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電路學

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Bode Plot

x axis : \(\omega\)
\(\omega\) 值算出 magnitude
y axis : \(dB = 20\times log_{10}(magnitude)\)

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

circuit-1.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
\(\tau=R_{TH}C\)
\(\tau=\dfrac{L}{R_{TH}}\)

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

\(Xcos(\omega t+\theta)=X\angle \theta\)

\(\frac{V_M\angle \theta_V}{I_M\angle \theta_i}=Z\angle \theta_Z=R+jX\)\(Z=\sqrt{R^2+X^2},\theta_Z=tan^{-1}\frac{X}{R}\) reverse:
\(R=\sqrt{Z}cos(\theta_Z)\) \(X=\sqrt{Z}sin(\theta_Z)\)

so \(\(j=1\angle 90^{\circ}\\ -j=1\angle -90^{\circ}\\ 1=1\angle 0^{\circ}\)\)

Z&Y

circuit-2.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

circuit-3.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

circuit-4.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

==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.

circuit-5.png

resonant

虛部為零時

RL : \(\omega_0=\dfrac{R}{L}\)
RLC : \(\omega_0=\dfrac{1}{\sqrt{LC}}\)

quality factor

circuit-6.png

series

circuit-7.png

\(|V_S|=Q|V_C|\\|V_S|=Q|V_L|\)

\(BW=\dfrac{\omega_0}{Q}=\dfrac{R}{L}\)

parallel

circuit-8.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:\(BW=\dfrac{\omega_0}{Q}=\dfrac{1}{RC}\)

circuit-9.png

\(\omega_{max}=\omega_0\sqrt{1-\dfrac{1}{2Q^2}}\)

parallel RLC with winding resistance

circuit-10.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. \(\dfrac{W}{2}\)\(W\)\(2W\) 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. \(0.1W\)\(\dfrac{W}{2}\)\(2W\)\(10W\) 算出真實的值,做修正

二次

image-20200604003630548 image-20200604003656882

\(\omega=\omega_0 → |H|=Q\)

image-20200604111258228

\(\dfrac{50}{(j\omega)^2(j\omega+0.5)}\)

  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

circuit-11.png

band pass filter

image-20200604113131828

image-20200604114626724

image-20200604114759684

image-20200604120823114

背背背

Ch13 Laplace Transform

表表(考試會給,不用背)

image-20200608023301413

image-20200618234510623

image-20200608132442011

==convolution 不教== (信號與系統)

complex

image-20200608152704947

image-20200613033231549

initial & final value theoreom

image-20200608160844141

image-20200608160818194

initial value (切換開關前) 還是要用之前的方法算

Ch14

C 之轉換

image-20200613001916993

image-20200613005917150

L 之轉換

image-20200613005934031

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