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A

Section A: Long Answer Questions

Attempt all / any as specified.

4 questions
1long14 marks

(a) With the help of a functional block diagram, describe the generalized configuration of a measurement system, clearly identifying the primary sensing element, variable-conversion element, variable-manipulation element, data-transmission element and data-presentation element. (b) Define accuracy, precision, resolution, sensitivity and dead zone as applied to a measuring instrument. (c) A pressure gauge has a range of 0–100 kPa and is guaranteed accurate to within ±1.5% of full-scale deflection. Determine the maximum possible error when the gauge reads 25 kPa and express it as a percentage of the actual reading.

(a) Generalized Configuration of a Measurement System

A measurement system can be represented by the following functional block diagram:

Measurand --> [Primary Sensing Element] --> [Variable-Conversion Element]
          --> [Variable-Manipulation Element] --> [Data-Transmission Element]
          --> [Data-Presentation Element] --> Observer
  • Primary sensing element: The element that first receives energy from the measured medium and produces an output that depends on the measurand (e.g. a thermocouple sensing temperature, a Bourdon tube sensing pressure).
  • Variable-conversion element: Converts the output of the sensing element into a more suitable form (often electrical) without changing the information content (e.g. a diaphragm displacement converted to capacitance change).
  • Variable-manipulation element: Changes the magnitude of the signal while keeping its nature the same (e.g. an amplifier that increases voltage level; signal conditioning).
  • Data-transmission element: Transmits the signal from one location to another (cables, wireless links, telemetry).
  • Data-presentation element: Conveys the measured value to a human observer or records it (analog meter, digital display, recorder).

(b) Definitions

  • Accuracy: The closeness of a measured value to the true (or accepted) value of the quantity, usually expressed as a percentage of full-scale or of the reading.
  • Precision: The degree of reproducibility/repeatability among independent measurements of the same quantity under the same conditions (consistency, not correctness).
  • Resolution: The smallest change in the measured quantity that the instrument can detect and indicate.
  • Sensitivity: The ratio of the change in output to the change in input, S=ΔoutputΔinputS = \dfrac{\Delta \text{output}}{\Delta \text{input}} (slope of the calibration curve).
  • Dead zone: The largest range of input values over which the instrument produces no detectable change in output (insensitivity around a region, often near zero).

(c) Numerical: Maximum Possible Error

The accuracy is ±1.5%\pm 1.5\% of full-scale deflection (FSD), with FSD =100= 100 kPa.

Maximum error=±1.5%×100 kPa=±1.5 kPa\text{Maximum error} = \pm 1.5\% \times 100\ \text{kPa} = \pm 1.5\ \text{kPa}

At a reading of 2525 kPa, expressed as a percentage of the actual reading:

%error=1.525×100=±6%\% \text{error} = \frac{1.5}{25} \times 100 = \pm 6\%

Result: Maximum possible error =±1.5= \pm 1.5 kPa, which is ±6%\pm 6\% of the actual reading. This illustrates that the relative error grows as the reading falls toward the low end of the scale.

measurement-systemserror-analysis
2long14 marks

(a) Explain the principle of operation of a resistance temperature detector (RTD) and a thermocouple, and compare them on the basis of range, linearity, sensitivity and the need for cold-junction compensation. (b) Design a Wheatstone-bridge signal-conditioning circuit suitable for a single active strain-gauge transducer. Derive the expression for the bridge output voltage in terms of the gauge factor and the applied strain, and explain how a half-bridge arrangement improves sensitivity and provides temperature compensation.

(a) RTD and Thermocouple

Resistance Temperature Detector (RTD): Operates on the principle that the electrical resistance of a pure metal (commonly platinum, Pt-100) increases almost linearly with temperature:

RT=R0(1+αT)R_T = R_0 \left(1 + \alpha T\right)

where R0R_0 is the resistance at 0C0^{\circ}\text{C} and α\alpha is the temperature coefficient of resistance. It is a passive sensor needing an excitation current and is read out using a bridge.

Thermocouple: Operates on the Seebeck effect — when two dissimilar metals are joined to form two junctions held at different temperatures, an EMF is generated proportional to the temperature difference:

E=S(ThotTcold)E = S\,(T_{hot} - T_{cold})

It is an active (self-generating) sensor.

