1
$\begingroup$

I have a time series of measurements that resembles the shape of an exponential function. The samples are a bit noisy and sometimes there is a weak sine like ripple signal ontop of it.

Simplified the function looks like this:

$$ y = a (1-e^{-bt + c}) + d $$

enter image description here

The data that I have lies between $t=$0 to 2, e.g. where the function is still curvy.

My goal is to find $a+d$, e.g. the value of the asymptote or the $y$ value of the graph above at infinite time.

My idea so far is to take the samples that I have, do least square fitting on the function above and then just solve it.

Q: Will this approach work or is there a easier way?

Q: I'm a bit lost how to do a least square fitting to a function like the one above. With a falling decaying exponential I would just take the log and least square fit a line, but how do I do this with my rising function?

Oh, one more thing: I don't need to do this as a one-off so I can't just plug the data into a general purpose solver. I have to implement this later for resource constrained microcontroller.

$\endgroup$
  • $\begingroup$ Are your sample points evenly spaced in time? aka Uniform Sampling. $\endgroup$ – Cedron Dawg Mar 14 '18 at 13:31
  • $\begingroup$ @CedronDawg They could be, but I consider dropping samples that I can identify as outliners or noise-bursts. $\endgroup$ – Nils Pipenbrinck Mar 15 '18 at 4:57
1
$\begingroup$

I think you are going to be stuck with an iterative approach.

$$ y = a (1-e^{-bt + c}) + d $$

First rewrite your equation like this:

$$ y = g + h e^{-bt} $$

Where:

$$ g = a + d $$

$$ h = -a e^c $$

You have a set of values $(t_n, y_n)$. Consider the $y$ values as a single vector $\vec Y $. Then pick an initial value for $b$ and construct a vector $\vec B$ using your $t$ values in $ e^{-bt} $. You now have a vector form of your problem:

$$ \vec Y = g \vec U + h \vec B $$

Where $\vec U$ is a vector of all ones.

You can solve for $g$ and $h$ by dotting $\vec Y$ with $\vec U$ and $\vec B$. This gives you a linear system of two equations two unknowns which is easily solvable.

$$ (\vec Y \cdot \vec U) = g (\vec U \cdot \vec U) + h (\vec B \cdot \vec U) $$

$$ (\vec Y \cdot \vec B) = g (\vec U \cdot \vec B) + h (\vec B \cdot \vec B) $$

Note that $ \vec U \cdot \vec U $ is your sample count and $ \vec Y \cdot \vec U $ is the sum of your $y$ values. Calculating $ \vec U \cdot \vec B $, $ \vec B \cdot \vec B $, and $ \vec Y \cdot \vec B $ can be done in single loop without actually having to store the vector values if you are memory constrained. Storing $ \vec B $ will save recalculation when you are measuring the fit.

To measure your fit, define the difference vector:

$$ \vec D = \vec Y - \left( g \vec U + h \vec B \right) $$

Your square fit is now:

$$ f = \vec D \cdot \vec D $$

Your problem now is to find the value of $b$ which minimizes $f$. There are several ways to do this. The fastest converging one might be to pick three values of b, calculate an f for each, fit a parabola to your $(b,f)$ points, find the minimum $b_*$. To iterate, choose three new tighter $b$'s around $b_*$. Repeat until good enough.

The value of $g$ is the answer you are looking for.

Hope this helps,

Ced


Followup

As I mentioned in my comment to Ben, this equation is overparameterized. My recommendation would be to get rid of $d$ by setting it to zero. This leaves

$$ a = g $$

$$ c = \ln( -h/g ) $$


As a differential equation, per r b-j:

$$ y = g + h e^{-bt+c} $$

$$ y' = h e^{-bt+c} (-b) $$

$$ y' = b ( g - y ) $$


Getting a good initial value of b:

There's nothing like sleeping on a problem. If your samples are evenly spaced in time then you can find a good initial value of b.

