Measurement and Signal Processing

[ozi-rock]

Hi guy's I'm currently studying for the repeat of the measurement and signals exam, I just can't figure this subject out but anyway I have a question if anyone can shed some light on, it's probably stupidly simple but here it is:

 

A liquid level sensor has an output range of 0 to 15cm. Use the calibration results given in the table to estimate the maximum hysteresis as a percentage of f.s.d.

 

Level h (cm) 0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15

Voltage output increase h 0 0.35 1.42 2.4 3.43 4.35 5.61 6.5 7.77 8.85 10.2

Voltage output decrease h 0.14 1.25 2.32 3.55 4.43 5.7 6.78 7.8 8.87 9.65 10.2

 

The answer given is 13.2%

 

I know %f.s.d = n max divided by Y max - Y min all multiplied by 100. so I got the largest difference between the input and output which I think is n max and I took Y min to be 0 and Y max to be 15. This left me with [2.19/(15-0)]*100 which = 14.6%.

 

Can anyone point out where I'm going wrong?

[lawcat]

The experiment goes like this (and it seems to me that this is not a lab, but a theoretical class, otherwise you would understand what is behind this science):

 

You are increasing liquid levels in a tube, and the tube contains some electromechanical sensors which sends out output, which is measured by a voltage measuring instrument.

 

You are calibrating the voltage reader instrument. (You do not care about liquid levels, but for the sake of symmetry in outputs, you are increasing the liquid levels by constant 1.5.) Based on those constant increases in liquid levels, you are measuring the deviation in voltage outputs as the liquid level goes up.

 

The instrument shows largest deviation from min to max voltage outputs in 7th iteration (5.7 -4.35). Thus, n max = 1.35. The maximum voltage output of the voltage measuring instrument for the trials is 10.2V, and the minimum is 0V.

 

Thus,

 

(5.7 -4.35) / (10.2-0) -> 13.2%

 

This instrument has standard deviation, or uncertainty, of up to 13.2% of the range of voltage outputs 10.2V. Having only single decimal point, and 13.2% deviation, this instrument is neither accurate nor precise.

[ozi-rock]

Thanks Lawcat, thats much clearer now. No your right this wasn't done with a lab it is one of our tutorial questions which I'm using to study. Thanks again.