Calibration line and nonlinearity of transducer

Method: fit $y = mx + c$ by least squares, then find the largest difference between the data and the line.

Least-squares fit

For $n = 6$ points: $\sum x = 30$, $\sum y = 15.04$, $\sum x^2 = 220$, $\sum xy = 110.10$ (units bar, V).

Here $\sum xy = 0 + 1.96 + 8.20 + 18.00 + 31.84 + 50.10 = 110.10$.

$$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{6(110.10) - 30(15.04)}{6(220) - 900} = \frac{660.6 - 451.2}{420} = \frac{209.4}{420} = 0.4986\ \text{V/bar}$$ $$c = \frac{\sum y - m\sum x}{n} = \frac{15.04 - 0.4986(30)}{6} = 0.0138\ \text{V}$$

So the line is $y = 0.4986\,x + 0.0138$.

Nonlinearity

Maximum deviation $= 0.0419$ V at 4 bar.

Full-scale output $= 5.01 - 0.02 = 4.99$ V.

$$\text{Nonlinearity} = \frac{0.0419}{4.99}\times 100 = 0.84\%\ \text{of FSO}$$

Answer: sensitivity (slope) $= 0.499$ V/bar, zero offset $= 0.0138$ V, maximum nonlinearity $= 0.84\%$ of FSO.