Comparison:

BasisRTDThermocouple
RangeModerate (200-200 to 850C\sim 850^{\circ}\text{C})Wide (up to 1700C\sim 1700^{\circ}\text{C})
LinearityGood, nearly linearNon-linear (needs polynomial correction)
SensitivityLower (small ΔR\Delta R)Low EMF (µV/°C) but usable
Cold-junction comp.Not requiredRequired (reference-junction compensation)

(b) Wheatstone-Bridge Signal Conditioning for a Single Active Strain Gauge

Connect one active strain gauge as one arm of a Wheatstone bridge; the other three arms are fixed precision resistors equal to the unstrained gauge resistance RR.

          +Vex
           |
      +-----+-----+
      |           |
     [R]        [R+ΔR]  <- active gauge
      |           |
   a  +---- Vo ---+  b
      |           |
     [R]         [R]
      |           |
      +-----+-----+
           |
          GND

For a balanced bridge excited by VexV_{ex}, with the active gauge changing by ΔR\Delta R, the output is:

Vo=Vex(R+ΔR2R+ΔR12)Vex4ΔRRV_o = V_{ex}\left(\frac{R+\Delta R}{2R+\Delta R} - \frac{1}{2}\right) \approx \frac{V_{ex}}{4}\cdot\frac{\Delta R}{R}

The gauge factor GG relates fractional resistance change to strain ε\varepsilon:

ΔRR=Gε\frac{\Delta R}{R} = G\,\varepsilon

Therefore:

VoVex4Gε\boxed{V_o \approx \frac{V_{ex}}{4}\,G\,\varepsilon}

Half-bridge improvement: Replace two opposite/adjacent arms with two active gauges — one in tension (+ΔR+\Delta R) and one in compression (ΔR-\Delta R). The differential outputs add:

VoVex2GεV_o \approx \frac{V_{ex}}{2}\,G\,\varepsilon

This doubles the sensitivity. Because both gauges experience the same ambient temperature, their temperature-induced resistance changes are equal and cancel in the bridge difference, giving inherent temperature compensation.

transducers-and-sensorssignal-conditioning
3long12 marks

(a) With a neat block diagram, explain the working of a successive-approximation type analog-to-digital converter (ADC) and state its advantages over the dual-slope (integrating) type. (b) Describe the architecture of a typical data-acquisition system (DAS), explaining the role of the multiplexer, sample-and-hold circuit, ADC and the need for anti-aliasing filtering. (c) A 10-bit ADC has a full-scale input of 5 V. Calculate its resolution and the quantization error.

(a) Successive-Approximation ADC (SAR)

Vin -->[ S/H ]--> [+ Comparator -] <-- Vdac
                        |
                        v
               [SAR Logic / Register] --> Digital Output
                        |
                        v
                      [DAC] --> Vdac (feedback)
        Clock --> SAR Logic

Working: The SAR register sets the MSB first and the DAC produces a trial voltage. The comparator checks whether VinV_{in} is above or below the DAC output. If Vin>VdacV_{in} > V_{dac} the bit is kept (1); otherwise it is cleared (0). The process repeats bit-by-bit down to the LSB. An nn-bit conversion is completed in exactly nn clock cycles by a binary search.

Advantages over dual-slope (integrating) type:

  • Much faster — fixed nn clocks vs. up to 2n2^{n} counts for dual-slope.
  • Higher conversion rate suited to data-acquisition with multiplexed channels.
  • Good balance of speed, resolution and cost. (Dual-slope is slower but offers excellent noise rejection and accuracy, used in DMMs.)

(b) Data-Acquisition System (DAS) Architecture

Sensors --> [Signal Conditioning] --> [Multiplexer] --> [S/H] --> [ADC] --> [Computer/MCU]
            (with anti-aliasing filters)
  • Multiplexer (MUX): Time-shares a single ADC among many input channels by sequentially selecting one analog channel at a time, reducing cost.
  • Sample-and-Hold (S/H): Captures and holds the instantaneous analog value constant during conversion so the input does not change while the ADC operates.
  • ADC: Converts the held analog voltage to a digital code for the processor.
  • Anti-aliasing filter: A low-pass filter placed before sampling to remove signal components above the Nyquist frequency, preventing high-frequency content from folding back (aliasing) into the band of interest.

