2
$\begingroup$

Is it possible to directly measure derivatives of higher order than 3? It can be jerk for instance. By "directly" I assume that we can measure a signal proportional to the jerk value without differentiating acceleration explicitly.

Intuitively I would say no, however I can not come up with any reasoning for it.

$\endgroup$
1
  • $\begingroup$ Seems like a pure semantics question to me -- it all depends on what you count as "direct". You can measure jerk by measuring the velocity of a strongly damped mass on a spring. Does that count as direct? $\endgroup$ – knzhou Jul 2 '18 at 14:33
2
$\begingroup$

It is generally not possible to measure any derivatives directly. Instead of measuring $\dot x = \tfrac{\partial x}{\partial t}$, what you do is either

  • Measure some other quantity which, according to theory, should be functionally related to the derivative of $x$. For instance, to determine acceleration, you might measure force instead and follow Newton's law.
  • Measure multiple values of $x$ itself, over some (generally, short) time span. Fit some smooth function to the data points, and calculate the derivative of that fitted function.
    The simplest implementation of this would be to simply take two measurements $x_0$ (at $t_0$) and $x_1$ (at $t_1$), do a linear interpolation $$x_{\mathrm{lin}}(t) = x_0 + (t-t_0)\cdot\frac{x_1-x_0}{t_1-t_0}$$ between them. The derivative of this is $\tfrac{x_1-x_0}{t_1-t_0}$, which can also be interpreted as the average velocity between the two points.

The first technique only works for derivatives that are linked to some other measurable quantity, which is often the case for first or second derivatives, less often for higher derivatives.

The interpolation technique can in principle be applied to any measurable quantity and any order of derivative, however you need to be careful. A simple approach would be to calculate the first derivative by linear interpolation as above, then the 2nd derivative by linear interpolation of the thus estimated 1st derivative, etc.. Unfortunately, this breaks horribly if there is any noise in the signal (which in practice it always is), because the difference between close neighbouring measure points gets very small but the noise level doesn't.

Another common attempt is fitting the entire measured curve with a polynomial. That makes it very easy to read off any derivative, but it only works well if the function has a globally simple shape, else you won't be able to fit it properly without going to very high polynomial degrees (which would make it unstable in particular at the outer bounds).

For serious applications, one will generally use more advanced interpolation to avoid the drawbacks of these simple techniques. This may involve splines and Fourier- or Legendre transforms, or the fitting of some specialised model with physical motivation.

$\endgroup$
5
  • $\begingroup$ Generically you are correct, but I would say there can be exceptions to directly measure derivatives. For example, velocity measurements using the (relativistic) doppler shift, or force measurements using spring compressions. $\endgroup$ – KF Gauss Jul 2 '18 at 15:58
  • 1
    $\begingroup$ @user157879 how are these not examples of the first method? $\endgroup$ – leftaroundabout Jul 2 '18 at 16:08
  • 1
    $\begingroup$ What about this? $\endgroup$ – HiddenBabel Jul 2 '18 at 16:33
  • $\begingroup$ @HiddenBabel an electronic differentiator doesn't really measure the derivative of voltage but the current floating into a capacitor. Or rather, the voltage across a resistor through which that current flows. That is, by Coulomb's and Ohm's laws, linked to the derivative of the input voltage, so this is again an example of the first method. $\endgroup$ – leftaroundabout Jul 2 '18 at 16:35
  • 4
    $\begingroup$ But in this case, nature does the conversion for you, so I say it's still a direct observation. Otherwise, no voltage measurement ever would be direct. I guess that's ok if you want to define it that way. $\endgroup$ – HiddenBabel Jul 2 '18 at 17:54

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.