Ch9 Sinusoids and Phasors

Introduction

Why sinusoid? 自然过程中很多都是这样的。

$v(t)=\overbrace{v_m}^{\mathrm{amplitude}}\sin/\cos(\underbrace{\overbrace{\omega}^{\mathrm{angular~frequency}} t+\overbrace{\Phi}^{\mathrm{phase}}}_{\mathrm{argument}})$

$\omega=2\pi f=\frac{2\pi}{T}$

Only cover steady-state response in Ch.10.



quite difficult to solve problem like this.

Time Domain $\Rightarrow $ ? Domain. Phasor Domain.

Sinusoids




需要比较幅值和相位角。



$v_1=-10\cos (\omega t+50^\circ)=-\sin(\omega t +50^\circ-90^\circ)$

$v_2$ leads $v_1$ by $30^\circ$. $v_1-v_2>0$ leads, $v_1-v_2<0$ lags.

$i_1=-4\sin(377t+25^\circ)=-4\cos (377t+25^\circ+90^\circ)$

$i_1-i_2=115^\circ+40^\circ=155^\circ$. $i_1$ leads $i_2$ by $155^\circ$.

Phasors

Expressions of complex number

Rectangular form: $\boldsymbol z=x+jy$ where $j=\sqrt{-1}.$, $x$ is real part and $y$ is imaginary part.

Polar or expoential form: \boldsymbol z=r\ang \varphi=re^{j\varphi}, where $r$ is magnitude, and $\varphi$ is phase.

From rectangular to polar.

$\forall x$ and $y$, $r=\sqrt{x^2+y^2},\varphi=\tan^{-1}\frac{y}{x}$.

$\forall r$ and $\varphi$, $x=r\cos \varphi,y=r\sin\varphi$. \boldsymbol z=x+jy=r\ang \varphi=re^{j\varphi}.


A complex number that represents amplitude and phase of a sinusoid, which is introduced to simplifying ac circuit analysis.

$v(t)=V_m\sin(\omega t+\varphi)$ can be written as

\boldsymbol V=V_m\ang \varphi

where $V_m$ is magnitude, and $\varphi$ is phase.

Euler’s identity and phasor representation

$e^{\pm j\varphi}=\cos\varphi\pm j\sin\varphi$

where $\cos \varphi=\operatorname{Re}(e^{j\varphi}),\sin \varphi=\operatorname{Im}(e^{j\varphi})$.


Sinor

展示了 cosine-sine diagram 和 Phasor diagram 的关系。




Summary

$v(t)=V_m\cos (\omega t+\varphi)$ can be taken as $v(t)=\operatorname{Re}(\bold V e^{j\omega t})$.

$v(t)=V_m\sin (\omega t+\varphi)$ can be taken as $v(t)=\operatorname{Im}(\bold V e^{j\omega t})$.

where \bold V=V_m e^{j\varphi}=VC_m \ang \varphi.




Phase 取决于象限。


Calculation of Phasors

Summation and subtraction of sinusoids using phasors.

Phasor domain $\Rightarrow $ Time domain $\Rightarrow$ Phasor domain.



$10\cos 2t=80i+0.24\frac{\mathrm d i}{\mathrm d t}+200\int_0^t i\mathrm d t$

$\Rightarrow$ Phasor:

10\ang 0^\circ=80 \bold I+0.24 j\omega \bold I+200\frac{\bold I}{j\omega}

$\omega=2$.

10\ang 0^\circ=(a+jb)\bold I \quad \bold I=I_m \ang \theta


-6\bold Ij+4\bold I+4\bold I/j=50\ang 75^\circ


\bold I=7.5 \ang 30^\circ \quad \bold V=120\ang 75^\circ

\bold V/\bold I=(120/7.5)\ang (75^\circ-30^\circ)=16\ang 45^\circ

Impedence.

Phasor Relationships for Circuit Elements

Circuit Elements: $R,L,C$.

Resistor

i(t)=I_m \cos (\omega t +\phi)\Rightarrow \bold I=I_m\ang \phi\Rightarrow v=iR=RI_m\cos (\omega t+\phi)\Rightarrow \bold V=RI_m\ang \phi.

这种情况下相位不变。


Inductor

\begin{aligned} &i(t)=I_m\cos (\omega t+\phi)\\ \Rightarrow &\bold I=I_m\ang \phi\\ \Rightarrow &v=L\frac{\mathrm d i}{\mathrm d t}=\omega LI_m\cos (\omega t+\phi+90^\circ) \text{~voltage leads current by 90}^\circ\\ \Rightarrow & \bold V=\omega L I_m \ang(\phi+90^\circ)=j\omega L\bold I~(e^{j90^\circ}=\cos 90^\circ+j\sin 90^\circ=j) \end{aligned}



Capacitor

$\bold V=\bold I/j\omega C,\bold I=j\omega C\bold V$


Summary


$\frac{\bold V}{\bold I}=R,j\omega L,\frac{1}{j\omega C}$

叫做阻抗 Impedence.









Take 110V ac source with 0 phase. \bold V_L=V_L \ang \phi, \phi=\arctan 85/110.


Forced response $v_0$:

$\bold V_0=j\omega L \bold I+\frac{1}{j\omega C} \bold I=0$

$\omega L=\frac{1}{\omega C}\Rightarrow \omega=\frac{1}{\sqrt{LC}}$

Impedance and Admittance

Impedance of elements. 阻抗

$Z=\frac{\bold V}{\bold I} (\Omega)$

Showing how circuit opposes sinusoidal current.

$\frac{\bold V}{\bold I}=R,j\omega L,\frac{1}{j\omega C}$

Admittance of elements. 导纳

$\frac{\bold I}{\bold V}=G,\frac{1}{j\omega L},j\omega C$




将交流电的分析方法转化为直流电的分析方式。Impedance 是复数，例如 $Z=3+j4$，代表感性阻抗。直流属于交流的特例，比如取 $\omega \to 0$，就可以得到 $i=0$.

Expressions of impedance

Rectangular form: $\bold Z=R\pm jX$ where $R:$ resistance（电 阻） $X$:reactance（电 抗）

Polar form: \bold Z=|\bold Z|\ang\theta where $|\bold Z|=\sqrt{R^2+X^2},\theta=\pm \tan^{-1} X/R$.

$+$: inductive or lagging, $-$ : capacitive or leading


Expressions of admittance

Rectangular form: $\bold Y=G+jB$ where $G$: conductance（电 导）$B$: susceptance （电 纳）

$G+jB=\frac{1}{R+jX}$

$G=\frac{R}{R^2+X^2},B=-\frac{X}{R^2+X^2}$

Nature of impedance



$\bold Z=5-j\frac{1}{\omega C}$

$\bold I=\frac{\bold V_s}{\bold Z},\bold V=-j\frac{1}{\omega C}\bold I$

Kirchhoff’s Laws in the Frequency Domain



Impedance Combinations

Voltage-division priciple and the series resistance still holds.






$Z_1=\frac{1}{j\omega C_1} \quad Z_2=8+j\omega L\quad Z_3=3+\frac{1}{j\omega C_2}\\ Z_{\mathrm{in}}=Z_1+Z_2//Z_3$


Keys: Problem given in time domain. Answer is in time domain.

\bold V_s=20\ang -15 ^\circ

$Z=60+\underbrace{\frac{1}{j4\cdot 10 \mathrm m}//j 4\cdot 5}_{Z_{\mathrm{eq}}}$

$\bold V_o=\frac{Z_{eq}}{Z_{eq}+60}\bold V_s$

计算器复数：https://zhuanlan.zhihu.com/p/68880025




Bridge Circuit.


其实是并联。