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SCIENCE · INVESTIGATE AND EXPLAIN

Infer a decay time separately from a voltage offset

A discharge trace can appear to approach a nonzero voltage because of a fixed instrument offset. The supplied true signal is 8exp(−t/2) V, but the indicated signal is Aexp(−t/τ)+o with A equal to instrument gain times true initial voltage. We deliberately keep the source generator visible for teaching while asking which parameters the observations alone identify.

Learning outcome: Compare offsets −1,0,+1 V at one gain and show which differences remain invariant. Then change gain and check that inferred τ remains the same while fitted amplitude changes. Freeze a third-second prediction and propose an independent reference to distinguish gain from initial voltage.

Starting recap

Use charge continuity, energy per charge, consistent SI units and an explicit circuit topology. Exponential responses are supplied algebraic models; initial state, source constraint and measurement loading must be declared.

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Try

With gain1.1 and offset .5 V, the indicated amplitude is 8.8 V, but differences retain decay factor exp(−.5)≈.606531 and recover τ=2 s. The fit gives offset .5 V and amplitude8.8 V, not independent proof of gain1.1 or true initial8 V.

Open the experiment

Predict

The exponential trace alone identifies amplitude A as:

Experiment

Compare offsets −1,0,+1 V at one gain and show which differences remain invariant. Then change gain and check that inferred τ remains the same while fitted amplitude changes. Freeze a third-second prediction and propose an independent reference to distinguish gain from initial voltage.

Infer a decay time without mistaking voltage offset for leakage

Worked teaching model · synthetic values · Synthetic discharge true V=8 exp(−t/2) V at exact t=0,1,2 s, indicated V=gain times true V plus offset. Three samples fit one exponential with constant offset. Gain and true initial voltage are not separately identifiable without an independent reference.

Reset keeps your prediction, captures and practice history.

A common gain and offset can conceal a true discharge

indicated voltage (V)time (s)069.30

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: offset exponential (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for A common gain and offset can conceal a true discharge

offset exponential

time (s)indicated voltage (V)
09.3
0.158.66414
0.38.07423
0.457.52694
0.67.0192
0.756.54815
0.96.11113
1.055.70569
1.25.32954
1.354.98058
1.54.65683
1.654.35647
1.84.07781
1.953.81929
2.13.57945
2.253.35694
2.43.15051
2.552.95899
2.72.78131
2.852.61647
32.46355
3.152.32167
3.32.19004
3.452.06792
3.61.95463
3.751.84952
3.91.75201
4.051.66155
4.21.57762
4.351.49975
4.51.42751
4.651.36049
4.81.29832
4.951.24063
5.11.18712
5.251.13747
5.41.09141
5.551.04868
5.71.00903
5.850.972249
60.938126

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Indication at zero seconds9.3V
Indication at one second5.83747V
Indication at two seconds3.73734V
Successive difference ratio0.606531ratio
Recovered time constant2s
Recovered indicated exponential amplitude8.8V
Recovered offset0.5V
Predicted indication at three seconds2.46355V

Differences remove a fixed offset; their ratio determines the decay factor under the single-exponential hypothesis.

With equally spaced times Delta t=1 s, successive differences have ratio exp(−Delta t/tau). The natural logarithm reverses the exponential: tau=−Delta t/ln(ratio). The fitted amplitude is gain times true initial voltage, so those two factors cannot be separated by this trace alone. A changed time constant might arise from capacitance, resistance or parallel leakage; an independent static resistance or capacitance reference is needed.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Differences remove a fixed offset; their ratio determines the decay factor under the single-exponential hypothesis.

With equally spaced times Delta t=1 s, successive differences have ratio exp(−Delta t/tau). The natural logarithm reverses the exponential: tau=−Delta t/ln(ratio). The fitted amplitude is gain times true initial voltage, so those two factors cannot be separated by this trace alone. A changed time constant might arise from capacitance, resistance or parallel leakage; an independent static resistance or capacitance reference is needed.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define topology, recipient and initial conditions

A discharge trace can appear to approach a nonzero voltage because of a fixed instrument offset. The supplied true signal is 8exp(−t/2) V, but the indicated signal is Aexp(−t/τ)+o with A equal to instrument gain times true initial voltage. We deliberately keep the source generator visible for teaching while asking which parameters the observations alone identify.

Derive the quantitative response

For readings at equally spaced times 0,1,2 s, let D01=R0−R1 and D12=R1−R2. The common offset cancels, and D12/D01=exp(−1/τ). The natural logarithm is the inverse of exp, so τ=−1/ln(D12/D01) seconds. Then A=D01/[1−exp(−1/τ)] and o=R0−A. This is an algebraic inversion of a supplied exponential, not a required differential-equation solution.

Check evidence and the limits of the model

Even an exact fit cannot separate gain from true initial voltage because only their product enters A. It also does not uniquely attribute a changed τ to capacitance: resistance, leakage or other unmodeled paths can alter the effective time constant. A fourth withheld time checks the single-exponential model, while an independent voltage reference or static resistance/capacitance observation tests different assumptions. Small differences and ratios near one can amplify reading errors, so a neat displayed result does not imply precise physical identification.

Worked original case

With gain1.1 and offset .5 V, the indicated amplitude is 8.8 V, but differences retain decay factor exp(−.5)≈.606531 and recover τ=2 s. The fit gives offset .5 V and amplitude8.8 V, not independent proof of gain1.1 or true initial8 V.

Investigation

Compare offsets −1,0,+1 V at one gain and show which differences remain invariant. Then change gain and check that inferred τ remains the same while fitted amplitude changes. Freeze a third-second prediction and propose an independent reference to distinguish gain from initial voltage.

Changed-case practice

Try the changed case before feedback

Equal-time voltage differences have ratio exp(−.25), and the sampling interval is 1 s. Find τ in seconds.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and evidence boundary

Original numerical cases and prose checked 30 October 2026 against the relevant primary explanation. All records are synthetic and all diagrams mathematical. No practical wiring, battery testing, capacitor handling, mains measurement or equipment safety rating is requested. Public feedback does not verify independent mastery; retain first predictions, contrasting recipes and criterion-based critique in the separate course evidence workspace.

Apply

The exponential trace alone identifies amplitude A as:

Public practice checks a specific claim. It does not establish independent mastery.

Which independent observation could reject this model rather than merely refit its parameters?

Continue

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