Measurements and Control Systems: SI Units, Errors and Uncertainty, Indicating Instruments, Potentiometer, Bridges, Q-Meter and Transfer Functions
2. Propagation of errors
When a result is computed from measured quantities, their errors combine. The limiting (worst-case) error assumes every error takes its most damaging sign; the root-sum-square (RSS) estimate assumes the errors are independent and random, which is more realistic for several sources and always smaller.
| Result | Limiting error | RSS (independent random) |
|---|---|---|
| R = x ± y | δR = δx + δy (absolute errors add) | σ_R = √(σ_x² + σ_y²) |
| R = xy or x/y | δR/R = δx/x + δy/y (relative errors add) | σ_R/R = √[(σ_x/x)² + (σ_y/y)²] |
| R = x^n | δR/R = n·δx/x | σ_R/R = n·σ_x/x |
| General R = f(x, y, …) | Σ|∂f/∂x_i|δx_i | √Σ(∂f/∂x_i · σ_i)² |
So power P = I²R with I known to ±1% and R to ±2% has a limiting error of 2 × 1 + 2 = ±4%, and an RSS error of √(2² + 2²) = ±2.83%. A subtraction of two nearly equal large readings is the dangerous case: the absolute errors add while the result shrinks, so the relative error explodes.
3. PMMC, moving-iron and electrodynamometer instruments
| Instrument | Deflecting torque | Scale and reading | Use |
|---|---|---|---|
| PMMC | T = NBAI, balanced by spring kθ | uniform; reads the average (dc) value | dc only; most sensitive; eddy-current damping from its aluminium former |
| Moving-iron (attraction or repulsion) | T = ½I² dL/dθ | non-uniform (square law); reads rms | ac and dc; robust, cheap ammeters and voltmeters |
| Electrodynamometer | T = I₁I₂ cos φ dM/dθ (fixed and moving coils) | reads rms, or true average power as a wattmeter | ac/dc transfer standard; wattmeter |
A PMMC movement of full-scale current I_m and resistance R_m becomes an ammeter of range I with a shunt R_sh = R_m I_m/(I − I_m) = R_m/(m − 1), m = I/I_m, and a voltmeter of range V with a series multiplier R_s = V/I_m − R_m. Its sensitivity is 1/I_m ohms per volt, and a low-sensitivity voltmeter loads a high-resistance circuit and reads low. For a waveform i = I₀ + I₁ sin ωt, a PMMC reads I₀, while a moving-iron or dynamometer ammeter reads √(I₀² + I₁²/2).
4. DC potentiometer, bridges and the Q-meter
The dc potentiometer compares an unknown emf with a known voltage drop along a uniform slide wire carrying a steady current. It is first standardised against a standard cell (the Weston cell, about 1.0186 V), and at balance the galvanometer shows null and no current is drawn from the source being measured — so the emf itself is read, free of the source’s internal resistance, and the galvanometer’s calibration does not matter, only its sensitivity.
| Bridge | Measures | Balance result |
|---|---|---|
| Wheatstone | medium R, about 1 Ω to a few MΩ | R₁R₄ = R₂R₃ (products of opposite arms equal) |
| Kelvin double | low R, below about 1 Ω | second ratio arms cancel lead and contact resistance |
| Maxwell inductance–capacitance | L of medium Q (1 < Q < 10) | L_x = R₂R₃C₄, R_x = R₂R₃/R₄, Q = ωC₄R₄ |
| Hay | L of high Q (Q > 10) | series R₄C₄ arm; balance depends on frequency |
| Anderson | L over a wide range, using a fixed standard capacitor | modified Maxwell bridge |
| Schering | C and its dissipation factor, including at high voltage | C_x = C₂R₄/R₃, D = ωC₄R₄ |
| Wien | frequency (or C) | f = 1/(2π√(R₁R₂C₁C₂)) |
The Q-meter drives a coil and a calibrated variable capacitor in series from a low-impedance oscillator of known output E, tunes to resonance, and reads Q = V_C/E directly. Its self-capacitance C_d is found by resonating at f₁ with C₁ and at nf₁ with C₂: C_d = (C₁ − n²C₂)/(n² − 1), which for n = 2 is (C₁ − 4C₂)/3.
