Math Bridge: Time Constant and -3 dB Bandwidth

A field team faces an unresolved physical question: Where τ and the -3 dB bandwidth come from They must answer it before changing sensor time constant on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is sensor time constant. The middle card applies this page's relationship. The green card is -3 db bandwidth. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Walk the arrows. The readouts use response = 1 - e^(-0.5/τ) and f_3dB = 1/(2πτ), the same formulas this page derives.

Derive the baseline in four named moves

1
Name the input. The chapter baseline for sensor time constant is 2.
2
Name the relationship. g(t) = g(0) x e^(-t/τ); response = 1 - e^(-t/τ); f_3dB = 1/(2πτ)
3
Substitute the chapter fixture. Set sensor time constant to 2. The page ledger gives -3 db bandwidth as 0.0796 Hz.
4
Read the result. Keep Hz beside the value. Use it only inside the technical boundary on this page.

Predict, then change sensor time constant

Try Predict the direction of -3 db bandwidth. Move one control, calculate, then check your prediction.

Observe The readouts use response = 1 - e^(-0.5/τ) and f_3dB = 1/(2πτ), the same formulas this page derives. Reset the control to 2 and compare -3 db bandwidth.

Explain Only sensor time constant moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only sensor time constant moves. Field effects named in the page's technical boundary stay fixed.

1. Begin with the physical story

Imagine the thermistor starts at 0% of a temperature step and the world suddenly moves to 100%. The sensor cannot jump there. Its thermal mass makes it close the gap gradually.

Phoebe: Keep your eye on the gap. At the beginning the gap is large, so the reading moves quickly. Near the end the gap is small, so it moves slowly. That single idea is the whole first-order model.

What does a derivative say?

A derivative is a rate: how fast something is changing at this instant. If x is the sensor reading and t is time, then dx/dt means “the reading's change per second.” It is the slope you would see if you zoomed into the response curve at one moment.

dx/dt = (x_final − x) / τ

Read that equation in words: the sensor's speed of change equals the gap still left to close, divided by the time constant.

Piece	Physical meaning	Unit check
x_final − x	The temperature gap the sensor has not caught yet.	Temperature
τ	How much the sensor's physics slows the catch-up.	Seconds
dx/dt	How fast the indicated temperature is changing now.	Temperature per second

2. Exponentials are “decay of the gap”

Let the remaining gap be g = x_final − x. A first-order sensor removes the same fraction of that gap over equal stretches of time. That is exponential decay.

g(t) = g(0) × e^(−t/τ); response = 1 − e^(−t/τ); f_3dB = 1/(2πτ)

The number e is about 2.718. You do not need to worship it or memorise its digits. It appears because continuous “fraction of what remains” change produces an exponential. At one time constant, t = τ, the gap multiplier is e⁻¹ ≈ 0.368. So 36.8% of the gap remains and 63.2% has been closed.

Phoebe: τ is not “time until finished.” It is the time needed to remove about 63% of the original gap. The last part takes longer because the remaining gap keeps shrinking.

TryMove τ from 0.5 s to 5 s and watch how much of the fixed 0.5 s event the sensor can show.

ObserveA larger τ makes the half-second response smaller and the −3 dB bandwidth lower.

ExplainThe readouts use response = 1 − e^(−0.5/τ) and f_3dB = 1/(2πτ), the same formulas this page derives.

Technical boundaries.

The slider keeps the event duration at the chapter's 0.5 s and varies only the ideal first-order time constant; a mounted probe still

a physical step test: Needs separate evidence

Use field evidence or a deeper model before release.

3. Derive the step response, one named move at a time

Assume the sensor starts at zero, x(0) = 0, and the final input x_final stays constant after the step.

Start with the physical lawdx/dt = (x_final − x) / τ

Separate the variablesMove every x term beside dx and move dt to the other side.dx / (x_final − x) = dt / τ

Integrate both sidesThe left integral is a logarithm; the minus sign appears because the derivative of x_final − x is −1.−ln(x_final − x) = t/τ + C

Collect the constantMultiply by −1 and rename the arbitrary constant.ln(x_final − x) = −t/τ + C₁

Undo the logarithmRaise e to both sides. The exponentiated constant becomes a new positive constant A.x_final − x = A e^(−t/τ)

Use the starting conditionAt t = 0 and x = 0, the equation says x_final = A × 1, so A = x_final.x_final − x = x_final e^(−t/τ)

Make x the subjectSubtract the remaining-gap term from x_final.x(t) = x_final [1 − e^(−t/τ)]

That last line is the chapter's step-response equation. Nothing new was assumed between the physical “close the gap” law and the exponential curve.

