something something i'm not responsible for lost marks. have fun.
| Name |
Symbol |
Value |
| Speed of light |
$c$ |
$3\times 10^8$ |
| Elementary charge |
$q_e$ |
$1.6022\times 10^{-19}$ |
| Magnetic constant |
$\mu_0$ |
$4\pi \times 10^{-7}$ |
$$
\begin{align}
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
\sin(\theta) &= \cos(\theta-90°)\\
\cos(\theta) &= \sin(\theta+90°)\\
\text{Euler's formula: } e^{j\theta} &= \cos(\theta)+j\sin(\theta)
\end{align}
$$
$$
\begin{align}
\text{Lumped model: } \lambda &= \frac{c}{f} \gggtr \text{dimension} \text{ (At least 10 times)}\\
i(t)&=\frac{dq}{dt} \Leftrightarrow q(t)=\int i(t)\cdot dt\\
P&=v\times i=i\times \frac{W}{q}\\
v&=\frac{W}{q} \Leftrightarrow W=v\times q = \int v\times i \cdot dt\\
\end{align}
$$
$$
\begin{align}
\text{Load line: } i_x &= -\frac{v_x}{R_T}+ \frac{v_t}{R_T}\\
\text{General: } A_v &= \frac{v_{out}}{v_{in}}\\
\text{Inverting: } A_v &= -\frac{R_f}{R_{in}}\\
\text{Non Inverting: } A_v &= 1+\frac{R_f}{R_{1}}\\
\text{Series of op amps total: } A_v &= (A_v)_1\times(A_v)_2\times \dots\times (A_v)_n
\end{align}
$$
| Inverting |
Non-inverting |
 |
 |
$$
\begin{align}
C &= \frac{q}{v}\\
i(t) = C\frac{dv}{dt} &\Leftrightarrow v(t) = \frac{1}{C}\int_0^t i(t)\cdot dt\\
\text{Series: }\frac{1}{C_T} &= \sum^N_{i=0}\frac{1}{C_i}\\
\text{Parallel: }C_T &= \sum^N_{i=0}C_i\\
\text{Energy: }E &= \frac{1}{2}Cv^2
\end{align}
$$
Differential equation solution
Where $v_s=v_\infty$:
$$
\begin{align}
\tau &= R\times C\\
v_C(t) &=
\begin{cases}
\begin{array}{lr}
v_0 & t\leq 0\\
v_\infty+(v_0-v_\infty)e^{-t/\tau} & t > 0
\end{array}
\end{cases}\\
& v_0 e^{-t/\tau} \text{ (Natural response, no input)}\\
& v_\infty\left(1-e^{-t/\tau}\right) \text{ (Forced response, input)}\\
i_C(t) &= \frac{v_s-v_C(t)}{R}=\frac{-(v_0-v_\infty)e^{-t/\tau}}{R}
\end{align}
$$
- When $t>0$, remove all independent sources, find equivalent resistance and capacitance, find $\tau$.
- Set C as open circuit, find initial capacitor voltage $v_0$ at $t=0$
- Set C as open circuit, find final capacitor voltage $v_\infty$ at $t\to\infty$
$$
\begin{align}
L &= \frac{\lambda}{i}\\
v(t)=L\frac{di}{dt} &\Leftrightarrow i(t)=\frac{1}{L}\int_0^tv\cdot dt\\
\text{Series: }L_T &= \sum^N_{i=0}L_i\\
\text{Parallel: }\frac{1}{L_T} &= \sum^N_{i=0}\frac{1}{L_i}\\
\text{Energy: }E &= \frac{1}{2}Li^2
\end{align}
$$
Differential equation solution
Where $v_s/R=i_\infty$:
$$
\begin{align}
\tau &= \frac{L}{R}\\
i(t) &=
\begin{cases}
\begin{array}{lr}
i_0 & t\leq 0\\
i_\infty+(i_0-i_\infty)e^{-t/\tau} & t > 0
\end{array}
\end{cases}\\
& i_0 e^{-t/\tau} \text{ (Natural response, no input)}\\
& i_\infty\left(1-e^{-t/\tau}\right) \text{ (Forced response, input)}\\
\end{align}
$$
- When $t>0$, remove all independent sources, find equivalent resistance and inductance, find $\tau$.
- Set L as short circuit, find initial inductor current $i_0$ at $t=0$
- 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 *
Current through capacitors in series is the same, so all capacitors have same charge stored $q$.
$$
\begin{align}
\text{Voltage drop over capacitor $i$: } v_{Di} &= v_T\frac{C_T}{C_i}\\
\text{Voltage divider: } v_i &= v_T\frac{C_T}{\frac{1}{C_i}+\frac{1}{C_{i+1}}+\dots+\frac{1}{C_N}}\\
\end{align}
$$
No voltage drop in steady state (Inductor is a short circuit)
Maximum power transfer in AC
$$
\begin{align}
\text{Condition: } &\overline{Z_L} = \overline{Z_S^*}\\
\text{Maximum power to load (50\%): } &2P_\text{avg}=P_\text{rms}\cdot\sqrt{2} = P_\text{max} = \frac{{|V_S|}^2}{4R_S}\\
& = \frac{{|V_L|}^2}{4R_L}\\
\text{Total maximum power: } &2P_\text{avg}=P_\text{rms}\cdot\sqrt{2} = P_\text{max} = \frac{{|V_S|}^2}{2R_S}
\end{align}
$$
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 $\bar{Z}_\text{load} = Z_\text{load}\angle\lambda = R+jX$:
$$
\let\lb=( \let\rb=) \def({\left\lb} \def){\right\rb} % Put \left(,\right) on (,)
\begin{align}
\text{PF from impedance: } \text{PF} &= \frac{\text{Re}(\bar{Z}\text{load})}{|\bar{Z}\text{load}|} = \frac{R}{\sqrt{R^2+X^2}} \
\text{PF from angles: } \text{PF} &= \cos(\lambda) = \cos(\arctan(\frac{X}{R}))
\end{align}
$$
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] |
$[0,-1)$ |
$[0,-1)$ |
$-1$ |
$$
\begin{align}
\text{Faraday's law: }\varepsilon &= -N\frac{d\varPhi}{dt}\\
\text{Ampere's law: }B &= \frac{\mu_0 I}{2\pi r}\\
\end{align}
$$
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:
$$
\begin{align*}
i_s &= i_p/n \\
&= \frac{V_p}{\frac{1}{n^2}\times \bar{Z}_{L}\times n} \\
\end{align*}\\
$$
$$
\begin{align}
i_s &=\frac{V_p\times n}{\bar{Z}_{L}}\\
\end{align}
$$
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
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 |
$(X+Y+\dots)'$ |
$(X\cdot Y\cdot \dots)'$ |
|
$=X'\cdot Y'\cdot\dots$ |
$=X'+Y'+\dots$ |
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,
$$
\begin{align*}
F(A,B) &= \sum\left(m_0,m_3\right)\\
&= A'\cdot B'+A\cdot B
\end{align*}
$$
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
$$
\begin{align*}
F(A,B) &= (A+B')\cdot(A'+B)\\
&= A\cdot A'+A\cdot B+B'\cdot A'+B'\cdot B &\text{(Distributivity)}\\
&= A\cdot B+B'\cdot A' &\text{(Complementarity)}\\
\end{align*}
$$
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