電路理論/相量/證明5

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

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

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

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

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

$\mathbb {G} =G_{m}e^{j*\phi }=G_{m}(cos(\phi )+j*sin(\phi ))=G_{m}cos(\phi )+jG_{m}sin(\phi )$