Phoebe guidesA field team has a real problem to settle: How response time sets a sampling floor They must decide what happens before they change sensor time to 90 percent in seconds on the device. Predict the direction first.
See the relationship first
The figure reads from left to right. The blue card is sensor time to 90 percent in seconds. The middle card uses this page's rule. The green card is corner frequency. Follow the arrows: set the input, use the rule, then read the result and its unit.
The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.
Derive the baseline in four moves
- 1
Name the input. The chapter baseline for sensor time to 90 percent in seconds is 30.
- 2
Name the rule. τ=t90/ln10; fc=1/(2πτ); fs,min=2fc
- 3
Put in the chapter value. Set sensor time to 90 percent in seconds to 30. The page rule gives corner frequency as 0.01222 Hz.
- 4
Read the result. Keep Hz next to the value. Use it only within the limits on this page.
Predict, then change sensor time to 90 percent in seconds
Try Predict what happens to corner frequency. Move one control, calculate, then check your idea.
Observe A slower response stretches τ, narrows physical bandwidth, and lengthens the mathematical sampling interval. Reset to 30 and compare corner frequency.
Explain Only sensor time to 90 percent in seconds moves here. The other chapter values stay fixed.
Check yourself
What should you do before you trust the result?
What does this small model leave out?
1. The sensing element has memory
A real humidity film cannot jump instantly to a new reading. In a first-order model, the remaining gap shrinks exponentially with time constant τ.
2. Recover τ from time-to-90%
At 90%Only 10%=0.1 of the gap remains.
Take logsln(0.1)=−t90/τ=−ln(10).
Make τ the subjectτ=t90/ln(10).
3. Move from τ to sampling
A first-order corner is fc=1/(2πτ). Nyquist requires fs≥2fc, so the longest interval at that floor is 1/(2fc).
This is a minimum mathematical floor, not a recommended control-loop rate.
4. Try the response-to-90%
TryMove between faster and slower sensing elements while keeping the chapter's ADC example fixed.
ObserveAt t90=30 s, τ=13.0 s, fc=0.0122 Hz, and the Nyquist floor is 0.0244 Hz: about one sample every 41 s.
ExplainA slower response stretches τ, narrows physical bandwidth, and lengthens the mathematical sampling interval.
A single first-order, isothermal response does not cover multistage diffusion, airflow, enclosure lag, humidity hysteresis, digital sensor conversion cadence, control latency, or anti-alias filter shape.
- Practical logging often samples faster than the bare Nyquist floor
- Needs separate evidence
Use field evidence or a deeper model before release.
5. Work the chapter's 30 seconds
The corresponding period 1/fc is about 82 s. That is a bandwidth scale, not a promise that every event repeats every 82 seconds.
6. Keep amplitude separate
The first line limits time resolution. The second limits voltage resolution. Neither one repairs the other.
7. Check yourself
Why divide t90 by ln(10)?
What sample interval matches the chapter's bare Nyquist floor?
Does a 10-bit ADC determine sensor bandwidth?
These are the chapter inputs, worked results, and named teaching assumptions.
- The catalog-typical 30 s humidity response
- Named teaching assumption
- 13.0 s time constant
- Time, interval, or service-life value
- 0.0122 Hz corner
- Frequency, sample rate, or event rate
- 0.0244 Hz sampling floor
- Frequency, sample rate, or event rate
- 41 s interval
- Time, interval, or service-life value
- 10-bit ADC
- Digital resolution or converter setting
- 3.3 V reference
- Voltage or voltage-step value
- 3.22 mV step
- Voltage or voltage-step value
- 62.0 dB ideal ceiling come from the chapter
- Gain, loss, margin, or level ratio
They are a teaching model, not a universal sensor data-sheet specification.