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CarboEurope workshop, Viterbo 2004Markus Reichstein Gap-filling algorithm Assumptions: NEE = NEE(R, T, VPD, t) +  NEE(R, T, VPD, t)  NEE(R+  R, T+ 

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Presentation on theme: "CarboEurope workshop, Viterbo 2004Markus Reichstein Gap-filling algorithm Assumptions: NEE = NEE(R, T, VPD, t) +  NEE(R, T, VPD, t)  NEE(R+  R, T+ "— Presentation transcript:

1 CarboEurope workshop, Viterbo 2004Markus Reichstein Gap-filling algorithm Assumptions: NEE = NEE(R, T, VPD, t) +  NEE(R, T, VPD, t)  NEE(R+  R, T+  VPD+  VPD, t+  t) The smaller  t and the more environmental constraints available the better Lag days Autokorrelation 01020

2 CarboEurope workshop, Viterbo 2004Markus Reichstein Gap-filling algorithm General type of approach same as Falge et al. (2001) Differences: –Dynamic averaging window size (as small as possible  better exploitation of temporal autocorrelation) –„Moving“ look-up table (  value to be filled always in the center of the class) –Combination of MDV and LUT methods

3 CarboEurope workshop, Viterbo 2004Markus Reichstein MDS method +Exploits meteorological drivers as much as available +Localized (exploits autocorrelation) +Fills all gaps +Gives tentative quality index +Yields error estimates for the flux +Easy to understand +Fast -Heuristic -Exhibits discontinuities with large gaps

4 Rg available with |dt| ≤ 7 days |dt| ≤ 1 hour |dt| ≤ 1 day (& same hour of day)Rg, T, VPD available with |dt| ≤ 21, 28,..., 140 daysRg available with |dt| ≤ 14, 21,..., 140 days |dt| ≤ 7, 14,... days Rg, T, VPD available with |dt| ≤ 14 days No Yes Fill with average of available values: f_met = 1; f_win=|dt|; f_qc=1; fqc_ok=1 Rg, T, VPD available with |dt| ≤ 7 days No Quality-controlled half-hourly data (storage, ustar,...) NEE present ? Yes f_met=0, f_qc=0; fqc_ok=1 Don‘t fill: f_met = 1; f_win=|dt|; f_qc=1; fqc_ok=1 Yes f_met = 3; f_win=|dt|; f_qc=1; fqc_ok=1 Yes f_met = 2; f_win=|dt|; f_qc=2; fqc_ok=0 Yes f_met = 3; f_win=|dt|; f_qc=2; fqc_ok=0 Yes f_met = 2; f_win=|dt|; f_qc=2 or 3; fqc_ok=0 Yes f_met = 1; f_win=|dt|; f_qc=2 or 3; fqc_ok=0 Yes f_met = 3; f_win=|dt|; f_qc= 3; fqc_ok=0 Yes


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