电路理论（英文班）知识汇总

提示

• 带单位，写 $t>0$ （如有）
• When solving Thevenin equialent/Norton equialent, determine the positive / negative voltage. If there is a voltage source across the resistor we want to obtain the Thevein equivalent(or there is a current source through the branch), we can only find $V_{th}$ or $I_{N}$ ,while the $R_{th}$ or $R_{N}$ is $0$ or $\infty$
• power is different from energy
• saturate 饱和
• DC 直流
• ideal transformer: When the voltage is not given in the problem, it can be set by yourself.
• fraction 分数 / 比例
• $w=2\pi f$
• Note the positive and negative of voltage and current!!!

DC circuits

• A branch ( Number $b$ )is a single two-terminal element in an electric circuit. A node( Number $n$ ) is the point of connection between two or more branches. A loop ( independent loops $l$ ) is a closed path in a circuit. ( $n-1$ is the number of tree branch.)

$$b=l+n-1$$

• Delta to Wye

$$R_{1}=\frac{R_{b}R_{c} }{R_{a}+R_{b}+R_{c} }$$

（自己在对面，两面夹击，除以所有）

• Wye to Delta

$$R_{a}=\frac{ {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1} }}{R_{1} }$$

（所有相乘，除以自己，得到对面）

• Nodal Analyses

$$\left[ \begin{matrix} {G_{1 1} } & {G_{1 2} } & {\cdots} & {G_{1 N} } \\ {G_{2 1} } & {G_{2 2} } & {\cdots} & {G_{2 N} } \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {G_{N 1} } & {G_{N 2} } & {\cdots} & {G_{N N} } \\ \end{matrix} \right] \left[ \begin{matrix} {v_{1} } \\ {v_{2} } \\ {\vdots} \\ {v_{N} } \\ \end{matrix} \right]=\left[ \begin{matrix} {i_{1} } \\ {i_{2} } \\ {\vdots} \\ {i_{N} } \\ \end{matrix} \right]$$

$G_{kk}$ Sum of the conductances connected to node k $G_{kj}=G_{jk}$ Negative of the sum of the conductances directly connectiong nodes k and j $v_{k}$ Voltage of node k $i_{k}$ Sum of all independent current sources directly connected to node k, current entering the node treated as positive

• Mesh Analyses

$$\left[ \begin{matrix} {R_{1 1} } & {R_{1 2} } & {\cdots} & {R_{1 N} } \\ {R_{2 1} } & {R_{2 2} } & {\cdots} & {R_{2 N} } \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {R_{N 1} } & {R_{N 2} } & {\cdots} & {R_{N N} } \\ \end{matrix} \right] \left[ \begin{matrix} {i_{1} } \\ {i_{2} } \\ {\vdots} \\ {i_{N} } \\ \end{matrix} \right]=\left[ \begin{matrix} {v_{1} } \\ {v_{2} } \\ {\vdots} \\ {v_{N} } \\ \end{matrix} \right]$$

$R_{kk}$ Sum of the resistances in mesh k $R_{kj}=R_{jk}$ Negative of the sum of the resistances in common with mesh $k$ and $j$ $i_{k}$ Unknown mesh current for mesh $k$ in the colockwise direction $v_{k}$ Sum taken clockwise of all independent voltage sources in mesh $k$ , voltage rise as positive(Current flows from the positive terminal)

• Super Node & Super Mesh
• Network theorems
• Superposition principle
• Source transformation
• Thevenins(open-circuit voltage)/Norton(short-circuit current) theorems

$$R_{N}=R_{Th}$$ $$I_{N}=\frac{V_{Th} }{R_{Th} }$$

• maximum power transfer:

$$R_{L}=R_{Th}$$

• Wheatstone bridge
• op amp equivalent circuit
• basic op amp circuits
• With an ideal op-amp, the output voltage changes immediately when the voltage at the input changes, and for this reason the current can potentially reach infinite values.
• current through a capacitor---like conductors---voltage can't change instantly

$$i=C \frac{dv}{dt}$$

• voltage across an inductor---like resistors---current can't change instantly

$$v=L \frac{di}{dt}$$

• energy stored in a capacitor $\frac{1}{2}Cv^2$; in an inductor $\frac{1}{2}Li^2$ ,energy stored is $\frac{1}{2}L(i(t)^2-i(0)^2)$
• Please note that Thevenin equivalent circuits do not generally exist for circuits involving capacitors and resistors.
• D-W/W-D of capacitor & inductor: find the inverse of capacitor
• Thevenin equivalent of circuits involving capacitors and resistors, we can get $C_{Th}$
• For capacitors: $v(0)^+=v(0)^-$
• For inductors: $i(0)^+=i(0)^-$
• Complete response:

$$Complete\ response=natural\ reponse+forced\ response=transient\ response+steady-state \ response$$