(c) Numerical: 10-bit ADC, Full Scale 5 V

Resolution (LSB step size):

Q=VFS2n=5210=510244.883 mVQ = \frac{V_{FS}}{2^{n}} = \frac{5}{2^{10}} = \frac{5}{1024} \approx 4.883\ \text{mV}

Quantization error: ±12\pm \tfrac{1}{2} LSB:

Quantization error=±Q2=±4.883 mV2±2.44 mV\text{Quantization error} = \pm\frac{Q}{2} = \pm\frac{4.883\ \text{mV}}{2} \approx \pm 2.44\ \text{mV}

Result: Resolution 4.88\approx 4.88 mV; maximum quantization error ±2.44\approx \pm 2.44 mV.

digital-instrumentsdata-acquisition
4long10 marks

(a) Explain the construction and working principle of a permanent-magnet moving-coil (PMMC) instrument. Derive the expression for the deflecting torque and explain why it can be used only for DC measurements. (b) Discuss the construction and operating principle of a seven-segment LED display and a liquid-crystal display (LCD), and compare them in terms of power consumption, viewing characteristics and suitability for portable digital instruments.

(a) PMMC (Permanent-Magnet Moving-Coil) Instrument

Construction: A rectangular coil of fine wire is wound on a light aluminium former and suspended (on jewelled pivots/springs) in the air gap of a permanent magnet with soft-iron pole pieces and a central soft-iron core, giving a strong radial uniform magnetic field. Control springs provide the restoring torque and also carry current to the coil; a pointer moves over a scale; damping is by eddy currents in the aluminium former.

Working / Deflecting torque: When current II flows, each side of the coil (length ll, NN turns) carries current in the field BB, producing a force F=NBIlF = NBIl on each side. With coil width bb, the deflecting torque is:

Td=NBIlb=NBIAT_d = N B I l b = N B I A

where A=lbA = l\,b is the coil area. Thus TdIT_d \propto I (the field BB is constant). At balance Td=Tc=KθT_d = T_c = K\theta, so:

θ=NBAKII\theta = \frac{N B A}{K}\,I \propto I

The deflection is directly proportional to current, giving a uniform (linear) scale.

Why DC only: Because the deflecting torque reverses direction when the current reverses. On AC the average torque over a cycle is zero, and the moving system (with finite inertia) cannot follow the rapid reversals, so the pointer reads zero. Hence PMMC instruments measure DC only.

(b) Seven-Segment LED vs. LCD

Seven-segment LED: Seven (plus a dot) light-emitting diodes arranged as bars; selectively forward-biasing segments forms digits 0–9. LEDs emit light (active emission) using injection electroluminescence in a forward-biased p-n junction.

LCD: Liquid-crystal molecules between polarizers and transparent electrodes (ITO). An applied AC field twists/aligns the molecules to modulate transmitted/reflected ambient or backlight light — the LCD does not emit light itself, it controls passage of light.

Comparison:

BasisSeven-segment LEDLCD
Power consumptionHigh (each lit segment draws current)Very low (µW range)
ViewingBright, visible in the dark, wide angleNeeds ambient light/backlight; limited viewing angle
Suitability (portable digital instruments)Less suitable on batteryPreferred — low power extends battery life

LCDs are therefore preferred for battery-powered portable digital instruments (DMMs, watches), while LEDs suit instruments needing bright displays readable in darkness.

analog-instrumentsdisplay-devices
B

Section B: Short Answer Questions

Attempt all / any as specified.

8 questions
5short6 marks

Explain the working principle of a Linear Variable Differential Transformer (LVDT) for displacement measurement. Sketch its output voltage versus core-displacement characteristic and comment on the phase reversal that occurs as the core moves through the null position.

LVDT (Linear Variable Differential Transformer)

Principle: An LVDT has one primary winding and two identical secondary windings (S1S_1, S2S_2) wound symmetrically on either side of the primary, with a movable ferromagnetic core linked to the object whose displacement is measured. The primary is excited by an AC voltage. The core position controls the mutual coupling between primary and each secondary.

