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---
title: Idiot's guide to ELEC4402 Communications Systems
author: peter
date: 2024-10-29 16:11:34 +0800
2024-11-02 04:39:24 +08:00
categories: [University, Guides] # 0-2 categories. Blogging | Electronics | Programming | Mechanical | SelfHosting
tags: [ELEC4402, communications systems, guide, formula sheet, datasheet, notes] # 0-\infty. systems | embedded | rf | microwave | electronics | solidworks | automation | tip
# image: assets/img/2024-10-29-Idiots-guide-to-ELEC/2024-10-29-Idiots-guide-to-ELEC/preview.png
math: true
---
2024-11-02 04:39:24 +08:00
These notes should provide a printable summary of all the formulas and procedures required to solve questions in the exam and tests. This unit allows you to bring infinite physical notes (except books borrowed from the UWA library) to all tests and the final exam. You can't rely on what material they provide in the test/exam, it is very _minimal_ to say the least. Hope this helps.
If you have issues or suggestions, [raise them on GitHub](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/issues/new). I accept [pull requests](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/pulls) for fixes or suggestions but the content must not be copyrighted under a non-GPL compatible license.
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## [Download the latest PDF revision 📄](https://raw.githubusercontent.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/refs/heads/master/README.pdf)
It is recommended to refer to use [the PDF copy](https://raw.githubusercontent.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/refs/heads/master/README.pdf) instead of this webpage for offline viewing/printing.
### License and information
Notes are open-source and licensed under the GNU GPL-3.0. **You must include the [full-text of the license](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/blob/master/COPYING.txt) and follow its terms when using these notes or any diagrams in derivative works** (but not when printing as notes)
Copyright (C) 2024 Peter Tanner
<details>
<summary>GPL copyright information</summary>
Copyright (C) 2024 Peter Tanner
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
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This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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You should have received a copy of the GNU General Public License
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</details>
## Other advice for this unit
### Get more exam papers on OneSearch
- You can access up to **6** more papers with this method (You normally only get the previous year's paper on LMS in week 12).
- Either [search "Communications" and filter by type "Examination Papers"](https://onesearch.library.uwa.edu.au/discovery/search?query=any%2Ccontains%2Ccommunications&tab=Everything&search_scope=MyInst_and_CI&vid=61UWA_INST%3AUWA&facet=rtype%2Cinclude%2Cexampaper&lang=en&offset=0)
- Or search old unit codes
- ELEC4301 Digital Communications and Networking
- ENGT4301 Digital Communications and Networking
- ELEC3302 Communications Systems
- Note that ELEC5501 Advanced Communications is a different unit.
<details>
<summary>Listing of examination papers on OneSearch</summary>
<ul>
<li>Communications Systems ELEC3302 Examination paper [2008 Supplementary]</li>
<li>Communications Systems ELEC4402 Examination paper [2014 Semester 2]</li>
<li>Communications Systems ELEC3302 Examination paper [2014 Semester 2]</li>
<li>Communications Systems ELEC3302 Examination paper [2008 Semester 1]</li>
<li>Digital Communications and Networking ENGT4301 Examination paper [2005 Supplementary]</li>
<li>Digital Communications and Networking ELEC4301 Examination paper [2009 Supplementary]</li>
</ul>
</details>
### Tests
- A lot of the unit requires you to learn processes and apply them. This is quite time consuming to do during the semester and the marking of the tests will destroy your wam if you do not know the process (especially compared to signal processing and signals and systems), I do not recommend doing this unit during thesis year.
- This formula sheet will attempt to condense all processes/formulas you may need in this unit.
- **You do not get given a formula sheet**, so you are entirely dependent on your own notes (except for some exceptions, such as the $\text{erf}(x)$ table). So bring good notes.
- Doing this unit after signal processing is a good idea.
## Printable notes begins on next page (in PDF)
<div style="page-break-after: always;"></div>
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## Fourier transform identities and properties
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| Time domain $x(t)$ | Frequency domain $X(f)$ |
| --------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \text{sinc}(fT)$ |
| $\text{sinc}(2Wt)$ | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ |
| $\exp(-at)u(t),\quad a>0$ | $\frac{1}{a + j2\pi f}$ |
| $\exp(-a\lvert t \rvert),\quad a>0$ | $\frac{2a}{a^2 + (2\pi f)^2}$ |
| $\exp(-\pi t^2)$ | $\exp(-\pi f^2)$ |
| $1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T$ | $T \text{sinc}^2(fT)$ |
| $\delta(t)$ | $1$ |
| $1$ | $\delta(f)$ |
| $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ |
| $\exp(j2\pi f_c t)$ | $\delta(f - f_c)$ |
| $\cos(2\pi f_c t)$ | $\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$ |
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| $\cos(2\pi f_c t+\theta)$ | $\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}$ |
| $\sin(2\pi f_c t)$ | $\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$ |
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| $\sin(2\pi f_c t+\theta)$ | $\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]$ |
| $\text{sgn}(t)$ | $\frac{1}{j\pi f}$ |
| $\frac{1}{\pi t}$ | $-j \text{sgn}(f)$ |
| $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ |
| $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ |
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| Time domain $x(t)$ | Frequency domain $X(f)$ | Property |
| ---------------------------------------- | ------------------------------------------------ | ----------------------------- |
| $g(t-a)$ | $\exp(-j2\pi fa)G(f)$ | Time shifting |
| $\exp(-j2\pi f_c t)g(t)$ | $G(f-f_c)$ | Frequency shifting |
| $g(bt)$ | $\frac{G(f/b)}{\|b\|}$ | Time scaling |
| $g(bt-a)$ | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)$ | Time scaling and shifting |
| $\frac{d}{dt}g(t)$ | $j2\pi fG(f)\quad$ | Differentiation wrt time |
| $tg(t)$ | $\frac{1}{2\pi}\frac{d}{df}G(f)\quad$ | Differentiation wrt frequency |
| $g^*(t)$ | $G^*(-f)$ | Conjugate functions |
| $G(t)$ | $g(-f)$ | Duality |
| $\int_{-\infty}^t g(\tau)d\tau$ | $\frac{1}{j2\pi f}G(f)+\frac{G(0)}{2}\delta(f)$ | Integration wrt time |
| $g(t)h(t)$ | $G(f)*H(f)$ | Time multiplication |
| $g(t)*h(t)$ | $G(f)H(f)$ | Time convolution |
| $ag(t)+bh(t)$ | $aG(f)+bH(f)$ | Linearity $a,b$ constants |
| $\int_{-\infty}^\infty x(t)y^*(t)dt$ | $\int_{-\infty}^\infty X(f)Y^*(f)df$ | Parseval's theorem |
| $E_x=\int_{-\infty}^\infty \|x(t)\|^2dt$ | $E_x=\int_{-\infty}^\infty \|X(f)\|^2df$ | Parseval's theorem |
| Description | Property |
| ----------------------------------- | ----------------- |
| $g(0)=\int_{-\infty}^\infty G(f)df$ | Area under $G(f)$ |
| $G(0)=\int_{-\infty}^\infty G(t)dt$ | Area under $g(t)$ |
$$
\begin{align*}
u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
\text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
