LaTeX公式测试

2026-01-05 · 修改于 2026-07-19· 884 字 · 约 3 分钟

分类：test

行内测试

勾股定理 $a^2 + b^2 = c^2$ 希腊字母与下标测试： $\omega_n = \sqrt{\frac{k}{m}}$ （无阻尼固有频率）

\frac{\partial^2 p}{\partial t^2} - c^2 \nabla^2 p = 0 \quad \Rightarrow \quad \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} + \frac{\partial^2 p}{\partial z^2} = \frac{1}{c^2} \frac{\partial^2 p}{\partial t^2}

\begin{aligned} M \ddot{x}(t) + C \dot{x}(t) + K x(t) &= F(t) \\ x(t) &= e^{-\zeta \omega_n t} \left( A \cos(\omega_d t) + B \sin(\omega_d t) \right) \end{aligned}

\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{K}{M} & -\frac{C}{M} \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{M} \end{bmatrix} u(t)

X(f) = \int_{-\infty}^{\infty} w(t) \cdot x(t) e^{-j 2\pi f t} dt \approx \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} kn}