The secondaries are connected in series opposition, so the output is:

Vo=VS1VS2V_o = V_{S1} - V_{S2}
  • At the null (centre) position, coupling to both secondaries is equal, VS1=VS2V_{S1}=V_{S2}, and Vo0V_o \approx 0.
  • When the core moves toward S1S_1, VS1>VS2V_{S1} > V_{S2} and VoV_o rises.
  • When the core moves toward S2S_2, VS2>VS1V_{S2} > V_{S1} and VoV_o rises with the opposite phase.

Output vs. core-displacement characteristic (described): The magnitude Vo|V_o| is a V-shaped curve with a minimum (ideally zero) at the null and rising linearly on both sides. If signed (phase-sensitive detection used), the characteristic is a straight line passing through the origin — positive for one direction, negative for the other.

|Vo|
  \        /
   \      /     (V-shape, min at null)
    \    /
     \  /
 -----\/------ displacement
      null

Phase reversal: As the core passes through the null, the dominant secondary changes from S1S_1 to S2S_2. Because the two secondaries are in opposition, the output voltage undergoes a 180180^{\circ} phase shift relative to the excitation. This phase reversal is used by a phase-sensitive demodulator to determine the direction of displacement (sign), which a simple magnitude reading cannot give.

transducers-and-sensors
6short6 marks

(a) Draw the circuit diagram of a three-op-amp instrumentation amplifier and write the expression for its overall gain. (b) State two reasons why an instrumentation amplifier is preferred over a simple difference amplifier in transducer signal conditioning.

(a) Three-Op-Amp Instrumentation Amplifier

 V1 --[+]A1                 
          \    R3   R4      
        Rg  >--[-]A3[+]----> Vout
          /    R3'  R4'      
 V2 --[+]A2                 
   (A1,A2 input buffers with gain-setting Rg; A3 = difference amp)

Stage 1: two non-inverting buffers (A1, A2) share a single gain resistor RgR_g with two equal resistors R1R_1. Stage 2 (A3) is a unity/scaled difference amplifier with resistors R2R_2 (input) and R3R_3 (feedback).

Overall gain:

VoutV2V1=(1+2R1Rg)R3R2\frac{V_{out}}{V_2 - V_1} = \left(1 + \frac{2R_1}{R_g}\right)\cdot\frac{R_3}{R_2}

The differential gain is set by the single resistor RgR_g, with the difference stage usually at unity (R3=R2R_3 = R_2), giving Av=1+2R1RgA_v = 1 + \dfrac{2R_1}{R_g}.

(b) Why preferred over a simple difference amplifier

  1. Very high input impedance at both inputs (signals go directly into op-amp non-inverting inputs), so it does not load the transducer/bridge; a simple difference amplifier presents finite, unequal input resistances that load the source and degrade accuracy.
  2. High common-mode rejection ratio (CMRR) with gain set by one resistor, and CMRR does not depend on tight matching of the input resistor ratios — making it ideal for amplifying small differential transducer signals riding on large common-mode voltages.
signal-conditioning
7short6 marks

Distinguish between systematic errors and random errors, giving one example of each. Explain how the standard deviation and probable error are used to express the uncertainty in a set of repeated measurements.

Systematic vs. Random Errors

Systematic (determinate) errors: Errors that have a definite cause and a consistent magnitude/sign, so they bias all readings in the same direction. They are reproducible and can in principle be calibrated out. Example: a zero-offset (incorrectly set zero) of an ammeter, or a meter with an incorrect calibration constant.

Random (indeterminate) errors: Errors due to many small, unpredictable fluctuations (noise, friction, observer variation) that cause scatter about the mean with no fixed sign. They cannot be eliminated but are reduced by averaging many readings. Example: fluctuating last-digit readings on a digital meter due to electrical noise.

Expressing Uncertainty

For nn repeated readings xix_i with mean xˉ\bar{x}:

Standard deviation measures the spread of the data:

σ=i=1n(xixˉ)2n1\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}

A small σ\sigma means high precision; it quantifies the dispersion of the random error about the mean.

