Example forward-difference
Consider the example of an ADCP with a beam angle of <math>20^{\circ}</math>, configured with a vertical bin size of 10 cm, recording profiles at 1 second intervals with a data segment length of 300 seconds. The Level 1 QC of the data identified that good data was typically returned from bins 1 to 30.
The velocity data from a single beam for a single data segment can therefore be visualised as:
Note: <math> b^{\prime}</math> is synonymous with <math> v^{\prime}</math> in this figure and the one below.
The square of the velocity difference between bins separated by <math>\delta</math> bins is then evaluated for each <math>t</math>. So for bin 4, <math>\delta=3</math> and <math>t=2</math>, we get:
- <math>\Delta^2(4,3,2) = \left[b^\prime(4,2) - b^\prime(7,2)\right]^2</math>
For bin 1 the squared velocity difference can be evaluated for <math>1\leqslant\delta\leqslant29</math>, whilst for bin 2 it is restricted to <math>1\leqslant\delta\leqslant28</math>, reducing by 1 with each bin, so that for bin 29, it can only be evaluated for <math>\delta=1</math> and there are no options for bin 30. This is summarised as follows:
The mean is then taken across the 300 profiles in the data segment i.e.
- <math>D(1,\delta) = \sum_{t=1}^{300}\Delta^2(1,\delta,t)</math>
Return to Forward-difference
