湖北工业大学电工电子实验教学示范中心.ppt

Exercise. In example 3, let gm=0.125S. Find the equivalent resistance seen by the capacitor and uC(t), t ≥0. + _ + _ + _ + _ 示例 已知图示电路中的电容原本充有 24V 电压，求K闭合后，电容电压和各支路电流随时间变化的规律。 解： 本题为求解一阶 RC 电路零输入响应问题 等效电路 t 0 则有 又由已知条件 利用并联分流，得 + _ + _ + _ 示例 t = 0 时 ，开关K由1→2，求电感电压和电流。 解： RL电路零输入响应问题 ① 一阶电路的零输入响应是由储能元件的初值引起的响应, 都是由初始值衰减为零的指数衰减函数。 ② 衰减快慢取决于时间常数 ? 。 ③ 同一电路中所有响应具有相同的时间常数。 ④ 一阶电路的零输入响应和初始值成正比，称为零输入线性。 小结 RC 电路 RL 电路 RC 电路 RL 电路 R 为与动态元件相连的一端口电路的等效电阻。 4. DC or step response of first-order circuits This section takes up the calculation of voltage and current responses when constant voltage or constant current sources are present. Linear Resistive circuit With independent sources L Linear Resistive circuit With independent sources C + _ Thevenin equivalent Thevenin equivalent C + _ + _ L + _ + _ C + _ + _ L + _ + _ Deriving the differential equation models characterizing each circuit’s voltage and current responses. By KVL and Ohm’ law, By KCL and Ohm’ law, Exercise. Constant differential equation models for the parallel RL and RC circuits of the figure. Note that these circuits are Norton equivalents of those in the figure. Again choose iL(t) as the response for the RL circuit and uC(t) as the response for the RC circuit. L + _ + _ C Answers: Observe that four differential equations have the same structure: where And the general formula for solving such a differential equation: for RC case for RL case where as long as x(t) is a capacitor voltage or inductor current, and f(τ)=F is a constant (nonimpulsive) forcing function. Which is valid for t≥t0. After some interpretation, this formula will serve as a basis for computing the response to RL and RC circuits driven by constant sources. if then for RC case for RL case Hence, the solution of the differential equation given constant or dc excitation becomes for RC case for RL case Example 8.5. For the circuit of the figure, suppose a 10V unit step excitation is applied at t=1 when it is found that the inductor cu