something something i'm not responsible for lost marks. have fun. # Constants | Name | Symbol | Value | | ----------------- | ------- | ----------------------- | | Speed of light | $c$ | $3\times 10^8$ | | Elementary charge | $q_e$ | $1.6022\times 10^{-19}$ | | Magnetic constant | $\mu_0$ | $4\pi \times 10^{-7}$ | # Basic math $$ \begin{align} x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \sin(\theta) &= \cos(\theta-90°)\\ \cos(\theta) &= \sin(\theta+90°)\\ \text{Euler's formula: } e^{j\theta} &= \cos(\theta)+j\sin(\theta) \end{align} $$ # Chapter 1 $$ \begin{align} \text{Lumped model: } \lambda &= \frac{c}{f} \gggtr \text{dimension} \text{ (At least 10 times)}\\ i(t)&=\frac{dq}{dt} \Leftrightarrow q(t)=\int i(t)\cdot dt\\ P&=v\times i=i\times \frac{W}{q}\\ v&=\frac{W}{q} \Leftrightarrow W=v\times q = \int v\times i \cdot dt\\ \end{align} $$ # Chapter 4 $$ \begin{align} \text{Load line: } i_x &= -\frac{v_x}{R_T}+ \frac{v_t}{R_T}\\ \text{General: } A_v &= \frac{v_{out}}{v_{in}}\\ \text{Inverting: } A_v &= -\frac{R_f}{R_{in}}\\ \text{Non Inverting: } A_v &= 1+\frac{R_f}{R_{1}}\\ \text{Series of op amps total: } A_v &= (A_v)_1\times(A_v)_2\times \dots\times (A_v)_n \end{align} $$ | Inverting | Non-inverting | | ---------------------------- | ---------------------------- | | ![](2022-06-07-14-07-20.png) | ![](2022-06-07-14-12-26.png) | # Chapter 5 ## Capacitor $$ \begin{align} C &= \frac{q}{v}\\ i(t) = C\frac{dv}{dt} &\Leftrightarrow v(t) = \frac{1}{C}\int_0^t i(t)\cdot dt\\ \text{Series: }\frac{1}{C_T} &= \sum^N_{i=0}\frac{1}{C_i}\\ \text{Parallel: }C_T &= \sum^N_{i=0}C_i\\ \text{Energy: }E &= \frac{1}{2}Cv^2 \end{align} $$ ### Differential equation solution Where $v_s=v_\infty$: $$ \begin{align} \tau &= R\times C\\ v_C(t) &= \begin{cases} \begin{array}{lr} v_0 & t\leq 0\\ v_\infty+(v_0-v_\infty)e^{-t/\tau} & t > 0 \end{array} \end{cases}\\ & v_0 e^{-t/\tau} \text{ (Natural response, no input)}\\ & v_\infty\left(1-e^{-t/\tau}\right) \text{ (Forced response, input)}\\ i_C(t) &= \frac{v_s-v_C(t)}{R}=\frac{-(v_0-v_\infty)e^{-t/\tau}}{R} \end{align} $$ 1. When $t>0$, remove all independent sources, find equivalent resistance and capacitance, find $\tau$. 2. Set C as open circuit, find initial capacitor voltage $v_0$ at $t=0$ 3. Set C as open circuit, find final capacitor voltage $v_\infty$ at $t\to\infty$ ## Inductor $$ \begin{align} L &= \frac{\lambda}{i}\\ v(t)=L\frac{di}{dt} &\Leftrightarrow i(t)=\frac{1}{L}\int_0^tv\cdot dt\\ \text{Series: }L_T &= \sum^N_{i=0}L_i\\ \text{Parallel: }\frac{1}{L_T} &= \sum^N_{i=0}\frac{1}{L_i}\\ \text{Energy: }E &= \frac{1}{2}Li^2 \end{align} $$ ### Differential equation solution Where $v_s/R=i_\infty$: $$ \begin{align} \tau &= \frac{L}{R}\\ i(t) &= \begin{cases} \begin{array}{lr} i_0 & t\leq 0\\ i_\infty+(i_0-i_\infty)e^{-t/\tau} & t > 0 \end{array} \end{cases}\\ & i_0 e^{-t/\tau} \text{ (Natural response, no input)}\\ & i_\infty\left(1-e^{-t/\tau}\right) \text{ (Forced response, input)}\\ \end{align} $$ 1. When $t>0$, remove all independent sources, find equivalent resistance and inductance, find $\tau$. 