y(t)=[y(0+)y（）]e.ppt

文档介绍：y(t)=[y(0+)y()]e+y()τty(0+)LRτ=RCτ=uC(0),iL(0)(t=0+时的等效电路)与输入无关,归结为求由电容元件或电感元件观察的输入电阻R归结为求解电阻电路(电容元件相当于开路,(1)y(0+)—待求电压或电流的初始值(2)y()—电路进入稳定状态后的待求电压或电流(3)τ—时间常数11-3-2求解一阶电路的三要素法11-3一阶电路的全响应电感元件相当于短路)uC(0+),iL(0+)(连续性)零输入响应、直流电源作用下的零状态响应和全响应例5图示电路开关S断开时处于稳定状态,t=0时将开关闭合,求开关闭合后的iL(t)和uL(t)。10mA1k+1k2k10V+uLiL1HS10mA1k1kiL(t0)5mA+2k2k10V+uL+10V(t=0+)y(t)=[y(0+)y()]e+y()τt11-3-2求解一阶电路的三要素法iL(0)=5mAiL(0+)=5mAuL(0+)=105=5V1k1k2k(求电感两端的输入电阻)iL=105e–1000tmAuL=5e–1000tVy(t)=[y(0+)y()]e+y()τt10mA1k+1k2k10V+uLiL1HS10mA1k+1k2k10ViL(t=)iL(0+)=5mAuL(0+)=5ViL()=5+5=10mAuL()=0=S11000y(t)=[y(0+)y()]e+y()τt11-3-2求解一阶电路的三要素法例6(见教材p286****题11-25)++12V2441H1FuiL(0)=(0)=6V++12V244iLuC(t<0)++uCu1u=u1uCiL(0+)=(0+)=6Vy(t)=[y(0+)y()]e+y()τt11-3-2求解一阶电路的三要素法++12V2441H1Fu++uCu1(t=)iL(0+)=(0+)=6VuC()=0u1()=8V1=S612=4Su1(0+)=9VuC(t)=6e(t)=8+e6tVu(t)=8+e6t6e++12V2441Fu++uCu1(t=0+)例7图示电路,开关断开时处于稳态,t=0时将开关闭合,求开关闭合后的响应uC、iL、i。++30V105F241y(t)=[y(0+)y()]e+y()τt11-3-2求解一阶电路的三要素法++30V10uCiL155(t0)iL(0)=1AuC(0)=20ViL(0+)=1AuC(0+)=20Vi1++30V105F241iL(0+)=1AuC(0+)=20Vi=i1iLiL()=0i1(0+)=4A/3uC()=18Vi1()=1==S101510+15241412===18+2e4tViL=e25tAi1=+e4tA152i=+e4te25tA152y(t)=[y(0+)y()]e+y()τt11-3-2求解一阶电路的三要素法—End1、相关概念一阶电路零输入响应零状态响应全响应时间常数uC(0)=0iL(0)=0第十一章电路暂态过程的时域分析y(t)=[y(0+)y()]e+y()τt2、求解一阶电路的三要素法y(0+):1)对换路前的电路求uC(0)、iL(0)2、求解一阶电路的三要素法2)由电容电压的连续性和电感电流的连续性iL(0+)=iL(0)uC(0+)=uC(0)3)求y(0+)y():对换路后的电路求稳态电压或稳态电流(电容以开路代替,电感以短路代替)第十一章电路暂态过程的时域分析y(t)=[y(0+)y()]e+y()τt(电路较为复杂时,画出t=0+的等效电路再求)y(0+)::=RC或=LR2)R为由电容元件两端或电感元件两端观察的1)与输入(电源)无关输入电阻2、求解一阶电路的三要素法第十一章电路暂态过程的时域分析y(t)=[y(0+)y()]e+y()τt