大学电路一阶电路的全响应问题
1个回答
开关闭合稳态后:
i(∞)=0A
4i(∞)=0V
Uc(∞)=1.5V*6/(9+6)=0.6V
开关断开后有：
电容初始值:Uc(0)=Uc(∞)=0.6V
i=CdUc(t)/dt
3CdUc(t)/dt+Uc(t)-4CdUc(t)/dt+6CdUc(t)/dt=0 (KVL方程)
5CdUc(t)/dt=Uc(t)
两边拉斯变换后有：
5C*L[dUc(t)/dt]=L[Uc(t)]
5C*{SL[Uc(t)]-Uc(0)}=L[Uc(t)]
5C*SL[Uc(t)]-5CUc(0)=L[Uc(t)]
5CUc(0)=5C*SL[Uc(t)]-L[Uc(t)]
5CUc(0)=(5C*S-1)L[Uc(t)]
L[Uc(t)]=5CUc(0)/(5C*S-1)
L[Uc(t)]=0.06/(0.1S-1)
则：
Uc(t)=(0.6e^10t)V
因为：i=CdUc(t)/dt
所以：
ui=-6i=-6CdUc(t)/dt
=-0.12dUc(t)/dt
=7.2t*e^(10t-1)V
