MEGR 3171 — Midterm 1 Study Guide

Exam Coverage

Module 1: Measurement performance metrics, statistical analysis, Kline-McClintock uncertainty, calibration, instrumentation amplifiers, Wheatstone bridges, active filters, and DAQ / Nyquist.
Module 2 (through Week 7): Strain gauges, rosette configurations, load cells, thermocouples with CJC, RTDs (Callendar-Van Dusen), thermistors (Steinhart-Hart), and radiation pyrometers.
Exam Format: Closed book, one 8.5 x 11” hand-written formula card permitted. Scientific calculator required.

Section 1: Measurement System Performance

Q1 — Conceptual Define accuracy and precision. An instrument always reads 5% above the true value but returns the same reading each time. Is it accurate? Is it precise?

Accuracy: closeness to true value. Precision: repeatability (consistency) of readings.

The instrument is NOT accurate (systematic 5% bias error). It IS precise (same reading each time = high repeatability). This is a classic example of a precise but inaccurate instrument.

Not accurate (5% bias). Highly precise (zero scatter). Can be corrected by calibration.

Q2 — ADC Resolution A 16-bit ADC has a ±10 V full-scale range. What is the voltage per count (step size)? What is the RMS quantization noise?

Full-scale range = 20 V. Step size = 20 / 2¹&sup6; = 20 / 65536 ≈ 0.3052 mV/count

RMS quantization noise = step / √12 = 0.3052 / 3.464 ≈ 0.0881 mV

Step size = 0.305 mV/count; RMS noise = 0.088 mV

Section 2: Statistics and Uncertainty

Q3 — Statistics Six measurements of a 50.00 N reference force: 49.82, 50.11, 49.95, 50.03, 49.88, 50.21 N. Compute: (a) mean, (b) standard deviation, (c) 95% CI using t_0.025,5 = 2.571.

(a) Mean = (49.82+50.11+49.95+50.03+49.88+50.21)/6 = 300.00/6 = 50.00 N

(b) Deviations: -0.18, +0.11, -0.05, +0.03, -0.12, +0.21. Sq: 0.0324, 0.0121, 0.0025, 0.0009, 0.0144, 0.0441. Sum = 0.1064. s² = 0.1064/5 = 0.02128. s = 0.1459 N.

Mean = 50.00 N, s = 0.146 N, 95% CI: [49.85, 50.15] N

Q4 — Kline-McClintock Power dissipated: P = I²·R. Current I = 3.00 ± 0.05 A, Resistance R = 10.0 ± 0.2 Ω. Find P and w_P.

P = (3.00)² × 10.0 = 90.0 W

∂P/∂I = 2IR = 2(3.00)(10.0) = 60.0 → (60.0 × 0.05)² = 9.0

∂P/∂R = I² = 9.00 → (9.00 × 0.2)² = 3.24

w_P = √(9.0 + 3.24) = √12.24 ≈ 3.50 W

P = 90.0 ± 3.5 W (3.9% uncertainty)

Section 3: Signal Conditioning and Filters

Q5 — INA Gain An INA129 has R = 25 kΩ. Calculate R_G needed for a gain of 500.

A = 1 + 2R/R_G → R_G = 2×25000/(500-1) = 50000/499 ≈ 100.2 Ω

R_G ≈ 100 Ω (use 100 Ω standard value)

Q6 — Wheatstone Bridge A full-bridge strain gauge circuit has V_ex = 10 V, G_f = 2.1, and an applied strain ε = 1200 µε. Two gauges in tension, two in compression. Calculate V_out.

Full-bridge with arms in tension (+ε) and compression (-ε) gives 4× the quarter-bridge output:

V_out = V_ex × G_f × ε = 10 × 2.1 × 1200×10&sup-⁶ = 10 × 0.00252 = 0.0252 V

V_out = 25.2 mV

Q7 — Aliasing A vibration signal contains frequencies up to 500 Hz. A DAQ system samples at 800 Hz. (a) Does aliasing occur? (b) At what frequency does the 500 Hz component alias?

(a) Nyquist = 800/2 = 400 Hz. Signal has 500 Hz component above 400 Hz. Yes, aliasing occurs.

(b) Alias frequency = |f_s - f_signal| = |800 - 500| = 300 Hz. The 500 Hz component appears as 300 Hz in the digitized data.

(a) Yes, aliasing occurs. (b) 500 Hz aliases to 300 Hz. Must sample at ≥1000 Hz with anti-aliasing filter.

Section 4: Sensor Systems

Q8 — Thermocouple A Type K thermocouple reads 12.207 mV at its terminals. The cold junction is at 20°C, which generates 0.798 mV (Type K, 0°C reference). Using the linear approximation 41 µV/°C, find the measurement junction temperature.

V_corrected = 12.207 + 0.798 = 13.005 mV

T ≈ 13.005 mV / 0.041 mV/°C ≈ 317.2°C

T ≈ 317°C

Q9 — Pt100 RTD A Pt100 RTD reads 119.40 Ω. Using the linear approximation R = 100(1 + 0.00385T), find the temperature. What connection method eliminates lead resistance error?

119.40 = 100(1 + 0.00385·T) → 1.1940 = 1 + 0.00385T → T = 0.194/0.00385 = 50.4°C

T ≈ 50.4°C. Use a 4-wire (Kelvin) connection to eliminate lead resistance error.

Q10 — FFT Interpretation True or False: A rectangular window applied to an FFT gives better amplitude accuracy than a Hanning window. Explain.

FALSE. The rectangular window has the worst spectral leakage. Unless the signal frequency falls exactly on an FFT bin (which requires the signal period to divide evenly into the record length), the rectangular window causes severe leakage that distorts amplitude measurements. The Hanning window greatly reduces leakage at a modest cost in frequency resolution. The flat-top window (not rectangular) gives the best amplitude accuracy.

False. Rectangular window has worst leakage → poor amplitude accuracy for general signals. Use Hanning (general) or flat-top (calibration).