make pdf link more obvious

README.md

@@ -1,5 +1,9 @@
 # Idiot's guide to ELEC4402 communication systems
 
+## [Download PDF 📄](/README.pdf)
+
+It is recommended to refer to use [the PDF copy](/README.pdf) instead of whatever GitHub renders.
+
 ## License and information
 
 Notes are open-source and licensed under the GNU GPL-3.0. **You must include the [full-text of the license](/COPYING.txt) and follow its terms when using these notes or any diagrams in derivative works** (but not when printing as notes)
@@ -25,26 +29,24 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
 
 </details>
 
-[Access a web-copy of the notes for printing](/README.html)
-
 I accept pull requests or suggestions but the content must not be copyrighted under a non-GPL compatible license.
 
 ## Fourier transform identities
 
 | **Time Function**                                                     | **Fourier Transform**                                                                                                                     |
-| --------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
-| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$                                           | $T \, \text{sinc}(fT)$                                                                                                                    |
+| --------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
+| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \text{sinc}(fT)$                                                                                                                       |
 | $\text{sinc}(2Wt)$                                                    | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$                                           |
-| $\exp(-at)u(t), \, a>0$                                                                                         | $\frac{1}{a + j2\pi f}$                                                                                                                   |
-| $\exp(-a\lvert t \rvert), \, a>0$                                                                               | $\frac{2a}{a^2 + (2\pi f)^2}$                                                                                                             |
+| $\exp(-at)u(t),\quad a>0$                                             | $\frac{1}{a + j2\pi f}$                                                                                                                   |
+| $\exp(-a\lvert t \rvert),\quad a>0$                                   | $\frac{2a}{a^2 + (2\pi f)^2}$                                                                                                             |
 | $\exp(-\pi t^2)$                                                      | $\exp(-\pi f^2)$                                                                                                                          |
-| $\begin{cases} 1 - \frac{\lvert t \rvert}{T}, & \lvert t \rvert < T \\ 0, & \lvert t \rvert \geq T \end{cases}$ | $T \, \text{sinc}^2(fT)$                                                                                                                  |
+| $1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T$              | $T \text{sinc}^2(fT)$                                                                                                                     |
 | $\delta(t)$                                                           | $1$                                                                                                                                       |
 | $1$                                                                   | $\delta(f)$                                                                                                                               |
 | $\delta(t - t_0)$                                                     | $\exp(-j2\pi f t_0)$                                                                                                                      |
 | $g(t-a)$                                                              | $\exp(-j2\pi fa)G(f)\quad\text{shift property}$                                                                                           |
 | $g(bt)$                                                               | $\frac{G(f/b)}{\|b\|}\quad\text{scaling property}$                                                                                        |
-| $g(bt-a)$                                                                                                       | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)\quad\text{shift \& scale}$                                                                |
+| $g(bt-a)$                                                             | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)\quad\text{shift and scale}$                                                               |
 | $\frac{d}{dt}g(t)$                                                    | $j2\pi fG(f)\quad\text{differentiation property}$                                                                                         |
 | $G(t)$                                                                | $g(-f)\quad\text{duality property}$                                                                                                       |
 | $g(t)h(t)$                                                            | $G(f)*H(f)$                                                                                                                               |
@@ -53,17 +55,19 @@ I accept pull requests or suggestions but the content must not be copyrighted un
 | $\cos(2\pi f_c t)$                                                    | $\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$                                                                                          |
 | $\sin(2\pi f_c t)$                                                    | $\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$                                                                                        |
 | $\text{sgn}(t)$                                                       | $\frac{1}{j\pi f}$                                                                                                                        |
-| $\frac{1}{\pi t}$                                                                                               | $-j \, \text{sgn}(f)$                                                                                                                     |
+| $\frac{1}{\pi t}$                                                     | $-j \text{sgn}(f)$                                                                                                                        |
 | $u(t)$                                                                | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$                                                                                               |
 | $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$                          | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ |
 
