数学知识:其他

求导

  • 链式法则适用(

\[\frac{\partial f(x)g(x)}{\partial x}=f'(x)g(x)+f(x)g'(x) \]

  • 偏导结果与变量形状相同

\[\frac{\partial f(x)}{\partial x_{m\times n}}=y_{m\times n} \]

  • 相乘时要满足可乘条件

\[a_{m\times n}b_{n\times p}c_{p\times q}=d_{m\times q} \]

向量求导

\[\dfrac{\partial x^Ta}{\partial x}=\dfrac{\partial a^Tx}{\partial x}=a \]

\[\dfrac{\partial x^Tx}{\partial x}=2x \]

\[\dfrac{\partial x^TAx}{\partial x}=Ax+A^Tx \]

\[\dfrac{\partial a^Txx^Tb}{\partial X}=ab^Tx+ba^Tx \]

矩阵求导

\[\dfrac{\partial a^TXb}{\partial X}=ab^T \]

\[\dfrac{\partial a^TX^Tb}{\partial X}=ba^T \]

\[\dfrac{\partial a^TXX^Tb}{\partial X}=ab^TX+ba^TX \]

\[\dfrac{\partial a^TX^TXb}{\partial X}=Xba^T+Xab^T \]

卷积

线性卷积

周期卷积

循环卷积

连续函数和离散函数的卷积

  1. 举例
    连续函数

\[f(t) = sin(2\pi t)\rm{rect}(\it{t}) \]

离散序列

\[a = (1, 1, 1, 1, 1, 1) \]

离散函数

\[g(t) = \sum\limits_{i=0}^{5}a_i\delta(t-i) \]

卷积

\[\begin{align*} f(t)*g(t)&=\sin(2\pi t)rect(t)*\sum\limits_{i=0}^{5}a_i\delta(t-i) \\ &=\sum\limits_{i=0}^{5}a_i\sin[2\pi*(t-i)]rect(t-i) \\ &=\sum\limits_{i=0}^{5}\sin[2\pi*(t-i)]rect(t-i) \end{align*} \]

仿真补充:!!!!!!!!!!!!!!!!

  1. 举例
    两个“离散的连续函数”卷积

\[f(t)=\cos(2\pi t)rect(t)*\sum\limits_{i=0}^{5}a_i\delta(t-i) \\ g(t)=f(-t) \]

卷积

\[\begin{align*} f(t)*g(t) = &[\cos(2\pi t)rect(t)*\sum\limits_{i=0}^{5}a_i\delta(t-i)] \\ &*[\cos(2\pi t)rect(t)*\sum\limits_{j=0}^{5}a_j\delta(t+j)] \\ =&[\cos(2\pi t)rect(t)*\cos(2\pi t)rect(t)] \\ &*[\sum\limits_{i=0}^{5}a_i\delta(t-i)*\sum\limits_{j=0}^{5}a_j\delta(t+j)] \\ =&\dfrac{1}{2}\cos(2\pi t)rect(\dfrac{t}{2})*\sum\limits_{i=0}^{5}\sum\limits_{j=0}^{5}a_ia_j\delta(t-i+j) \\ =&\dfrac{1}{2}\sum\limits_{i=0}^{5}\sum\limits_{j=0}^{5}a_ia_j\cos[2\pi(t-i+j)]rect(\dfrac{t-i+j}{2}) \end{align*} \]

傅里叶变换

FT

\[f(t)=\sum\limits_{i=0}^{N-1}f(x_i)e^{-j\pi t} \]

DFT

FFT

内积、外积、点乘、叉乘

结果

向量(同维)内积后得数值,外积(不同维)得矩阵,点乘(同维)得数值,叉乘(同维)得同维向量。

参考:https://blog.csdn.net/Dust_Evc/article/details/127502272

posted @ 2024-01-11 13:47  工大鸣猪  阅读(6)  评论(0编辑  收藏  举报