# Linearize Shifted Exponential Decay

I'm just having trouble linearizing a quick function I'm graphing. The function is of a shifted exponential decay from a capacitor : $V(t) = V_0e^{-t/\tau}+V_1$. I previously had the function without the $V_1$ addition shifting it which just simply needed a natural logarithm on both sides, but I'm not sure what to do from here.

For the graph, we have a list of known values for V and t, $V_0$ and $\tau$ are the unknowns we have to find. $V_1$ is assumed to be known, but we have to use the graph to guess as to it's value. I can't take the natural log of both sides because that gets me nowhere, there's no rule for (x + b) in a natural log, and I don't think I can just subtract $V_1$ from either side because I wouldn't have my known V values to graph anymore.

Any help to this would be greatly appreciated.

• Does it help to consider $V(t)-V(0)$, or $V(t)-V(t_0)$ for some other useful $t_0$? That would eliminate $V_1$. – MPW Sep 13 '17 at 20:53
• @MPW If I were to use $V(t) - V(0)$ I would just be replacing $V_1$ in my equation with $V_0$. Maybe you're getting at something I'm not following. – DavisM12 Sep 13 '17 at 22:59

The model $$V(t) = V_0\,e^{-\frac t \tau}+V_1$$ is nonlinear with respect to its parameters.

You can make it linear if you know either $\tau$ or $V_1$. Admitting that you have data covering a large range, probably the simplest is to use for $V_1$ a value which is "just" below the smallest value of $V$ and then, define $W(t_i)=V(t_i)-V_1^{(guess)}$ and, as usual, take logarithms to get $\log(V_0)$ and $\frac 1 \tau$ from a linear regression.

This being done, you have all the elements to start the nonlinear regression and get the three parameters.

From $V(t)$ for $t=0$ we have $$V(0)=V_0+V_1\quad\Longrightarrow\quad V_0=V(0)-V_1$$

From the value of $V(t)$ for $t=1$ we have $$\frac{V(1)-V_1}{V_0}=\mathrm e^{-1/\tau}\quad\Longrightarrow\quad \tau=\frac{1}{\log\left(\frac{V_0}{V(1)-V_1}\right)}$$

$-V_0/\tau$ is the angular coefficient of the line tangent to the exponential function at $t=0$.

The equation of the linearized function is $$V_l(t)=V_0-V_0\frac{t}{\tau}+V_1$$ So the vertical axis intecerpt is for $t=0$, i.e. $V_l (0)=V_0+V_1$, and the orizontal axis intercept is when $V_l(t)=0$, i.e. for $t=\tau\frac{V_0+V_1}{V_0}$. Note that for $t=\tau$ we have $V_l(\tau)=V_1$.

• I'm not entirely sure I follow or where I would use this to create a linearized function of the original one I had. The outcome from this needs to be something that can then be plotted and give a usable slope and y-intercept. – DavisM12 Sep 13 '17 at 22:57
• I've added the linearized function – alexjo Sep 13 '17 at 23:09
• @DavisM12 Picture added – alexjo Sep 14 '17 at 13:37