\text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\
\text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
\end{align*}
$$
### Fourier transform of continuous time periodic signal
Required for some questions on **sampling**:
<!-- Transform a continuous time-periodic signal $x(t)=\sum_{n=-\infty}^\infty x_p(t-nT_s)$ with period $T_s$: -->
Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
$$
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
$$
Calculate $C_n$ coefficient as follows from $x_p(t)$:
<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
$$
\begin{align*}
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
\end{align*}
$$
### $\text{rect}$ function
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![rect](/assets/img/2024-10-29-Idiots-guide-to-ELEC/rect.drawio.svg)
### Bessel function
$$
\begin{align*}
\sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
J_n(\beta)&=(-1)^nJ_{-n}(\beta)
\end{align*}
$$
### White noise
$$
\begin{align*}
R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
G_w(f)&=\frac{N_0}{2}\\
N_0&=kT\\
G_y(f)&=|H(f)|^2G_w(f)\\
G_y(f)&=G(f)G_w(f)\\
\end{align*}
$$
### WSS
$$
\begin{align*}
\mu_X(t) &= \mu_X\text{ Constant}\\
R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
\end{align*}
$$
### Ergodicity
$$
\begin{align*}
\braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
\braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\
E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\
\end{align*}
$$
| Type | Normal | Mean square sense |
| ----------------------------------- | ------------------------------------------------------- | ----------------------------------------------------------- |
| ergodic in mean | $$\lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X$$ | $$\lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0$$ |
| ergodic in autocorrelation function | $$\lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau)$$ | $$\lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0$$ |
Note: **A WSS random process needs to be both ergodic in mean and autocorrelation to be considered an ergodic process**
### Other identities
$$
\begin{align*}
f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\
a(f*g) &= (af)*g \quad\text{Convolution associative}\\
\sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right)
\end{align*}
$$
### Other trig
$$
\begin{align*}
\cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\
e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\
e^{-j\alpha}+e^{j\alpha}&=2 \cos(\alpha)\\
\cos(-A)&=\cos(A)\\
\sin(-A)&=-\sin(A)\\
\sin(A+\pi/2)&=\cos(A)\\
\sin(A-\pi/2)&=-\cos(A)\\
\cos(A-\pi/2)&=\sin(A)\\
\cos(A+\pi/2)&=-\sin(A)\\
\int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\
\end{align*}
$$
$$
\begin{align*}
\cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\
\sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\
\cos(A)\cos(B) &= \frac{1}{2} (\cos (A-B)+\cos (A+B)) \\
\cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\
\sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\
\end{align*}
$$
$$
\begin{align*}
\cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
\cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
\sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\
\sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
\cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
\cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
\end{align*}
$$
## IQ/Complex envelope
Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $e^{j\theta}$ form.
### Convert complex envelope representation to time-domain representation of signal
$$
\begin{align*}
g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\
&=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\
&=A(t)\cos(2\pi f_c t+\phi(t))\\
A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\
\phi(t)&\quad\text{Phase}\\
g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\
g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\
\end{align*}
$$
### For transfer function
$$
\begin{align*}
h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
&=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\
\Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))}
\end{align*}
$$
## AM
### CAM
$$
\begin{align*}
m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
m_a&=k_a A_m \quad\text{ (Symmetrical $m(t)$)}\\
x(t)&=A_c\cos(2\pi f_c t)\left[1+k_a m(t)\right]=A_c\cos(2\pi f_c t)\left[1+m_a m(t)/A_c\right], \\
&\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\
P_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\
P_x &=\frac{1}{4}{m_a}^2{A_c}^2\\
\eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
B_T&=2f_m=2B
\end{align*}
$$
$B_T$: Signal bandwidth
$B$: Bandwidth of modulating wave
Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
### DSB-SC
$$
\begin{align*}
x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
B_T&=2f_m=2B
\end{align*}
$$
## FM/PM
$$
\begin{align*}
s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
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s(t) &= A_c\cos(\theta_i(t))=A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_{-\infty}^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\
s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\quad\text{FM single tone}\\
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f_i(t) &= \frac{1}{2\pi}\frac{d}{dt}\theta_i(t)\quad\text{Instantaneous frequency from instantaneous phase}\\
\Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
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\Delta f&=\max_t(f_i(t))- \min_t(f_i(t))\quad\text{Maximum frequency deviation}\\
\beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\
D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}\\
\end{align*}
$$
### Bessel form and magnitude spectrum (single tone)
$$
\begin{align*}
s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
\end{align*}
$$
### FM signal power
$$
\begin{align*}
P_\text{av}&=\frac{ {A_c}^2}{2}\\
P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
\text{band\_index}&=0\implies f_c+0f_m\\
\text{band\_index}&=1\implies f_c+1f_m,\dots\\
\end{align*}
$$
### Carson's rule to find $B$ (98% power bandwidth rule)
$$
\begin{align*}
B &= 2Mf_m = 2(\beta + 1)f_m\\
&= 2(\Delta f+f_m)\\
&= 2(k_f A_m+f_m)\\
&= 2(D+1)W_m\\
B &= \begin{cases}
2(\Delta f+f_m) & \text{FM, sinusoidal message}\\
2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
\end{cases}\\
\end{align*}
$$
### Complex envelope
$$
\begin{align*}
s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
\tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
\end{align*}
$$
### Band
| Narrowband | Wideband |
| ------------- | ------------- |
| $D<1,\beta<1$ | $D>1,\beta>1$ |
## Power, energy and autocorrelation
$$
\begin{align*}
G_\text{WGN}(f)&=\frac{N_0}{2}\\
G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\
G_x(f)&=\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\text{ (PSD)}\\
G_x(f)&=\mathfrak{F}[R_x(\tau)]\text{ (WSS)}\\
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P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df\quad\text{For zero mean}\\
P_x&={\sigma_x}^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\quad\text{For zero mean}\\
P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\
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E_x&=\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\quad\text{Parseval's theorem}\\
R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
\end{align*}
$$
##
## Noise performance
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Coherent detection system.
