Đâu là công thức tổng quát nhất tính góc pha của A ( e j ω )

$\text{a}\text{r}\text{g}\left[A\left(e{}^{j\omega }\right)\right]=\left\{2k+\frac{1}{2}\left[1-\frac{A\left(e{}^{j\omega }\right)}{\left|A\left(e{}^{j\omega }\right)\right|}\right]\right\}\pi$

$\{\begin{array}{c}A\left(e{}^{j\omega }\right)\ge 0\to \text{a}\text{r}\text{g}\left[A\left(e{}^{j\omega }\right)\right]=2k\pi \\ A\left(e{}^{j\omega }\right)<0\to \text{a}\text{r}\text{g}\left[A\left(e{}^{j\omega }\right)\right]=\left(2k+1\right)\pi \end{array}$

$\left|A\left(e{}^{j\omega }\right)\right|=\left|X\left(e{}^{j\omega }\right)\right|$

$\varphi \left[A\left(e{}^{j\omega }\right)\right]=\phi \left[X\left(e{}^{j\omega }\right)\right]$