2. Set L as short circuit, find initial inductor current $i_0$ at $t=0$ 3. Set L as short circuit, find final inductor current $i_\infty$ at $t\to\infty$ ## Voltage drop in DC for capacitor and inductor at steady state ``` CAPACITOR: INDUCTOR: v_T _ v_T _ | <- V_1 | <- V_1 C1 = ) <- V_D1 L1 3 ) <- V_D1 | | C2 = C2 3 | | ... ... | | CN = LN 3 | | GND * GND * ``` ## Capacitor Current through capacitors in series is the same, so all capacitors have same charge stored $q$. $$ \begin{align} \text{Voltage drop over capacitor $i$: } v_{Di} &= v_T\frac{C_T}{C_i}\\ \text{Voltage divider: } v_i &= v_T\frac{C_T}{\frac{1}{C_i}+\frac{1}{C_{i+1}}+\dots+\frac{1}{C_N}}\\ \end{align} $$ ## Inductor No voltage drop in steady state (Inductor is a short circuit) # Chapter 7 ## Maximum power transfer in AC $$ \begin{align} \text{Condition: } &\overline{Z_L} = \overline{Z_S^*}\\ \text{Maximum power to load (50\\\%): } &2P_\text{avg}=P_\text{rms}\cdot\sqrt{2} = P_\text{max} = \frac{{|V_S|}^2}{4R_S}\\ & = \frac{{|V_L|}^2}{4R_L}\\ \text{Total maximum power: } &2P_\text{avg}=P_\text{rms}\cdot\sqrt{2} = P_\text{max} = \frac{{|V_S|}^2}{2R_S} \end{align} $$ ![](2022-06-01-16-41-13.png) ## Complex Power Where $\bar{V}=V\angle\theta$ and $\bar{I}=I\angle\phi$: $$ \let\lb=( \let\rb=) \def\({\left\lb} \def\){\right\rb} % Put \left(,\right) on \(,\) \begin{align} \text{Complex [VA]: }\bar{S} &= \bar{V}_\text{rms}\times \bar{I}_\text{rms}^* = \frac{\bar{V}\times \bar{I}^*}{2} = \frac{VI}{2}\angle\(\theta-\phi\)\\ \text{Apparent [VA]: } |\bar{S}|\\ \text{Real [W]: } P &= |\bar{S}| \cos\(\theta-\phi\) = \text{Re}\(\bar{S}\)\\ \text{Reactive [VAR]: } Q &= |\bar{S}| \sin\(\theta-\phi\) = \text{Im}\(\bar{S}\)\\ Q &= P\tan\(\arccos\(\text{PF}\)\)\\ \text{Power Factor (PF): } \text{PF} &= \frac{P}{|\bar{S}|} = \frac{P}{\sqrt{P^2+Q^2}}\\ \text{PF from angles: } \text{PF} &= \cos\(\theta-\phi\) = \cos\(\arctan\(\frac{Q}{P}\)\)\\ \end{align} $$ and where $\bar{Z}\_\text{load} = Z_\text{load}\angle\lambda = R+jX$: $$ \let\lb=( \let\rb=) \def\({\left\lb} \def\){\right\rb} % Put \left(,\right) on \(,\) \begin{align} \text{PF from impedance: } \text{PF} &= \frac{\text{Re}\(\bar{Z}_\text{load}\)}{|\bar{Z}_\text{load}|} = \frac{R}{\sqrt{R^2+X^2}} \\ \text{PF from angles: } \text{PF} &= \cos\(\lambda\) = \cos\(\arctan\(\frac{X}{R}\)\) \end{align} $$ ### Types of power factors Where $\bar{S}=|\bar{S}|\angle\varphi$: $$ \varphi = \arctan\left(\frac{Q}{P}\right)$$ | | Lagging | Leading | Unity | | ----------- | -------------- | ------------- | ------------ | | Voltage | Current behind | Current ahead | In phase | | Load type | Inductive | Capacitive | Resistive | | $Q$ | $Q>0$ | $Q<0$ | $Q=0$ | | $\varphi$ | $\varphi>0°$ | $\varphi<0°$ | $\varphi=0°$ | | PF [Load] | $[0,1)$ | $[0,1)$ | $1$ | | PF [Source] | $[0,-1)$ | $[0,-1)$ | $-1$ | # Chapter 8 $$ \begin{align} \text{Faraday's law: }\varepsilon &= -N\frac{d\varPhi}{dt}\\ \text{Ampere's law: }B &= \frac{\mu_0 I}{2\pi r}\\ \end{align} $$ ## Transformer Step up: $n>1$\ Step down: $n<1$ $$ \begin{align} \frac{V_s}{V_p} &= \frac{N_s}{N_p}=\frac{i_p}{i_s}=n\\ \bar{Z}_{in} &= \frac{1}{n^2}\bar{Z}_L \end{align} $$ Derived from equation 42: $$ \begin{align*} i_s &= i_p/n \\ &= \frac{V_p}{\frac{1}{n^2}\times \bar{Z}_{L}\times n} \\ \end{align*}\\ $$ $$ \begin{align} i_s &=\frac{V_p\times n}{\bar{Z}_{L}}\\ \end{align} $$ ## Motor Note - this section on motors is a bit sketchy, best to refer to slides! For permanent motors, define permanent torque constant $k_{TP}=k_T\varPhi$\ and define permanent armature constant $k_{aP} = k_a\varPhi$ Note, back emf should oppose $v_a$ and $i_a$ $$ \begin{align} \text{Back emf: }e_b=k_a\times \varPhi\times \omega_m = k_{aP} \times\omega_m\\ \end{align} $$ For ideal motor, torque and armature constants are the same: $k_a=k_T$, $k_{aP}=k_{TP}$ $$ \begin{align} \text{Heat dissipated: } P_e &= e_b\times i_a = k_{aP} \times \omega_m \times i_a\\ \text{Mechanical power: } P_m &= \omega_m\times T_L = k_{TP}\times \omega_m \times i_a\\ \end{align} $$ Define $p$ as number of magnetic poles and $M$ as the number of parallel paths in armature winding. $$ \begin{align} \text{Constants for ideal motor: } k_a &= k_T = \frac{pN}{2\pi M} \end{align} $$ ### Most important motor equations to solve For permanent magnet DC motor in DC steady state:\ Define viscous frictional damping coefficient $b$ and load torque $T_L$ - If $b$ is not defined, assume no damping (??? todo - check) $$ \begin{align} &\begin{cases} 0 &= v_a - i_a R_a - k_{aP} \omega_m &= v_a - i_a R_a - e_b \\ k_{TP} i_a &= T_L + b\times \omega_m \end{cases}\\ \end{align} $$ Define total resistance $R = R_\text{armature} + R_\text{source}$\ These are derived from the previous equations: $$ \begin{align} &\text{Analog speed control (Voltage): } T = \frac{k_{TP}}{R}v_s - \frac{k_{TP}k_{aP}}{R} \omega_m\\ &\text{Analog speed control (Current): } T = \frac{k_{TP}R_S}{R}i_s - \frac{k_{TP}k_{aP}}{R} \omega_m \end{align} $$ ``` + v_a | s R_a | 3 L_a | (M) | I_a | V - GND ``` # Chapter 9 ## Axioms and theorems Operations are also commutative, associative. | Name | 1 | 2 | | --------------- | ----------------------------- | ---------------------------- | | Identity | $X+0=X$ | $X\cdot 1 = X$ | | Null | $X+1=1$ | $X\cdot 0=0$ | | Idempotency | $X+X=X$ | $X\cdot X=X$ | | Involution | $(X')'=X$ | | | Complementarity | $X+X'=1$ | $X\cdot X'=0$ | | Uniting | $X\cdot Y + X\cdot Y'=X$ | $(X+Y)(X+Y')=X$ | | Absorption | $X+X\cdot Y=X$ | $X\cdot(X+Y)=X$ | | Absorption | $(X+Y')\cdot Y=X\cdot Y$ | $(X\cdot Y')+Y=X+Y$ | | Distributivity | $X\cdot Y+X\cdot Z = X (Y+Z)$ | $X+(Y\cdot Z)=(X+Y)(X+Z)$ | | Factoring | $(X+Y)\cdot(X'+Z)$ | $X\cdot Y+X'\cdot Z$ | | | $=X\cdot Z+X'\cdot Y$ | $=(X+Z)\cdot(X'+Y)$ | | Consensus | $X\cdot Y+Y\cdot Z+X'\cdot Z$ | $(X+Y)\cdot(Y+Z)\cdot(X'+Z)$ | | | $=X\cdot Y+X'\cdot Z$ | $=(X+Y)\cdot(X'+Z)$ | | DeMorgan's | $(X+Y+\dots)'$ | $(X\cdot Y\cdot \dots)'$ | | | $=X'\cdot Y'\cdot\dots$ | $=X'+Y'+\dots$ | ## Sum of products A min term is the intersection of the inverse of the inputs, or the NOR of the inputs. | $\dotsc$ | $A$ | $B$ | $C$ | minterm | | -------- | ------- | ------- | ------- | ------------------------------------ | | $\dotsc$ | $0$ | $0$ | $0$ | $m_0=\dotsc\cdot A'\cdot B'\cdot C'$ | | $\dotsc$ | $0$ | $0$ | $1$ | $m_1=\dotsc\cdot A'\cdot B'\cdot C$ | | $\dotsc$ | $0$ | $1$ | $0$ | $m_2=\dotsc\cdot A'\cdot B\cdot C'$ | | $\dotsc$ | $\dots$ | $\dots$ | $\dots$ | $\dotsc$ | Sum of products is the sum of the minterms when $F(A,B,C,\dots)$ is $1$ (**TRUE**).\ Example | $A$ | $B$ | $F(A,B)$ | minterm | | --- | --- | -------- | ---------------- | | $0$ | $0$ | $1$ | $m_0=A'\cdot B'$ | | $0$ | $1$ | $0$ | $m_1=A'\cdot B$ | | $1$ | $0$ | $0$ | $m_2=A\cdot B'$ | | $1$ | $1$ | $1$ | $m_3=A\cdot B$ | In the example, $$ \begin{align*} F(A,B) &= \sum\left(m_0,m_3\right)\\ &= A'\cdot B'+A\cdot B \end{align*} $$ ## Product of sums A max term is the union of the inverse of the inputs, or the NAND of the inputs. | $\dotsc$ | $A$ | $B$ | $C$ | maxterm | | -------- | ------- | ------- | ------- | -------------------- | | $\dotsc$ | $0$ | $0$ | $0$ | $M_0=\dotsc+A+B +C$ | | $\dotsc$ | $0$ | $0$ | $1$ | $M_1=\dotsc+A+B +C'$ | | $\dotsc$ | $0$ | $1$ | $0$ | $M_2=\dotsc+A+B'+C$ | | $\dotsc$ | $\dots$ | $\dots$ | $\dots$ | $\dotsc$ | Product of sums is the product of the maxterms when $F(A,B,C,\dots)$ is $0$ (**FALSE**).\ Example | $A$ | $B$ | $F(A,B)$ | maxterm | | --- | --- | -------- | ----------- | | $0$ | $0$ | $1$ | $M_0=A +B$ | | $0$ | $1$ | $0$ | $M_1=A +B'$ | | $1$ | $0$ | $0$ | $M_2=A'+B$ | | $1$ | $1$ | $1$ | $M_3=A'+B'$ | In the example, $$ \begin{align*} F(A,B) &= \prod\left(M_1,M_2\right)\\ &= (A+B')\cdot(A'+B) \end{align*} $$ Product of sums is equal to sum of products $$ \begin{align*} F(A,B) &= (A+B')\cdot(A'+B)\\ &= A\cdot A'+A\cdot B+B'\cdot A'+B'\cdot B &\text{(Distributivity)}\\ &= A\cdot B+B'\cdot A' &\text{(Complementarity)}\\ \end{align*} $$