$$
\begin{align*}
y(t) &= m(t)\cos(2\pi f_c t+\theta) = m(t)\cos^2(2\pi f_c t+\theta)\text{ After IF mixing.}\\
&= m(t)\frac{1}{2}(1+\cos(4 f_c t+2\theta))\\
&= \frac{1}{2}m(t)+\frac{1}{2}\cos(4 f_c t+2\theta)\\
\implies S_u(f) &= \left(\frac{1}{2}\right)^2S_u(f)\quad\text{After LPF.}
\end{align*}
$$
Use formualas from previous section, [Power, energy and autocorrelation](#power-energy-and-autocorrelation).\
Use these formulas in particular:
$$
\begin{align*}
G_\text{WGN}(f)&=\frac{N_0}{2}\\
G_x(f)&=|H(f)|^2G_w(f)&\text{Note the square in $|H(f)|^2$}\\
P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df&\text{Often perform graphical integration}\\
\end{align*}
$$
$$
\begin{align*}
\text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\
\text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\
\text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\
\text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio}
\end{align*}
$$
## Sampling
$$
\begin{align*}
t&=nT_s\\
T_s&=\frac{1}{f_s}\\
x_s(t)&=x(t)\delta_s(t)=x(t)\sum_{n\in\mathbb{Z}}\delta(t-nT_s)=\sum_{n\in\mathbb{Z}}x(nT_s)\delta(t-nT_s)\\
X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\
\end{align*}
$$
### Procedure to reconstruct sampled signal
Analog signal $x'(t)$ which can be reconstructed from a sampled signal $x_s(t)$: Put $x_s(t)$ through LPF with maximum frequency of $f_s/2$ and minimum frequency of $-f_s/2$. Anything outside of the BPF will be attenuated, therefore $n$ which results in frequencies outside the BPF will evaluate to $0$ and can be ignored.
Example: $f_s=5000\implies \text{LPF}\in[-2500,2500]$
Then iterate for $n=0,1,-1,2,-2,\dots$ until the first iteration where the result is 0 since all terms are eliminated by the LPF.
TODO: Add example
Then add all terms and transform $\bar X_s(f)$ back to time domain to get $x_s(t)$
### Fourier transform of continuous time periodic signal (1)
Required for some questions on **sampling**:
<!-- Transform a continuous time-periodic signal $x(t)=\sum_{n=-\infty}^\infty x_p(t-nT_s)$ with period $T_s$: -->
Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
$$
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
$$
Calculate $C_n$ coefficient as follows from $x_p(t)$:
<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
$$
\begin{align*}
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
\end{align*}
$$
<!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ -->
### Nyquist criterion for zero-ISI
Do not transmit more than $2B$ samples per second over a channel of $B$ bandwidth.
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![By Bob K - Own work, CC0, https://commons.wikimedia.org/w/index.php?curid=94674142](/assets/img/2024-10-29-Idiots-guide-to-ELEC/Nyquist_frequency_&_rate.svg)
### Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to `assets/img/2024-10-29-Idiots-guide-to-ELEC/sampling.png`)
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Cannot add directly due to copyright!
2024-10-29 16:38:03 +08:00
<!-- ![sampling](copyrighted_assets/img/2024-10-29-Idiots-guide-to-ELEC/sampling.png) -->
<!-- ![sampling](/assets/img/2024-10-29-Idiots-guide-to-ELEC/sampling.png) -->
## Quantizer
$$
\begin{align*}
\Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\
\end{align*}
$$
### Quantization noise
$$
\begin{align*}
e &:= y-x\quad\text{Quantization error}\\
\mu_E &= E[E] = 0\quad\text{Zero mean}\\
{\sigma_E}^2&=E[E^2]-0^2=\int_{-\Delta/2}^{\Delta/2}e^2\times\left(\frac{1}{\Delta}\right) de\quad\text{Where $E\thicksim 1/\Delta$ uniform over $(-\Delta/2,\Delta/2)$}\\
\text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\
\text{SQNR(dB)}&=10\log_{10}(\text{SQNR})
\end{align*}
$$
### Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to `assets/img/2024-10-29-Idiots-guide-to-ELEC/quantizer.png`)
2024-10-29 16:38:03 +08:00
Cannot add directly due to copyright!