-| **Function Name**    | **Formula**                                                                                             |
-| -------------------- | ------------------------------------------------------------------------------------------------------- |
-| Unit Step Function   | $u(t) = \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}$                     |
-| Signum Function      | $\text{sgn}(t) = \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}$                    |
-| sinc Function        | $\text{sinc}(2Wt) = \frac{\sin(2\pi W t)}{2\pi W t}$                                                    |
-| Rectangular Function | $\text{rect}(t) = \Pi(t) = \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}$ |
-| Convolution          | $g(t)*h(t)=(g*h)(t)=\int_\infty^\infty g(\tau)h(t-\tau)d\tau$                                           |
+```math
+\begin{align*}
+    u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
+    \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
+    \text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\
+    \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
+    g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
+\end{align*}
+```
 
 ### Fourier transform of continuous time periodic signal
 
@@ -73,21 +77,21 @@ Required for some questions on **sampling**:
 
 Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
 
-$$
+```math
 X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
-$$
+```
 
 Calculate $C_n$ coefficient as follows from $x_p(t)$:
 
 <!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
 
-$$
+```math
 \begin{align*}
     % C_n&=X_p(nf_s)\\
     C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
        &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
 \end{align*}
-$$
+```
 
 ### $\text{rect}$ function
 
@@ -95,16 +99,16 @@ $$
 
 ### Bessel function
 
-$$
+```math
 \begin{align*}
-    \sum_{n\in\Z}{J_n}^2(\beta)&=1\\
+    \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
     J_n(\beta)&=(-1)^nJ_{-n}(\beta)
 \end{align*}
-$$
+```
 
 ### White noise
 
-$$
+```math
 \begin{align*}
 R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
 G_w(f)&=\frac{N_0}{2}\\
@@ -112,27 +116,27 @@ N_0&=kT\\
 G_y(f)&=|H(f)|^2G_w(f)\\
 G_y(f)&=G(f)G_w(f)\\
 \end{align*}
-$$
+```
 
 ### WSS
 
-$$
+```math
 \begin{align*}
     \mu_X(t) &= \mu_X\text{ Constant}\\
     R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
     E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
 \end{align*}
-$$
+```
 
 ### Ergodicity
 
-$$
+```math
 \begin{align*}
     \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
     \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\
     E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\
 \end{align*}
-$$
+```
 
 | Type                                | Normal                                                  | Mean square sense                                           |
 | ----------------------------------- | ------------------------------------------------------- | ----------------------------------------------------------- |
@@ -143,17 +147,17 @@ $$
 
 ### Other identities
 
-$$
+```math
 \begin{align*}
     f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\
     a(f*g) &= (af)*g \quad\text{Convolution associative}\\
     \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right)
 \end{align*}
-$$
+```
 
 ### Other trig
 
-$$
+```math
 \begin{align*}
     \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\
     e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\
@@ -164,11 +168,11 @@ $$
     \sin(A-\pi/2)&=-\cos(A)\\
     \cos(A-\pi/2)&=\sin(A)\\
     \cos(A+\pi/2)&=-\sin(A)\\
-    \int_{x\in\R}\text{sinc}(A x) &= \frac{1}{|A|}\\
+    \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\
 \end{align*}
-$$
+```
 
-$$
+```math
 \begin{align*}
     \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\
     \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\
@@ -176,9 +180,9 @@ $$
     \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\
     \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\
 \end{align*}
-$$
+```
 
-$$
+```math
 \begin{align*}
     \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
     \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
@@ -187,7 +191,7 @@ $$
     \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
     \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
 \end{align*}
-$$
+```
 
 ## IQ/Complex envelope
 
@@ -195,7 +199,7 @@ Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $
 
 ### Convert complex envelope representation to time-domain representation of signal
 
-$$
+```math
 \begin{align*}
 g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\
 &=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\
@@ -205,23 +209,23 @@ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\
 g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\
 g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\
 \end{align*}
-$$
+```
 
 ### For transfer function
 
-$$
+```math
 \begin{align*}
 h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
 &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\
 \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))}
 \end{align*}
-$$
+```
 
