SME
Measurements & Instrumentation · Problem type
First-order sensor step response
Use this for “A thermometer with τ = 4 s goes from 20 °C into 100 °C water. What does it read after 10 s?”
Chart
y(t) = 100 + (20 − 100)exp(−t/τ), τ = 4 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 10 s: y = 93.43 °C.
This is an example. Type your own values to start a new problem.
y (°C)
=
yf (°C)
+
(
y₀ (°C)
−
yf (°C)
)
×
e
−
÷
f
Share reached
calculated · %
Formulas
f
=
100
×
(1
−
e
−
t
÷
τ
)
91.7915
=
100
×
(1
−
e
−10 ÷ 4
)
y
=
yf
+
(
y₀
−
yf)
×
e
−
t
÷
τ
93.4332
=
100
+
(20
−
100)
×
e
−10 ÷ 4
Assumptions
•
A first-order sensor: one energy store, so it lags by one time constant τ.
•
The measured value jumps at t = 0 and then holds.
•
After 3τ it has made 95% of the change, after 5τ more than 99%.
Step-by-step
WE KNOW
t = 10 s, τ = 4 s, y₀ = 20 °C, yf = 100 °C
FIND
share reached (f), reading (y)
Step 1 · Find share reached (f)
Use f = 100 × (1 − e−t ÷ τ)
The share of the step still to go falls by the factor e each time constant; the rest is made.
f
=
100
×
(1
−
e
−10 ÷ 4
)
f
=
100
×
(1
−
0.08208)
f
=
100
×
0.9179
f
=
91.7915%
Step 2 · Find reading (y)
Use y = yf + (y₀ − yf) × e−t ÷ τ
The
gap
to
the
final
value
shrinks
by
the
factor
e
−
t
÷
τ
;
add
what
is
left
of
it
to
yf.
y
=
100
+
(20
−
100)
×
e
−10 ÷ 4
y
=
100
−
80
×
0.082085
y
=
100
−
6.5668
y
=
93.4332
°C
Check
Put the numbers back in. Both sides should match.
91.7915
=
100
×
(1
−
e
−10 ÷ 4
)
✓
93.4332
=
100
+
(20
−
100)
×
e
−10 ÷ 4
✓
Topic check
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