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ACs processing. TAO-cruise August 2006. total signal. 0.2 m m interpolated values. The 0.2 m m signal is removed from the total signal using the 0.2 m m interpolated values. 1. Total vs 0.2 m m signal. 0.2 m m signal. Present also in 0.2 m m measurements.
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ACs processing TAO-cruise August 2006
total signal 0.2 mm interpolated values The 0.2 mm signal is removed from the total signal using the 0.2 mm interpolated values 1. Total vs 0.2mm signal 0.2 mm signal
Present also in 0.2 mm measurements 2. Spectral discontinuity correction
Q: Which is the correct one, LPS or SPS? Longer Portion of the Spectrum A: For ap we don’t care because we know that ap(NIR)≈0. Shorter Portion of the Spectrum 2. Spectral discontinuity correction
2. Spectral discontinuity correction • Correction method • assume that the LPS is the correct one Longer Portion of the Spectrum
2. Spectral discontinuity correction • Correction method • assume that the LPS is the correct one • using the first 2 ls of the LPS, predict the value of ap at the last l of the SPS
2. Spectral discontinuity correction • Correction method • assume that the LPS is the correct one • using the first 2 ls of the LPS, predict the value of ap at the last l of the SPS • compute the difference between the predicted and observed values of ap at the last l of the SPS • subtract such difference from the SPS
2. Spectral discontinuity correction • Case of cp • There is no l at which we a-priori know the value of cp • Arbitrarily assume that LSP is correct • Apply correction as for ap • Evaluate the bias introduced by such arbitrary assumption
2. Spectral discontinuity correction • Bias evaluation (cp) • NOTE: For this data set, cp(750) (minimal values) ranges from 0.02 to 0.08 m-1 • For most of the cp data the discontinuity is not important • When it is important, our arbitrary assumption introduces biases of the order of 0.0007/0.02=4% (conservative estimate) • There is no preferential direction for the shift
2. Spectral discontinuity correction • Importance of this correction for ap • NOTE: For this data set, ap(676) ranges from 0.001 to 0.005 m-1 • For most of the ap data the discontinuity is not important • When it is important, the discontinuity can be of the order of 0.0003/0.005=6% to 0.0003/0.001=30% of ap(676) • There is no preferential direction for the shift
3. Interpolation to common l • Because the ACs produces ap and cp spectra with different wavelength centers, we interpolated each spectrum to have equally spaced values every 2 nm from 400 to 750 nm
4. Correction for residual T-dependence • In theory by subtracting the 0.2 mm interpolated signal from the total signal, we should not need any TS-correction of our ap spectra • However, because the ap that we are measuring have really low values in the red region (0.001-0.005 m-1), even small DTs between two consecutive 0.2 mm filtrations could introduce relatively large biases in ap
from Sullivan et al. 2006 4. Correction for residual T-dependence • What ranges of DT and DS are we observing between 1-hr distant data points?
4. Correction for residual T-dependence • What ranges of DT and DS are we observing between 1-hr distant data points? ignore DS
note that part of the b-corr depends on DT 4. Correction for residual T-dependence • b(l) is independent of DT (and DS) 2. look at Zaneveld’s scattering correction #3: scattering correction
4. Correction for residual T-dependence How to get DT? • Assume that there exists a spectral region (lref) where: • Set lref=710:740 nm and NIR=730 nm • For each spectrum, find DT that minimizes: where: d) Apply T-correction to ap using the derived DT
4. Correction for residual T-dependence no correction
4. Correction for residual T-dependence b-correction
4. Correction for residual T-dependence Tb-correction
5. Compute cpT • The temperature-corrected cp is finally calculated as:
7. CD content • processed ACs data merged with PAR, lat, long, SST and salinity data • Sullivan et al. 2006 table interpolated every 2 nm (instrument specific)
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