2023-05-23

Two-point model for determining the coefficient g

This note is in addition to this paper.
It details the formula for determining the coefficient \(g\) using a two-point model.

The problem is this.
There is a plot of current \(I\) versus time \(t\), shown in Figure 1.
We can measure two points on this graph: 2 and 3.
It is also known that the

*τ*time is approximately half of the descending part of this graph, and the required coefficient can be expressed through this value: \[g = K^{\large {\tau / (t - \tau)}} \tag{1}\] Here: \(K\) is a known coefficient depending on the measurements on the graph.
If we knew exactly

*τ*, then this problem would not arise, and we would calculate \(g\) using formula (1). But this parameter greatly affects the result even with a small error, and in the schema it is only roughly defined. In this case, we consider that*τ*is unknown, and the coefficient we need is determined by two points on the graph.
Fig.1. Graph of current
I versus time t |

For a two-point model, you need to prepare the appropriate formulas.
First we output

*τ*: \[\tau = {t \over 1 + \ln(K) / \ln(g)} \tag{2}\] Since we believe that this parameter is unknown to us, but the data of two measurement points are known, we can rewrite this formula as follows: \[\tau = {t_2 \over 1 + \ln(K_2) / \ln(g)} = {t_3 \over 1 + \ln(K_3) / \ln(g)} \tag{3}\] Here: \(t_2, K_2, t_3, K_3\) - measurement data for two points. Where do we get the desired coefficient: \[ \ln g = {t_2 \ln(K_3) - t_3 \ln(K_2) \over t_3 - t_2} \tag{4}\] The same formula can be written in another way: \[ g = { K_3^{\large {t_2 \over t_3 - t_2}} \over K_2^{\large {t_3 \over t_3 - t_2}} } \tag{5}\] Thus, bypassing the*τ*, we got the desired result. At the same time, we can now find the*τ*parameter itself with maximum accuracy if we substitute expression (4) into formula (2): \[ \tau = {t_2 \ln(K_3) - t_3 \ln(K_2) \over \ln(K_3) - \ln(K_2)} \tag{6}\] Q.E.D.