12．已知G(s)=G1(s)G2(s)，且已分别测试得到：
G1(jω)的幅频特性|G1(jω)|=A1(ω) ，相频∠G1(jω)=ψ1(ω)
G2(Jω)的幅频特性|G2(jω)|=2 ，相频∠G2(jω)=-0.1ω
则
A G(jω)=2A1(ω).e-j0.1ωψ1(ω)
B G(jω)=[2+A1(ω)].ej[ψ1(ω)-0.1ω]
C G(jω)=2A1(ω).ej[ψ1(ω)-0.1ω]
D G(jω)=[2+A1(ω)].E-j0.1ωψ1(ω)