 ## AM
 
 ### CAM
 
-$$
+```math
 \begin{align*}
     m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
     m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
@@ -233,7 +237,7 @@ $$
     \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
     B_T&=2f_m=2B
 \end{align*}
-$$
+```
 
 $B_T$: Signal bandwidth
 $B$: Bandwidth of modulating wave
@@ -242,16 +246,16 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
 
 ### DSB-SC
 
-$$
+```math
 \begin{align*}
     x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
     B_T&=2f_m=2B
 \end{align*}
-$$
+```
 
 ## FM/PM
 
-$$
+```math
 \begin{align*}
     s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
     s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\
@@ -260,34 +264,30 @@ $$
     \Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
     D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}
 \end{align*}
-$$
+```
 
 ### Bessel form and magnitude spectrum (single tone)
 
-$$
+```math
 \begin{align*}
     s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
 \end{align*}
-$$
-
-<!-- ### Bessel repr.
-
-$$s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m) t)$$ -->
+```
 
 ### FM signal power
 
-$$
+```math
 \begin{align*}
     P_\text{av}&=\frac{{A_c}^2}{2}\\
     P_\text{band\_index}&=\frac{{A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
     \text{band\_index}&=0\implies f_c+0f_m\\
     \text{band\_index}&=1\implies f_c+1f_m,\dots\\
 \end{align*}
-$$
+```
 
 ### Carson's rule to find $B$ (98% power bandwidth rule)
 
-$$
+```math
 \begin{align*}
 B &= 2Mf_m = 2(\beta + 1)f_m\\
     &= 2(\Delta f+f_m)\\
@@ -298,7 +298,7 @@ B &= \begin{cases}
     2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
 \end{cases}\\
 \end{align*}
-$$
+```
 
 #### $\Delta f$ of arbitrary modulating signal
 
@@ -306,7 +306,7 @@ Find instantaneous frequency $f_\text{FM}$.
 
 $M$: Number of **pairs** of significant sidebands
 
-$$
+```math
 \begin{align*}
 s(t)&=A_c\cos(\theta_\text{FM}(t))\\
 f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\
@@ -317,17 +317,17 @@ W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\
 D &= \frac{\Delta f}{W_m}\\
 B_T &= 2(D+1)W_m
 \end{align*}
-$$
+```
 
 ### Complex envelope
 
-$$
+```math
 \begin{align*}
     s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
     s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
     \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
 \end{align*}
-$$
+```
 
 ### Band
 
@@ -337,7 +337,7 @@ $$
 
 ## Power, energy and autocorrelation
 
-$$
+```math
 \begin{align*}
     G_\text{WGN}(f)&=\frac{N_0}{2}\\
     G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
@@ -350,24 +350,24 @@ $$
     E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\
     R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
 \end{align*}
-$$
+```
 
 ##
 
 ## Noise performance
 
-$$
+```math
 \begin{align*}
     \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\
     \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\
     \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\
     \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio}
 \end{align*}
-$$
+```
 
 ## Sampling
 
-$$
+```math
 \begin{align*}
     t&=nT_s\\
     T_s&=\frac{1}{f_s}\\
@@ -375,7 +375,7 @@ $$
     X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
     B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\
 \end{align*}
-$$
+```
 
 ### Procedure to reconstruct sampled signal
 
@@ -397,21 +397,21 @@ Required for some questions on **sampling**:
 
 Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
 
-$$
+```math
 X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
-$$
+```
 
 Calculate $C_n$ coefficient as follows from $x_p(t)$:
 
 <!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
 
-$$
+```math
 \begin{align*}
     % C_n&=X_p(nf_s)\\
     C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
        &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
 \end{align*}
-$$
+```
 
 <!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ -->
 
@@ -428,15 +428,15 @@ Do not transmit more than $2B$ samples per second over a channel of $B$ bandwidt
 
 ## Quantizer
 
-$$
+```math
 \begin{align*}
     \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\
 \end{align*}
-$$
+```
 
 ### Quantization noise
 
-$$
+```math
 \begin{align*}
     e &:= y-x\quad\text{Quantization error}\\
     \mu_E &= E[E] = 0\quad\text{Zero mean}\\
@@ -444,7 +444,7 @@ $$
     \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\
     \text{SQNR(dB)}&=10\log_{10}(\text{SQNR})
 \end{align*}
-$$
+```
 
 ### Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to `images/quantizer.png`)
 
@@ -455,7 +455,7 @@ $$
 
 ![binary_codes](images/Line_Codes.drawio.svg)
 
-$$
+```math
 \begin{align*}
     R_b&\rightarrow\text{Bit rate}\\
     D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\
@@ -469,7 +469,7 @@ $$
     G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\
     G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b
 \end{align*}
-$$
+```
 
 ## Modulation and basis functions
 
@@ -479,25 +479,27 @@ $$
 
 #### Basis functions
 
-$$
+```math
 \begin{align*}
     \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
 \end{align*}
-$$
+```
 
 #### Symbol mapping
 
-$$b_n:\{1,0\}\to a_n:\{1,0\}$$
+```math
+b_n:\{1,0\}\to a_n:\{1,0\}
+```
 
 #### 2 possible waveforms
 
-$$
+```math
 \begin{align*}
     s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\
     s_1(t)&=0\\
     &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{{A_c}^2}{2}\times T_b + 0)=\frac{{A_c}^2}{4}T_b$}
 \end{align*}
-$$
+```
 
 Distance is $d=\sqrt{2E_b}$
 
@@ -505,25 +507,27 @@ Distance is $d=\sqrt{2E_b}$
 
 #### Basis functions
 
-$$
+```math
 \begin{align*}
     \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
 \end{align*}
-$$
+```
 
 #### Symbol mapping
 
-$$b_n:\{1,0\}\to a_n:\{1,\color{lime}-1\color{white}\}$$
+```math
+b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}
+```
 
 #### 2 possible waveforms
 
-$$
+```math
 \begin{align*}
     s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\
     s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\
     &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{{A_c}^2}{2}\times T_b + \frac{{A_c}^2}{2}\times T_b)=\frac{{A_c}^2}{2}T_b$}
 \end{align*}
-$$
+```
 
 Distance is $d=2\sqrt{E_b}$
 
@@ -531,45 +535,45 @@ Distance is $d=2\sqrt{E_b}$
 
 #### Basis functions
 
-$$
+```math
 \begin{align*}
     T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\
     \varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\
     \varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\
 \end{align*}
-$$
+```
 
 ### 4 possible waveforms
 
-$$
+```math
 \begin{align*}
     s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\
     s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\
     s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\
     s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\
 \end{align*}
-$$
+```
 
 Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows:
 
-$$
+```math
 \begin{align*}
     s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
           &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
           &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
 \end{align*}
-$$
+```
 
 #### Signal
 
-$$
+```math
 \begin{align*}
     \text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\
     I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\
     Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\
     x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)]
 \end{align*}
-$$
+```
 
 ### Example of waveform
 
@@ -615,20 +619,20 @@ Remember that $T=2T_b$
 
 Find transfer function $h(t)$ of matched filter and apply to an input:
 
-$$
+```math
 \begin{align*}
     h(t)&=s_1(T-t)-s_2(T-t)\\
     h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\
     s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\
     n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output}
 \end{align*}
-$$
+```
 
 ### 2. Bit error rate
 
 Bit error rate (BER) from matched filter outputs and filter output noise
 
-$$
+```math
 \begin{align*}
     % H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right)
 
@@ -643,7 +647,7 @@ $$
     \text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\
     \text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\
 \end{align*}
-$$
+```
 
