电路理论/相量/证明1

$g(t)=G_{m}cos(\omega t)$ 起点

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

$g(t)=G_{m}\operatorname {Re} (e^{j*0}e^{j\omega t})$

$g(t)=\operatorname {Re} (G_{m}e^{j0}e^{j\omega t})$

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

$\mathbb {G} =G_{m}e^{j0}=G_{m}(cos(0)+j*sin(0))=Gm$