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Let's $q(t)\in \mathbb{R}$ represents a noisy raw signal, the filter is $$ \begin{align} \dot{x}_1 &= x_2 + g\gamma_1(q-x_1) \\ \dot{x}_2 &= g^2\gamma_2(q-x_1) \end{align} $$ where $x_1$ represents the smoothed version of $q$ and $x_2$ represents an estimate for $\dot{q}$ and $g$ is the filter's gain. The constants $\gamma_{1,2}$ are positive constants. What is the name of this filter and how it works?

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  • $\begingroup$ What is the filter's input, and what is its output? Where is the independent variable (time)? What is $q$ and what is its "smoothed version"? $\endgroup$
    – MBaz
    Apr 12, 2022 at 22:00
  • $\begingroup$ @MBaz the raw signal $q(t)$ is the input. The outputs are $x_{1,2}$. The smoothed version is preventing higher frequencies. $\endgroup$
    – CroCo
    Apr 12, 2022 at 22:01
  • $\begingroup$ Do you mean that the outputs are the $x$s with dots? And, if there are two outputs, don't you have two filters? $\endgroup$
    – MBaz
    Apr 12, 2022 at 22:04
  • $\begingroup$ @MBaz the outputs are $x_{1,2}$ and because they are coupled I believe this is a one filter but I'm not sure since I'm asking what is this filter? $\endgroup$
    – CroCo
    Apr 12, 2022 at 22:06
  • $\begingroup$ @MBaz you could simulate it by adding some noise to $q(t)$ and you will see it is working good (i.e. you can integrate $\dot{x}_{1,2}$ to obtain $x_{1,2}$). $\endgroup$
    – CroCo
    Apr 12, 2022 at 22:15

1 Answer 1

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I am afraid there is no special designation for the parameters you indicate.

Your expression describes a Second Order Band Pass Filter, which is clearer by taking its State Space form, and making some notation and variable changes: $$ \dot{\left[\begin{array}{cc}x_1 \\x_2 \end{array}\right]} =\left[\begin{array}{cc}-a_1 & 1\\-a_2 & 0\end{array}\right] \left[\begin{array}{cc}x_1 \\x_2 \end{array}\right] +\left[\begin{array}{cc}a_1 \\a_2 \end{array}\right] u $$

After taking into Laplace Domain and doing some matrices, we have exposed the Transfer Functions: $$ \left[\begin{array}{cc}x_1 \\x_2 \end{array}\right] ={1 \over s^2+a_1s+a_2}\left[\begin{array}{cc}s & 1\\-a_2 & s+a_1\end{array}\right] \left[\begin{array}{cc}a_1 \\a_2 \end{array}\right] u\\ =\left[\begin{array}{cc}{sa_1+a_2 \over s^2+a_1s+a_2}\\{sa_2 \over s^2+a_1s+a_2}\end{array}\right] u $$

As you see, the roots of $s^2+a_1s+a_2$ describe the poles of your Second Order System. Inspecting the Discriminant $\Delta$, you can check the poles will always be stable for $\gamma_i$ positive. If $\Delta=g^2(\gamma_1^2-4\gamma_2)>0$ the poles will be real, and the sytem will be Overdamped.

If $\Delta=g^2(\gamma_1^2-4\gamma_2)<0$, the poles will be complex, the system will be Underdamped, and we can calculate natural frequency $\omega$ and damping $\zeta$: $$ \omega=g\sqrt{\gamma_2}\\ \zeta={\gamma_1 \over 2 \sqrt{\gamma_2}} $$

The zeros for $x_1$ and $x_2$ are $g\gamma_1/\gamma_2$ and $0$ respectively. And since we have only a Order 2, care should be taken to determine the cut frequencies for the bandpass.

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    $\begingroup$ but why $x_2$ represents the first derivative? First derivative in Laplace transform is only $s$ but I see in your answer it is a polynomial of second degree. $\endgroup$
    – CroCo
    Apr 16, 2022 at 23:30
  • $\begingroup$ Exactly. The algebra is not lying, there is not a derivative in there. Unless you start adding some additional conditions, such as $u(t)$ moves very slowly in order their frequency components are very small with respect to the poles frequencies, then $s^2+a_1s+a_2 \approx a_2$ and only so, under that specific consideration, you might say, $$x_2(t)\approx {sa_2 \over a_2}u(t)={d \over dt}u(t)$$ $\endgroup$
    – Brethlosze
    Apr 17, 2022 at 18:33
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    $\begingroup$ so it is an estimate not first derivative so to speak, right? $\endgroup$
    – CroCo
    Apr 17, 2022 at 18:35
  • $\begingroup$ Though there is no reason in here to estimate a derivative, since having the signal, there is nothing preventing you to simply compute it, but yes. $\endgroup$
    – Brethlosze
    Apr 17, 2022 at 18:39
  • $\begingroup$ interesting, I've seen this filter is being used in robotics with relatively high frequency where encoders provide noisy measurements. $\endgroup$
    – CroCo
    Apr 17, 2022 at 18:42

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