Second-Order Differential Equation of the Spring-Mass-Damper System
Consider a system with the transfer function given by equation (1). This is the standard form transfer function of a second-order lag element.
\zeta represents the damping ratio, \omega_n represents the natural angular frequency, and y denotes the position of the mass, the u is the force as input.
damping ratio: 減衰比
natural angular frequency: 固有角振動数
\begin{align}
G(s)= \frac{\omega_n^2}{s^2+ 2\zeta\omega_ns + \omega_n^2}
\end{align}Let’s say Y(s) is output and U is input, the form (1) is denoted as (2).
\begin{align}
(s^2+ 2\zeta\omega_ns + \omega_n^2)Y(s) = \omega_n^2U(s)
\end{align}Performing the inverse Laplace transform yields the following time-domain description. In this equation, \frac{d^2y}{dt^2} represents the second derivative of the position y with respect to time t.
\begin{align}
\frac{d^2y(t)}{dt^2} + 2\zeta\omega_n\frac{dy(t)}{dt} + \omega_n^2y(t) = \omega_n^2u(t)
\end{align}This formula is seen as Second-Order Differential Equation of the Spring-Mass-Damper System.
This differential equation can also be described using linear algebra.
\begin{align}
\frac{dx_1(t)}{dt}=x_2(t)\\
\frac{d}{dt}\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix}
=\begin{bmatrix}
0 & 1 \\
-\omega_n^2 & - 2\zeta\omega_n \\
\end{bmatrix}
\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix}
+ \begin{bmatrix}
0 \\
\omega_n^2 \\
\end{bmatrix} u(t) \\
y(t) =\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix} \\
\end{align}we can use another form using (7), (8)
\begin{align}
\omega_n^2 \coloneqq \frac{k}{m} \\
2\zeta\omega_n \coloneqq \frac{c}{m} \\
\frac{dx_2(t)}{dt}=\frac{1}{m}(-c\frac{dy(t)}{dt} - ky(t) + u(t) ) \\
\frac{d}{dt}\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix}
=\begin{bmatrix}
0 & 1 \\
-\frac{k}{m} & - \frac{c}{m} \\
\end{bmatrix}
\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix}
+ \begin{bmatrix}
0 \\
\frac{1}{m} \\
\end{bmatrix} u(t) \\
y(t) =\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1(t) \\
x_2(t) \\
\end{bmatrix} \\
\end{align}Using state equations and output equations (observation equations), the following can be described simply. This formulation is used in Kamlan Filter.
\begin{align}
\frac{d(t)}{dt} x(t)=Ax(t) + bu(t)\\
y(t) = c^Tx(t)+du(t)
\end{align}The solution of state equations
\begin{align}
\frac{d}{dt} x(t)=Ax(t) + bu(t)\\
x(0) = x_0
\end{align}\begin{align}
u(t) \coloneqq 0 \\
\frac{d}{dt} x(t)=Ax(t) \\
x(t) = e^{At} x_0 \\
e^{At} = I + At + \frac{1}{2!}A^2t^2 + \ldots + \frac{1}{k!}A^kt^k
\end{align}You can get e using inverse laplace transform,
\begin{align}
e^{At} = \mathcal{L}[(sI-A)^{-1}]
\end{align}