<!-- ![quantizer](copyrighted_assets/img/2024-10-29-Idiots-guide-to-ELEC/quantizer.png) -->
<!-- ![quantizer](/assets/img/2024-10-29-Idiots-guide-to-ELEC/quantizer.png) -->
## Line codes
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![binary_codes](/assets/img/2024-10-29-Idiots-guide-to-ELEC/Line_Codes.drawio.svg)
$$
\begin{align*}
R_b&\rightarrow\text{Bit rate}\\
D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\
A&\rightarrow m_a\\
V(f)&\rightarrow\text{Pulse shape}\\
V_\text{rectangle}(f)&=T\text{sinc}(fT\times\text{DutyCycle})\\
G_\text{MunipolarNRZ}(f)&=\frac{(M^2-1)A^2D}{12}|V(f)|^2+\frac{(M-1)^2}{4}(DA)^2\sum_{l=-\infty}^{\infty}|V(lD)|^2\delta(f-lD)\\
G_\text{MpolarNRZ}(f)&=\frac{(M^2-1)A^2D}{3}|V(f)|^2\\
G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right), \text{NB}_0=R_b\\
G_\text{polarNRZ}(f)&=\frac{A^2}{R_b}\text{sinc}^2\left(\frac{f}{R_b}\right)\\
G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\
G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b
\end{align*}
$$
## Modulation and basis functions
![Constellation diagrams](./assets/img/2024-10-29-Idiots-guide-to-ELEC/Constellation.drawio.svg)
### BASK
#### Basis functions
$$
\begin{align*}
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
\end{align*}
$$
#### Symbol mapping
$$
b_n:\{1,0\}\to a_n:\{1,0\}
$$
#### 2 possible waveforms
$$
\begin{align*}
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\
s_1(t)&=0\\
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$}
\end{align*}
$$
Distance is $d=\sqrt{2E_b}$
### BPSK
#### Basis functions
$$
\begin{align*}
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
\end{align*}
$$
#### Symbol mapping
$$
b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}
$$
#### 2 possible waveforms
$$
\begin{align*}
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\
s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$}
\end{align*}
$$
Distance is $d=2\sqrt{E_b}$
### QPSK ($M=4$ PSK)
#### Basis functions
$$
\begin{align*}
T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\
\varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\
\varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\
\end{align*}
$$
### 4 possible waveforms
$$
\begin{align*}
s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\
s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\
s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\
s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\
\end{align*}
$$
Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows:
$$
\begin{align*}
s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
&=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
&=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
\end{align*}
$$
#### Signal
$$
\begin{align*}
\text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\
I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\
Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\
x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)]
\end{align*}
$$
### Example of waveform
<details>
<summary>Code</summary>
<pre><code>
tBitstream[bitstream_, Tb_, title_] :=
Module[{timeSteps, gridLines, plot},
timeSteps =
Flatten[Table[{(n - 1) Tb, bitstream[[n]]}, {n, 1,
Length[bitstream]}] /. {t_, v_} :> { {t, v}, {t + Tb, v}}, 1];
gridLines = {Join[
Table[{n Tb, Dashed}, {n, 1, 2 Length[bitstream], 2}],
Table[{n Tb, Thin}, {n, 0, 2 Length[bitstream], 2}]], None};
plot =
Labeled[ListLinePlot[timeSteps, InterpolationOrder -> 0,
PlotRange -> Full, GridLines -> gridLines, PlotStyle -> Thick,
Ticks -> {Table[{n Tb,
Row[{n, "\!\(\*SubscriptBox[\(T\), \(b\)]\)"}]}, {n, 0,
Length[bitstream]}], {-1, 0, 1}},
LabelStyle -> Directive[Bold, 12],
PlotRangePadding -> {Scaled[.05]}, AspectRatio -> 0.1,
ImageSize -> Large], {Style[title, "Text", 16]}, {Right}]];
tBitstream[{0, 1, 0, 0, 1, 0, 1, 1, 1, 0}, 1, "Bitstream Step Plot"]
tBitstream[{-1, -1, -1, -1, 1, 1, 1, 1, 1, 1}, 1, "I(t)"]
tBitstream[{1, 1, -1, -1, -1, -1, 1, 1, -1, -1}, 1, "Q(t)"]
</code></pre>
</details>
Remember that $T=2T_b$
| | |
| ----------------------- | ----------------------------------------------------------------------- |
| $b_n$ | ![QPSK bits](/assets/img/2024-10-29-Idiots-guide-to-ELEC/qpsk-bits.svg) |
| $I(t)$ (Odd, 1st bits) | ![QPSK bits](/assets/img/2024-10-29-Idiots-guide-to-ELEC/qpsk-it.svg) |
| $Q(t)$ (Even, 2nd bits) | ![QPSK bits](/assets/img/2024-10-29-Idiots-guide-to-ELEC/qpsk-qt.svg) |
## Matched filter
### 1. Filter function
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Find transfer function $h(t)$ of matched filter and apply to an input:\
Note that $x(T-t)$ is equivalent to horizontally flipping $x(t)$ around $x=T/2$.
$$
\begin{align*}
h(t)&=s_1(T-t)-s_2(T-t)\\
h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\
