From Sensor Reading to Calibrated Quantity
A sensor usually produces a raw reading, such as a voltage, resistance, or digital count. A calibration model converts that reading into the quantity of interest. For example, a voltage may be converted into temperature, or a resistance into concentration.
Even after calibration, the input reading is not perfectly exact. If the raw reading has a small measurement error, the calibrated output changes as well. The derivative of the calibration model describes how strongly the output responds to a small change in its input. It therefore gives a direct way to estimate how input uncertainty affects the calibrated result.
In this tutorial, the calibration model itself is treated as known. We estimate the output uncertainty caused by error in the measured input; uncertainty in the calibration coefficients is not included.
What the Derivative Means in a Calibration
The value \(f'(x_0)\) is the local sensitivity of the calibrated output to changes in the raw reading near \(x_0\). Its units are output units divided by input units. For example, if \(x\) is measured in volts and \(y\) in degrees Celsius, then \(f'(x_0)\) is measured in degrees Celsius per volt.
The signed estimate \(f'(x_0)h\) preserves the direction of change. If it is positive, the small input error is estimated to raise the calibrated output; if it is negative, the output is estimated to fall. An uncertainty magnitude is nonnegative, so its first-order estimate uses the absolute value \(|f'(x_0)|\delta\).
The central calculation is short, but its interpretation matters: use the derivative at the nominal measured input \(x_0\), multiply by the input error or its uncertainty magnitude, and keep the units visible.
Why the Linear Estimate Works
The derivative is defined by the limiting ratio of output change to input change. That definition says that, for sufficiently small input changes, the output change is well approximated by the derivative times the input change.
Equivalently, the difference between the quotient and \(f'(x_0)\) tends to zero. Write that difference as \(r(h)\), so \(r(h)\to 0\) as \(h\to 0\). Multiplying the quotient equation by \(h\) gives
The term \(f'(x_0)h\) is the linear part of the output change. The remaining term \(h\,r(h)\), divided by \(h\), tends to zero as \(h\) tends to zero. Thus the linear part provides the first-order estimate for small input errors. This reasoning is local: it does not claim that the estimate is an exact bound for every possible input error.
A Practical Calculation Procedure
For a sensor reading with a stated input uncertainty, organize the calculation so the nominal calibrated result and its estimated uncertainty remain distinct.
Write \(y=f(x)\), with raw sensor input \(x\) and calibrated output \(y\).
Use \(x_0\) for the reported input and \(\delta\) for the magnitude of its uncertainty.
Compute \(f(x_0)\) and \(f'(x_0)\).
Estimate the output uncertainty as \(|f'(x_0)|\delta\), and state the output units.
Confirm that the derivative uses output units per input unit, and that the reported uncertainty is a magnitude.
Worked Examples
Worked Example: Linear Voltage-to-Temperature Calibration
A temperature sensor is calibrated by \(T(V)=100(V-0.50)\), where voltage \(V\) is measured in volts and calibrated temperature \(T\) is in degrees Celsius. The measured voltage is \(V_0=1.20\ \text{V}\), with uncertainty \(\delta V=0.01\ \text{V}\). Find the nominal temperature and estimate its uncertainty.
First evaluate the calibration curve at the measured voltage:
Differentiate the calibration model with respect to voltage:
The estimated temperature uncertainty is the sensitivity magnitude times the voltage uncertainty:
Thus the calibrated reading is \(70\ ^\circ\text{C}\), with a first-order estimated uncertainty of \(1\ ^\circ\text{C}\). Because this calibration curve is linear, the output change is exactly \(100\) times the voltage change: input changes of \(+0.01\ \text{V}\) and \(-0.01\ \text{V}\) produce temperature changes of \(+1\ ^\circ\text{C}\) and \(-1\ ^\circ\text{C}\). This checks the estimate.
Worked Example: Quadratic Pressure Calibration
A pressure sensor has calibration model \(P(V)=2V+0.5V^2\), where \(V\) is in volts and \(P\) is in kilopascals. The measured voltage is \(V_0=2.00\ \text{V}\), with uncertainty \(\delta V=0.02\ \text{V}\). Estimate the calibrated pressure and its uncertainty.
