# square root extraction

L

#### Luke

i have used the formula 4mA + (4*square root of linear o/p - 4mA. but i'm getting the wrong answers. where am i going wrong

B

#### Bruce Durdle

You need to normalise the input signal first.

If you are talking about a 4 to 20 mA signal:

This has a span of (20-4) = 16, so the normalised value is (I - LRV)/span = (I - 4)/16. This will have a range of 0 to 1.

Now take the square root of this value = R.

Now convert R back to the signal range of 4 to 20 mA: I = (R * span) + LRV = (R * 16) + 4 mA.

So if the input current is 17 mA:<pre>
Normalise this to get (17 - 4)/16 = 13/16 = 0.8125

Take the square root: SQRT(.8125) = 0.9014

Convert back to mA: (.9014 * 16) + 4 = 14.42 + 4 = 18.42 mA.</pre>

For any other range, replace the 4 and 20 values with the appropriate
levels.

Cheers,
Bruce

C

#### curt wuollet

Units perhaps? Should be in Amps, .004, etc.

Regards
cww

R

#### Ray R. Powell

For square root extraction for a flowrate dealing with D/P this is a good definition to remember: The square root of the percentage of the calibration range is the actual percentage of the calculated flowrate.