s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\
n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output}
\end{align*}
$$
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### 2. Bit error rate of matched filter
Bit error rate (BER) from matched filter outputs and filter output noise
$$
\begin{align*}
Q(x)&=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Leftrightarrow\text{erf}\left(\frac{x}{\sqrt{2}}\right)=1-2Q(x)\\
E_b&=d^2=\int_{-\infty}^\infty|s_1(t)-s_2(t)|^2dt\quad\text{Energy per bit/Distance}\\
T&=1/R_b\quad\text{$R_b$: Bitrate}\\
E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
P(\text{W})&=10^{\frac{P(\text{dB})}{10}}\\
P_\text{RX}(W)&=P_\text{TX}(W)\cdot10^{\frac{P_\text{loss}(\text{dB})}{10}}\quad \text{$P_\text{loss}$ is expressed with negative sign e.g. "-130 dB"}\\
\text{BER}_\text{MatchedFilter}&=Q\left(\sqrt{\frac{d^2}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{2N_0}}\right)\\
\text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\
\text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\
\end{align*}
$$
<div style="page-break-after: always;"></div>
## Value tables for $\text{erf}(x)$ and $Q(x)$
### $Q(x)$ function
| $x$ | $Q(x)$ | $x$ | $Q(x)$ | $x$ | $Q(x)$ | $x$ | $Q(x)$ |
| ------ | ---------- | ------ | ----------------------- | ------ | ------------------------ | ------ | ------------------------ |
| $0.00$ | $0.5$ | $2.30$ | $0.010724$ | $4.55$ | $2.6823 \times 10^{-6}$ | $6.80$ | $5.231 \times 10^{-12}$ |
| $0.05$ | $0.48006$ | $2.35$ | $0.0093867$ | $4.60$ | $2.1125 \times 10^{-6}$ | $6.85$ | $3.6925 \times 10^{-12}$ |
| $0.10$ | $0.46017$ | $2.40$ | $0.0081975$ | $4.65$ | $1.6597 \times 10^{-6}$ | $6.90$ | $2.6001 \times 10^{-12}$ |
| $0.15$ | $0.44038$ | $2.45$ | $0.0071428$ | $4.70$ | $1.3008 \times 10^{-6}$ | $6.95$ | $1.8264 \times 10^{-12}$ |
| $0.20$ | $0.42074$ | $2.50$ | $0.0062097$ | $4.75$ | $1.0171 \times 10^{-6}$ | $7.00$ | $1.2798 \times 10^{-12}$ |
| $0.25$ | $0.40129$ | $2.55$ | $0.0053861$ | $4.80$ | $7.9333 \times 10^{-7}$ | $7.05$ | $8.9459 \times 10^{-13}$ |
| $0.30$ | $0.38209$ | $2.60$ | $0.0046612$ | $4.85$ | $6.1731 \times 10^{-7}$ | $7.10$ | $6.2378 \times 10^{-13}$ |
| $0.35$ | $0.36317$ | $2.65$ | $0.0040246$ | $4.90$ | $4.7918 \times 10^{-7}$ | $7.15$ | $4.3389 \times 10^{-13}$ |
| $0.40$ | $0.34458$ | $2.70$ | $0.003467$ | $4.95$ | $3.7107 \times 10^{-7}$ | $7.20$ | $3.0106 \times 10^{-13}$ |
| $0.45$ | $0.32636$ | $2.75$ | $0.0029798$ | $5.00$ | $2.8665 \times 10^{-7}$ | $7.25$ | $2.0839 \times 10^{-13}$ |
| $0.50$ | $0.30854$ | $2.80$ | $0.0025551$ | $5.05$ | $2.2091 \times 10^{-7}$ | $7.30$ | $1.4388 \times 10^{-13}$ |
| $0.55$ | $0.29116$ | $2.85$ | $0.002186$ | $5.10$ | $1.6983 \times 10^{-7}$ | $7.35$ | $9.9103 \times 10^{-14}$ |
| $0.60$ | $0.27425$ | $2.90$ | $0.0018658$ | $5.15$ | $1.3024 \times 10^{-7}$ | $7.40$ | $6.8092 \times 10^{-14}$ |
| $0.65$ | $0.25785$ | $2.95$ | $0.0015889$ | $5.20$ | $9.9644 \times 10^{-8}$ | $7.45$ | $4.667 \times 10^{-14}$ |
| $0.70$ | $0.24196$ | $3.00$ | $0.0013499$ | $5.25$ | $7.605 \times 10^{-8}$ | $7.50$ | $3.1909 \times 10^{-14}$ |
| $0.75$ | $0.22663$ | $3.05$ | $0.0011442$ | $5.30$ | $5.7901 \times 10^{-8}$ | $7.55$ | $2.1763 \times 10^{-14}$ |
| $0.80$ | $0.21186$ | $3.10$ | $0.0009676$ | $5.35$ | $4.3977 \times 10^{-8}$ | $7.60$ | $1.4807 \times 10^{-14}$ |
| $0.85$ | $0.19766$ | $3.15$ | $0.00081635$ | $5.40$ | $3.332 \times 10^{-8}$ | $7.65$ | $1.0049 \times 10^{-14}$ |
| $0.90$ | $0.18406$ | $3.20$ | $0.00068714$ | $5.45$ | $2.5185 \times 10^{-8}$ | $7.70$ | $6.8033 \times 10^{-15}$ |
| $0.95$ | $0.17106$ | $3.25$ | $0.00057703$ | $5.50$ | $1.899 \times 10^{-8}$ | $7.75$ | $4.5946 \times 10^{-15}$ |
| $1.00$ | $0.15866$ | $3.30$ | $0.00048342$ | $5.55$ | $1.4283 \times 10^{-8}$ | $7.80$ | $3.0954 \times 10^{-15}$ |
| $1.05$ | $0.14686$ | $3.35$ | $0.00040406$ | $5.60$ | $1.0718 \times 10^{-8}$ | $7.85$ | $2.0802 \times 10^{-15}$ |
| $1.10$ | $0.13567$ | $3.40$ | $0.00033693$ | $5.65$ | $8.0224 \times 10^{-9}$ | $7.90$ | $1.3945 \times 10^{-15}$ |
| $1.15$ | $0.12507$ | $3.45$ | $0.00028029$ | $5.70$ | $5.9904 \times 10^{-3}$ | $7.95$ | $9.3256 \times 10^{-16}$ |
| $1.20$ | $0.11507$ | $3.50$ | $0.00023263$ | $5.75$ | $4.4622 \times 10^{-9}$ | $8.00$ | $6.221 \times 10^{-16}$ |
| $1.25$ | $0.10565$ | $3.55$ | $0.00019262$ | $5.80$ | $3.3157 \times 10^{-9}$ | $8.05$ | $4.1397 \times 10^{-16}$ |
| $1.30$ | $0.0968$ | $3.60$ | $0.00015911$ | $5.85$ | $2.4579 \times 10^{-9}$ | $8.10$ | $2.748 \times 10^{-16}$ |
| $1.35$ | $0.088508$ | $3.65$ | $0.00013112$ | $5.90$ | $1.8175 \times 10^{-9}$ | $8.15$ | $1.8196 \times 10^{-16}$ |
| $1.40$ | $0.080757$ | $3.70$ | $0.0001078$ | $5.95$ | $1.3407 \times 10^{-9}$ | $8.20$ | $1.2019 \times 10^{-16}$ |