Evaluate the nominal pressure:
Differentiate the model and evaluate its derivative at the measured voltage:
Therefore the first-order output uncertainty estimate is
The result is \(6.00\ \text{kPa}\), with estimated uncertainty \(0.08\ \text{kPa}\). To check the calculation against this particular model, evaluate the calibration curve at the two ends of the input uncertainty interval. At \(V=1.98\ \text{V}\),
This is \(5.9202-6.00=-0.0798\ \text{kPa}\) from the nominal output. At \(V=2.02\ \text{V}\),
This is \(6.0802-6.00=0.0802\ \text{kPa}\). These changes are close to the first-order estimates \(-0.08\ \text{kPa}\) and \(+0.08\ \text{kPa}\); the small differences arise because the calibration curve is curved.
Worked Example: Calibration with Negative Local Sensitivity
A sensor’s calibrated temperature is modeled by \(T(I)=30-4I+0.2I^2\), where current \(I\) is in milliamperes and \(T\) is in degrees Celsius. At \(I_0=5.00\ \text{mA}\), the input uncertainty is \(\delta I=0.05\ \text{mA}\). Find the nominal output and estimate its uncertainty.
The nominal output is
The derivative and its value at the nominal input are
The signed first-order output change corresponding to an input change \(h\) is approximately \((-2\ ^\circ\text{C}/\text{mA})h\). For the uncertainty magnitude, use the absolute value:
The calibrated output is \(15\ ^\circ\text{C}\), with estimated uncertainty \(0.10\ ^\circ\text{C}\). The negative derivative indicates that a small increase in current corresponds locally to a decrease in calibrated temperature. As a check, the two endpoint inputs give
and
The changes from \(15\ ^\circ\text{C}\) are \(+0.1005\ ^\circ\text{C}\) and \(-0.0995\ ^\circ\text{C}\), close to the estimated magnitude \(0.10\ ^\circ\text{C}\), with the expected direction reversed as current increases.
Common Pitfall: Mixing Up the Units or the Direction
A calibration model specifies which quantity is a function of which. If \(T=f(V)\), then the relevant sensitivity is \(dT/dV\), not \(dV/dT\). These derivatives have reciprocal units and generally different numerical values. Also, the uncertainty calculation uses a magnitude: a negative derivative conveys direction, but does not make an uncertainty negative.
It is also important to retain the units during the multiplication. In the corrected calculation, milliamperes cancel, leaving degrees Celsius, as required for an output uncertainty.
Interpreting the Estimate
The derivative-based estimate is useful because a calibration curve can have different sensitivities at different readings. A fixed input uncertainty may correspond to a small output uncertainty in one region and a larger one in another. Evaluating the derivative at the actual nominal reading captures this local variation.
The result is a first-order estimate, not automatically a guaranteed worst-case bound for a nonlinear calibration. It describes the output change predicted by the local linear approximation. Keep the nominal calibrated value \(f(x_0)\) separate from the uncertainty estimate \(|f'(x_0)|\delta\), and report both with their units. When the derivative is negative, report the uncertainty as a positive magnitude and use the derivative’s sign only when describing the direction of a particular input change.
Check Your Understanding
For each question, identify the raw input, calibrated output, and relevant derivative before calculating.
- A calibration model is \(y=5x\), with \(x_0=3.0\) and input uncertainty \(0.02\). What are the nominal output and first-order output uncertainty?
- If \(y=f(x)\) is measured in grams and \(x\) in volts, what are the units of \(f'(x)\)?
- Why does an uncertainty estimate use \(|f'(x_0)|\delta\) rather than \(f'(x_0)\delta\)?
- For \(y=x^2\) at \(x_0=4\), with input uncertainty \(0.1\), calculate the first-order output uncertainty.
- In one sentence, explain why the derivative-based estimate is local rather than necessarily an exact bound.
