电路理论/相量/证明8

$g(t)=G_{m}sin(\omega t-\phi )$

$g(t)=G_{m}cos(\omega t-\phi -{\frac {\pi }{2}})$

$g(t)=G_{m}\operatorname {Re} (e^{j(\omega t-\phi -{\frac {\pi }{2}})})$

$g(t)=G_{m}\operatorname {Re} (e^{j*(-\phi -{\frac {\pi }{2}})}e^{j\omega t})$

$g(t)=\operatorname {Re} (G_{m}e^{j*(-\phi -{\frac {\pi }{2}})}e^{j\omega t})$

$g(t)=\operatorname {Re} (\mathbb {G} e^{j\omega t})$

$\mathbb {G} =G_{m}e^{j*(-\phi -{\frac {\pi }{2}})}=G_{m}(cos(-\phi -{\frac {\pi }{2}})+j*sin(-\phi -{\frac {\pi }{2}}))$

$=G_{m}cos(-\phi -{\frac {\pi }{2}})+jG_{m}sin(-\phi -{\frac {\pi }{2}})$

$=G_{m}(cos(-\phi )cos({\frac {\pi }{2}})+sin(-\phi )sin({\frac {\pi }{2}}))+jG_{m}(sin(-\phi )cos({\frac {\pi }{2}})-cos(-\phi )sin({\frac {\pi }{2}}))$

$=G_{m}sin(-\phi )-jG_{m}cos(-\phi )$

$=-G_{m}sin(\phi )-jG_{m}cos(\phi )$