Quan hệ giữa phổ biên pha với phần thực và phần ảo của tín hiệu như thế nào?

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=-\text{a}\text{c}\text{t}\text{g}\frac{j\text{I}\text{m}\left[X\left(e{}^{j\omega }\right)\right]}{\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]}=\phi \left(\omega \right)$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=\text{a}\text{r}\text{c}\text{t}\text{g}\frac{\text{I}\text{m}\left[X\left(e{}^{j\omega }\right)\right]}{\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]}=\phi \left(\omega \right)$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=\text{a}\text{c}\text{t}\text{g}\frac{j\text{I}\text{m}\left[X\left(e{}^{j\omega }\right)\right]}{\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]}=\phi \left(\omega \right)$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=-\text{a}\text{c}\text{t}\text{g}\frac{\text{I}\text{m}\left[X\left(e{}^{j\omega }\right)\right]}{\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]}=\phi \left(\omega \right)$