积化和差

$\cos \alpha \cos \beta = \frac{1}{2}[\cos (\alpha + \beta ) + \cos (\alpha - \beta )]$ $\sin \alpha \cos \beta = \frac{1}{2}[\sin (\alpha + \beta ) + \sin (\alpha - \beta )]$ $\sin \alpha \sin \beta = \frac{1}{2}[\cos (\alpha - \beta ) - \cos (\alpha + \beta )]$ $\cos \alpha \sin \beta = \frac{1}{2}[\sin (\alpha + \beta ) - \sin (\alpha - \beta )]$

自相关与互相关

${R_X}({t_2},{t_1}) = R_X^*({t_1},{t_2})$ ${R_{XY}}({t_2},{t_1}) = R_{YX}^*({t_1},{t_2})$ ${R_X}({t_2},{t_1}) = {R_X}(\tau ),\quad \tau = {t_1} - {t_2}$

矩阵微分

$Y$、 $B$和 $R$均代表矩阵， $z$和 $a$代表向量，上标T表示转置， $*$表示共轭，H表示共轭转置。
$\frac{{\partial {Y^{\mathop{\rm T}\nolimits} }B}}{{\partial B}} = Y$ $\frac{{\partial {B^{\mathop{\rm T}\nolimits} }Y}}{{\partial B}} = Y$

规律总结： “前面”为转置，对“不转置”求导，结果为“另一个不转置”

$\frac{{\partial {B^{\mathop{\rm T}\nolimits} }{Y^{\mathop{\rm T}\nolimits} }YB}}{{\partial B}} = 2{Y^{\mathop{\rm T}\nolimits} }YB$ $\frac{{\partial {B^{\mathop{\rm T}\nolimits} }B}}{{\partial B}} = 2B$ $\frac{{\partial {B^{\mathop{\rm T}\nolimits} }WB}}{{\partial B}} = WB + {W^{\mathop{\rm T}\nolimits} }B$
特别地， ${{W^{\mathop{\rm T}\nolimits} } = W}$时，
$\frac{{\partial {B^{\mathop{\rm T}\nolimits} }WB}}{{\partial B}} = WB + {W^{\mathop{\rm T}\nolimits} }B = 2WB$

若 ${\nabla _{{z^*}}}$表示对向量 ${{z^*}}$进行微分， ${\nabla _z}$表示对向量 ${{z}}$进行微分，则
${\nabla _{{z^*}}}({a^{\rm{H}}}z){\rm{ = }}{\bf{0}}$ ${\nabla _{{z^*}}}({z^{\rm{H}}}a){\rm{ = }}a$ ${\nabla _{{z^*}}}({z^{\rm{H}}}Rz){\rm{ = }}Rz$ ${\nabla _z}({a^{\rm{H}}}z) = {a^*}$ ${\nabla _z}({z^{\rm{H}}}a) = {\bf{0}}$ ${\nabla _z}({z^{\rm{H}}}Rz) = {R^{\rm{T}}}{z^{\rm{*}}}{\rm{ = (}}{R^{\rm{H}}}z{)^{\rm{*}}}$

一个计算小技巧

已知基向量两两正交

$\int_{ - \infty }^\infty {{f_m}(t)f_n^*(t)dt} = \delta (m - n)$

${s(t)}$可由基向量线性组合近似

$\hat s(t) = \sum\limits_{k = 1}^K {{s_k}{f_k}(t)}$

由误差与基向量正交，有

$\left\langle {s(t) - \hat s(t),{f_n}(t)} \right\rangle = 0 \Rightarrow {s_n} = \left\langle {s(t),{f_n}(t)} \right\rangle$

则误差的二范数为

${\varepsilon _e} = \int_{ - \infty }^\infty {(s(t) - \hat s(t)){{(s(t) - \hat s(t))}^*}dt}$

$= \int_{ - \infty }^\infty {|s(t){|^2}dt} - \int_{ - \infty }^\infty {\sum\limits_{k = 1}^K {{s_k}{f_k}(t)} \cdot {s^*}(t)dt}- \left\langle {s(t) - \hat s(t),{{\hat s}^*}(t)} \right\rangle$

$= \int_{ - \infty }^\infty {|s(t){|^2}dt} - \sum\limits_{k = 1}^K {{s_k}{{[\int_{ - \infty }^\infty {s(t)f_k^*(t)dt} ]}^*}}$

$= \int_{ - \infty }^\infty {|s(t){|^2}dt} - \sum\limits_{k = 1}^K {{s_k}s_k^*}$

$= \int_{ - \infty }^\infty {|s(t){|^2}dt} - \sum\limits_{k = 1}^K {|{s_k}{|^2}}$