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# 電路學 powered by $\LaTeX$ ## 共同網頁 http://cc.ee.ntu.edu.tw/~ntueecircuit/ ## Ch5 ### Superposition **step 1** make independent current source 斷路 (i=0) **step 2** make independent voltage source 短路 (v=0) ### Thevenin & Norton 打開 terminal → 算 V~OC~ → 算 R~TH~ ==算 R~TH~ or 時 independent voltage source 短路 independent current source 斷路== if have only dependent souces : 在 terminal 自由接上一個 voltage/current source if have both independent & dependent sources : 打開 terminal → 算 V~OC~ → terminal 接上電線算 I~SC~ → $R_{TH}=\frac{V_{OC}}{I_{SC}}$ $ ### source exchange ![](https://i.imgur.com/MRqLp2f.png) ## Ch6 ### Capacitance $Q=CV \\i=C\frac{dV_C}{dt}$ ### Inductance $\phi=Li \\V_L=L\frac{di_L}{dt}$ ## Ch7 ### first order **step 1** $v(t) = K_1 + K_2e^{-\frac{t}{\tau}}$ $i(t) = K_1 + K_2e^{-\frac{t}{\tau}}$ **step 2** in _t = 0_ 電容形同斷路 電感形同短路 **step 3** in _t = 0+_ 電容形同 independent voltage source $v_C(0+)=v_C(0-)$ 電感形同 independent current source $i_L(0+)=i_L(0-)$ **step 4** in _t=$\infty$_ (_>5$\tau$_) 電容形同斷路 電感形同短路 **step 5** $R_{TH}$ 為以電容/電感為 terminal 的等效電阻 $\tau=\frac{1}{R_{TH}C}$ $\tau=\frac{L}{R_{TH}}$ ### second order $s^2Ke^{st}+2\zeta \omega_0sKe^{st}+{\omega_0}^2Ke^{st}$ $s^2+2\zeta \omega_0s+{\omega_0}^2=0$ V(t)'s characteristic : $s^2+\frac{1}{RC}s+\frac{1}{LC}=0$ i(t)'s characteristic : $s^2+\frac{L}{R}s+\frac{1}{LC}=0$ (這兩個可直接從題目帶 KVL KCL 推得) overdamped : 相異實根 $x(t)=K_1e^{s_1t}+K_2e^{s_2t}$ underdamped : 虛根 $s=\alpha\pm\beta i$ $x(t)=e^{\alpha t}(K_1cos(\beta t)+K_2sin(\beta t))$ critically damped : 重根 $x(t)=K_1e^{st}+K_2te^{st}$ ## Ch8 ==**注意正負號!!!**== ### x+jy 公式 if $x+jy=re^{j\theta}$ $$ \left\{ \begin{array}{lr} r=\sqrt{x^2+y^2} \\ \theta=\tan^{-1}{\dfrac{y}{x}}\\ x=rcos\theta ,y=rsin\theta\\ \dfrac{1}{e^{j\theta}}=e^{-j\theta} \end{array} \right. $$ ### phasor <!-- \usepackage{steinmetz} \phase{a}$ --> $Xcos(\omega t+\theta)=X\angle \theta$ $\frac{V_M\angle \theta_V}{I_M\angle \theta_i}=Z\angle \theta_Z=R+jX$ → $Z=\sqrt{R^2+X^2},\theta_Z=tan^{-1}\frac{X}{R}$ reverse: $R=\sqrt{Z}cos(\theta_Z)$ $X=\sqrt{Z}sin(\theta_Z)$ so $$ \left\{ \begin{array}{ln} j=1\angle 90^{\circ}\\ -j=1\angle -90^{\circ}\\ 1=1\angle 0^{\circ} \end{array} \right. $$ ### Z&Y ![](https://i.imgur.com/D7ON7T1.png) $$ \left\{ \begin{array}{lr} Z_R=R \\ Z_L=j\omega L \\ Z_C=\frac{j}{\omega C}=\frac{1}{j\omega C} \\ \end{array} \right. $$ $Y=\dfrac{1}{Z}$(單位: S (siemens)) 並聯 → 相加 串聯 → 如同電阻並聯 ## Ch9 $cos(\theta_i)cos(\theta_V)=\frac{1}{2}[cos(\theta_i-\theta_V)-cos(\theta_i+\theta_V)]$ $rms = \dfrac{max}{\sqrt{2}}$ 電器 120V 是 rms ### Max average power ![](https://i.imgur.com/5JdGAdl.png) ### Power Factor $P = V_{rms}I_{rms}cos(\theta_V-\theta_i)$ $pf=\dfrac{P}{V_{rms}I_{rms}}=cos(\theta_z)$ ### Complex Power ![](https://i.imgur.com/5mGK9QJ.png) $S=V_{rms}I^*_{rms}\\=V_{rms}\angle\theta_VI_{rms}\angle-\theta_i\\=V_{rms}I_{rms}\angle(\theta_V-\theta_i)\\=P+Qj$ $$ \left\{ \begin{array}{ln} P=V_{rms}I_{rms}cos(\theta_V-\theta_i)\\ Q=V_{rms}I_{rms}sin(\theta_V-\theta_i)\\ \end{array} \right. $$ $$ \left\{ \begin{array}{ln} C:\theta_V-\theta_i=-90^{\circ}→leading\\ R:\theta_V-\theta_i=0^{\circ}\\ L:\theta_V-\theta_i=90^{\circ}→lagging \end{array} \right. $$ S單位:VA P單位:W Q單位:VARs ## Ch12 ==20200529-1 有 3 題習題講解 沒看== ==注意算 transfer function 時不一定直接是電阻相除,可能有並聯的情況== e.g. <img src="https://i.loli.net/2020/06/03/fYrITBm42CP6qS9.png" width=300 style="float:left"> ![](https://i.imgur.com/TFOWUds.png) ### resonant 虛部為零時 RL : $\omega_0=\dfrac{R}{L}$ RLC : $\omega_0=\dfrac{1}{\sqrt{LC}}$ ### quality factor ![](https://i.imgur.com/o0BZip2.png) #### series ![ ](https://i.imgur.com/LfQocNW.png) $|V_S|=Q|V_C|\\|V_S|=Q|V_L|$ $BW=\dfrac{\omega_0}{Q}=\dfrac{R}{L}$ #### parallel ![](https://i.imgur.com/3TaScgq.png) $|I_S|=Q|I_C|\\|I_S|=Q|I_L|$ $BW=\dfrac{\omega+0}{Q}=\dfrac{1}{RC}$ ### Bandwidth $BW=\dfrac{\omega_0}{Q}$ series:$BW=\dfrac{\omega_0}{Q}=\dfrac{R}{L}$ parallel:$BW=\dfrac{\omega_0}{Q}=\dfrac{1}{RC}$ ![](https://i.imgur.com/AxdbbMH.png) $\omega_{max}=\omega_0\sqrt{1-\dfrac{1}{2Q^2}}$ ### parallel RLC with winding resistance ![](https://i.imgur.com/UKyMBm2.png) R~w~ = 電感內電阻 $R_{par} = \dfrac{L}{CR_w}$ ### Bode Plot zeros : 分子 poles : 分母 #### 一次 ##### dB 弄成這種形式 $\dfrac{100(j\omega+100)}{(j\omega+1)(j\omega+10)(j\omega+50)}$ $j\omega+x$ 1. 若為分子 → 在 x 前 +0,x 後 +20dB/decade (x 為轉折頻率) 2. 若為分母 → 在 x 前 -0,x 後 -20dB/decade 3. 算出各區間的斜率 4. 帶值算出最左邊的點的 magnitude,再用 20log(magnitude) 換成 dB 5. 從那個點依各區間斜率劃出整個 bode plot 6. 在 $\dfrac{W}{2}$ 、$W$ 、$2W$ 算出真實的值,做修正 (W為轉折頻率) ##### phase $j\omega+x$ 1. 若在分子,則 +45$^{\circ}$/decade, $omega = x$ 時通過 45$^{\circ}$ ,直到 90$^{\circ}$ 2. 