vector standard deviation
I'm investigating the sensitivity of some modelling software to data noise.
This has involved generating a 'clean' current data set of purely tidal currents (a tidal ellipse) with spring-neap variation through a 15 day record. I have then added random, normally distributed noise, with specific standard deviations for each test scenario (s.d.=0, 1, 2 & 3cm/s). The noise is added simultaneously to both the x & y components of each data record.
Is this a fair simulation NDP data? or, is the estimated velocity prescision returned by DEPLOY.exe applicable to both vector components or only to the scalar speed?

This has involved generating a 'clean' current data set of purely tidal currents (a tidal ellipse) with spring-neap variation through a 15 day record. I have then added random, normally distributed noise, with specific standard deviations for each test scenario (s.d.=0, 1, 2 & 3cm/s). The noise is added simultaneously to both the x & y components of each data record.
Is this a fair simulation NDP data? or, is the estimated velocity prescision returned by DEPLOY.exe applicable to both vector components or only to the scalar speed?
Dear Jeremy
The noise in the instrument originates in the beam velocities. The noise then transforms into the XYZ velocities as specified by the transformation matrix in such a way that if:
Vx = a1V1 + a2V2 + a3V3
Vy = b1V1 + b2V2 + b3V3
Vz = c1V1 + c2V2 + c3V3
then the standard deviation in Vx is:
sigmaVx=sigmaVi * sqrt(a1*a1 + a2*a2 + a3*a3)
where sigmaVi is the standard deviation in the beam velocities (it should be the same for all three beams).
Your question is then if
sqrt(a1*a1 + a2*a2 + a3*a3) = sqrt(b1*b1 + b2*b2 + b3*b3)
and the answer is yes
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'> - for all of our instruments where the beams are symmetric.
- Atle Lohrmann
The noise in the instrument originates in the beam velocities. The noise then transforms into the XYZ velocities as specified by the transformation matrix in such a way that if:
Vx = a1V1 + a2V2 + a3V3
Vy = b1V1 + b2V2 + b3V3
Vz = c1V1 + c2V2 + c3V3
then the standard deviation in Vx is:
sigmaVx=sigmaVi * sqrt(a1*a1 + a2*a2 + a3*a3)
where sigmaVi is the standard deviation in the beam velocities (it should be the same for all three beams).
Your question is then if
sqrt(a1*a1 + a2*a2 + a3*a3) = sqrt(b1*b1 + b2*b2 + b3*b3)
and the answer is yes
'> - for all of our instruments where the beams are symmetric.- Atle Lohrmann
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