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Chapter 3 · 4 hours

Static Characteristics of Measurement System

Practice questions

Practice questions and answers

4 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 3+3 marks

(a) Differentiate between accuracy and precision with a suitable example.
(b) Define sensitivity, threshold, resolution and tolerance of a measuring instrument.

Answer

(a) Accuracy and precision

Accuracy is the closeness of a measured value to the true value. Precision is the closeness of repeated readings of the same quantity to each other (repeatability).

PointAccuracyPrecision
MeaningNearness to true valueNearness of readings to one another
Error typeAffected by systematic errorAffected by random error
Expressed as% of full scale or % of readingStandard deviation, repeatability
Improved byCalibrationBetter design, averaging
Possible alone?Average of scattered readings can be accurateReadings can be precise yet wrong

Example: a target shot. Shots all clustered far from the bullseye are precise but not accurate. Shots scattered evenly around the bullseye are accurate on average but not precise. A voltmeter reading 4.98, 4.99, 4.98 V for a true 5.50 V is precise but inaccurate.

(b) Definitions

  • Sensitivity: ratio of the change in output to the change in input, S=Δqo/ΔqiS = \Delta q_o/\Delta q_i (slope of the calibration curve). Example: 5 mV/bar.
  • Threshold: the smallest input change from zero that produces a detectable output.
  • Resolution: the smallest change in input (from a non-zero value) that the instrument can detect. For a digital meter it is one least-count digit.
  • Tolerance: the maximum permitted departure from a specified value, usually given as ±\pm value or ±%\pm\%; it is the permissible error, not the instrument's inherent accuracy.
  • Practice · 6 marks

Write short notes on (a) hysteresis, (b) dead space, and (c) linearity of a measuring instrument. Explain the terms independent, terminal and least-squares linearity with a sketch.

Answer

(a) Hysteresis

Hysteresis is the difference in output for the same input depending on whether the input is increasing or decreasing. It is caused by friction, backlash, elastic after-effect and magnetic hysteresis. It is quoted as the maximum difference between the up-scale and down-scale outputs, as a percentage of full-scale output.

 Output |           ___--- up
        |       _--/ _--- down
        |    _-/ _--/   <- hysteresis band
        |  _/ _-/
        | /_-/
        +------------------ Input

(b) Dead space

Dead space (dead band) is the range of input over which there is no change in output, even though the input changes, usually because of backlash, friction or a preload. The instrument responds only after the input leaves this band. Hysteresis is not the same: it exists at all points, while dead space is a flat zone.

(c) Linearity

Linearity is the closeness of the calibration curve to a straight line. Nonlinearity is the maximum deviation from the reference line, expressed as % of full-scale output. The reference line can be defined in three ways.

  • Terminal (end-point) linearity: a line joining the zero and full-scale points. It is simple but gives large deviations.
  • Independent linearity: two parallel lines enclose the curve with minimum spacing; the line midway is the reference. Nonlinearity is half the spacing.
  • Least-squares linearity: the line that minimizes ∑(yi−mxi−c)2\sum (y_i - mx_i - c)^2. It gives the smallest overall error and is the most used.
 Output |        . . .  actual curve
        |     .  /  <- best straight line
        |   . /
        | ./
        +----------------- Input
  • Practice · 8 marks

A pressure transducer was calibrated with the following data.
Pressure (bar)0246810
Output (V)0.020.982.053.003.985.01
Fit the least-squares straight line, find the sensitivity and the zero offset, and calculate the maximum nonlinearity as a percentage of full-scale output (FSO).

Answer

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

Least-squares fit

For n=6n = 6 points: ∑x=30\sum x = 30, ∑y=15.04\sum y = 15.04, ∑x2=220\sum x^2 = 220, ∑xy=149.57\sum xy = 149.57 (units bar, V).

x (bar)y (V)Fitted (V)Deviation (V)
00.020.0138+0.0062
20.981.0110-0.0310
42.052.0082+0.0419
63.003.0053-0.0052
83.984.0024-0.0224
105.014.9996+0.0105
m=n∑xy−∑x∑yn∑x2−(∑x)2=6(149.57)−30(15.04)6(220)−900=446.22420=0.4986 V/barm = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{6(149.57) - 30(15.04)}{6(220) - 900} = \frac{446.22}{420} = 0.4986\ \text{V/bar} c=∑y−m∑xn=15.04−0.4986(30)6=0.0138 Vc = \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+0.0138y = 0.4986\,x + 0.0138.

Nonlinearity

Maximum deviation =0.0419= 0.0419 V at 4 bar.

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

Nonlinearity=0.04194.99×100=0.84% of FSO\text{Nonlinearity} = \frac{0.0419}{4.99}\times 100 = 0.84\%\ \text{of FSO}

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

  • Practice · 2+2 marks

(a) Explain sensitivity to disturbance, with zero drift and sensitivity drift (scale-factor drift) as the two forms.
(b) A pressure transducer has a sensitivity of 5 mV/bar at 20 ∘^\circC. The zero drift is 0.2 mV/∘^\circC and the sensitivity drift is 0.01 mV/bar per ∘^\circC. Find the output when it reads 6 bar at 40 ∘^\circC.

Answer

(a) Sensitivity to disturbance

Environmental inputs such as ambient temperature, supply voltage, vibration or humidity can change the output of an instrument even when the measurand is constant. These are called disturbances (interfering and modifying inputs). Two effects are described.

  • Zero drift (zero shift): the whole calibration line moves up or down by a constant amount, so the output at zero input changes. It is quoted per unit disturbance, e.g. mV/∘^\circC.
  • Sensitivity drift (scale-factor drift): the slope of the calibration line changes, so the error grows with input. Quoted as, e.g., (mV/bar) per ∘^\circC.
 Output |      /  <- sensitivity drift (slope change)
        |    //
        |  ///_/  <- zero drift (parallel shift)
        |_/_/
        +---------------- Input

(b) Numerical

Temperature change ΔT=40−20=20 ∘\Delta T = 40 - 20 = 20\ ^\circC.

Nominal output =5×6=30= 5 \times 6 = 30 mV.

Zero drift =0.2×20=4= 0.2 \times 20 = 4 mV.

Sensitivity drift effect =(0.01×20)×6=1.2= (0.01\times 20)\times 6 = 1.2 mV.

Vo=30+4+1.2=35.2 mVV_o = 30 + 4 + 1.2 = 35.2\ \text{mV}

Answer: output =35.2= 35.2 mV, an error of +5.2+5.2 mV (17%) due to temperature.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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