4. Turn τ into the chapter's settling numbers

Divide the response by the final value to get the fraction completed:

fraction completed = x(t)/x_final = 1 − e^(−t/τ)

Time	Substitution	Completed	Chapter meaning
1τ = 2 s	1 − e⁻¹	0.632 = 63.2%	One time constant is still far from settled.
3τ = 6 s	1 − e⁻³	0.950 = 95.0%	This gives the chapter's roughly 6-second 95% point.
5τ = 10 s	1 − e⁻⁵	0.993 = 99.3%	This is the chapter's “about 99%” milestone.

5. Derive the −3 dB bandwidth from the same model

A time response and a frequency response are two views of the same sensor. Start from the first-order transfer function used by the chapter's model:

Write the transfer functionH(s) = 1 / (1 + sτ)

Ask about a sinusoidFor frequency response, replace s with jω, where ω is angular frequency in radians per second.H(jω) = 1 / (1 + jωτ)

Take the magnitudeThe size of 1 + jωτ is √[1 + (ωτ)²].|H(jω)| = 1 / √[1 + (ωτ)²]

Define the −3 dB point−3 dB in amplitude means the output is 1/√2 ≈ 0.707 of the low-frequency amplitude.1 / √[1 + (ωτ)²] = 1 / √2

Square both sides1 / [1 + (ωτ)²] = 1/2

Cross-multiply2 = 1 + (ωτ)²

Subtract 1 and take the positive rootFrequency is non-negative, so use ωτ = 1.ω = 1/τ

Convert radians per second to hertzBecause ω = 2πf, divide by 2π.f_3dB = 1 / (2πτ)

Phoebe: The −3 dB point is a marker, not a brick wall. Frequencies above it are not magically zero; they are increasingly attenuated. It also is not, by itself, an anti-alias guarantee.

6. Work every chapter-specific number

Bandwidth of the 2-second thermistor

f_3dB = 1 / (2 × π × 2 s) = 1 / 12.566 s = 0.0796 Hz

That is below 0.1 Hz, exactly as the chapter states.

What 100 samples per second buys

100 samples/s × 2 s = 200 samples per τ

100 samples/s × 6 s = 600 samples by the 95% point

Those samples describe the same slow climb in fine detail. They do not make the thermistor reach the final temperature sooner.

What a half-second spike looks like

fraction seen = 1 − e^(−0.5/2) = 1 − e⁻⁰·²⁵ = 0.221 = 22.1%

Even in the ideal first-order model, after 0.5 seconds the probe has shown only about 22.1% of the step. The event-rate comparison tells the same story:

event rate = 1 / 0.5 s = 2 Hz

2 Hz / 0.0796 Hz = 25.1

The half-second event's simple rate check is about 25 times the sensor's −3 dB bandwidth. The honest design response is a faster sensor or a changed requirement, not pretending that a faster ADC recovers the peak.

7. Check yourself

Try each question before revealing the answer. Writing one line of working is enough.

1. In plain language, what does dx/dt = (x_final − x)/τ say?

Answer: The reading changes faster when the remaining gap is large and slower when the gap is small; τ sets how quickly that catch-up happens.

2. For τ = 2 s, what percentage of the step is complete after 2 s?

Answer: One τ has passed, so 1 − e⁻¹ = 0.632: about 63.2%.

3. Why does the chapter use about 6 s for the 95% point?

Answer: Three time constants have passed: 3 × 2 s = 6 s, and 1 − e⁻³ = 0.950.

4. Calculate the −3 dB bandwidth for τ = 2 s.

Answer: 1/(2πτ) = 1/(2 × π × 2) = 0.0796 Hz, below 0.1 Hz.

5. Does sampling at 100 samples/s make the half-second spike visible at full size?

Answer: No. The sensor itself reaches only 22.1% of the step in 0.5 s. Sampling faster records more points on that attenuated response.

Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

Every numerical input on this page comes from the companion chapter: τ = 2 s: Time, interval, or service-life value
100 samples/s: Frequency, sample rate, or event rate
a 0.5 s spike: Time, interval, or service-life value
3τ ≈ 95%: Percentage, ratio, or gain
5τ ≈ 99% milestones: Percentage, ratio, or gain
the “below 0.1 Hz” bandwidth statement: Frequency, sample rate, or event rate
22.1%: Percentage, ratio, or gain
0.0796 Hz: Frequency, sample rate, or event rate
600-sample: Device, payload, or sample count
200-sample-per-τ: Device, payload, or sample count
2 Hz: Frequency, sample rate, or event rate
25.1×: Percentage, ratio, or gain

The first-order model is an approximation; a real mounted probe must still be step-tested.