How to fit an exponential decay curve to data (with a worked example)

How to · Updated 2026-11-10

Fit y = a·exp(−x/t) + c to measured data by least squares, read the time constant with its uncertainty, and check the fit with residuals and chi-square.

To fit an exponential decay, choose the model y = a·exp(−x/t) + c and let a least-squares fit find the starting height a, the time constant t and the level c that the data settle towards. The curve fitter does this in your browser: paste x and y (and the uncertainty in y if you know it), pick Exponential decay with offset, and read each parameter with its uncertainty, the chi-square test and a residual plot.

What you need

• Pairs of x and y values, such as time and temperature, or time and counts. Ten or more points along the whole fall work best.
• The uncertainty in each y value, if you have it. A thermometer read to ± 0.5 °C, or √N for counted events, are typical. Without it the fit still runs, but the chi-square test is skipped.
• A rough idea of the shape. Does the curve level off above zero? Then you need the offset c. Does it head all the way to zero? Then a plain exponential may do.

If your data exist only as a printed graph, read the points off it first with the graph digitiser.

Step by step

1. Paste your data into the Data box: one row per point, columns separated by tabs, commas or spaces, with an optional header row. You can also open a CSV, text or Excel file.
2. Check the x column, y column and Uncertainty in y menus. Choose “None (unweighted)” for the last if you have no uncertainties.
3. Pick Exponential decay with offset from the Model menu. The formula box fills with a*exp(-x/t) + c, and the Starting values box gets guesses taken from your data. If a fit goes wrong, press Guess starting values again or type better ones.
4. Read the parameter table. Written gives each parameter rounded with its uncertainty, ready for a report; the last column gives the 95 % interval.
5. Look at the second plot. The residuals (data minus curve, divided by σ when weighted) should scatter without a pattern, mostly between the dashed lines at ± 2.
6. Download the fitted values and residuals as a CSV, or the report as an image, PDF or Word file.

The fit chooses the parameters that make the weighted sum of squared misses as small as possible:

χ² = Σ [(yᵢ − f(xᵢ)) ÷ σᵢ]²

Here f(x) = a·exp(−x/t) + c is the curve and σᵢ is the uncertainty of point i. With t inside the exponential there is no one-step formula, so the tool searches for the minimum (Levenberg-Marquardt) from the guesses. The reduced chi-square is χ² divided by the degrees of freedom (number of points minus number of parameters). It is close to 1 when the model fits and the σ values are honest.

A worked example

A mug of hot water cools on a desk. Its temperature, read every two minutes and then less often, is (made-up numbers, each ± 0.5 °C):

time_min  temp_C  sigma
0   81.3  0.5
2   71.4  0.5
4   64.2  0.5
6   57.9  0.5
8   51.5  0.5
10  47.2  0.5
12  42.6  0.5
14  40.1  0.5
16  36.8  0.5
18  34.2  0.5
20  32.6  0.5
24  29.0  0.5
30  26.1  0.5

With Exponential decay with offset, the fit gives:

• a = 59.95 ± 0.66 °C (how far above the final level the water started)
• t = 11.90 ± 0.34 min, with a 95 % interval of 11.14 to 12.67 min
• c = 21.20 ± 0.71 °C (the level it is heading for: room temperature)
• χ² = 4.63 with 10 degrees of freedom, reduced χ² = 0.463, p = 0.91

The headline calls this consistent with the uncertainties; a reduced chi-square this far below 1 hints that ± 0.5 °C was generous.

Now switch the model to Straight line. The reduced chi-square jumps to 142, and the residuals bend in a clear U shape: a line is simply the wrong model. Exponential without the offset does better but still gives a reduced chi-square near 30, because it forces the water to cool towards 0 °C.

The correlation table shows t and c at −0.93: a longer time constant with a lower room temperature fits almost as well. Logging for longer would separate them.

Check your answer

• Look at the ends. At x = 0 the curve should be near a + c: 59.95 + 21.20 = 81.15 °C, close to the first reading of 81.3 °C. Far out it should approach c.
• Use the time constant. After one t the excess above c falls to 1/e, about 37 %. At 12 min the excess is 42.6 − 21.2 = 21.4 °C, which is 36 % of 59.95 °C. That matches.
• Half-life. If you need it, it is t × ln 2 = 11.90 × 0.693 = 8.25 min.
• Residuals beat R². All three models above have R² above 0.88. Only the residual plot and chi-square show which one is right.

Questions people ask

Why not just take the log and fit a straight line? That only works when c is zero or known, and taking logs distorts the uncertainties, so the small values near the end get too much weight. A direct nonlinear fit avoids both problems.

My fit says it “did not settle”. What now? Give starting values near the truth: a as the drop you see, t as the time to fall about two thirds of the way, c as the final level. Then try again.

How do I carry t into another formula? Use t = 11.90 ± 0.34 min as an input with standard uncertainty in the uncertainty propagation calculator. If the same formula also uses c, remember the two are correlated, which that calculator does not include.