Probable error (PE): the value such that half the readings lie within ±PE\pm PE of the mean. For a normal distribution:

PE=0.6745σPE = 0.6745\,\sigma

Thus the result is stated as xˉ±σ\bar{x} \pm \sigma (or xˉ±PE\bar{x} \pm PE), giving the range within which the true value is expected to lie due to random error.

error-analysismeasurement-systems
8short6 marks

With a block diagram, explain the operation of a digital frequency counter. Explain the meaning of the term 'gate time' and describe how it affects the resolution and accuracy of the frequency measurement.

Digital Frequency Counter

Input --> [Amplifier/Schmitt Shaper] --> [AND/Main Gate] --> [Decade Counter] --> [Display]
                                            ^
                                            |
[Crystal Oscillator] --> [Time-Base Divider] --> [Gate Control Flip-Flop]

Operation: The unknown signal is amplified and shaped (Schmitt trigger) into clean pulses. A highly stable crystal oscillator drives a chain of decade dividers (the time base) which produces a precise gating interval. The gate-control flip-flop opens the main AND gate for exactly this interval, allowing the input pulses to pass to a decade counter. The number of pulses counted during the gate interval equals the frequency:

f=N (counts)tg (gate time)f = \frac{N \ (\text{counts})}{t_g \ (\text{gate time})}

The count is latched and shown on the digital display, then the counter resets for the next cycle.

Gate Time

Gate time (tgt_g) is the precisely controlled interval (e.g. 0.1 s, 1 s, 10 s) during which the main gate is held open to count input pulses.

Effect on resolution and accuracy:

  • A longer gate time counts more pulses, improving resolution (more significant digits) and reducing the relative effect of the ±1\pm 1 count quantization error, since that error is ±1/N\pm 1/N of the reading.
  • For a 1 s gate, the displayed count directly equals the frequency in Hz with 1 Hz resolution; a 10 s gate gives 0.1 Hz resolution.
  • A shorter gate time gives faster updates but coarser resolution and larger relative error.
  • Overall accuracy is also limited by the time-base (crystal) accuracy and the ±1\pm1 count uncertainty; long gate times favour accuracy for low frequencies, while a period-measurement mode is preferred for very low frequencies.
digital-instruments
9short6 marks

(a) Explain how a basic d'Arsonval PMMC movement is converted into a multi-range DC ammeter using shunt resistors. (b) A 50 µA, 2 kΩ movement is to be used to measure currents up to 1 mA. Calculate the required shunt resistance.

(a) Multi-Range DC Ammeter using Shunts

A d'Arsonval PMMC movement carries only a small full-scale current ImI_m. To measure a larger current II, a low-resistance shunt RshR_{sh} is connected in parallel with the movement (coil resistance RmR_m) so that most of the current bypasses the delicate movement:

I=Im+IshI = I_m + I_{sh}

Since the shunt and movement share the same voltage:

ImRm=IshRsh=(IIm)RshI_m R_m = I_{sh} R_{sh} = (I - I_m) R_{sh} Rsh=ImRmIIm\boxed{R_{sh} = \frac{I_m R_m}{I - I_m}}

For a multi-range ammeter, several shunts (or an Ayrton/universal shunt) are selected by a range switch to give different full-scale currents.

(b) Numerical

Given: Im=50 μA=50×106I_m = 50\ \mu\text{A} = 50\times10^{-6} A, Rm=2 kΩ=2000 ΩR_m = 2\ \text{k}\Omega = 2000\ \Omega, full-scale I=1 mA=1×103I = 1\ \text{mA} = 1\times10^{-3} A.

Rsh=ImRmIIm=(50×106)(2000)(1×103)(50×106)R_{sh} = \frac{I_m R_m}{I - I_m} = \frac{(50\times10^{-6})(2000)}{(1\times10^{-3}) - (50\times10^{-6})} =0.10.95×103=105.26 Ω= \frac{0.1}{0.95\times10^{-3}} = 105.26\ \Omega

Result: Required shunt resistance Rsh105.3 ΩR_{sh} \approx 105.3\ \Omega.

analog-instruments
10short6 marks

Explain the working principle of a capacitive transducer for the measurement of liquid level. Discuss why such transducers require high-frequency excitation and a charge-amplifier or AC bridge for signal conditioning.

