电路理论/相量/证明6

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

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

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

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

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

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

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

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

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

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