Measurements · Module 3 of 10
Measurement System Dynamic Response
Instruments take time to respond, and reading them too soon gives the wrong answer. This module models measurement systems as first and second order and quantifies their lag through the time constant and natural frequency.
Readiness check
This module models how instruments respond in time. Tick only what you can do closed-notes.
- Evaluate a natural logarithm.
- Recall that a decaying exponential settles toward a final value.
- Take a square root of a ratio.
- Recall the mass-spring natural frequency idea.
- Multiply by 2π to convert rad/s to Hz.
The core idea
A measurement system is modelled by its order. A first-order system responds to a step as 1 − e−t/τ, reaching 63.2 percent of the change in one time constant τ. A second-order system adds inertia, with a natural frequency ωn = √(k/m). The lag these describe is the dynamic error.
first order: y/y∞ = 1 − e−t/τtime to a fraction: t = τ ln(1/(1−fraction))second order: ωn = √(k/m)No real instrument responds instantly; it takes time to follow a changing input, and that lag is a measurable, predictable error. The simplest useful model is first order, governed by a single time constant τ. Following a sudden step change, a first-order system approaches its new value as 1 − e−t/τ, so after one time constant it has covered 63.2 percent of the change, after about three constants 95 percent, and after five about 99 percent. A thermometer plunged into hot water behaves this way. To reach a chosen fraction of the change takes t = τ ln(1/(1−fraction)), so demanding 90 percent takes τ ln(10), about 2.3 time constants. Systems with mass and stiffness are second order, adding a natural frequency ωn = √(k/m) and a damping ratio ζ that together decide whether the response overshoots and rings. The gap between what the instrument reads and the true instantaneous input is the dynamic error, and keeping it small means choosing a system fast enough, small τ or high ωn, for the signal you are measuring. This is why a slow sensor cannot faithfully record a fast event.
The skills, taught in order
Five skills quantify how fast an instrument can be trusted.
3.1 System order
A zero-order system responds instantly and scales by its sensitivity; a first-order system has one time constant; a second-order system has mass and stiffness, so it can overshoot. Identifying the order sets which model applies.
3.2 The first-order time constant
The time constant τ is the single number that governs a first-order response. In one τ the system covers 63.2 percent of a step, and the smaller τ is, the faster and more faithful the instrument.
| Time elapsed | Fraction of step reached |
|---|---|
| 1 τ | 63.2% |
| 3 τ | 95.0% |
| 5 τ | 99.3% |
The universal first-order settling table. Waiting about five time constants is the usual rule for a settled reading.
3.3 Dynamic error and time lag
While the system settles, its reading differs from the true input; that difference is the dynamic error, and the delay is the time lag. Both shrink as the input changes more slowly relative to τ.
3.4 Second-order response
A second-order system is set by its natural frequency ωn = √(k/m) and damping ratio ζ. Light damping lets it overshoot and ring near ωn; heavy damping makes it sluggish. Good instruments are damped near critical.
3.5 Frequency response and bandwidth
An instrument passes low frequencies faithfully but attenuates and delays those near or above its characteristic frequency. The band it reproduces accurately is its bandwidth, which must cover the signal's spectrum.
Engineering connection: a thermocouple with a 5 s time constant cannot track a temperature that swings every second; you either pick a faster probe or correct for the known lag.
Worked example 1: first-order settling time
A thermometer behaves as a first-order system with a time constant of 5 s. How long after a step change does it reach 90 percent of the change?
- ProblemFind the time to 90 percent for the thermometer in Figure 1.
- Given / findτ = 5 s, target fraction 0.90. Find the time.
- AssumptionsIdeal first-order response to a step input.
- Modelt = τ ln(1/(1 − fraction)).
- Equationst = 5 × ln(1/0.10)= 5 × ln 10
- Solvet = 5 × 2.303 = 11.5 s.
- CheckAt one τ (5 s) it is at 63.2 percent, and 90 percent should take a bit over two constants, which 11.5 s is.
- ConclusionThe thermometer needs about 11.5 s to read within 10 percent of the true step, so a faster event would be badly distorted.
Worked example 2: second-order natural frequency
A force sensor behaves as a second-order system with stiffness k = 2000 N/m and effective mass m = 0.5 kg. Find its natural frequency in rad/s and in Hz.
- ProblemFind the natural frequency of the sensor in Figure 2.
- Given / findk = 2000 N/m, m = 0.5 kg. Find ωn and fn.
- AssumptionsLinear second-order system; natural frequency taken undamped.
- Modelωn = √(k/m); fn = ωn/2π.
- Equationsωn = √(2000/0.5) = √4000fn = ωn/2π
- Solveωn = 63.2 rad/s; fn = 63.2/6.283 = 10.06 Hz.
- Check√4000 ≈ 63.2, and dividing by about 6.28 gives near 10 Hz, consistent.
- ConclusionThe sensor resonates near 10 Hz, so it can faithfully measure forces well below that, but signals near 10 Hz would be amplified by resonance.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Assuming instant response | Reading a transient too early | "How many time constants have passed?" | Wait about five τ for a settled reading. |
| τ as full response time | Thinking one τ means done | "Is 63 percent enough?" | One τ gives only 63.2 percent of the change. |
| Slow sensor on fast signal | Amplitude flattened, phase lagged | "Is the sensor faster than the signal?" | Match bandwidth to the signal's spectrum. |
| Ignoring resonance | Readings amplified near ωn | "Is the signal near the natural frequency?" | Keep the signal well below ωn or damp it. |
Practice ladder
A first-order sensor has τ = 2 s. How long to reach 95 percent of a step?
Show answer
t = 2 × ln(1/0.05) = 2 × 3.00 = 6.0 s (about three time constants).
A first-order instrument has τ = 0.5 s. Find the time to reach 99 percent.
Show answer
t = 0.5 × ln(1/0.01) = 0.5 × 4.605 = 2.30 s.
A second-order system has k = 5000 N/m and m = 2 kg. Find ωn and fn.
Show answer
ωn = √(5000/2) = √2500 = 50 rad/s; fn = 50/2π = 7.96 Hz.
You must record a temperature that changes noticeably every 2 s. What thermocouple time constant would you specify, and why?
What good work looks like
To follow a 2 s change with small dynamic error, τ should be much smaller than 2 s, ideally about 0.2 s or less so several time constants fit within each change. A good answer ties τ to the input timescale and notes the settling-time rule.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Model one instrument as first or second order, estimate its time constant or natural frequency, and judge whether it can follow a real signal.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What fraction is reached in one time constant?
63.2 percent of a step change.
2. Write the time to a given fraction.
t = τ ln(1/(1 − fraction)).
3. Write the natural frequency.
ωn = √(k/m).
4. What is dynamic error?
The gap between the reading and the true input while the system settles.
5. Why keep signals below ωn?
Near resonance a lightly damped system amplifies and distorts the reading.
Textbook mapping
This module follows Figliola and Beasley, Theory and Design for Mechanical Measurements, 5th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| The general measurement system model | Figliola and Beasley, Section 3.2, General Model for a Measurement System |
| First and second-order systems | Figliola and Beasley, Section 3.3, Special Cases of the General System Model |
| Transfer functions and phase | Figliola and Beasley, Sections 3.4 and 3.5 |
Section numbers refer to Figliola and Beasley, 5th edition. Any edition with the same chapter titles is equivalent for study.