# How much data is required to reliably fit an exponential function?

I have some test data for a battery (pulsed discharge curves), and I am attempting to fit a second-order RC model to those curves.

In many cases it looks like I can get a good fit, but the longest RC time constant I get can range from ~0.75 hours to ~5 hours. The test data I have is only the first 30 minutes of the discharge step response. Is that enough of a duration to adequately fit for a time constant that might be a few hours long?

My feeling is no... but where is the boundary between having enough test data to fit an exponential curve with some lengthy time constant?

EDIT FOR MORE CONTEXT: Here is a wikipedia link to an RC circuit, https://en.wikipedia.org/wiki/RC_circuit.

You can see an image of the RC exponential curve here, https://en.wikipedia.org/wiki/RC_circuit#/media/File:Series_RC_capacitor_voltage.svg.

My question is essentially, how much of the curve do you need to know (starting from t=0) to fit the exponential.

• What are RC curves? Can you please phrase your question in terms someone not familliar with your domain knowledge may be able to understand?> Jul 27, 2020 at 23:15
• edited for more context Jul 27, 2020 at 23:27
• are you interested in predicted error, or are you more interested in the precision of the estimates of various coefficients? Jul 28, 2020 at 0:19
• i'm interested in the precision of the estimates of the various coefficients. Jul 28, 2020 at 0:43
• @aosborme. The wording of your question is not understandable. What function do you want to fit ? What are the measured variables ? What are the parameters to evaluate ? Can't you post an example of data ? Jul 30, 2020 at 8:43

This isn't an answer so much as a lengthy comment. I will edit as I get more information.

Because you're modelling an exponential phenomenon, you can do linear regression on the log outcomes. This allows you to leverage the rich theory of linear regression.

In OLS, the variability of the coefficients is given by

$$\operatorname{Var}\left(\hat{\beta}_{j}\right)=\frac{\sigma_{y \mid \mathbf{x}}^{2}}{(n-1) \sigma_{x_{j}}^{2}}$$

Here $$\sigma_{x_{j}}^{2}$$ is the sample variance of the predictor (in your case, time), $$n$$ is the sample size, and $$\sigma_{y \mid \mathbf{x}}^{2}$$ is the noise in the outcome. There are then 3 ways to increase precision:

• Use an outcome with lower noise