• $\tau$ : for RC, $\tau=RC$ , for $RL$ , $\tau=\frac{L}{R}$
• Singularity functions:
  • unit step function $u(t)$ : $u ( t )=\left\{\begin{matrix} {0,} & {\qquad t < 0} \\ {1,} & {\qquad t > 0} \\ \end{matrix} \right.$
  • unit impulse function: $\delta( t )=\begin{cases} {0,} & {\qquad t < 0} \\ {\mathrm{U n d e f i n e d},} & {\qquad t=0} \\ {0,} & {\qquad t > 0} \\ \end{cases}$ ，Meaningful only when integrating
  • unit ramp function: $r ( t )=\left\{\begin{matrix} {0,} & {\qquad t \leq0} \\ {t,} & {\qquad t \geq0} \\ \end{matrix} \right.$
• wave representation: not to have negative coefficients in singularity functions and Note the coefficients on the axes
• $\tau=RC / \frac{L}{R}$
• step response(short-cut approach)

$$x ( t )=x ( \infty)+[ x ( t_{0}^{+} )-x ( \infty) ] e^{-( t-t_{0} ) / \tau}$$

• Short-cut approach can only be applied to voltage across capacitor.
• Sometimes can't use this method(in op amp), so we need to use straightforward approach.
• finding initial values: looking at voltage across the capacitors and current through the inductors, and get other values
• source free RLC series:
  • quadratic equation: $s^2+\frac{R}{L}s+\frac{1}{LC}=0$
  • natural frequencies:
    • $s_{1}=-\alpha+\sqrt{ \alpha^2-w_{0}^2 }$
    • $s_{2}=-\alpha-\sqrt{ \alpha^2-w_{0}^2 }$
  • neper frequency: $\alpha=\frac{R}{2L}$ .
  • resonant frequency / undamped natural frequency: $w_{0}=\frac{1}{\sqrt{ LC } }$
  • $s^2+2as+w_{0}^2=0$
  • $i(t)=A_{1}e^{ s_{1}t }+A_{2}e^{ s_{2}t }$ , and $A_{1}$, $A_{2}$ determined from the initial values( $i(0)$ and $\frac{di(0)}{dt}$)
• three types of solutions
  • overdamped case $\alpha>w_{0}$ ,two roots are negative and real
    • decays and approaches zero as it increases
    • $i(t)=A_{1}e^{ s_{1}t }+A_{2}e^{ s_{2}t }$
  • critically damped case $\alpha=w_{0}$ , $s_{1}=s_{2}=-\alpha$
    • $i(t)=(A_{2}+A_{1}t)e^{ -\alpha t }$
  • underdamped case $\alpha<w_{0}$
    • $s_{1}=-\alpha+jw_{d}$ , $s_{2}=-\alpha-jw_{d}$
    • $w_{d}=\sqrt{ w_{0}^2-\alpha^2 }$
    • $i(t)=e^{ -\alpha t }(B_{1}\cos w_{d}t+B_{2}\sin w_{d}t)$
• source free RLC parallel
  • $\alpha=\frac{1}{2RC}$
  • $w_{0}=\frac{1}{\sqrt{ LC } }$
  • solving the constant by getting $v(0)$ and $\frac{dv(0)}{dt}$
• step response: $v_{ss} / i_{ss}(t)=v / i(\infty)=V_{s} / I_{s}$
  • $x(t)=x_{t}(t)+x_{ss}(t)$
• Simple method: Simplify as much as possible to RLC series or RLC parallel or directly use Thevenins theroem.

AC circuits

Fundamentals

• $V_{m}$ amplitude
• $\cos(wt+\phi)\to \angle \phi$
• $\sin(wt+\phi)\to \angle{\phi-90^\circ}$
• $\frac{dv}{dt} \Leftrightarrow jwV$
• $\int v \, dt \Leftrightarrow \frac{V}{jw}$
• inductor: $V=jwLI$ , and $I$ lags $V$
• capacitor: $V=\frac{I}{jwC}$ , and $V$ lags $I$
• $Y=G+jB=\frac{1}{R+jX}$
• other: $Y-\Delta$ , Kirchhoff's law is same as DC circuits
• superposition theorem: different frequencies

Power analysis

Be careful to distinguish between amplitude and RMS values!