5. Basics of control systems: the transfer function
The transfer function of a linear time-invariant system is the ratio of the Laplace transform of its output to that of its input with all initial conditions zero, G(s) = Y(s)/U(s). Its denominator is the characteristic polynomial; the roots of that polynomial are the poles and those of the numerator the zeros. A system is stable when every pole lies in the left half of the s-plane.
- Series blocks multiply: G₁G₂. Parallel blocks add: G₁ + G₂.
- A negative-feedback loop with forward path G and feedback H has closed-loop transfer function T = G/(1 + GH); the loop gain GH appears in the denominator.
- Feedback makes T nearly 1/H when GH is large, so the result is set by the feedback element and becomes insensitive to drift in G — why precision instruments and physiological regulators both rely on it.
- For unity feedback the steady-state error to a step is 1/(1 + K_p), with K_p = lims→0 G(s); one integrator in G (type 1) drives it to zero.
Unity feedback around G(s) = K/[s(s + a)] gives the closed-loop characteristic equation s² + as + K = 0, a standard second-order system with ω_n = √K and ζ = a/(2√K): raising the gain speeds the response but lowers the damping. The same loop structure describes the baroreceptor reflex, where arterial pressure is sensed, compared with a set point in the brainstem and corrected through heart rate and vessel tone.
Key takeaways
- Systematic error limits accuracy and needs calibration; random error limits precision and falls as σ/√n; precision index h = 1/(σ√2).
- Sums add absolute errors, products add relative errors, powers multiply them; RSS is smaller than the limiting error.
- PMMC: uniform scale, dc average, R_sh = R_m/(m − 1), R_s = V/I_m − R_m; MI and dynamometer: square law, rms.
- Potentiometer draws no current at balance; Kelvin for low R, Maxwell for medium-Q L, Hay for high-Q L, Schering for C and loss.
- T = G/(1 + GH) for negative feedback; poles in the left half-plane mean stability.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which of these is NOT an SI base unit?
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Answer: A — volt
The seven base units are metre, kilogram, second, ampere, kelvin, mole and candela. The volt is derived, W/A = kg·m²·s⁻³·A⁻¹. The ampere is the base electrical unit, which is why the volt is not.In the Gaussian law of errors written as y = (h/√π)e−h²x², the precision index h is related to the standard deviation σ by
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Answer: A — h = 1/(σ√2)
Matching the exponent with −x²/(2σ²) gives h² = 1/(2σ²), so h = 1/(σ√2): the smaller the scatter, the larger the index. 0.6745σ is the probable error, a different quantity with the units of the reading.Five readings of a reference voltage are 98, 100, 102, 100 and 100 mV. Their sample standard deviation (in mV), to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.414
The mean is 100 mV and the squared deviations sum to 4 + 0 + 4 + 0 + 0 = 8, so s = √(8/(5 − 1)) = √2 = 1.414 mV. Dividing by n instead of n − 1 gives √1.6 = 1.265 mV, the population formula applied to a sample.Power is computed as P = I²R with I = 2 A ± 1% and R = 50 Ω ± 2%, the two errors being independent and random. The root-sum-square relative uncertainty in P (in %), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.83
The I² term contributes 2 × 1% = 2% and R contributes 2%, so the RSS uncertainty is √(2² + 2²) = √8 = 2.83%. Adding them gives the limiting error 4%; forgetting the square on I gives √(1² + 2²) = 2.24%.A PMMC movement has full-scale current 1 mA and resistance 100 Ω. The shunt resistance (in Ω) needed to make it a 0–1 A ammeter, to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.1001
m = 1/0.001 = 1000, so R_sh = R_m/(m − 1) = 100/999 = 0.1001 Ω; the shunt carries 999 mA while the meter carries 1 mA at the same voltage, 0.1 V. Dividing by m instead of m − 1 gives 0.1 Ω, a small but real error in a precision shunt.The same 1 mA, 100 Ω PMMC movement is to become a 0–10 V voltmeter. The series multiplier resistance (in Ω) is ____.
Numerical answer — type the value.