Capacitive Transducer for Liquid-Level Measurement

Principle: A capacitive level sensor uses two concentric/parallel electrodes (or a single insulated probe and the metal tank wall) forming a capacitor. As the liquid level rises, the dielectric between the electrodes partly changes from air (εr1\varepsilon_r \approx 1) to the liquid (εr>1\varepsilon_r > 1), so the effective permittivity and hence the capacitance increases in proportion to the level:

C=ε0[εr,liqh+(Hh)](2π)ln(b/a)C = \frac{\varepsilon_0\,[\varepsilon_{r,liq}\,h + (H-h)]\,(2\pi)}{\ln(b/a)}

For concentric cylinders of inner/outer radii a,ba,b, total height HH and liquid height hh. Thus CC varies linearly with level hh; measuring CC gives the level. (For conductive liquids an insulated probe is used and capacitance is measured to the liquid surface.)

Why High-Frequency Excitation and Charge Amplifier / AC Bridge

  • The transducer capacitance is small (pF range), so its impedance ZC=12πfCZ_C = \dfrac{1}{2\pi f C} is very high at low frequency. Using a high excitation frequency lowers this impedance to a measurable level, increases the signal current, and improves sensitivity and signal-to-noise ratio.
  • A capacitor passes no DC, so a changing (AC) excitation is necessary to produce any output at all.
  • A charge amplifier (op-amp with capacitive feedback) converts the tiny charge change into a usable output voltage independent of cable capacitance, while an AC bridge balances the unknown CC against a reference, giving an output proportional to the capacitance change and rejecting stray effects. These conditioners are needed because the high source impedance makes the sensor very sensitive to leakage, stray capacitance and noise.
transducers-and-sensorssignal-conditioning
11short4 marks

Define the terms aliasing and Nyquist rate in the context of sampled data systems. State the sampling theorem and explain, with a simple example, the consequence of sampling a signal below the Nyquist rate.

Aliasing and Nyquist Rate

  • Aliasing: The phenomenon by which a high-frequency signal component, when sampled too slowly, becomes indistinguishable from (masquerades as) a lower-frequency component in the sampled data, producing a false (alias) frequency.
  • Nyquist rate: The minimum sampling rate equal to twice the highest frequency present in the signal, fN=2fmaxf_N = 2f_{max}, required to reconstruct it without aliasing.

Sampling Theorem

Shannon/Nyquist sampling theorem: A continuous band-limited signal containing no frequency higher than fmaxf_{max} can be completely reconstructed from its samples if it is sampled at a rate fs>2fmaxf_s > 2 f_{max}.

Example of Under-Sampling

Suppose a 700 Hz700\ \text{Hz} sine wave is sampled at fs=1000 Hzf_s = 1000\ \text{Hz} (below the Nyquist rate of 1400 Hz1400\ \text{Hz}). The signal appears as an alias at:

fsf=1000700=300 Hz|f_s - f| = |1000 - 700| = 300\ \text{Hz}

So the recorded data shows a spurious 300 Hz300\ \text{Hz} tone instead of the true 700 Hz700\ \text{Hz}, and the original signal cannot be recovered. This is why an anti-aliasing low-pass filter must remove components above fs/2f_s/2 before sampling.

data-acquisitiondisplay-devices
12short4 marks

Differentiate between active and passive transducers, and between primary and secondary transducers, giving one practical example of each category.

Active vs. Passive Transducers

  • Active (self-generating) transducer: Produces its own electrical output (voltage/current) directly from the measured energy, needing no external power supply. Example: thermocouple (generates EMF from temperature), piezoelectric crystal, photovoltaic cell.
  • Passive transducer: Does not generate energy itself; it changes a passive electrical parameter (R, L or C) in response to the measurand and requires external excitation to give an output. Example: strain gauge / RTD (resistance change), LVDT, capacitive sensor.

Primary vs. Secondary Transducers

  • Primary transducer: The first element that directly senses the physical quantity and converts it into a mechanical or intermediate signal. Example: a Bourdon tube that converts pressure into a mechanical displacement.
  • Secondary transducer: Receives the output of the primary element and converts that intermediate (usually mechanical) signal into a usable electrical output. Example: an LVDT coupled to the Bourdon tube that converts the displacement into a proportional voltage.

In the Bourdon-tube + LVDT pressure gauge, the Bourdon tube is the primary transducer and the LVDT is the secondary transducer.

measurement-systems

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