• instantaneous power: $p(t)=v(t)i(t)$
• average power: $P=\frac{1}{2}V_{m}I_{m}\cos(\theta_{v}-\theta_{i})$
  • $\theta_{v}=\theta_{i}$
  • $P=\frac{1}{2}V_{m}I_{m}=\frac{1}{2}I_{m}^2R=\frac{1}{2}\mid I\mid^2R$
• maximum average power transfer (so a factor of $\frac{1}{2}$ precedes the equation)
  • $Z_{L}=Z^*_{Th}$
  • 
• when $X_{L}=0$, $R_{L}=\mid Z_{Th}\mid$
• rms: $X_{rms}=\sqrt{ \frac{1}{T}\int _{0}^T \,x^2 dt }$
  • $I_{rms}=\frac{I_{m} }{\sqrt{ 2 } }$
  • $V_{rms}=\frac{V_{m} }{\sqrt{ 2 } }$
  • if $V / I$ has constant like $c+d\sin \dots$ , the $V_{rms} / I_{r ms}=\sqrt{ c^2+\frac{d^2}{2} }$
• apparent power(VA):
  • $S=\frac{1}{2}VI^*$
  • $S=V_{rms}I^*_{rms}$
  • $S=I^2_{rms}Z=\frac{V^2_{rms} }{Z^*}$ （注意什么时候带共轭什么时候不带共轭）
  • When answering the question, S should be written in the form $P+jQ$ .
  • $Q$ is the reactive power
• power factor $pf=\frac{P}{S}=\cos(\theta_{v}-\theta_{i})$
  • leading: current leads voltage
  • lagging: voltage leads current
  • power factor correction
    • phasor diagram:
    • power factor correction
    • $C=\frac{Q_{c} }{w^2V_{r ms} }=\frac{P(\tan\theta_{1}-\tan\theta_{2})}{wV^2_{rms} }$
    • $Q_{L}=\frac{V^2_{r ms} }{X_{L} }$
• reactive power: $Q=Im(S)=I^2_{r ms} X=V_{r ms}I_{r ms}\sin(\theta_{v}-\theta_{i})$
  • $Q=0$ resistive loads
  • $Q<0$ capacitive loads(leading pf)
  • $Q>0$ inductive loads(lagging pf)
  • power triangle: using $Q$ , not pf, that means, leading/lagging depends on $Q$ in the figure
• $|Z|=\sqrt{ R^2+X^2 }$

Three-Phase Circuits

• in three-phase circuits, $V_{p}$ is rms value of the phase voltages
• phase sequence: time order in which voltage pass through their respective maximum values
  • positive: $V_{an},V_{bn},V_{cn}$ counterclockwise, a leads b leads c
    • abc sequence
  • negative: $V_{an},V_{bn},V_{cn}$ clockwise
    • acb sequence
• phase xx: line-to-neutual xx
• line xx: line-to-line xx
• $Y$ connection of the source:
  • $V_{L}=\sqrt{ 3 }V_{P}$
  • $V_{ab}=\sqrt{ 3 }V_{an}\angle 30^\circ$
• $\Delta$ connection of the load:
  • $I_{L}=\sqrt{ 3 }I_{P}$
  • $I_{a}=\sqrt{ 3 }I_{AB}\angle -30^\circ$
• power (remind to $\times3$ )
  • $S=3S_{p}=3V_{p}I_{p}^*=3I_{p}^2Z_{p}=\frac{3V_{p}^2}{Z_{p}^*}$
  • $\mathbf{Z_{p} }=Z_{p}\angle\theta$
  • $\mathbf{S}=3V_{p}I_{p}\angle\theta=\sqrt{ 3 }V_{L}I_{L}\angle \theta$
• power loss of transmission lines in three-phase system and single-phase system
  • single-phase: $I_{L}=\frac{P_{L} }{V_{L} }$
    • $P_{loss}=2I_{L}^2R=2R \frac{P_{L}^2}{V_{L} }$ (2 wires)
  • three-phase: $I_{L}=\frac{P_{L} }{\sqrt{ 3 }V_{L} }$
    • $P_{loss}=3I_{L}^2R=R \frac{P_{L}^2}{V_L^2}$
• no need to get conjuction when doing $\Delta-Y$ .

Magnetically Coupled Circuits

Be careful with the dot convention!!!

• voltage ratings of transformers are usually specified in rms values
• mutual voltage(eg)
  • $v_{1}=M_{12} \frac{di_{2} }{dt}$
• $M=k\sqrt{ L_{1}L_{2} }$
• dot convention
  • enter the dotted terminal $\to$ positive at the dotted terminal
  • leave the dotted terminal $\to$ negatifve at the dotted terminal
• induction in series
  • aiding connection: $L=L_{1}+L_{2}+2M$
  • opposing connection: $L=L_{1}+L_{2}-2M$
• Transformer
  • From two KVL: $Z_{in}=R_{1}+jwL_{1}+ \frac{ {w^2M^2} }{R_{2}+jwL_{2}+Z_{L} }$
  • $Z_{R}=\frac{w^2M^2}{R_{2}+jwL_{2}+Z_{L} }$
• I-V relationships

$$\left[ \begin{matrix} { {\bf V}_{1} } \\ { {\bf V}_{2} } \\ \end{matrix} \right]=\left[ \begin{matrix} {j \omega L_{1} } & {j \omega M} \\ {j \omega M} & {j \omega L_{2} } \\ \end{matrix} \right] \left[ \begin{matrix} { {\bf I}_{1} } \\ { {\bf I}_{2} } \\ \end{matrix} \right]$$

