ACs processing TAO-cruise August 2006. 1. Total vs 0.2 m signal total signal 0.2 m signal The 0.2 ...

ACs processing

TAO-cruise August 2006

1. Total vs 0.2m signal

total signal

0.2 m signal

The 0.2 m signal is removed from the total signal using the 0.2 m interpolated values

0.2 m interpolated values

2. Spectral discontinuity correction

Present also in 0.2 m measurements


Longer Portion of the Spectrum

Shorter Portion of

the Spectrum

Q: Which is the correct one, LPS or SPS?

A: For ap we don’t care because we know that ap(NIR)≈0.


Correction methoda) assume that the LPS is

the correct one

Longer Portion of the Spectrum



the correct oneb) using the first 2 s of

the LPS, predict the value of ap at the last of the SPS



the correct oneb) using the first 2 s of

the LPS, predict the value of ap at the last of the SPS

c) compute the difference between the predicted and observed values of ap at the last of the SPS

d) subtract such difference from the SPS


Case of cp

a) There is no at which we a-priori know the value of cp

b) Arbitrarily assume that LSP is correct

c) Apply correction as for ap

d) Evaluate the bias introduced by such arbitrary assumption


Bias evaluation (cp)

a) NOTE: For this data set, cp(750) (minimal values) ranges from 0.02 to 0.08 m-1

b) For most of the cp data the discontinuity is not important

c) When it is important, our arbitrary assumption introduces biases of the order of 0.0007/0.02=4% (conservative estimate)

d) There is no preferential direction for the shift


Importance of this correction for ap

a) NOTE: For this data set, ap(676) ranges from 0.001 to 0.005 m-1

b) For most of the ap data the discontinuity is not important

c) 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)

d) There is no preferential direction for the shift

3. Interpolation to common • 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 m 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 Ts between two consecutive 0.2 m filtrations could introduce relatively large biases in ap

• What ranges of T and S are we observing between 1-hr distant data points?


from Sullivan et al. 2006


• What ranges of T and S are we observing between 1-hr distant data points?

ignore S


1. b() is independent of T (and S)

)()(

),()(),()()( b

NIRb

TNIRaNIRaTaaa wp

wpTbp

)(Tpa

scattering correction

note that part of the b-corr depends on T)(NIRaTp

2. look at Zaneveld’s scattering correction #3:


ref

)()(

),()(),()( refrefref

bNIRb

TNIRaNIRaTaa wp

wp

How to get T?a) Assume that there exists a spectral region (ref) where:

0)()(

),()(),()( refrefref

b

NIRb

TNIRaNIRaTaa wp

wp

b) Set ref=710:740 nm and NIR=730 nm

c) For each spectrum, find T that minimizes:

TTa Sw )(),( refref where:

d) Apply T-correction to ap using the derived T

)()(

)(b

NIRb

NIRap

)()(

)()( b

NIRb

NIRaa pp

)(Tbpa

)(pa



no correction


b-correction


Tb-correction

5. Compute cpT

• The temperature-corrected cp is finally calculated as:

pTbp

Tp bac

6. Biofouling

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)

Some highlights

-125ºW -140ºW

125º 140º

biofouling

ACs processing TAO-cruise August 2006. 1. Total vs 0.2 m signal total signal 0.2 m signal The 0.2 ...

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