 760_20180508121341_calibrationinfrared.png 760_20180508121357_infraredca.png 760_20180508121525_comparison.png 760_20180508121538_overlap.png 760_20180508121949_721319689.jpg
Comments related to this report
YuhangZhao - 11:17, Wednesday 09 May 2018 (763)

Here I attach the rms integration of the four error signal curves.

EleonoraCapocasa - 08:04, Monday 14 May 2018 (773)

The error signals for green and infrared account for the closed loop laser frequency noise filtered by the pole of the cavity  (which is different for green and IR).

In the fist attached plot, the error spectra has been divided for the corrispondig pole in order to go back to the close loop laser frequency noise.

freq nois =  err sig * ( sqrt( 1+(f/f0)^2)     with f0 = 55 Hz for IR and 1.45 kHz for green

The two curves obtained shoud be coincindents. The discrepancy (about a factor 2.5) suggests that there is maybe an issue with the calibration.

YuhangZhao - 21:45, Wednesday 16 May 2018 (780)

Corresponding to the comment of Eleonora, the bandwidth of filter cavity for infrared is 114Hz but not 55Hz. Then I think we can explain the result (almost).

bandwidth=FSR/Finesse=500000/4355=114

Matteo Barsuglia - 05:58, Friday 18 May 2018 (782)

I think there is a factor 2 missing in the formula: the pole of the cavity is FSR/(2*F) = 500000/(2*4355) = 57 Hz