| $1.45$ | $0.073529$ | $3.75$ | $8.8417 \times 10^{-5}$ | $6.00$ | $9.8659 \times 10^{-10}$ | $8.25$ | $7.9197 \times 10^{-17}$ |
| $1.50$ | $0.066807$ | $3.80$ | $7.2348 \times 10^{-5}$ | $6.05$ | $7.2423 \times 10^{-10}$ | $8.30$ | $5.2056 \times 10^{-17}$ |
| $1.55$ | $0.060571$ | $3.85$ | $5.9059 \times 10^{-5}$ | $6.10$ | $5.3034 \times 10^{-10}$ | $8.35$ | $3.4131 \times 10^{-17}$ |
| $1.60$ | $0.054799$ | $3.90$ | $4.8096 \times 10^{-5}$ | $6.15$ | $3.8741 \times 10^{-10}$ | $8.40$ | $2.2324 \times 10^{-17}$ |
| $1.65$ | $0.049471$ | $3.95$ | $3.9076 \times 10^{-5}$ | $6.20$ | $2.8232 \times 10^{-10}$ | $8.45$ | $1.4565 \times 10^{-17}$ |
| $1.70$ | $0.044565$ | $4.00$ | $3.1671 \times 10^{-5}$ | $6.25$ | $2.0523 \times 10^{-10}$ | $8.50$ | $9.4795 \times 10^{-18}$ |
| $1.75$ | $0.040059$ | $4.05$ | $2.5609 \times 10^{-5}$ | $6.30$ | $1.4882 \times 10^{-10}$ | $8.55$ | $6.1544 \times 10^{-18}$ |
| $1.80$ | $0.03593$ | $4.10$ | $2.0658 \times 10^{-5}$ | $6.35$ | $1.0766 \times 10^{-10}$ | $8.60$ | $3.9858 \times 10^{-18}$ |
| $1.85$ | $0.032157$ | $4.15$ | $1.6624 \times 10^{-5}$ | $6.40$ | $7.7688 \times 10^{-11}$ | $8.65$ | $2.575 \times 10^{-18}$ |
| $1.90$ | $0.028717$ | $4.20$ | $1.3346 \times 10^{-5}$ | $6.45$ | $5.5925 \times 10^{-11}$ | $8.70$ | $1.6594 \times 10^{-18}$ |
| $1.95$ | $0.025588$ | $4.25$ | $1.0689 \times 10^{-5}$ | $6.50$ | $4.016 \times 10^{-11}$ | $8.75$ | $1.0668 \times 10^{-18}$ |
| $2.00$ | $0.02275$ | $4.30$ | $8.5399 \times 10^{-6}$ | $6.55$ | $2.8769 \times 10^{-11}$ | $8.80$ | $6.8408 \times 10^{-19}$ |
| $2.05$ | $0.020182$ | $4.35$ | $6.8069 \times 10^{-6}$ | $6.60$ | $2.0558 \times 10^{-11}$ | $8.85$ | $4.376 \times 10^{-19}$ |
| $2.10$ | $0.017864$ | $4.40$ | $5.4125 \times 10^{-6}$ | $6.65$ | $1.4655 \times 10^{-11}$ | $8.90$ | $2.7923 \times 10^{-19}$ |
| $2.15$ | $0.015778$ | $4.45$ | $4.2935 \times 10^{-6}$ | $6.70$ | $1.0421 \times 10^{-11}$ | $8.95$ | $1.7774 \times 10^{-19}$ |
| $2.20$ | $0.013903$ | $4.50$ | $3.3977 \times 10^{-6}$ | $6.75$ | $7.3923 \times 10^{-12}$ | $9.00$ | $1.1286 \times 10^{-19}$ |
| $2.25$ | $0.012224$ | | | | | | |
Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
### $\text{erf}(x)$ function
| $x$ | $\text{erf}(x)$ | $x$ | $\text{erf}(x)$ | $x$ | $\text{erf}(x)$ |
| ------ | --------------- | ------ | --------------- | ------ | --------------- |
| $0.00$ | $0.00000$ | $0.75$ | $0.71116$ | $1.50$ | $0.96611$ |
| $0.05$ | $0.05637$ | $0.80$ | $0.74210$ | $1.55$ | $0.97162$ |
| $0.10$ | $0.11246$ | $0.85$ | $0.77067$ | $1.60$ | $0.97635$ |
| $0.15$ | $0.16800$ | $0.90$ | $0.79691$ | $1.65$ | $0.98038$ |
| $0.20$ | $0.22270$ | $0.95$ | $0.82089$ | $1.70$ | $0.98379$ |
| $0.25$ | $0.27633$ | $1.00$ | $0.84270$ | $1.75$ | $0.98667$ |
| $0.30$ | $0.32863$ | $1.05$ | $0.86244$ | $1.80$ | $0.98909$ |
| $0.35$ | $0.37938$ | $1.10$ | $0.88021$ | $1.85$ | $0.99111$ |
| $0.40$ | $0.42839$ | $1.15$ | $0.89612$ | $1.90$ | $0.99279$ |
| $0.45$ | $0.47548$ | $1.20$ | $0.91031$ | $1.95$ | $0.99418$ |
| $0.50$ | $0.52050$ | $1.25$ | $0.92290$ | $2.00$ | $0.99532$ |
| $0.55$ | $0.56332$ | $1.30$ | $0.93401$ | $2.50$ | $0.99959$ |
| $0.60$ | $0.60386$ | $1.35$ | $0.94376$ | $3.00$ | $0.99998$ |
| $0.65$ | $0.64203$ | $1.40$ | $0.95229$ | $3.30$ | $0.999998$\*\* |
| $0.70$ | $0.67780$ | $1.45$ | $0.95970$ | | |
\*\*The value of $\text{erf}(3.30)$ should be $\approx0.999997$ instead, but this value is quoted in the formula table.
<div style="page-break-after: always;"></div>
### Receiver output shit
$$
\begin{align*}
r_o(t)&=\begin{cases}
s_{o1}(t)+n_o(t) & \text{code 1}\\
s_{o2}(t)+n_o(t) & \text{code 0}\\
\end{cases}\\
n&: \text{AWGN with }\sigma_o^2\\
\end{align*}
$$
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<!--
$$
\begin{align*}
G_x(f)
\end{align*}
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``` -->
## ISI, channel model
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### Raised cosine (RC) pulse
![Raised cosine pulse](/assets/img/2024-10-29-Idiots-guide-to-ELEC/RC.drawio.svg)
$$
0\leq\alpha\leq1
$$
⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method.
### Nyquist criterion for zero ISI
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<!-- P_r(kT)=\begin{cases}
1 & k=0\\
0 & k\neq0
\end{cases} -->
$$
\begin{align*}
D &> 2W\quad\text{Use $W$ from table below depending on modulation scheme.}\\
B_\text{Nyquist} &= \frac{W}{1+\alpha}\\
\alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
\end{align*}
$$
### Nomenclature
$$
\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*}
$$
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### Bandwidth $W$ and bit error rate of modulation schemes
To solve this type of question:
