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LaTeX公式测试

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）

复杂公式测试

∂2p∂t2−c2∇2p=0⇒∂2p∂x2+∂2p∂y2+∂2p∂z2=1c2∂2p∂t2\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}

公式对齐

Mx¨(t)+Cx˙(t)+Kx(t)=F(t)x(t)=e−ζωnt(Acos⁡(ωdt)+Bsin⁡(ωdt))\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}

矩阵与线性代数

[x˙x¨]=[01−KM−CM][xx˙]+[01M]u(t)\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)

傅里叶变换

X(f)=∫−∞∞w(t)⋅x(t)e−j2πftdt≈∑n=0N−1x[n]e−j2πNknX(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}