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Entry  Wed Jun 9 11:46:01 2021, Anchal, Paco, Summary, AUX, Xend Green Laser PDH OLTF measurement image-6f2923a3-01ce-4d04-bc53-d8db0238e195.jpgimage-72223f4b-3b74-4574-a7ad-de6628a2c5e9.jpgX_Green_ARM_PDH_OLTF.pdf
    Reply  Thu Jun 10 14:01:36 2021, Anchal, Summary, AUX, Xend Green Laser PDH OLTF measurement loop algebra AUX_PDH_LOOP.pdf
       Reply  Mon Jun 14 18:57:49 2021, Anchal, Update, AUX, Xend is unbearably hot. Green laser is loosing lock in 10's of seconds XAUX_PDH_Err_In_ASD.pdfXAUX_PZT_Out_Mon_ASD.pdf
       Reply  Tue Jun 15 15:26:43 2021, Anchal, Paco, Summary, AUX, Xend Green Laser PDH OLTF measurement loop algebra, excitation at control point AuxPDHloop.pdf
          Reply  Fri Jun 18 10:07:23 2021, Anchal, Paco, Summary, AUX, Xend Green Laser PDH OLTF with coherence XEND_PDH_OLTF_with_Coherence.pdfBeta_Amp.pdf
Message ID: 16202     Entry time: Tue Jun 15 15:26:43 2021     In reply to: 16197     Reply to this: 16213
Author: Anchal, Paco 
Type: Summary 
Category: AUX 
Subject: Xend Green Laser PDH OLTF measurement loop algebra, excitation at control point 

Attachment 1 shows the case when excitation is sent at control point i.e. the PZT output. As before, free running laser noise \eta in units of Hz/rtHz is added after the actuator and I've also shown shot noise being added just before the detector.

Again, we have a access to three output points for measurement. \alpha right at the output of mixer (the PDH error signal), \beta the feedback signal to be applied by uPDH box (PZT Mon) and \gamma the output of the summing box SR560.

Doing loop algebra as before, we get:

\large \alpha = \frac{\eta}{K(s) A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} + \frac{\chi}{C(s) K(s) A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} - \frac{\nu_e}{K(s) } \frac{G_{OL}(s)}{1 - G_{OL}(s)}

\large \beta = \frac{\eta}{A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} + \frac{\chi}{C(s) A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} - \nu_e \frac{G_{OL}(s)}{1 - G_{OL}(s)}

\large \gamma= \frac{\eta}{A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} + \frac{\chi}{C(s) A(s)} \frac{G_{OL}(s)}{1 - G_{OL}(s)} - \nu_e \frac{1}{1 - G_{OL}(s)}

So measurement of \large G_{OL}(s) can be done by

\large G_{OL}(s) \approx \frac{\beta}{\gamma}

  • For frequencies, where \large G_{OL}(s) is large enough, to have an SNR of 100, we need that ratio of \large \nu_e to integrated noise is 100.
  • Assuming you are averaging for 'm' number of cycles in your swept sine measurement, time of integration for the noise signal would be \large \frac{m}{f}where f is the frequency point of the seeping sine wave.
    • This means, the amplitude of integrated laser frequency noise at either \large \beta or \large \gamma would be \large \sqrt{\left(\frac{\eta(f)}{A(f)}\right)^2\frac{f}{m}} = \frac{\eta(f) \sqrt{f}}{A(f)\sqrt{m}}
    • Therefore, signal to laser free running noise ratio at f would be \large S = \frac{\nu_eA(f)\sqrt{m}}{\eta(f) \sqrt{f}}.
    • This means to keep a constant SNR of S, we need to shape the excitation amplitude as \large \nu_e \sim S \frac{\eta(f) \sqrt{f}}{A(f)\sqrt{m}}
    • Putting in numbers for X end Green PDH loop, laser free-running frequency noise ASD is 1e4/f Hz/rtHz, laser PZT actuation is 1MHz/V, then for 10 integration cycles and SNR of 100, we get: \large \nu_e \sim 100 \times \frac{10^4 \sqrt{f}}{f \times10^6 \sqrt{10}} = \frac{30\, mV}{\sqrt{f}}
  • Assuming you are averaging for a constant time \large \tau in swept sine measurement, then the amplitude of integrated laser free noise would be \large \sqrt{\left(\frac{\eta(f)}{A(f)}\right)^2 \frac{1}{\tau}} = \frac{\eta(f) }{A(f)\sqrt{\tau}}
    • In this case, signal to laser free-running noise ratio at f would be \large S = \frac{\nu_eA(f)\sqrt{\tau}}{\eta(f)}
    • This means to keep a constant SNR of S, we need to shape the excitation amplitude as \large \nu_e \sim S\frac{\eta(f)}{A(f)\sqrt{\tau}}
    • Again putting in numbers as above and integration time of 1s, we need an excitation amplitude shape \large \nu_e \sim 100 \times \frac{10^4 }{f \times10^6 \sqrt{1}} = \frac{1\, V}{f}

This means at 100 Hz, with 10 integration cycles, we should have needed only 3 mV of excitation signal to get an SNR of 100. However, we have been unable to get good measurements with even 25 mV of excitation. We tried increasing the cycles, that did not work either.

This post is to summarize this analysis. We need more tests to get any conclusions.

Attachment 1: AuxPDHloop.pdf  655 kB  Uploaded Tue Jun 15 16:29:50 2021  | Hide | Hide all
AuxPDHloop.pdf
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