22-Agric-A5 Principles of Instrumentation · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. 04-Agric-A5 Principles of Instrumentation, National Exams December 2015 — a three-hour open-book exam; any non-communicating calculator is permitted. Questions 1 and 2 are compulsory (20 marks each); candidates then choose any three (3) of Questions 3-7 (20 marks each) for a 100-mark paper. All seven questions are worked here.
Reference texts. E.O. Doebelin, Measurement Systems: Application and Design, 5th ed. (calibration, standards, static/dynamic sensor characteristics, second-order step response); J.P. Bentley, Principles of Measurement Systems, 4th ed. (accuracy vs. precision, error propagation, signal conditioning); P. Horowitz and W. Hill, The Art of Electronics, 3rd ed. (bridge circuits, instrumentation amplifiers, CMRR, ADC architectures); J. Fraden, Handbook of Modern Sensors: Physics, Designs, and Applications, 5th ed. (thermistors, strain gages, photodetectors); R.W. Fox, A.T. McDonald and P.J. Pritchard, Introduction to Fluid Mechanics, 7th ed. (orifice and venturi metering).
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
a) Given. $R_{ref}$-$R_{th}$ divider, $V_{out}=V_{ex}\cdot R_{th}/(R_{ref}+R_{th})$; the thermistor's resistance at 30°C (from the table) is $R_{th}(30)=80\,000\ \Omega$. Find. The $R_{ref}$ that maximizes measurement sensitivity around a 30°C operating point.
Approach. Sensitivity is $|dV_{out}/dR_{th}|=V_{ex}R_{ref}/(R_{ref}+R_{th})^2$; maximizing this over $R_{ref}$ at a fixed operating $R_{th}$ gives $R_{ref}=R_{th}$ — the matched-source condition, exactly analogous to maximum power transfer.
An 80 kΩ reference resistor gives the largest possible $dV_{out}/dR_{th}$ — and hence the largest signal per degree of temperature change — right at the 30°C "warm day" operating point the question specifies; moving further from $R_{ref}=R_{th}$ in either direction reduces sensitivity even though $V_{out}$ itself keeps changing monotonically with $R_{ref}$.
b) Given. $R_{ref}=100\,000\ \Omega$, $V_{ex}=5\ \text{V}$; table values bracketing 20.5°C: $R_{th}(20)=125\,500\ \Omega$, $R_{th}(21)=119\,800\ \Omega$. Find. Approximate sensitivity (V per °C) of the measurement at 20.5°C.
Approach. Get $dR_{th}/dT$ from the two adjacent table points (their midpoint IS 20.5°C), get $dV_{out}/dR_{th}$ from the divider-sensitivity formula, and chain-rule the two together.
| Quantity | Result |
|---|---|
| $R_{ref}$ for max. sensitivity at 30°C | $80\ \text{k}\Omega$ |
| $R_{th}(20.5^\circ\text{C})$ (interpolated) | $122\,650\ \Omega$ |
| $|dV_{out}/dR_{th}|$ at $R_{ref}=100\ \text{k}\Omega$ | $1.009\times10^{-5}\ \text{V}/\Omega$ |
| Sensitivity $S(20.5^\circ\text{C})$ | $\approx0.0575\ \text{V}/^\circ\text{C}$ |
c) Both $R_{th}(T)$ and $V_{out}(R_{th})$ are strongly NON-LINEAR (convex) functions of temperature over a 20°C span, so by Jensen's inequality the time-average of a non-linear function of a fluctuating quantity is NOT the same as the function evaluated at the average of that quantity. A concrete check using the table's own data: for temperature swinging between 20°C and 40°C (a 20°C span, mean 30°C, matching part (a)), the naive average resistance is $$\bar{R}=\dfrac{R_{th}(20)+R_{th}(40)}{2}=\dfrac{125\,500+52\,190}{2}=88\,845\ \Omega,$$ which is about 11% HIGHER than the true value $R_{th}(30^\circ\text{C})=80\,000\ \Omega$ at the actual mean temperature — because $R_{th}(T)$ curves upward (convex) as $T$ falls, the resistance spends "more of its swing" at the high-resistance, low-temperature end than a straight-line average would suggest. Converting that biased average resistance back through the calibration curve gives an APPARENT temperature of only about 27.7°C — roughly 2.3°C too cold, purely from averaging the wrong (non-linear) quantity, with no actual sensor error at all. The same bias, though smaller, survives even if the OUTPUT VOLTAGE is averaged instead of the resistance: with $R_{ref}=100\ \text{k}\Omega$, $V_{ex}=5\ \text{V}$ from part (b), $V_{out}(20)=2.783\ \text{V}$ and $V_{out}(40)=1.715\ \text{V}$ average to $2.249\ \text{V}$, against the TRUE mean-temperature value $V_{out}(30^\circ\text{C})=2.222\ \text{V}$ — a smaller but still real bias (implied apparent temperature $\approx29.5^\circ\text{C}$, about 0.5°C low), because $V_{out}(T)$ is a milder (though still non-linear) function of $T$ than $R_{th}(T)$ itself. Averaging EITHER raw electrical quantity therefore gives a temperature reading that is systematically biased relative to the true time-average temperature.
d) Convert every individual sample to TEMPERATURE first (via the full, non-linear calibration curve), and only THEN average the resulting temperature values — never average the raw resistance or raw voltage samples and convert afterward. Temperature is the physically additive quantity whose time-average is actually wanted; $R$ and $V_{out}$ are merely non-linear proxies for it, and Jensen's-inequality bias (part c) only enters when averaging is performed on the non-linear proxy instead of on the linear quantity of interest. In practice this means sampling the thermistor circuit at a rate fast enough (the same Nyquist-adequacy requirement already discussed in Q2a) to resolve how $T(t)$ actually varies, converting each sample through the calibration curve in software, and computing $T_{avg}=(1/\Delta t)\int T(t)\,dt$ (a running sum in a data-acquisition system) on the already-linearized temperature values.