 <div style="page-break-after: always;"></div>
 
@@ -673,7 +677,7 @@ $$
 
 | $x$    | $Q(x)$     | $x$    | $Q(x)$                  | $x$    | $Q(x)$                   | $x$    | $Q(x)$                   |
 | ------ | ---------- | ------ | ----------------------- | ------ | ------------------------ | ------ | ------------------------ |
-| $0.00$ | $0.5       | $2.30$ | $0.010724$              | $4.55$ | $2.6823 \times 10^{-6}$  | $6.80$ | $5.231 \times 10^{-12}$  |
+| $0.00$ | $0.5$      | $2.30$ | $0.010724$              | $4.55$ | $2.6823 \times 10^{-6}$  | $6.80$ | $5.231 \times 10^{-12}$  |
 | $0.05$ | $0.48006$  | $2.35$ | $0.0093867$             | $4.60$ | $2.1125 \times 10^{-6}$  | $6.85$ | $3.6925 \times 10^{-12}$ |
 | $0.10$ | $0.46017$  | $2.40$ | $0.0081975$             | $4.65$ | $1.6597 \times 10^{-6}$  | $6.90$ | $2.6001 \times 10^{-12}$ |
 | $0.15$ | $0.44038$  | $2.45$ | $0.0071428$             | $4.70$ | $1.3008 \times 10^{-6}$  | $6.95$ | $1.8264 \times 10^{-12}$ |
@@ -726,7 +730,7 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
 
 ### Receiver output shit
 
-$$
+```math
 \begin{align*}
     r_o(t)&=\begin{cases}
         s_{o1}(t)+n_o(t) & \text{code 1}\\
@@ -734,14 +738,14 @@ $$
     \end{cases}\\
     n&: \text{AWGN with }\sigma_o^2\\
 \end{align*}
-$$
+```
 
 <!--
-$$
+```math
 \begin{align*}
     G_x(f)
 \end{align*}
-$$ -->
+``` -->
 
 ## ISI, channel model
 
@@ -751,20 +755,22 @@ TODO:
 
 ### Nomenclature
 
-$$
+```math
 \begin{align*}
     D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\
     T&\rightarrow\text{Symbol Duration}\\
     M&\rightarrow\text{Symbol set size}\\
     W&\rightarrow\text{Bandwidth}\\
 \end{align*}
-$$
+```
 
 ### Raised cosine (RC) pulse
 
 ![Raised cosine pulse](images/RC.drawio.svg)
 
-$$0\leq\alpha\leq1$$
+```math
+0\leq\alpha\leq1
+```
 
 ⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method.
 
@@ -773,50 +779,43 @@ To solve this type of question:
 1. Use the formula for $D$ below
 2. Consult the BER table below to get the BER which relates the noise of the channel $N_0$ to $E_b$ and to $R_b$.
 
-<!-- $$
-V_\text{RC}(f)=\begin{cases}
-    T & 0\le|f|\le(1-\alpha)/2T\\
-    \frac{T}{2}\left(1+\cos\left[\right]\right)
-\end{cases}
-$$ -->
-
 | Linear modulation ($M$-PSK, $M$-QAM)                | NRZ unipolar encoding                              |
 | --------------------------------------------------- | -------------------------------------------------- |
-| $W=B_\text{\color{lime}abs-abs}$                    | $W=B_\text{\color{lime}abs}$                       |
+| $W=B_\text{\color{green}abs-abs}$                   | $W=B_\text{\color{green}abs}$                      |
 | $W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ | $W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ |
 | $D=\frac{W\text{ symbol/s}}{1+\alpha}$              | $D=\frac{2W\text{ symbol/s}}{1+\alpha}$            |
 
 #### Symbol set size $M$
 
-$$
+```math
 \begin{align*}
     D\text{ symbol/s}&=\frac{2W\text{ Hz}}{1+\alpha}\\
     R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\
     M\text{ symbol/set}&=2^k\\
     E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
 \end{align*}
-$$
+```
 
 ### Nyquist stuff
 
-#### Condition for 0 ISI
+#### Condition for 0 ISI TODO:
 
-$$
+```math
 P_r(kT)=\begin{cases}
     1 & k=0\\
     0 & k\neq0
 \end{cases}
-$$
+```
 
 #### Other
 
-$$
+```math
 \begin{align*}
     \text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\
     \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
     T&=1/D
 \end{align*}
-$$
+```
 
 <div style="page-break-after: always;"></div>
 
@@ -854,7 +853,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems
 
 ### Entropy for discrete random variables
 
-$$
+```math
 \begin{align*}
     H(x) &\geq 0\\
     H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\
@@ -866,7 +865,7 @@ $$
     H(x|y) &= H(x,y)-H(y)\\
     H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\
 \end{align*}
-$$
+```
 
 Entropy is **maximized** when all have an equal probability.
 
