# What is the name of this filter

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?

• 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"?
– MBaz
Apr 12, 2022 at 22:00
• @MBaz the raw signal $q(t)$ is the input. The outputs are $x_{1,2}$. The smoothed version is preventing higher frequencies. Apr 12, 2022 at 22:01
• Do you mean that the outputs are the $x$s with dots? And, if there are two outputs, don't you have two filters?
– MBaz
Apr 12, 2022 at 22:04
• @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? Apr 12, 2022 at 22:06
• @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 $x_{1,2}$). Apr 12, 2022 at 22:15

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.

• 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. Apr 16, 2022 at 23:30
• 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, $$x_2(t)\approx {sa_2 \over a_2}u(t)={d \over dt}u(t)$$ Apr 17, 2022 at 18:33
• so it is an estimate not first derivative so to speak, right? Apr 17, 2022 at 18:35
• 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. Apr 17, 2022 at 18:39
• interesting, I've seen this filter is being used in robotics with relatively high frequency where encoders provide noisy measurements. Apr 17, 2022 at 18:42