# 相量

## 定义

${\displaystyle A\cdot \cos(\omega t+\theta )={\frac {A\cdot e^{j(\omega t+\theta )}}{2}}+{\frac {A\cdot e^{-j(\omega t+\theta )}}{2}}}$ [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${\displaystyle {\frac {\omega }{2\pi }}}$ 。）

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

{\displaystyle {\begin{aligned}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\{Ae^{j(\theta +{\tfrac {\pi }{2}})}\cdot e^{j\omega t}\right\}\end{aligned}}}

## 运算法则

### 与常数（标量）相乘

{\displaystyle {\begin{aligned}\operatorname {Re} \{(Ae^{j\theta }\cdot Be^{j\phi })\cdot e^{j\omega t}\}&=\operatorname {Re} \{(ABe^{j(\theta +\phi )})\cdot e^{j\omega t}\}\\&=AB\cos(\omega t+(\theta +\phi ))\end{aligned}}}

### 微分和积分

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

${\displaystyle {\frac {d\ v_{C}(t)}{dt}}+{\frac {1}{RC}}v_{C}(t)={\frac {1}{RC}}v_{S}(t)}$

${\displaystyle v_{S}(t)=V_{P}\cdot \cos(\omega t+\theta ),\,}$

{\displaystyle {\begin{aligned}v_{S}(t)&=\operatorname {Re} \{V_{s}\cdot e^{j\omega t}\}\\\end{aligned}}}
${\displaystyle v_{C}(t)=\operatorname {Re} \{V_{c}\cdot e^{j\omega t}\},}$

${\displaystyle j\omega V_{c}+{\frac {1}{RC}}V_{c}={\frac {1}{RC}}V_{s}}$

${\displaystyle V_{c}={\frac {1}{1+j\omega RC}}\cdot (V_{s})={\frac {1-j\omega RC}{1+(\omega RC)^{2}}}\cdot (V_{P}e^{j\theta })\,}$

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

${\displaystyle v_{C}(t)={\frac {1}{\sqrt {1+(\omega RC)^{2}}}}\cdot V_{P}\cos(\omega t+\theta -\phi (\omega ))}$

### 加法

{\displaystyle {\begin{aligned}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{aligned}}}

${\displaystyle A_{3}^{2}=(A_{1}\cos {\theta _{1}}+A_{2}\cos {\theta _{2}})^{2}+(A_{1}\sin {\theta _{1}}+A_{2}\sin {\theta _{2}})^{2},}$
${\displaystyle \theta _{3}=\arctan {\left({\frac {A_{1}\sin {\theta _{1}}+A_{2}\sin {\theta _{2}}}{A_{1}\cos {\theta _{1}}+A_{2}\cos {\theta _{2}}}}\right)},}$

${\displaystyle A_{3}^{2}=A_{1}^{2}+A_{2}^{2}-2A_{1}A_{2}\cos(180^{\circ }-\Delta \theta ),=A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}\cos(\Delta \theta ),}$

${\displaystyle A_{1}\angle \theta _{1}+A_{2}\angle \theta _{2}=A_{3}\angle \theta _{3}.}$

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

## 脚注

1. ^
• j虚数单位${\displaystyle j^{2}=-1}$ ）。
• 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${\displaystyle {\frac {d}{dt}}(e^{j\omega t})=j\omega e^{j\omega t}}$ 得出，表明複指数是微分运算的本征函数
3. ^ 证明：

${\displaystyle {\frac {d\ \operatorname {Re} \{V_{c}\cdot e^{j\omega t}\}}{dt}}+{\frac {1}{RC}}\operatorname {Re} \{V_{c}\cdot e^{j\omega t}\}={\frac {1}{RC}}\operatorname {Re} \{V_{s}\cdot e^{j\omega t}\}}$

(式1)

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

${\displaystyle {\frac {d\ \operatorname {Im} \{V_{c}\cdot e^{j\omega t}\}}{dt}}+{\frac {1}{RC}}\operatorname {Im} \{V_{c}\cdot e^{j\omega t}\}={\frac {1}{RC}}\operatorname {Im} \{V_{s}\cdot e^{j\omega t}\}}$

(式2)

更显而易见的关系如下述方程所示：

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

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

${\displaystyle j\omega V_{c}\cdot e^{j\omega t}+{\frac {1}{RC}}V_{c}\cdot e^{j\omega t}={\frac {1}{RC}}V_{s}\cdot e^{j\omega t}}$
${\displaystyle \left(j\omega V_{c}+{\frac {1}{RC}}V_{c}={\frac {1}{RC}}V_{s}\right)\cdot e^{j\omega t}}$
${\displaystyle j\omega V_{c}+{\frac {1}{RC}}V_{c}={\frac {1}{RC}}V_{s}}$

证毕。

## 参考文献

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