@@ -874,64 +873,63 @@ Entropy is **maximized** when all have an equal probability.
 
 TODO: Cut out if not required
 
-$$
+```math
 \begin{align*}
     h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx
 \end{align*}
-$$
+```
 
 ### Mutual information
 
 Amount of entropy decrease of $x$ after observation by $y$.
 
-$$
+```math
 \begin{align*}
     I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\
 \end{align*}
-$$
+```
 
 ### Channel model
 
 Vertical, $x$: input\
 Horizontal, $y$: output
 
-$$
+```math
 \mathbf{P}=\left[\begin{matrix}
     p_{11} & p_{12} &\dots & p_{1N}\\
     p_{21} & p_{22} &\dots & p_{2N}\\
     \vdots & \vdots &\ddots & \vdots\\
     p_{M1} & p_{M2} &\dots & p_{MN}\\
 \end{matrix}\right]
-$$
+```
 
-$$
+```math
 \begin{array}{c|cccc}
-    P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\
-    \hline
+    P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline
     x_1 & p_{11} & p_{12} & \dots & p_{1N} \\
     x_2 & p_{21} & p_{22} & \dots & p_{2N} \\
     \vdots & \vdots & \vdots & \ddots & \vdots \\
     x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\
 \end{array}
-$$
+```
 
 Input has probability distribution $p_X(a_i)=P(X=a_i)$
 
-Channel maps alphabet $\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}$
+Channel maps alphabet $`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$
 
 Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
 
-$$
+```math
 \begin{align*}
     p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\
         &= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\
        [\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}
 \end{align*}
-$$
+```
 
 #### Fast procedure to calculate $I(y;x)$
 
-$$
+```math
 \begin{align*}
     &\text{1. Find }H(x)\\
     &\text{2. Find }[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] = [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}\\
@@ -940,7 +938,7 @@ $$
     &\text{5. Find }H(x|y)=H(x,y)-H(y)\\
     &\text{6. Find }I(y;x)=H(x)-H(x|y)\\
 \end{align*}
-$$
+```
 
 ### Channel types
 
@@ -951,7 +949,7 @@ $$
 
 #### Channel capacity of weakly symmetric channel
 
-$$
+```math
 \begin{align*}
     C &\to\text{Channel capacity (bits/channels used)}\\
     N &\to\text{Output alphabet size}\\
@@ -959,38 +957,38 @@ $$
     C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\
     R &< C \text{ for error-free transmission}
 \end{align*}
-$$
+```
 
 #### Channel capacity of an AWGN channel
 
-$$
+```math
 y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
-$$
+```
 
-$$
+```math
 C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)
-$$
+```
 
 #### Channel capacity of a bandwidth AWGN channel
 
 Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition.
 
-$$
+```math
 \begin{align*}
     P_s&\to\text{Bandwidth limited average power}\\
     y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\
     C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\
     C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W)
 \end{align*}
-$$
+```
 
 ## Channel code
 
 |                  |                                   |                                                                                                                              |
-| ---------------- | --------------------------------- | -------------------------------------------------------------------------------------------------------------------------- |
+| ---------------- | --------------------------------- | ---------------------------------------------------------------------------------------------------------------------------- |
 | Hamming weight   | $w_H(x)$                          | Number of `'1'` in codeword $x$                                                                                              |
 | Hamming distance | $d_H(x_1,x_2)=w_H(x_1\oplus x_2)$ | Number of different bits between codewords $x_1$ and $x_2$ which is the hamming weight of the XOR of the two codes.          |
-| Minimum distance | $d_\text{min}$                    | **IMPORTANT**: $x\neq\bold{0}$, excludes weight of all-zero codeword. For a linear block code, $d_\text{min}=w_\text{min}$ |
+| Minimum distance | $d_\text{min}$                    | **IMPORTANT**: $x\neq\textbf{0}$, excludes weight of all-zero codeword. For a linear block code, $d_\text{min}=w_\text{min}$ |
 
 ### Linear block code
 
@@ -1011,7 +1009,7 @@ For a linear block code, $d_\text{min}=w_\text{min}$
 
 Each generator vector is a binary string of size $n$. There are $k$ generator vectors in $\mathbf{G}$.
 
