大学电路一阶电路的全响应问题
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/08/29 13:04:50
大学电路一阶电路的全响应问题
![](http://img.wesiedu.com/upload/9/a0/9a0b6710f9b8329e2a124caedcdb8be4.jpg)
![](http://img.wesiedu.com/upload/9/a0/9a0b6710f9b8329e2a124caedcdb8be4.jpg)
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
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