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Entry  Wed Jan 30 15:07:32 2019, awade, anchal, HowTo, FSS, Shot noise in FSS loops? 
    Reply  Fri Feb 1 14:59:01 2019, awade, anchal, HowTo, FSS, Noise tallies CTN_FSS_south_blockdiagram.pdf
       Reply  Fri Feb 1 20:20:11 2019, awade, HowTo, FSS, Modifying North FSS EOM path 
          Reply  Tue Feb 5 11:40:02 2019, awade, HowTo, FSS, Modifying South FSS boost 
             Reply  Tue Feb 5 20:39:52 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis FSS_Modified_Analysis.pdfFSS_Mod_Analysis.zip
                Reply  Wed Feb 6 19:13:06 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis FSS_Modified_Analysis.pdf
                   Reply  Wed Feb 6 21:30:24 2019, awade, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis EOM_Vrms_FunctionOfLowerFrequencyBoundOfTF.pdfTTFSS_lisomodel_FSS_Mod_Analysis.ipynb.zip
                      Reply  Thu Feb 7 10:33:27 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis 
Message ID: 2307     Entry time: Thu Feb 7 10:33:27 2019     In reply to: 2306
Author: anchal 
Type: Summary 
Category: FSS 
Subject: Complete model of TTFSS box with TF and crossover analysis 

where ms is EOM phase slope (15 mrad/V) and V_EOM is the applied actuation voltage. As you wrote, that leads to

\delta \tilde f(f) = \frac{m_s}{2\pi} (-i 2 \pi f) \tilde V_{EOM}(f) = -i m_s f\tilde V_\textrm{EOM}(f) \quad [\textrm{Hz}/\textrm{V}]


This actually has units of Hz/Hz. Note, that \delta \tilde{f}(f) is just Fourier transform of frequency actuation, so it is unitless. I used this to get the transfer function of EOM actuation, from applied signal in V to resulting actuation in Hz, which is the prefactor of -\iota m_s f above having units of Hz/V.


If we assume that the EOM is taking 100% of the load for canceling laser frequency noise at a given frequency then it follows that the applied voltage to exactly cancel laser frequency noise is 

\delta \tilde V_\textrm{EOM}(f) = \frac{S^\textrm{Laser}_f (f)}{\delta \tilde f(f)} = i \frac{10^4}{m_s} \frac{1}{f^2}\quad [\textrm{V}/\sqrt{\textrm{Hz}}]

So, if EOM is taking the whole load of frequency noise, actuation signal ASD would be:

\delta \tilde{V_{EOM}}(f) = \frac{TF_{EOMpath}}{-\iota m_s f} S_f^{laser}(f) \quad \left[ V/\sqrt{Hz} \right]

But obviously, this is wrong because this just assume that we are not feeding back the actuation signal at all. So instead, I assumed that if everything is 'ideally' working and we actually have the frequency noise suppressed by a teamwork of PZT and EOM, the incoming noise signal to EOM path's transfer function is:

\frac{1}{1+TF_{PZTpath}(f) + TF_{EOMpath}(f)} S_f^{laser}\quad \left[ Hz/\sqrt{Hz}\right]

So, the actuation signal generated for EOM by EOM path's transfer function in an ideally working loop is:

\delta\tilde{V_{EOM}}(f )=\frac{TF_{EOMpath}(f)}{-\iota m_sf}\frac{1}{1-TF_{PZTpath}(f) - TF_{EOMpath}(f)} S_f^{laser}\quad \left[ V/\sqrt{Hz}\right]

The error in the last plot in PSL:2305. is that I forgot to divide by -\iota m_sf since transfer functions in the code are from Hz to Hz. So I think this is the real error.


One thing to note is that the above estimates are an upper bound assuming that the EOM is taking all of the load down to that frequency point and that the PZT path isn't fighting or out of phase with EOM.  To correctly compute the load on the EOM you are going to have to break down the EOM only portion of the loop from the laser frequency to the point of voltage injected into the EOM.  This can be done by effectively nesting the PZT loop into the round trip gain in a way similar to that described in Josh Smith's Thesis section 2.6.2.  Finding the actuation signal should be similar to finding the PLL actuation signal, at this point in the loop it is the G/(1-G)/A copy of the sensor noise.  In the high gain regime the applied EOM control signal should just be the laser frequency  divided by the EOM frequency slope. Of course you can compute for G_EOM OLG to get a true value with a bunch of algebra.

But on reading section 2.6.2 of Josh Smith's Thesis (which btw has a typo in Eq. 2.33 and 2.34), I did the thing of nesting the PZT path with the plant. So as per Eq. 2.31:

TF'_{PZTpath,roundtrip} = \frac{1}{1 - TF_{PZTpath}(f)}

is the transfer fucntion of Plant for the simplified loop with just EOM as the actuator. Then the actuation signal ASD would be (note S_f^{laser}(f) is free running laser frequency noise ASD):

\delta \tilde{V_{EOM}}(f) = \frac{TF_{EOMpth}(f)}{-\iota m_s f}\frac{TF'_{PZTpath,rountrip}(f)}{1-TF'_{PZTpath,rountrip}(f)TF_{EOMpath}(f)}S_f^{laser}(f)

which turns out to:

\delta \tilde{V_{EOM}}(f) = \frac{TF_{EOMpth}(f)}{-\iota m_s f}\frac{1}{1-TF_{PZTpath}(f)-TF_{EOMpath}(f)}S_f^{laser}(f)

Which is exactly the same as above. But still my model has this error. I'll fix it and post it soon. For readers in the future, the last set of equations are the only correct equations to the best of my knowledge.

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