# 相量

## 定义

$A\cdot \cos(\omega t + \theta) = \frac{A\cdot e^{j(\omega t + \theta)}}{2}+\frac{A\cdot e^{-j(\omega t + \theta)}}{2}$[注 1]
（其中A和θ分別表波的振幅以及相位，而其頻率f則定義為$\frac{\omega}{2\pi}$。）

\begin{align} A\cdot \cos(\omega t + \theta) &= \operatorname{Re} \left\{ A\cdot e^{j(\omega t + \theta)}\right\} \\ &= \operatorname{Re} \left\{ A e^{j\theta} \cdot e^{j\omega t}\right\} \\ \end{align}

\begin{align} A\cdot \cos(\omega t + \theta) &= A\cdot \sin(\omega t + \theta + \frac{\pi}{2}) \\ &= \operatorname{Im} \left\{ A\cdot e^{j(\omega t + \theta+\tfrac{\pi}{2})}\right\} \\ &= \operatorname{Im} \left\{ A e^{j(\theta+\tfrac{\pi}{2})} \cdot e^{j\omega t}\right\} \end{align}

## 运算法则

### 与常数（标量）相乘

\begin{align} \operatorname{Re}\{(A e^{j\theta} \cdot B e^{j\phi})\cdot e^{j\omega t} \} &= \operatorname{Re}\{(AB e^{j(\theta+\phi)})\cdot e^{j\omega t} \} \\ &= AB \cos(\omega t +(\theta+\phi)) \end{align}

### 微分和积分

\begin{align} \operatorname{Re}\left\{\frac{d}{dt}(A e^{j\theta} \cdot e^{j\omega t})\right\} &= \operatorname{Re}\{A e^{j\theta} \cdot j\omega e^{j\omega t}\} \\ &= \operatorname{Re}\{A e^{j\theta} \cdot e^{j\tfrac{\pi}{2}} \omega e^{j\omega t}\} \\ &= \operatorname{Re}\{\omega A e^{j(\theta + \tfrac{\pi}{2})} \cdot e^{j\omega t}\} \\ &= \omega A\cdot \cos(\omega t + \theta + \frac{\pi}{2}) \end{align}

$\frac{d\ v_C(t)}{dt} + \frac{1}{RC}v_C(t) = \frac{1}{RC}v_S(t)$

$v_S(t) = V_P\cdot \cos(\omega t + \theta),\,$

\begin{align} v_S(t) &= \operatorname{Re} \{V_s \cdot e^{j\omega t}\} \\ \end{align}
$v_C(t) = \operatorname{Re} \{V_c \cdot e^{j\omega t}\},$

$j \omega V_c + \frac{1}{RC} V_c = \frac{1}{RC}V_s$

$V_c = \frac{1}{1 + j \omega RC} \cdot (V_s) = \frac{1-j\omega R C}{1+(\omega R C)^2} \cdot (V_P e^{j\theta})\,$

$\frac{1}{\sqrt{1 + (\omega RC)^2}}\cdot e^{-j \phi(\omega)}\,$，其中$\phi(\omega) = \arctan(\omega RC)\,$。（简化的极坐标形式为：$\frac{1}{\sqrt{1 + (\omega RC)^2}} \angle -\arctan(\omega RC)$

$v_C(t) = \frac{1}{\sqrt{1 + (\omega RC)^2}}\cdot V_P \cos(\omega t + \theta- \phi(\omega))$

### 加法

\begin{align} A_1 \cos(\omega t + \theta_1) + A_2 \cos(\omega t + \theta_2) &= \operatorname{Re} \{A_1 e^{j\theta_1}e^{j\omega t}\} + \operatorname{Re} \{A_2 e^{j\theta_2}e^{j\omega t}\} \\ &= \operatorname{Re} \{A_1 e^{j\theta_1}e^{j\omega t} + A_2 e^{j\theta_2}e^{j\omega t}\} \\ &= \operatorname{Re} \{(A_1 e^{j\theta_1} + A_2 e^{j\theta_2})e^{j\omega t}\} \\ &= \operatorname{Re} \{(A_3 e^{j\theta_3})e^{j\omega t}\} \\ &= A_3 \cos(\omega t + \theta_3), \end{align}

$A_3^2 = (A_1 \cos{\theta_1}+A_2 \cos{\theta_2})^2 + (A_1 \sin{\theta_1}+A_2 \sin{\theta_2})^2,$
$\theta_3 = \arctan{\left(\frac{A_1 \sin{\theta_1} + A_2 \sin{\theta_2}}{A_1 \cos{\theta_1} + A_2 \cos{\theta_2}}\right)},$

$A_3^2 = A_1^2 + A_2^2 - 2 A_1 A_2 \cos(180^\circ - \Delta\theta), = A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta\theta),$

$A_1 \angle \theta_1 + A_2 \angle \theta_2 = A_3 \angle \theta_3.$

$\cos(\omega t) + \cos(\omega t + \frac{2\pi}{3}) + \cos(\omega t +\frac{4\pi}{3}) = 0\,$

## 脚注

1. ^
• j虚数单位$j^2 = -1$）。
• 虚数单位用j表示是電機工程學中的用法，而在数学中则一般用i表示虚数单位。
2. ^ 可由$\frac{d}{dt}(e^{j \omega t}) = j \omega e^{j \omega t}$得出，表明複指数是微分运算的本征函数
3. ^ 证明：

$\frac{d\ \operatorname{Re} \{V_c \cdot e^{j\omega t}\}}{dt} + \frac{1}{RC}\operatorname{Re} \{V_c \cdot e^{j\omega t}\} = \frac{1}{RC}\operatorname{Re} \{V_s \cdot e^{j\omega t}\}$

(式1)

由於对所有$t\,$，更清楚地说是所有$t-\frac{\pi}{2\omega },\,$，上式均成立，因此下式同样成立：

$\frac{d\ \operatorname{Im} \{V_c \cdot e^{j\omega t}\}}{dt} + \frac{1}{RC}\operatorname{Im} \{V_c \cdot e^{j\omega t}\} = \frac{1}{RC}\operatorname{Im} \{V_s \cdot e^{j\omega t}\}$

(式2)

更显而易见的关系如下述方程所示：

$\frac{d\ \operatorname{Re} \{V_c \cdot e^{j\omega t}\}}{dt} = \operatorname{Re} \left\{ \frac{d\left( V_c \cdot e^{j\omega t}\right)}{dt} \right\} = \operatorname{Re} \left\{ j\omega V_c \cdot e^{j\omega t} \right\}$
$\frac{d\ \operatorname{Im} \{V_c \cdot e^{j\omega t}\}}{dt} = \operatorname{Im} \left\{ \frac{d\left( V_c \cdot e^{j\omega t}\right)}{dt} \right\} = \operatorname{Im} \left\{ j\omega V_c \cdot e^{j\omega t} \right\}$

将以上二式代入式1式2，然後令式2乘以$j\,$，最後将式1和乘$j\,$後的式2相加，得到：

$j\omega V_c \cdot e^{j\omega t} + \frac{1}{RC}V_c \cdot e^{j\omega t} = \frac{1}{RC}V_s \cdot e^{j\omega t}$
$\left(j\omega V_c + \frac{1}{RC}V_c = \frac{1}{RC}V_s\right) \cdot e^{j\omega t}$
$j\omega V_c + \frac{1}{RC}V_c = \frac{1}{RC}V_s$

证毕。

## 参考文献

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