1
$\begingroup$

A continuous measurement gives a function $m(t)$. The measurement corresponds physically to the integration of a function $f(t)$ over a defined interval $[t-t_0,t]$ ($t_0$ being a constant)

$$m(t)=\int_{t-t_0}^tf(t)dt$$

By numerically calculating continuously the derivative of the measurement $m(t)$, is it possible to find the function $f(t)$?

1 Answers 1

0

No.

For example $$g(t)=f(t) + \cos\left(2\pi \dfrac{t}{t_0}\right)$$ will produce the same $m(t)$ and $m'(t)$ and there are many other examples with a fluctuating component of period $t_0$ or with a period which divides $t_0$

As another example $$h(t)=f(t)+k$$ will produce an $m(t)$ which is $k t_0$ larger but the same $m'(t)$

  • 0
    We are only interested in relative variations of $f(t)$. In practice $f(t)$ is a physical dimension which nominal value $n_0$ is known. So in average $$\bar{f}(t)\arrowvert_0^\infty=n_0$$. – 2017-02-12
  • 0
    Therefore the constant should not be a problem. Following your reasoning, if $f(t)$ is decomposed in Fourier Series, all periods multiples of $t_0$ will be "lost" by the integration. – 2017-02-12