THEORY EXAMINATION (SEM–IV) 2016-17 SENSOR & INSTRUMENTATION

Course: B.Tech (EE / EN / ECE / EI / IC / AEI)
Subject Code: EE405
Subject Title: Sensor & Instrumentation
Exam Type: Theory
Duration: 3 Hours
Maximum Marks: 100

SECTION – A (10 × 2 = 20 Marks)

Short conceptual questions testing fundamental understanding

\text{Resolution} = \frac{5V}{2^8 - 1} = 19.6 \text{ mV/step} \] | | (i) | **Sample & Hold Circuit** | Captures analog signal and holds it constant during A/D conversion; prevents distortion due to sampling lag. | | (j) | **Pressure Sensors** | Include strain gauge, piezoelectric, capacitive, and resonant sensors — each converts pressure to an electrical signal. | --- ## ⚙️ **SECTION – B (5 × 10 = 50 Marks)** Descriptive and numerical questions covering transducers, signal processing, and display systems:contentReference[oaicite:2]{index=2}. --- ### **(a) Active High-Pass Filter** - **Transfer Function:** \[ H(s) = \frac{sRC}{1 + sRC}

Cutoff Frequency:

• fc=12πRCf_c = \frac{1}{2\pi RC}fc​=2πRC1​

Output: Amplified high-frequency signal; attenuates low frequencies.

(b) Hall Effect Transducer

Principle: When current-carrying conductor placed in magnetic field experiences transverse voltage.

Applications: Magnetic field strength, displacement, current measurement.

(c) Frequency Division Multiplexing (FDM)

Multiple analog signals transmitted simultaneously using distinct frequency bands.

Block Diagram Includes: Signal source → Modulator → Bandpass filter → Multiplexer → Channel → Demultiplexer → Receiver.

Used in telemetry and communication systems.

(d) LCD Working

Field Effect Type: Electric field aligns liquid crystal molecules to control light transmission.

Light Scattering Type: Unaligned molecules scatter light; alignment removes scattering.

Advantages: Low power, compact, good visibility.

(e) Strain Gauge Problem

Given:
R=120Ω,G=2,E=200 GN/m2,σ=140 MN/m2R = 120 \Omega, G = 2, E = 200 \text{ GN/m}^2, \sigma = 140 \text{ MN/m}^2R=120Ω,G=2,E=200 GN/m2,σ=140 MN/m2

Strain=σE=7×10−4\text{Strain} = \frac{\sigma}{E} = 7 \times 10^{-4}Strain=Eσ​=7×10−4 ΔR=GR×Strain=2×120×7×10−4=0.168 Ω\Delta R = G R \times \text{Strain} = 2 × 120 × 7×10^{-4} = 0.168 \, \OmegaΔR=GR×Strain=2×120×7×10−4=0.168Ω

Voltage Change measured through Wheatstone bridge proportional to ΔR.

(f) DAC Conversion Techniques

Weighted Resistor DAC

R–2R Ladder Network

Sigma–Delta Converter

Flash (Parallel) DAC
Each balances between speed, cost, and resolution.

(g) Level Meters

Types: Capacitive, resistive, ultrasonic, radar, and optical.

Advantages: Non-contact (ultrasonic), precise (radar); Disadvantages: Cost, sensitivity to temperature/humidity.

(h) Transducer Classification

SECTION – C (2 × 15 = 30 Marks)

Analytical and practical problem-based questions

Q3. Thermocouple

Working Principle: Based on Seebeck effect — EMF generated due to temperature difference between two dissimilar metals.

Common Types and Ranges:

Advantages: Rugged, inexpensive, wide range.
Disadvantages: Nonlinear output, reference junction compensation needed.

Q4. Strain Gauge – Gauge Factor Derivation

G=ΔR/Rε=1+2μ+Δρ/ρεG = \frac{\Delta R / R}{\varepsilon} = 1 + 2\mu + \frac{\Delta \rho / \rho}{\varepsilon}G=εΔR/R​=1+2μ+εΔρ/ρ​

where μ\muμ = Poisson’s ratio.
Numerical Example:
Given: G=2,σ=500 kg/cm2,E=2×106 kg/cm2G = 2, \sigma = 500 \text{ kg/cm}^2, E = 2×10^6 \text{ kg/cm}^2G=2,σ=500 kg/cm2,E=2×106 kg/cm2

ε=σE=2.5×10−4\varepsilon = \frac{\sigma}{E} = 2.5×10^{-4}ε=Eσ​=2.5×10−4 ΔRR=Gε=2(2.5×10−4)=5×10−4\frac{\Delta R}{R} = G\varepsilon = 2(2.5×10^{-4}) = 5×10^{-4}RΔR​=Gε=2(2.5×10−4)=5×10−4 

Q5. Branch-Specific Questions

The question varies by department

Summary

This Sensor & Instrumentation (EE405) paper comprehensively tests: