commit 275e45f20d10563609c5f44b9df014c93cf7cc3f Author: Peter Date: Tue Jun 7 03:18:09 2022 +0800 stuff diff --git a/2022-05-31-15-46-20.png b/2022-05-31-15-46-20.png new file mode 100644 index 0000000..59c4271 Binary files /dev/null and b/2022-05-31-15-46-20.png differ diff --git a/2022-05-31-15-52-51.png b/2022-05-31-15-52-51.png new file mode 100644 index 0000000..35e8723 Binary files /dev/null and b/2022-05-31-15-52-51.png differ diff --git a/2022-05-31-15-53-03.png b/2022-05-31-15-53-03.png new file mode 100644 index 0000000..37b958f Binary files /dev/null and b/2022-05-31-15-53-03.png differ diff --git a/2022-06-01-16-41-13.png b/2022-06-01-16-41-13.png new file mode 100644 index 0000000..56342a2 Binary files /dev/null and b/2022-06-01-16-41-13.png differ diff --git a/FORMULA.html b/FORMULA.html new file mode 100644 index 0000000..51c05ef --- /dev/null +++ b/FORMULA.html @@ -0,0 +1,286 @@ + + + + + + + + + + + + +

Chapter 1

+

i(t)=dqdt⇔q(t)=∫i(t)⋅dtP=v×i=i×Wqv=Wq⇔W=v×q=∫v×i⋅dt +\begin{align} + 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

+

Load line: ix=−vxRT+vtRTGeneral: Av=voutvinInverting: Av=−RfRinNon Inverting: Av=1+RfRin +\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_{in}}\\ +\end{align} +

+ +

Inverting

+
+ +

Non-inverting

+
+ +

Chapter 5

+

Capacitor

+

C=qvi(t)=Cdvdt⇔v(t)=1C∫0ti(t)⋅dtSeries: 1CT=∑i=0N1CiParallel: CT=∑i=0NCiEnergy: E=12Cv2 +\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 vs=v∞v_s=v_\infty:

+

τ=R×CvC(t)={v0t≤0v∞+(v0−v∞)e−t/τt>0v0e−t/τ (Natural response, no input)v∞(1−e−t/τ) (Forced response, input)iC(t)=vs−vC(t)R=−(v0−v∞)e−t/τR +\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. Remove all independent sources, find equivalent resistance and capacitance, find τ\tau.
    2. +
  2. Set C as open circuit, find initial capacitor voltage v0v_0 at t=0t=0
    3. +
  3. Set C as open circuit, find final capacitor voltage v∞v_\infty at t→∞t\to\infty
    4. +
+

Inductor

+

L=λiv(t)=Ldidt⇔i(t)=1L∫0tv⋅dtSeries: LT=∑i=0NLiParallel: 1LT=∑i=0N1LiEnergy: E=12Li2 +\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 vs/R=i∞v_s/R=i_\infty:

+

τ=LRi(t)={i0t≤0i∞+(i0−i∞)e−t/τt>0i0e−t/τ (Natural response, no input)i∞(1−e−t/τ) (Forced response, input) +\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} +

+ +

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 qq.

+

Voltage drop over capacitor i: vDi=vTCTCiVoltage divider: vi=vTCT1Ci+1Ci+1+⋯+1CN\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

+

Condition: ZL‾=ZS∗‾Maximum power to load (50%): 2Pavg=Prms⋅2=Pmax=∣VS∣24RS=∣VL∣24RLTotal maximum power: 2Pavg=Prms⋅2=Pmax=∣VS∣22RS +\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} +

+ +

+

Complex Power

+

Where Vˉ=V∠θ\bar{V}=V\angle\theta and Iˉ=I∠ϕ\bar{I}=I\angle\phi:

+

Complex [VA]: Sˉ=Vˉrms×Iˉrms∗=Vˉ×Iˉ∗2=VI2∠(θ−ϕ)Apparent [VA]: ∣S∣Real [W]: P=∣S∣cos⁡(θ−ϕ)=Re(Sˉ)Reactive [VAR]: Q=∣S∣cos⁡(θ−ϕ)=Im(Sˉ)Q=Ptan⁡(arccos⁡(Power factor))Power Factor=P∣S∣=cos⁡(arctan⁡(QP)) +\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]: } |S|\\ + \text{Real [W]: } P &= |S| \cos(\theta-\phi) = \text{Re}(\bar{S})\\ + \text{Reactive [VAR]: } Q &= |S| \cos(\theta-\phi) = \text{Im}(\bar{S})\\ + Q &= P\tan(\arccos(\text{Power factor}))\\ + \text{Power Factor} &= \frac{P}{|S|} = \cos\left(\arctan\left(\frac{Q}{P}\right)\right) +\end{align} +

+ +

Where Sˉ=∣S∣∠φ\bar{S}=|S|\angle\varphi

+

φ=arctan⁡(QP) \varphi = \arctan\left(\frac{Q}{P}\right)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
LaggingLeading
VoltageCurrent behindCurrent ahead
Load typeInductiveCapacitive
QQQ>0Q>0Q<0Q<0
φ\varphiφ>0\varphi>0φ<0\varphi<0
+

Chapter 8

+

Constants

+

μ0=4π×10−7 +\mu_0=4\pi \times 10^{-7} +

+ +

Faraday’s law: ε=−NdΦdtAmpere’s law: B=μ0I2πr +\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>1n>1
+Step down: n<1n<1

+

VsVp=NsNp=ipis=nZˉin=1n2ZˉL +\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} +

+ +

Motor

+

For permanent motors, define permanent torque constant kTP=kTΦk_{TP}=k_T\varPhi

+

T=kT×Φ×ia=kTP×iaPmech=ωmechT=ωmech×kTP×ia +\begin{align} + T &= k_T\times\varPhi \times i_a = k_{TP} \times i_a\\ + P_\text{mech} &= \omega_\text{mech} T = \omega_\text{mech} \times k_{TP}\times i_a +\end{align} +

+ +

Back emf

+

Define for permanent motors, define permanent armature constant kaP=kaΦk_{aP} = k_a\varPhi
+Note, back emf should oppose vav_a and iai_a

+

eb=ka×Φ×ωmech=kaP×ωmech +\begin{align} + e_b=k_a\times \varPhi\times \omega_\text{mech} = k_{aP} \times\omega_\text{mech}\\ +\end{align} +

+ +

Summary

+

For ideal motor, torque and armature constants are the same: ka=kTk_a=k_T
+Define pp as number of magnetic poles and MM as the number of parallel paths in armature winding.

+

Power dissipated: Pe=eb×ia=kaP×ωmech×iaConstants for ideal motor: ka=kT=pN2πM +\begin{align} + \text{Power dissipated: } P_e &= e_b\times i_a = k_{aP} \times \omega_\text{mech} \times i_a\\ + \text {Constants for ideal motor: } k_a &= k_T = \frac{pN}{2\pi M} + +\end{align} +

+ +

For permanent magnet DC motor in DC steady state:
+Define viscous frictional damping coefficient bb and load torque TLT_L

+

{0=va−iaRa−kaPωmech=va−iaRa−ebkTPia=TL+b×ωmechAnalog speed control (Voltage): T=kTPRvs−kTPkaPRωmechAnalog speed control (Current): T=kTPRSRis−kTPkaPRωmech +\begin{align} + &\begin{cases} + + 0 &= v_a - i_a R_a - k_{aP} \omega_\text{mech} = v_a - i_a R_a - e_b \\ + k_{TP} i_a &= T_L + b\times \omega_\text{mech} + + \end{cases}\\ + + &\text{Analog speed control (Voltage): } T = \frac{k_{TP}}{R}v_s - \frac{k_{TP}k_{aP}}{R} \omega_\text{mech}\\ + + &\text{Analog speed control (Current): } T = \frac{k_{TP}R_S}{R}i_s - \frac{k_{TP}k_{aP}}{R} \omega_\text{mech} +\end{align} +

+ + + + \ No newline at end of file diff --git a/FORMULA.md b/FORMULA.md new file mode 100644 index 0000000..22ca625 --- /dev/null +++ b/FORMULA.md @@ -0,0 +1,416 @@ +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 + +$$ +\begin{align} + \text{Load line: } i_x &= -\frac{v_x}{R_T}+ \frac{v_t}{R_T}\\ + \text{General: } A_v &= \frac{v_{out}}{v_{in}}\\ + \text{Inverting: } A_v &= -\frac{R_f}{R_{in}}\\ + \text{Non Inverting: } A_v &= 1+\frac{R_f}{R_{in}}\\ + \text{Series of op amps total: } A_v &= (A_v)_1\times(A_v)_2\times \dots\times (A_v)_n +\end{align} +$$ + +## Inverting + +![](2022-05-31-15-52-51.png) + +## Non-inverting + +![](2022-05-31-15-53-03.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)$$ + +| | Lagging | Leading | Unity | +| ----------- | -------------- | ------------- | ------------ | +| Voltage | Current behind | Current ahead | In phase | +| Load type | Inductive | Capacitive | Resistive | +| $Q$ | $Q>0$ | $Q<0$ | $Q=0$ | +| $\varphi$ | $\varphi>0°$ | $\varphi<0°$ | $\varphi=0°$ | +| PF [Load] | $[0,1)$ | $[0,1)$ | $1$ | +| PF [Soruce] | $[0,-1)$ | $[0,-1)$ | $-1$ | + +# Chapter 8 + +$$ +\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*} +$$ diff --git a/FORMULA.pdf b/FORMULA.pdf new file mode 100644 index 0000000..22fd119 Binary files /dev/null and b/FORMULA.pdf differ