在分母則 -45$^{\circ}$/decade,直到 -90$^{\circ}$ 3. (若只有 $j\omega$ (x=0) 則分子 → 一直 90$^{\circ}$;分母 → 一直 -90$^{\circ}$) 4. 把漸進線疊起來 5. 在 $0.1W$、$\dfrac{W}{2}$ 、$2W$ 、$10W$ 算出真實的值,做修正 #### 二次 ![image-20200604003630548](https://i.loli.net/2020/06/04/dkZcwy1Y57lUPXJ.png) ![image-20200604003656882](https://i.loli.net/2020/06/04/KZOmbtUvnLg8d6W.png) $\omega=\omega_0 → |H|=Q$ ![image-20200604111258228](https://i.loli.net/2020/06/04/3gP4nGVOJtcNEhz.png) $\dfrac{50}{(j\omega)^2(j\omega+0.5)}$ 1. 規則大致如一次,但 $j\omega^2$ 變成是 $\pm$ 40dB/decade 2. 帶值算算出最左側 magnitude 之後,用 40log(magnitude) 換成 dB #### phase ### 12.5 filter network ![](https://i.imgur.com/xD9CZed.png) #### band pass filter ![image-20200604113131828](https://i.loli.net/2020/06/04/DjOZxlTqJAgvHau.png) ![image-20200604114626724](https://i.loli.net/2020/06/04/Fhi31HA5ytnDzQe.png) ![image-20200604114759684](https://i.loli.net/2020/06/04/18ni6QckRsXvhKy.png) ![image-20200604120823114](https://i.loli.net/2020/06/04/gZA7wE4ux3KXc2O.png) <p style="color:red;font-size:44px">背背背</p> ## Ch13 Laplace Transform ### 表表(考試會給,不用背) ![image-20200608023301413](https://i.loli.net/2020/06/08/AVQ2DP76wsbgkKI.png) ![image-20200618234510623](https://i.loli.net/2020/06/18/ZAY9rMDhGe2an64.png) ![image-20200608132442011](https://i.loli.net/2020/06/08/WGLFYIZH5xov1XN.png) ==convolution 不教== (信號與系統) <!-- {dqw|dwqed} --> <!-- so i just realized i can use atoms markdown editor and Typora as previewer damn --> ### complex ![image-20200608152704947](https://i.loli.net/2020/06/08/YwfvTRegnUXI4yj.png) ![image-20200613033231549](https://i.loli.net/2020/06/13/PHsDZMhe9WloXmc.png) ### initial & final value theoreom ![image-20200608160844141](https://i.loli.net/2020/06/08/uiNREhOmobQn5KU.png) ![image-20200608160818194](https://i.loli.net/2020/06/08/CDWA7YLcNVwv8K6.png) initial value (切換開關前) 還是要用之前的方法算 ## Ch14 ### C 之轉換 ![image-20200613001916993](https://i.loli.net/2020/06/13/TD2FsIfXBiZS9V7.png) ![image-20200613005917150](https://i.loli.net/2020/06/13/a1es4QopDCfHwKO.png) ### L 之轉換 ![ ](https://i.loli.net/2020/06/13/GUxu2WNyoPw7rsH.png) ![image-20200613005934031](https://i.loli.net/2020/06/13/7ZhRCzO9IcUnbKe.png) ### stable? pole(分母的根) 1. 在複數平面的左邊(i.e. 實部為負) → stable 2. 在複數平面的右邊(i.e. 實部為正) → unstable ![image-20200619011957342](https://i.loli.net/2020/06/19/vNGSosAUnaIXPum.png)

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