Example forward-difference

Consider the example of an ADCP with a beam angle of [math]\displaystyle{ 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]\displaystyle{ b^{\prime} }[/math] is synonymous with [math]\displaystyle{ v^{\prime} }[/math] in this figure and the one below.

The square of the velocity difference between bins separated by [math]\displaystyle{ \delta }[/math] bins is then evaluated for each [math]\displaystyle{ t }[/math]. So for bin 4, [math]\displaystyle{ \delta=3 }[/math] and [math]\displaystyle{ t=2 }[/math], we get:

[math]\displaystyle{ \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]\displaystyle{ 1\leqslant\delta\leqslant29 }[/math], whilst for bin 2 it is restricted to [math]\displaystyle{ 1\leqslant\delta\leqslant28 }[/math], reducing by 1 with each bin, so that for bin 29, it can only be evaluated for [math]\displaystyle{ \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]\displaystyle{ D(1,\delta) = \sum_{t=1}^{300}\Delta^2(1,\delta,t) }[/math]

---

Return to Forward-difference