• change to T circuit: (using mesh method)

$$\left[ \begin{array} {c} { {\mathrm{V}_{1} }} \\ { {\mathrm{V}_{2} }} \\ \end{array} \right]=\left[ \begin{array} {c c} { {j \omega( L_{a}+L_{c} )} } & { {j \omega L_{c} }} \\ { {j \omega L_{c} }} & { {j \omega( L_{b}+L_{c} )} } \\ \end{array}\right] \left[ \begin{array} {c} { {\mathrm{I}_{1} }} \\ { {\mathrm{I}_{2} }} \\ \end{array} \right]$$ $$L_{a}=L_{1}-M$$ $$L_{b}=L_{2}-M$$ $$L_{c}=M$$

• change to $\Pi$ circuit (using nodal method)

$$\left[ \begin{array} {c} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{array} \right]=\left[ \begin{array} {c c} { {\frac{1} {j \omega L_{A} }+\frac{1} {j \omega L_{C} }} } & { {-\frac{1} {j \omega L_{C} }} } \\ { {-\frac{1} {j \omega L_{C} }} } & { {\frac{1} {j \omega L_{B} }+\frac{1} {j \omega L_{C} }} } \\ \end{array} \right] \left[ \begin{array} {c} { {\mathbf{V}_{1} }} \\ { {\mathbf{V}_{2} }} \\ \end{array} \right]$$ $$L_{A}={\frac{L_{1} L_{2}-M^{2} } {L_{2}-M} } \, \qquad L_{B}={\frac{L_{1} L_{2}-M^{2} } {L_{1}-M} }$$ $$L_{C}=\frac{L_{1} L_{2}-M^{2} } {M}$$

• ideal transformers
  • $\frac{V_{2} }{V_{1} }=\frac{N_{2} }{M_{1} }=n$
  • $v_{1}$ , $v_{2}$ both positive at dotted terminal-- $+n$ otherwise-- $-n$
  • $I_{1}$ , $I_{2}$ both enter or leave-- $-n$ otherwise-- $+n$
  • $S_{1}=S_{2}$
  • reflected impedance $Z_{in}=\frac{Z_{L} }{n^2}$

Two-Port Networks

• current: always in the two-port, transmission: $-I$
• impedance $[z]$ , admittance $[y]$ ,hybrid $[h]$ , inverse hybrid $[g]$ , transmission $[T]$ , inverse transmission $[t]$

$$\left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ { {\mathbf{V}_{2} }} \end{matrix} \right]=[ \mathbf{z} ] \left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{matrix} \right], \qquad\left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{matrix} \right]=[ \mathbf{y} ] \left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ {\mathbf{V}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ {\mathbf{I}_{2} } \\ \end{matrix} \right]=[ \mathbf{h} ] \left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ {\mathbf{V}_{2} } \\ \end{matrix} \right]$$ $$\left[ \begin{matrix} {\mathbf{I}_{1} } \\ {\mathbf{V}_{2} } \\ \end{matrix} \right]=\left[ \mathbf{g} \right] \left[ \begin{matrix} {\mathbf{V}_{1} } \\ {\mathbf{I}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} {\mathbf{V}_{1} } \\ {\mathbf{I}_{1} } \\ \end{matrix} \right]=\left[ \mathbf{T} \right] \left[ \begin{matrix} {\mathbf{V}_{2} } \\ {-\mathbf{I}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} {\mathbf{V}_{2} } \\ {\mathbf{I}_{2} } \\ \end{matrix} \right]=\left[ \mathbf{t} \right] \left[ \begin{matrix} {\mathbf{V}_{1} } \\ {-\mathbf{I}_{1} } \\ \end{matrix} \right]$$

• reciprocal
  • $z_{12}=z_{21}$
  • $y_{12}=y_{21}$
  • $h_{12}=-h_{21}$
  • $g_{12}=-g_{21}$
  • $\Delta_{T}=1$
  • $\Delta_{t}=1$
  • has dependent sources are not reciprocal
• relationship

$$[ {\bf y} ]=[ {\bf z} ]^{-1}, \qquad[ {\bf g} ]=[ {\bf h} ]^{-1}, \qquad[ {\bf t} ] \neq[ {\bf T} ]^{-1}$$

• problems
  • determine parameters
  • Thevenin equivalent-solve quadratic equation

Interesting problems

• 7.50 $t=0$ ，$R_{3}$ is shorted
• when $i_{0}=0$, the source can be look like a wire.
• 
• 13.47 using mesh method.