1. Use the formula for $D$ below
2. Consult the BER table below to get the BER which relates the noise of the channel $N_0$ to $E_b$ and to $R_b$.
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| Linear modulation | Half |
| --------------------------------------------------- | -------------------------------------------------- |
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| BPSK, QPSK, $M$-PSK, $M$-QAM, ASK, FSK | $M$-PAM, PAM |
| RZ unipolar, Manchester | NRZ Unipolar, NRZ Polar, Bipolar RZ |
| $W=B_\text{\color{green}abs-abs}$ | $W=B_\text{\color{green}abs}$ |
| $W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ | $W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ |
| $D=\frac{W\text{ symbol/s}}{1+\alpha}$ | $D=\frac{2W\text{ symbol/s}}{1+\alpha}$ |
$$
\begin{align*}
R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\
M\text{ symbol/set}&=2^k\\
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T\text{ s/symbol}&=1/(D\text{ symbol/s})\\
E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
\end{align*}
$$
### Table of bandpass signalling and BER
| **Binary Bandpass Signaling** | **$B_\text{null-null}$ (Hz)** | **$B_\text{abs-abs}\color{red}=2B_\text{abs}$ (Hz)** | **BER with Coherent Detection** | **BER with Noncoherent Detection** |
| --------------------------------- | -------------------------------- | ---------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------- |
| ASK, unipolar NRZ | $2R_b$ | $R_b (1 + \alpha)$ | $Q\left( \sqrt{E_b / N_0} \right)$ | $0.5\exp(-E_b / (2N_0))$ |
| BPSK | $2R_b$ | $R_b (1 + \alpha)$ | $Q\left( \sqrt{2E_b / N_0} \right)$ | Requires coherent detection |
| Sunde's FSK | $3R_b$ | | $Q\left( \sqrt{E_b / N_0} \right)$ | $0.5\exp(-E_b / (2N_0))$ |
| DBPSK, $M$-ary Bandpass Signaling | $2R_b$ | $R_b (1 + \alpha)$ | | $0.5\exp(-E_b / N_0)$ |
| QPSK/OQPSK (**$M=4$, PSK**) | $R_b$ | $\frac{R_b (1 + \alpha)}{2}$ | $Q\left( \sqrt{2E_b / N_0} \right)$ | Requires coherent detection |
| MSK | $1.5R_b$ | $\frac{3R_b (1 + \alpha)}{4}$ | $Q\left( \sqrt{2E_b / N_0} \right)$ | Requires coherent detection |
| $M$-PSK ($M > 4$) | $2R_b / \log_2 M$ | $\frac{R_b (1 + \alpha)}{\log_2 M}$ | $\frac{2}{\log_2 M} Q\left( \sqrt{2 \log_2 M \sin^2 \left( \pi / M \right) E_b / N_0} \right)$ | Requires coherent detection |
| $M$-DPSK ($M > 4$) | $2R_b / \log_2 M$ | $\frac{R_b (1 + \alpha)}{2 \log_2 M}$ | | $\frac{2}{\log_2 M} Q\left( \sqrt{4 \log_2 M \sin^2 \left( \pi / (2M) \right) E_b / N_0} \right)$ |
| $M$-QAM (Square constellation) | $2R_b / \log_2 M$ | $\frac{R_b (1 + \alpha)}{\log_2 M}$ | $\frac{4}{\log_2 M} \left( 1 - \frac{1}{\sqrt{M}} \right) Q\left( \sqrt{\frac{3 \log_2 M}{M - 1} E_b / N_0} \right)$ | Requires coherent detection |
| $M$-FSK Coherent | $\frac{(M + 3) R_b}{2 \log_2 M}$ | | $\frac{M - 1}{\log_2 M} Q\left( \sqrt{(\log_2 M) E_b / N_0} \right)$ | |
| Noncoherent | $2M R_b / \log_2 M$ | | | $\frac{M - 1}{2 \log_2 M} 0.5\exp({-(\log_2 M) E_b / 2N_0})$ |
Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems
### PSD of modulated signals
| Modulation | $G_x(f)$ |
| ---------- | ------------------------------------------------------------------------------------------------- |
| Quadrature | $\color{red}\frac{ {A_c}^2}{4}[G_I(f-f_c)+G_I(f+f_c)+G_Q(f-f_c)+G_Q(f+f_c)]$ |
| Linear | $\color{red}\frac{\|V(f)\|^2}{2}\sum_{l=-\infty}^\infty R(l)\exp(-j2\pi l f T)\quad\text{What??}$ |
### Symbol error probability
- Minimum distance between any two point
- Different from bit error since a symbol can contain multiple bits
## Information theory
### Entropy for discrete random variables
$$
\begin{align*}
H(x) &\geq 0\\
H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\
H(x,y) &= -\sum_{x_i\in A_x}\sum_{y_i\in A_y} p_{XY}(x_i,y_i)\log_2(p_{XY}(x_i,y_i)) \quad\text{Joint entropy}\\
H(x,y) &= H(x)+H(y) \quad\text{Joint entropy if $x$ and $y$ independent}\\
H(x|y=y_j) &= -\sum_{x_i\in A_x} p_X(x_i|y=y_j) \log_2(p_X(x_i|y=y_j)) \quad\text{Conditional entropy}\\
H(x|y) &= -\sum_{y_j\in A_y} p_Y(y_j) H(x|y=y_j) \quad\text{Average conditional entropy, equivocation}\\
H(x|y) &= -\sum_{x_i\in A_x}\sum_{y_i\in A_y} p_X(x_i,y_j) \log_2(p_X(x_i|y=y_j))\\
H(x|y) &= H(x,y)-H(y)\\
H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\
\end{align*}
$$
Entropy is **maximized** when all have an equal probability.
### Differential entropy for continuous random variables
TODO: Cut out if not required
$$
\begin{align*}
h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx
\end{align*}
$$
### Mutual information
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![Mutual information](/assets/img/2024-10-29-Idiots-guide-to-ELEC/MutualInformation.drawio.svg)
Amount of entropy decrease of $x$ after observation by $y$.
$$
\begin{align*}
I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\
\end{align*}
$$
### Channel model
Vertical, $x$: input\
Horizontal, $y$: output
$$
\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]
$$
$$
\begin{array}{c|cccc}
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}
$$
Input has probability distribution $p_X(a_i)=P(X=a_i)$
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Channel maps alphabet $`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$
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Output has probability distribution $p_Y(b_j)=P(y=b_j)$
$$
\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*}
$$
#### Fast procedure to calculate $I(y;x)$
$$
\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}\\
&\text{3. Multiply each row in $\textbf{P}$ by $p_X(a_i)$ since $p_{XY}(x_i,y_i)=P(y_i|x_i)P(x_i)$}\\
&\text{4. Find $H(x,y)$ using each element from (3.)}\\
&\text{5. Find }H(x|y)=H(x,y)-H(y)\\
&\text{6. Find }I(y;x)=H(x)-H(x|y)\\
\end{align*}
$$
### Channel types
| Type | Definition |
| ----------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------- |
| Symmetric channel | Every row is a permutation of every other row, Every column is a permutation of every other column. $\text{Symmetric}\implies\text{Weakly symmetric}$ |
| Weakly symmetric | Every row is a permutation of every other row, Every column has the same sum |
#### Channel capacity of weakly symmetric channel
$$
\begin{align*}
C &\to\text{Channel capacity (bits/channels used)}\\
N &\to\text{Output alphabet size}\\
\mathbf{p} &\to\text{Probability vector, any row of the transition matrix}\\
C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\
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R_b &< C \text{ for error-free transmission}
\end{align*}
$$
#### Channel capacity of an AWGN channel
$$
y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
$$
$$
C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)
$$
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#### Channel capacity of a bandwidth limited AWGN channel
$$
\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)\\
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C&=W\log_2(1+\text{SNR})\\
\text{SNR}&=P_s/(N_0 W)
\end{align*}
$$
#### Shannon limit
$$
\begin{align*}
R_b &< C\\
\implies R_b &< W\log_2\left(1+\frac{P_s}{N_0 W}\right)\quad\text{For bandwidth limited AWGN channel}\\
\frac{E_b}{N_0} &> \frac{2^\eta-1}{\eta}\quad\text{SNR per bit required for error-free transmission}\\
\eta &= \frac{R_b}{W}\quad\text{Spectral efficiency (bit/(s-Hz))}\\
\eta &\gg 1\quad\text{Bandwidth limited}\\
\eta &\ll 1\quad\text{Power limited}
\end{align*}
$$
## Channel code
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Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition.