-$$
+```math
 \begin{align*}
 \mathbf{g}_i&=[\begin{matrix}
     g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1}
@@ -1029,11 +1027,11 @@ $$
     g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\
 \end{matrix}\right]
 \end{align*}
-$$
+```
 
 A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$:
 
-$$
+```math
 \begin{align*}
 \mathbf{m}&=[\begin{matrix}
     m_{0}& \dots & m_{n-2} & m_{k-1}
@@ -1041,13 +1039,13 @@ $$
 \color{darkgray}\mathbf{m}&\color{darkgray}=[101001]\quad\text{Example for $k=6$}\\
 \mathbf{x} &= \mathbf{m}\mathbf{G}=m_0\mathbf{g}_0+m_1\mathbf{g}_1+\dots+m_{k-1}\mathbf{g}_{k-1}
 \end{align*}
-$$
+```
 
 ### Systemic linear block code
 
 Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits.
 
-$$
+```math
 \begin{align*}
 \mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[
     \begin{array}{c|c}
@@ -1070,13 +1068,13 @@ $$
 \mathbf{x} &= \mathbf{m}\mathbf{G}= \mathbf{m} \begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\begin{array}{c|c}[\mathbf{mI}_k & \mathbf{mP}]\end{array}=\begin{array}{c|c}[\mathbf{m} & \mathbf{b}]\end{array}\\
 \mathbf{b} &= \mathbf{m}\mathbf{P}\quad\text{Parity bits of $\mathbf{x}$}
 \end{align*}
-$$
+```
 
 #### Parity check matrix $\mathbf{H}$
 
 Transpose $\mathbf{P}$ for the parity check matrix
 
-$$
+```math
 \begin{align*}
 \mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\
 &=\left[
@@ -1103,7 +1101,7 @@ $$
 \end{matrix}\end{array}\right]\\
 \mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid}
 \end{align*}
-$$
+```
 
 #### Procedure to find parity check matrix from list of codewords
 
@@ -1114,30 +1112,28 @@ $$
 
 Example:
 
-$$
+```math
 \begin{array}{cccc}
-    x_1 & x_2 & x_3 & x_4 & x_5 \\
-    \hline
+    x_1 & x_2 & x_3 & x_4 & x_5 \\\hline
     \color{magenta}1&\color{magenta}0&1&1&0\\
     \color{magenta}0&\color{magenta}1&1&1&1\\
     \color{magenta}0&\color{magenta}0&0&0&0\\
     \color{magenta}1&\color{magenta}1&0&0&1\\
 \end{array}
-$$
+```
 
 Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
 
-$$
-\begin{align*}
+```math
 \begin{align*}
+\begin{aligned}
     x_3 &= x_1\oplus x_2\\
     x_4 &= x_1\oplus x_2\\
     x_5 &= x_2\\
-\end{align*}
+\end{aligned}
 \implies\textbf{P}&=
-\begin{array}{c|ccc}
-    & x_1 & x_2 \\
-    \hline
+\begin{array}{c|cc}
+    & x_1 & x_2 \\\hline
     x_3&1&1&\\
     x_4&1&1&\\
     x_5&0&1&\\
@@ -1156,7 +1152,7 @@ $$
     0 & 0 & 1\\
 \end{matrix}\end{array}\right]
 \end{align*}
-$$
+```
 
 #### Error detection and correction
 
@@ -1166,6 +1162,6 @@ $$
 
 ## CHECKLIST
 
-- Transfer function in complex envelope form ($\tilde{h}(t)$) should be divided by two.
+- Transfer function in complex envelope form $\tilde{h}(t)$ should be divided by two.
 - Convolutions: do not forget width when using graphical method
 - todo: add more items to check