The two secondaries of a linear variable differential transformer (LVDT) showed a magnitude of 2 V (RMS) for zero displacement position of the core. It is noted that the phase of one of the secondaries has a deviation of one degree from the expected phase. Other than this deviation, the LVDT is ideal.
If the differential output sensitivity of the LVDT is 1 mV (RMS)/1 µm, the output for zero displacement is __________ µm (rounded off to one decimal place).
ANSWER
The output for zero displacement is about 34.9 µm (equivalent).
Official answer: 34.5 to 35.5Source: Official answer key, GATE 2023 IN (IIT Kanpur)
At the null position of an ideal LVDT the two secondary voltages have equal magnitude and equal phase, so the series-opposition (differential) output is zero. Here the only defect is that one secondary is out of phase.
Take each secondary magnitude as , with and , so .
The differential output is the phasor difference: .
Substituting: .
This residual (quadrature) voltage is nearly out of phase with the in-phase signal, so moving the core cannot cancel it. It remains at the null position as the output for zero displacement.
Using the sensitivity of , the equivalent displacement is .
KEY IDEA
A phase error between the two secondaries cannot be nulled by core position. The differential output is the phasor difference, , which leaves a residual quadrature voltage at zero displacement. Dividing by the sensitivity gives the equivalent displacement.
WATCH OUT FOR
Treating the zero-displacement output as exactly zero because the magnitudes are equal. Subtracting magnitudes instead of phasors gives 0. Another slip is using (one secondary against the other as reference) or converting the angle in degrees directly instead of radians.
From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 19 Sensing principles questions, with notes