| | | |
| ---------------- | --------------------------------- | ---------------------------------------------------------------------------------------------------------------------------- |
| Hamming weight | $w_H(x)$ | Number of `'1'` in codeword $x$ |
| Hamming distance | $d_H(x_1,x_2)=w_H(x_1\oplus x_2)$ | Number of different bits between codewords $x_1$ and $x_2$ which is the hamming weight of the XOR of the two codes. |
| Minimum distance | $d_\text{min}$ | **IMPORTANT**: $x\neq\textbf{0}$, excludes weight of all-zero codeword. For a linear block code, $d_\text{min}=w_\text{min}$ |
### Linear block code
Code is $(n,k)$
$n$ is the width of a codeword
$2^k$ codewords
A linear block code must be a subspace and satisfy both:
1. Zero vector must be present at least once
2. The XOR of any codeword pair in the code must result in a codeword that is already present in the code table.
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3. $d_\text{min}=w_\text{min}$ (Implied by (1) and (2).)
### Code generation
Each generator vector is a binary string of size $n$. There are $k$ generator vectors in $\mathbf{G}$.
$$
\begin{align*}
\mathbf{g}_i&=[\begin{matrix}
g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1}
\end{matrix}]\\
\color{darkgray}\mathbf{g}_0&\color{darkgray}=[1010]\quad\text{Example for $n=4$}\\
\mathbf{G}&=\left[\begin{matrix}
\mathbf{g}_0\\
\mathbf{g}_1\\
\vdots\\
\mathbf{g}_{k-1}\\
\end{matrix}\right]=\left[\begin{matrix}
g_{0,0}& \dots & g_{0,n-2} & g_{0,n-1}\\
g_{1,0}& \dots & g_{1,n-2} & g_{1,n-1}\\
\vdots & \ddots & \vdots & \vdots\\
g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\
\end{matrix}\right]
\end{align*}
$$
A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$:
$$
\begin{align*}
\mathbf{m}&=[\begin{matrix}
m_{0}& \dots & m_{n-2} & m_{k-1}
\end{matrix}]\\
\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*}
$$
### Systemic linear block code
Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits.
$$
\begin{align*}
\mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[
\begin{array}{c|c}
\begin{matrix}
1 & 0 & \dots & 0\\
0 & 1 & \dots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0& 0 & \dots & 1\\
\end{matrix}
&
\begin{matrix}
p_{0,0}& \dots & p_{0,n-2} & p_{0,n-1}\\
p_{1,0}& \dots & p_{1,n-2} & p_{1,n-1}\\
\vdots & \ddots & \vdots & \vdots\\
p_{k-1,0}& \dots & p_{k-1,n-2} & p_{k-1,n-1}\\
\end{matrix}\end{array}\right]\\
\mathbf{m}&=[\begin{matrix}
m_{0}& \dots & m_{n-2} & m_{k-1}
\end{matrix}]\\
\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*}
$$
#### Parity check matrix $\mathbf{H}$
Transpose $\mathbf{P}$ for the parity check matrix
$$
\begin{align*}
\mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\
&=\left[
\begin{array}{c|c}
\begin{matrix}
{\textbf{p}_0}^\text{T} & {\textbf{p}_1}^\text{T} & \dots & {\textbf{p}_{k-1}}^\text{T}
\end{matrix}
&
\mathbf{I}_{n-k}\end{array}\right]\\
&=\left[
\begin{array}{c|c}
\begin{matrix}
p_{0,0}& \dots & p_{0,k-2} & p_{0,k-1}\\
p_{1,0}& \dots & p_{1,k-2} & p_{1,k-1}\\
\vdots & \ddots & \vdots & \vdots\\
p_{n-1,0}& \dots & p_{n-1,k-2} & p_{n-1,k-1}\\
\end{matrix}
&
\begin{matrix}
1 & 0 & \dots & 0\\
0 & 1 & \dots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0& 0 & \dots & 1\\
\end{matrix}\end{array}\right]\\
\mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid}
\end{align*}
$$
#### Procedure to find parity check matrix from list of codewords
1. From the number of codewords, find $k=\log_2(N)$
2. Partition codewords into $k$ information bits and remaining bits into $n-k$ parity bits. The information bits should be a simple counter (?).
3. Express parity bits as a linear combination of information bits
4. Put coefficients into $\textbf{P}$ matrix and find $\textbf{H}$
Example:
$$
\begin{array}{cccc}
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*}
\begin{aligned}
x_3 &= x_1\oplus x_2\\
x_4 &= x_1\oplus x_2\\
x_5 &= x_2\\
\end{aligned}
\implies\textbf{P}&=
\begin{array}{c|cc}
& x_1 & x_2 \\\hline
x_3&1&1&\\
x_4&1&1&\\
x_5&0&1&\\
\end{array}\\
\textbf{H}&=\left[
\begin{array}{c|c}
\begin{matrix}
1&1\\
1&1\\
0&1\\
\end{matrix}
&
\begin{matrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{matrix}\end{array}\right]
\end{align*}
$$
#### Error detection and correction
**Detection** of $s$ errors: $d_\text{min}\geq s+1$
**Correction** of $u$ errors: $d_\text{min}\geq 2u+1$
## CHECKLIST
- 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