First derivative analog filter

Question

I'm reading about fault detection via signal processing in time domain. One possibility is to check that first derivative of the signal is in some predefined bounds. The text says that to obtain the first derivative of the output signal $y(t)$, I can use this analog filter: $$F(p)=\frac{p}{\frac{1}{\tau}p+1}$$ for some sufficiently large $\tau$. Why is that? I don't know much about analog filters. I only know that Laplace transform of the first derivative of a function $f(t)$ with zero initial conditions is $pF(p)$. I also understand that $X(p)=\frac{1}{\frac{1}{\tau}p+1}$ is Laplace transform of exponential decay function $e^{-\tau t}$. Could you please explain to me why the filter needs to be designed like that and how does it work physically?

Thank you.

My wrong reasoning: If $u(t)$ is the input signal and $y(t)$ is the rate of that signal, then $$ y(t) = u(t)'. \\ $$ The transfer function would then be $$ \begin{align} Y &= pU \\ F &= \frac{Y}{U} = p. \end{align} $$

Answer 1

I will answer this with the understanding that $p$ is the Laplace variable (which I will call $s$).

As you stated, differentiation in the time domain corresponds to multiplication by $s$ in the Laplace domain. So it seems that we could get what we want by just choosing $F(s)=s$; then the output is the derivative of the input, i.e.,

$$y(t) = \frac{du}{dt}$$.

So why the $\tau s+1$ in the denominator?

This corresponds to a lowpass filter with bandwidth $1/\tau$. The output of $F(s) = s/(\tau s+1)$ is the derivative of the input signal after it has been smoothed by a lowpass filter with the desired bandwidth (or alternatively, this is the same as taking the derivative of the input signal and then smoothing it with the lowpass filter). This prevents the derivative operator from picking up high-frequency components of the input signal and amplifying noise.

I don't know much about fault detection, but I can give a simple example. Say you're trying to detect when $u'(t)$ exceeds a fixed threshold. If you had no lowpass filter, small-amplitude (but high-frequency) noise in your measurement $u$ could trigger your detector even when it really shouldn't. But by setting $\tau$ you can filter out components of $u$ above a certain frequency ($1/\tau$) so they don't affect your derivative estimate.