Calibrating a Voltage Sensor with a Modulus Response
A pressure-rig technician is validating a sensor whose output voltage v, in volts, rises with test time t, in minutes, according to v(t) = 3t/(t + 1) for t ≥ 0. The display does not show voltage directly; it shows a processed bar reading R(v) = |v − 2| + 1, a modulus function chosen so that the display never shows less than 1 bar. During a 40-minute run the logger records a reading of 1.5 bar at two separate moments, and the operator, assuming a single reading should correspond to a single time, logs the run as a sensor fault. Before the rig is stripped down, the engineer must decide whether the double reading is a genuine fault or an expected consequence of the functions involved. The safety case also needs the display reading written directly as a function of time, together with the maximum reading the display can reach, and the time at which the sensor voltage passes through 2 V, since that is the point where the display behaviour changes.
1. Write R(v(t)) as a function of t alone, in piecewise form if necessary, and state its range.
Read the worked answer
First find the range of v(t). For t ≥ 0, v(t) = 3t/(t + 1) = 3 − 3/(t + 1), which increases from 0 towards 3 but never reaches 3. So 0 ≤ v < 3.
Split R at its turning point v = 2, i.e. where 3t/(t + 1) = 2, giving 3t = 2t + 2, so t = 2.
For 0 ≤ t ≤ 2 (so v ≤ 2): R = (2 − v) + 1 = 3 − v = 3 − 3t/(t + 1) = 3/(t + 1). At t = 0 this is 3; at t = 2 it is 1.
For t > 2 (so v > 2): R = (v − 2) + 1 = v − 1 = 3t/(t + 1) − 1 = (2t − 1)/(t + 1). At t = 2 this is 3/3 = 1; as t → ∞ it tends to 2 without reaching it.
So R(t) = 3/(t + 1) for 0 ≤ t ≤ 2, and R(t) = (2t − 1)/(t + 1) for t > 2.
Range: the first branch contributes [1, 3]; the second contributes (1, 2). The union is [1, 3], so the display can show anything from 1 bar up to 3 bar, with 3 bar only at t = 0.
2. The log shows a reading of 1.5 bar at two different times. Find both times algebraically and explain why this is not evidence of a faulty sensor. State what restriction on t would make the reading reversible.
Read the worked answer
Solve R(t) = 1.5 on each branch.
Branch 0 ≤ t ≤ 2: 3/(t + 1) = 1.5 gives t + 1 = 2, so t = 1, which lies in [0, 2].
Branch t > 2: (2t − 1)/(t + 1) = 1.5 gives 2t − 1 = 1.5t + 1.5, so 0.5t = 2.5 and t = 5, which is in the branch.
Check: at t = 1, v = 3/2 = 1.5 V and |1.5 − 2| + 1 = 1.5 bar. At t = 5, v = 15/6 = 2.5 V and |2.5 − 2| + 1 = 1.5 bar.
So the two readings occur because the modulus folds voltages symmetrically about 2 V: 1.5 V and 2.5 V are both 0.5 V from 2 V and therefore display identically. R(v) is not one-to-one on [0, 3), so a repeated reading is expected behaviour, not a fault.
To make the reading reversible, restrict the voltage to one side of the fold: either 0 ≤ v ≤ 2 (equivalently 0 ≤ t ≤ 2, where R falls from 3 to 1) or 2 ≤ v < 3 (t ≥ 2, where R rises from 1 towards 2). On either restricted piece the function is monotonic and hence invertible.
3. Find the inverse of v(t) and hence the time at which the sensor voltage reaches 2 V. Comment on the behaviour of the display at that instant.
Read the worked answer
Let v = 3t/(t + 1). Then v(t + 1) = 3t, so vt + v = 3t, giving t(3 − v) = v and t = v/(3 − v). This is defined for 0 ≤ v < 3, matching the range of v(t), and reproduces t = 0 at v = 0.
Voltage 2 V: t = 2/(3 − 2) = 2 minutes, consistent with the branch point found earlier.
Display at t = 2: taking the limit from the first branch, R → 3/(2 + 1) = 1; from the second branch, R → (2(2) − 1)/(2 + 1) = 1. The two formulas agree at t = 2, so R(t) is continuous there. The reading reaches its minimum value of 1 bar at t = 2 and then rises again towards 2 bar.
Implication for the safety case: the display passes through 1 bar once, at t = 2 min, and any value between 1 and 2 bar is displayed twice (once before and once after t = 2). Time must therefore be used to disambiguate the reading, or the modulus removed from the signal path.