# 低通RC电路对方波的响应



## 问题与解决方案

RC电路中电容器上的电压 $V_C(s)$ 和输入电压 $V_i(s)$ 在 $s$ 域中的关系方程已确定。
$V_i(s) - R \; C \; s \; V_C(s) - V_C(s) = 0$       (I)

$\displaystyle v_i(t) = V_0 \sum_{n=0}^{n=\infty} \left\{ u(t - n\;T)- u (t-(n+1/2)\;T) \right\}$

$\displaystyle \mathscr{L} \{ v_i (t) \} = V_0 \sum_{n=0}^{n=\infty} \left\{ \mathscr{L} \{ u(t - n\;T) \} - \mathscr{L} \{ u (t-(n+1/2)\;T) \} \right\}$

$\dfrac{e^{-\alpha s }}{s}$

$\displaystyle V_i(s) = V_0 \sum_{n=0}^{n=\infty} \left\{ \dfrac{ e^{-n\;T\;s}}{s} - \dfrac {e^{-(n+1/2)\;T \; s}}{s} \right\}$

$\displaystyle V_0 \sum_{n=0}^{n=\infty} \left\{ \dfrac{ e^{-n\;T\;s}}{s} - \dfrac {e^{-(n+1/2)\;T \; s}}{s} \right\} = R \; C \; s \; V_C(s) + V_C(s)$

$\displaystyle V_C(s) = \dfrac{V_0}{s(R\;C\;s + 1)} \sum_{n=0}^{n=\infty} \left\{ e^{-n\;T\;s} - e^{-(n+1/2)\;T s} \right\}$       (II)

$\dfrac{V_0}{s(R\;C s + 1)} = \dfrac{V_0}{s} - \dfrac{R\;C V_0}{R\;C s + 1}$

$\dfrac{V_0}{s(R\;C s + 1)} = V_0 \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right)$

$\displaystyle V_C(s) = V_0 \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \sum_{n=0}^{n=\infty} \left\{ e^{-n\;T\;s} - e^{-(n+1/2)\;T \; s} \right\}$

$\displaystyle V_C(s) = V_0 \sum_{n=0}^{n=\infty} \left\{ e^{-n\;T\;s} \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \\\\ \quad \quad \quad \quad - e^{-(n+1/2)\;T s} \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \right\}$

$\displaystyle v_C(t) = \mathscr{L^{-1}} \left\{ V_c(s) \right\}$
$\displaystyle = V_0 \sum_{n=0}^{n=\infty} \left\{ \mathscr{L^{-1}} \left\{ e^{-n\;T\;s} \left (\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \right\} \\\\ \quad \quad \quad \quad - \mathscr{L^{-1}} \left\{ e^{-(n+1/2)\;T \; s} \left (\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \right\} \right\}$

$\mathscr{L^{-1}} ( e^{-\tau s} F(s) )$

$\mathscr{L^{-1}} ( e^{-\tau s} F(s) ) = u(t- \tau) f(t - \tau)$，其中 $f(t)$ 是 $F(s)$ 的逆拉普拉斯变换

$\displaystyle \mathscr{L^{-1}} \left\{ \dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right\}$
$\displaystyle = \mathscr{L^{-1}} \left\{ \dfrac{1}{s} \right\} - \mathscr{L^{-1}} \left\{ \dfrac{1}{s + \dfrac{1}{R\;C}} \right\}$
$= u(t) - u(t) e^{-\frac{t}{R\;C}} = u(t)(1 - e^{-\frac{t}{R\;C}} )$

$\displaystyle v_C(t) = \displaystyle V_0 \sum_{n=0}^{n=\infty} \left \{ u(t-nT) \; \left(1 - e^{- \dfrac{t - n \; T}{R \;C} } \right) \\\\ \quad \quad \quad \quad - u(t-(n+1/2)T) \; \left(1 - e^{-\dfrac{ t - (n + 1/2) T}{\; R \; C} } \right) \right\}$

$R\;C = 200 \times 5 \times 10^{-3} = 1$ 秒

a) $T = 15 RC = 15$ 秒
b) $T = 10 RC = 10$ 秒
c) $T = 5 RC = 5$ 秒
d) $T = 2 RC = 2$ 秒

RC电路对阶跃电压的响应