正弦量与相量

Sinusoids and Phasors

正弦量

辅助角公式 (The auxiliary angle formula):

$$A\cos (\omega t)+B\sin (\omega t)=C\cos (\omega t-\theta )$$

其中

$$C=\sqrt{A^2+B^2},\theta ={\tan }^{-1}\left(\frac{B}{A}\right)$$

前提为频率相同

相量

复数的基本形式：

$$\begin{array}{rlr}z& =x+jy& \text{Rectangular form}\\ z& =r\mathrm{\angle }\varphi & \text{Polar form}\\ z& =r(\cos \varphi +j\sin \varphi )& \text{Trigonometric form}\\ z& =r{e}^{j\varphi }& \text{Exponential form}\end{array}$$

其中

$$r=\sqrt{x^2+y^2},\varphi ={\tan }^{-1}\left(\frac{y}{x}\right)$$

复数的基本代数运算：

Operation	Result
Addition	$z_1+z_2=(x_1+x_2)+j(y_1+y_2)$
Subtraction	$z_1-z_2=(x_1-x_2)+j(y_1-y_2)$
Multiplication	$z_1{z}_2=r_1{r}_2\mathrm{\angle }({\varphi }_1+{\varphi }_2)$
Division	${\displaystyle \frac{z_1}{z_2}=\frac{r_1}{r_2}\mathrm{\angle }({\varphi }_1-{\varphi }_2)}$
Reciprocal	${\displaystyle \frac{1}{z}=\frac{1}{r}\mathrm{\angle }-\varphi }$
Square Root	${\displaystyle \sqrt{z}=\sqrt{r}\mathrm{\angle }\frac{\varphi }{2}}$
Conjugate	$z^{\ast }=x-jy=r\mathrm{\angle }-\varphi =r{e}^{-j\varphi }$

时域 (Time domain)：

$$\begin{array}{rl}v(t)& =V_m\cos (\omega t+\varphi )\\ & =\text{Re}\left(V_m{e}^{j(\omega t+\varphi )}\right)\\ & =\text{Re}\left(V_m{e}^{j\varphi }{e}^{j\omega t}\right)\\ & =\text{Re}\left(\mathbf{V}{e}^{j\omega t}\right)\end{array}$$

相量域 (Phasor domain)：

$$\mathbf{V}=V_m\mathrm{\angle }\varphi$$

微积分运算：

$$\begin{array}{rl}\frac{\mathrm{d}v(t)}{\mathrm{d}t}& ⟺j\omega \mathbf{V}\\ \int v(t)\mathrm{d}t& ⟺\frac{\mathbf{V}}{j\omega }\end{array}$$

注：积分运算忽略了积分常数，因为在交流稳态分析中，我们假设没有直流分量。

阻抗与导纳

Impedance and Admittance

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms (Ω).

$$\mathbf{V}=\mathbf{Z}\mathbf{I}$$$${\mathbf{Z}}_{eq}={\mathbf{Z}}_1+{\mathbf{Z}}_2+\cdots +{\mathbf{Z}}_N$$

The admittance Y is the reciprocal of impedance, measured in siemens (S).

$$\mathbf{I}=\mathbf{Y}\mathbf{V}$$$${\mathbf{Y}}_{eq}={\mathbf{Y}}_1+{\mathbf{Y}}_2+\cdots +{\mathbf{Y}}_N$$

基尔霍夫定律

For KVL,

$$\sum \mathbf{V}=0$$

For KCL,

$$\sum \mathbf{I}=0$$