Time constant of exponential decay

Hello Matlab,

I have those two equations of exponential decay with time constant of the first one tu1=3800 sec. I plot those graphs and then from the graph, when I find the 36% decay of the initial value, I read different value tu2=5397. Can anyone help me what I am missing here?

y0=2;

tu1=3800;
tu2=1200;
t=0:1:10000;

y=y0*exp(-t/tu1)+ 0.4;
y2=(y0/2)*exp(-t/tu2)+0.5;

plot(t,y,t,y2)
legend('1','2')

[Walter Roberson]

That does not appear to me to be 36% decay; it looks more like 60-ish percent decay.

But I am not clear as to what your question is?

[friet]

In short my question is:

tu1=3800;
y=y0*exp(-t/tu1)+ 0.4;

I have ploted the above equation and when I calculate the time constant from the plot(i.e., 36% of the original value, (0.36*2.4=0.88)) the time is 5397 instead of 3800.

[John D'Errico]

What you are doing with that 36% point only makes sense if you have a PURE exponential decay. You don't.

Answer 1

[John D'Errico]

As Walter pointed out, you don't have a purely exponential model. Yet, this is way easier than you think to solve. Two lines of code to recover the time constant. ;-)

Presume the model is a negative exponential, plus an unknown constant, sampled at an arithmetic sequence in t.

dt = t(2) - t(1);
dy = diff(y);

So assume the model was

y(t) = y0*exp(-t/tc) + yinf

Then when we used diff, it kills off the unknown additive constant. (i.e., the asymptote as t-->infinity.) We will get something like

y0*(exp(-(t + dt)/tc) - exp(-t/tc)) = y0*exp(-t/tc)*(1 - exp(-dt/tc))

Here, dt is quite small compared to tc, but both dt and tc are fixed, thus not a function of t. We don't know either y0 or tc, YET. Take the log.

log(diff(y)) = log(y0*(1-exp(-dt/tc))) - t/tc

Again, the first term is constant, thus not a function of t. While we don't know what that constant is, we don't care, as this part is targeted at solving for tc. The second term is linear in t.

We solve for tc using polyfit.

P1 = polyfit(t(2:end),log(-diff(y)),1)
P1 =
  -0.00026316      -7.5495

tc = -1/P1(1)
tc =
         3800

Yup. That looks familiar.

To recover the unknowns from the second series,

P2 = polyfit(t(2:end),log(-diff(y2)),1)
P2 =
  -0.00083333      -7.0897

tc2 = -1/P2(1)
tc2 =
         1200

Ok. That really was pretty easy. Can we now recover the constants y0 and the asymptote at infinity? Easy peasy too. Just one more line. Oddly enough, we use polyfit once more.

polyfit(exp(-t/tc),y,1)
ans =
          2          0.4

For the second curve...

polyfit(exp(-t/tc2),y2,1)
ans =
          1          0.5

I think you should recognize each of those coefficients for each dataset.

Such a versatile tool, that polyfit. We used it twice, fitting a straight line fit each time, yet we solved a nonlinear curve fitting problem, recovering not only the rate constant, but the asymptote at infinity as well as the multiplicative constant out front.

Kind of pretty. (Yes, I played a bit fast and loose with the statistics here.)

[friet]

Thanks I have a conceptual question If you have a function of y(t) = y0*exp(-t/tc) + yinf where yinf is non- zero constant, is your time constant is tc? Because when i plot my equation, i dont see tc to be 36.7% decay of the original value. Thanks for the help with fitting.

[John D'Errico]

Again, that 36% decay point is not relevant when the asymptote is non-zero. So what you are doing is meaningless here.

For example, suppose you added some arbitrary constant to that curve? The 36% point will be wildly different. In fact, if you added a large enough constant to y, then you will never even see a 36% decay point at all. This should help to convince you that what you are trying to do is in fact meaningless. Sorry, but it is.

I think you learned somewhere (from someone) that you can do this computation, using a specific percent decay to establish a time constant. What you do not seem to understand is it is NOT applicable to the problem you have posed. PERIOD. PERIOD. You learned a trick that applies to a different problem. But that does not mean it applies to every possible problem.

[friet]

Thank you so much for explanation. If you have a graph with non-zero asymptote, is there any way to know the time constant?

[John D'Errico]

Let me explain what you are trying to do.

Given a PURE exponential decay curve.

y = y0*exp(-t/tc)

Assume that you locate the 36% decay point on the curve, so the value t_36 such that

y(t_36)/y(t0) = 0.36

See that this corresponds to the ratio

0.36 = exp(-t_36/tc)/exp(-t0/tc)

So y0 goes away. Taking the log of both sides, we have

log(0.36) = -(t_36 - t0)/tc

So you can recover the time constant tc as

tc = -(t_36 - t0)/log(0.36)

But, we know that

log(0.36)
ans =
    -1.0217

So log(0.36) is darn near -1. Actually, the 36.788% point is what you might want, to be mathematically pedantic.

exp(-1)
ans =
      0.36788

But if we assume that log(.36) is close enough to -1, then we have a simple estimate for the time constant as simply:

tc = t0 - t_36

But that ONLY applies to the very simple case of an absolutely pure exponential decay.

What you are trying to do completely invalidates what I just did above.

[John D'Errico]

For the case of a curve with a non-zero asymptote, I gave it to you already! READ MY ANSWER. There I showed you a very simple code to compute that time constant, requiring all of two lines of MATLAB code. In fact, I also gave you a complete derivation thereof.

Answer 2

[Walter Roberson]

The time constant tu1=3800 applies to a pure exponential. You do not have a pure exponential: you have an exponential plus a constant. If you look at 3800 you will see 0.88 plus the constant.

[friet]

So at t=3800, i got 1.136. and this is not equal to 0.88+0.5=1.38.