The simplest involves using three evenly spaced points at time $t_A$, $t_B$, and $t_C$.

$$ t_B = t_A + s $$ $$ t_C = t_B + s $$

Calculate the following ratio:

$$ r = \frac{ y_B - y_A }{ y_C - y_B } $$

The value of b can be now be solved for:

$$ r = \frac{ ( g + h e^{-bt_B} ) - ( g + h e^{-bt_A} ) }{ ( g + h e^{-bt_C} ) - ( g + h e^{-bt_B} ) } $$ $$ r = \frac{ h e^{-b( t_A + s ) } - h e^{-bt_A} }{ h e^{-b( t_B + s ) } - h e^{-bt_B} } $$ $$ r = \frac{ e^{-b t_A }( e^{-bs} + 1 ) }{ e^{-b t_B }( e^{-bs} + 1 ) } $$ $$ r = e^{-b ( t_A - t_B ) } $$ $$ b = \frac{ \ln( r ) } { t_B - t_A } $$

In order to mitigate the effects of noise, more points can be used. Another formula would be:

$$ r = \frac{ y_1 - y_2 + y_3 - y_4 + y_5 - y_6 }{ y_7 - y_8 + y_9 - y_{10} + y_{11} - y_{12} } $$

$$ b = \frac{ \ln( r ) } { t_7 - t_1 } $$

Any number of formulas can be created. The rules are that the coefficients have to add up to zero and the time spacing has to be the same in the numerator and the denominator. Also, you want to stay on the steeper part of the curve so the differences in values are larger compared to your noise level.

With enough sample values used, the value of $b$ might be good enough that the first calculation of $g$ will be accurate enough that no iterations are needed.

$\endgroup$
  • $\begingroup$ Why not use "ln" on each side of the equation y=g+he−bt? Then you could use a linear fit? $\endgroup$ – Ben Mar 13 '18 at 13:44
  • 1
    $\begingroup$ @Ben, the "g+" throws that approach off as mentioned by the OP. Also that would give a least squares fit in log space, not in the data space. BTW, a linear least squares fit is the same approach I gave except B would be simply the values of t. $\endgroup$ – Cedron Dawg Mar 13 '18 at 14:02
  • $\begingroup$ Yeah you're right. My bad. $\endgroup$ – Ben Mar 13 '18 at 14:38
  • $\begingroup$ @Ben, No worries. It's also interesting that although a+d can be solved for, the individual values of a, d, and c can't be tied down. An increase in a can be compensated for by a decrease in d and a shift in c. Ergo, there is no absolute time scale. $\endgroup$ – Cedron Dawg Mar 13 '18 at 15:09
  • $\begingroup$ This is easier than I thought. Haven't tried it yet but I'll give it a test-run tomorrow. Like the fact that the solution is lean on the storage side of things. $\endgroup$ – Nils Pipenbrinck Mar 14 '18 at 2:24
1
$\begingroup$

i actually had this as a problem for my master's thesis 4 decades ago. it was before i was doing anything in audio. the application was predicting the equilibrium pressure of a very slow osmosis experiment (where the time constant was about a minute).

this was for a process ostensibly obeying this physical equation:

$$ \frac{d}{dt}P(t) = k\big(P_\text{eq}-P(t)\big) $$

where $P(t)$ is measured and $k$ and $P_\text{eq}$ are unknown.

so, from memory, here is the continuous-time result:

$$ P_\text{eq} = \frac{2 \int_{t_0}^{t} P^2(u) \, du - (P(t)+P(t_0)) \int_{t_0}^{t} P(u) \, du }{2\int_{t_0}^{t} P(u) \, du - (P(t)+P(t_0))(t-t_0) } $$

which is intended to minimize the magnitude of:

$$ \frac{d}{dt}P(t) - k\big(P_\text{eq}-P(t)\big) $$

so i think it was meant to minimize:

$$ \int\limits_{t_0}^{t} \left( \frac{d}{du}P(u) - k\big(P_\text{eq}-P(u)\big) \right)^2 du $$

take the derivatives with respect to the two unknowns, $k$ and $P_\text{eq}$, set the derivatives to zero (to find the minimum) solve two equations and two unknowns.

$\endgroup$
  • $\begingroup$ Thanks a lot! I have to admit it's been decades since I've dealed with differential equations. I have to refresh my math to see if this works for me. $\endgroup$ – Nils Pipenbrinck Mar 14 '18 at 2:26
  • $\begingroup$ the integrals for the $P_\text{eq}$ calculation can be turned into Riemann summations for the purpose discrete-time input, $P[n]$ instead of $P(t)$. $\endgroup$ – robert bristow-johnson Mar 14 '18 at 3:16
  • $\begingroup$ BTW, i implemented this in a "resource-constrained micro[processor]" called the Motorola MC6800. i was able to do this because the sample rate was very slow. about 10 samples per second. i had a 12-bit D/A which i doubled in use as a successive-approx A/D. i remember the terms in the numerator were 40-bits and in the denominator were 24-bits, all done with 8-bit bytes. i had to write my own multiplication (Booth's alg) and division routines. the result was 16 bits going into that 12-bit D/A. it worked good. $\endgroup$ – robert bristow-johnson Mar 14 '18 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.