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Answer: 9900
The total resistance must be V/I_m = 10/0.001 = 10 000 Ω, so R_s = 10 000 − 100 = 9900 Ω, giving a sensitivity of 1000 Ω/V. Quoting 10 000 Ω forgets that the coil’s own resistance is already part of the path.A current i = 5 + 5 sin ωt A flows through a PMMC ammeter. The meter reads
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Answer: A — 5 A
The PMMC torque follows the instantaneous current and the pointer settles at its average, which is the dc component, 5 A. 6.12 A = √(25 + 12.5) is the rms value a moving-iron meter would show, and 10 A is the peak.A current i = 3 + 4 sin ωt A flows through a moving-iron ammeter. Its reading (in A), to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 4.123
A moving-iron meter reads rms: √(3² + 4²/2) = √(9 + 8) = √17 = 4.123 A. Adding the dc and the ac rms (3 + 2.83 = 5.83 A) is the classic slip — rms values of different frequencies combine in quadrature — and √(9 + 16) = 5 A uses the ac peak instead of its rms.A Maxwell inductance–capacitance bridge balances with R₂ = 1 kΩ, R₃ = 1 kΩ, and a 0.5 µF capacitor C₄ in parallel with R₄ = 10 kΩ. The unknown inductance (in H) is ____.
Numerical answer — type the value.
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Answer: 0.5
L_x = R₂R₃C₄ = 10³ × 10³ × 0.5 × 10⁻⁶ = 0.5 H, and R_x = R₂R₃/R₄ = 100 Ω, so at ω = 1000 rad/s Q = ωC₄R₄ = 5, within the medium-Q range this bridge suits. Multiplying by R₄ as well gives 5000 H, which confuses the inductance balance with the Q expression.The Schering bridge is used chiefly to measure
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Answer: A — capacitance and dielectric loss (dissipation factor)
The Schering bridge compares a capacitor with a loss-free standard and yields both C_x and its dissipation factor D = ωC₄R₄, even at high voltage. Low resistance is the Kelvin double bridge, high-Q inductance is Hay’s bridge and frequency is Wien’s.In a Q-meter a coil resonates at frequency f with 460 pF and at 2f with 100 pF. Its self-capacitance (in pF) is ____.
Numerical answer — type the value.
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Answer: 20
f ∝ 1/√[L(C + C_d)], so doubling f needs (C₁ + C_d) = 4(C₂ + C_d), giving C_d = (C₁ − 4C₂)/3 = (460 − 400)/3 = 20 pF. Using C₁ − 2C₂ (treating f as proportional to 1/C) gives 260 pF.A forward path G(s) = 10/(s + 1) is closed with unity negative feedback. The dc gain of the closed loop, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.909
T(s) = G/(1 + G) = 10/(s + 11), so T(0) = 10/11 = 0.909, and the closed-loop time constant is 1/11 s against the open loop’s 1 s. Taking the open-loop dc gain 10 ignores the feedback; 10/9 comes from using 1 − G for negative feedback.Which statements about the dc potentiometer are correct?
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Answer: A — at balance it draws no current from the emf being measured; B — it is standardised against a standard cell; D — its accuracy does not depend on the galvanometer’s calibration
It is a null method: at balance no current flows in the galvanometer branch, so the source’s internal drop vanishes and the galvanometer only has to detect zero, not measure. The working current is standardised with a standard cell. It is a dc instrument; ac potentiometers are a separate class.Which statements about resistance bridges are correct?
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Answer: A — the Kelvin double bridge is meant for resistances below about 1 Ω; B — the Kelvin bridge’s second set of ratio arms removes the effect of lead and contact resistance; C — the Wheatstone bridge balances when the products of opposite arms are equal
Below about 1 Ω the leads and contacts are comparable to the unknown, and the Kelvin bridge’s inner ratio arms cancel the connecting link’s resistance. The Wheatstone balance R₁R₄ = R₂R₃ is the equality of opposite-arm products. Neither is an inductance bridge — that is Maxwell, Hay or Anderson.A unity-feedback system has G(s) = K/[s(s + 2)]. For the closed loop to have damping ratio ζ = 0.5, K must be
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Answer: A — 4
The characteristic equation is s² + 2s + K = 0, so ω_n = √K and 2ζω_n = 2, i.e. ζ = 1/√K. ζ = 0.5 needs √K = 2, K = 4 (ω_n = 2 rad/s). K = 1 gives ζ = 1, critical damping, and K = 2 gives ζ = 0.707.