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# Idiot's guide to ELEC4402 communication systems
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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.
## [Download PDF 📄](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 whatever GitHub renders.
### License and information
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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)
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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
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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.
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- To be honest, you select (relevant) questions from these for class test study as well.
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< 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.
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- Try to practice questions from the exam papers on OneSearch mentioned above before class test, since you don't get past questions for the tests.
### Final exam 2024
As quoted from the lectures:
- No short questions
- 10 calculation questions
- From exam front page: Any printed or written material (e.g. dictionaries, textbooks, statutes, cases, legislation, annotations, lecture notes, tutorials with solutions, notes) except UWA Library books. Photocopied or printed eBook of a textbook is permitted.
- So yes you can still bring as many papers as you want to the final, unlike in signals and systems/signal processing where the final only allowed two pages.
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**[https://www.petertanner.dev/posts/Idiots-guide-to-ELEC4402-Communications-Systems/](https://www.petertanner.dev/posts/Idiots-guide-to-ELEC4402-Communications-Systems/)**
**Notes are open-source and licensed under the [GNU GPL-3.0 ](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/blob/master/COPYING.txt ). Suggest any corrections or changes on [GitHub ](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems ).**
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## Fourier transform identities and properties
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| Time domain $x(t)$ | Frequency domain $X(f)$ |
| ---------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \text{sinc}(fT)$ |
| $\text{sinc}(2Wt)$ | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ |
| $\exp(-at)u(t),\quad a>0$ | $\frac{1}{a + j2\pi f}$ |
| $\exp(-a\lvert t \rvert),\quad a>0$ | $\frac{2a}{a^2 + (2\pi f)^2}$ |
| $\exp(-\pi t^2)$ | $\exp(-\pi f^2)$ |
| $1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T \quad \text { tri }( t / T )$ | $ T \text { sinc }^ 2 ( fT )$ |
| $\delta(t)$ | $1$ |
| $1$ | $\delta(f)$ |
| $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ |
| $\exp(j2\pi f_c t)$ | $\delta(f - f_c)$ |
| $\cos(2\pi f_c t)$ | $\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$ |
| $\cos(2\pi f_c t+\theta)$ | $\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}$ |
| $\sin(2\pi f_c t)$ | $\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$ |
| $\sin(2\pi f_c t+\theta)$ | $\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]$ |
| $\text{sgn}(t)$ | $\frac{1}{j\pi f}$ |
| $\frac{1}{\pi t}$ | $-j \text{sgn}(f)$ |
| $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ |
| $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ |
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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 $g(t)$ |
```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}\\
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\text{tri}(t/T) & = \begin{cases} 1 - \frac{|t|}{T}, & \lvert t\rvert < T \\ 0 , & \lvert t \rvert \geq T \end { cases }= \frac { 1 }{ T } \Pi ( t / T )* \Pi ( t / T )& \text { Triangle Function } \\
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g(t)*h(t)=(g*h)(t)& =\int_\infty^\infty g(\tau)h(t-\tau)d\tau& \text{Convolution}\\
\end{align*}
```
<!-- MATH END -->
### 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 $T_s$:
```math
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
```
<!-- MATH END -->
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*}
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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 $T_s$}
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\end{align*}
```
<!-- MATH END -->
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### Shape functions
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### Random processes examples
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$$
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\begin{align*}
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& \text{Example: separate RV from expression}\\
X(t) & = A\cos(2\pi f_c t)\quad A\thicksim \mathcal{N}(\mu=5,\sigma^2=1)\\
\implies E[X(t)] & = E[A\cos(2\pi f_c t)] = E[A]\cos(2\pi f_c t) = 5\cos(2\pi f_c t)\\
& \text{Example: random phase}\\
X(t) & = B\cos(2\pi f_c t+\theta)\quad \theta\thicksim \mathcal{U}(0,2\pi)\\
\implies E[X(t)] & = E[B\cos(2\pi f_c t+\theta)] = B\int_0^{2\pi}\underbrace{\frac{1}{2\pi}}_{\text{uniform}}\cos(2\pi f_c t+\theta)d\theta=0
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\end{align*}
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$$
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### Wide sense stationary (WSS)
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Two conditions for WSS:
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| Constant mean | Autocorrelation only dependent on time difference |
| ---------------------------------- | ------------------------------------------------- |
| $\mu_X(t) = \mu_X\text{ Constant}$ | $R_{XX}(t_1,t_2)=R_X(t_1-t_2)=R_X(\tau)$ |
| $\mu_X(t)=E[X(t)]$ | $E[X(t_1)X(t_2)]=E[X(t)X(t+\tau)]$ |
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### 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*}
```
<!-- MATH END -->
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| 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 | $\lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau)$ | $\lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0$ |
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Note: **A WSS random process needs to be both ergodic in mean and autocorrelation to be considered an ergodic process**
### 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*}
```
<!-- MATH END -->
### 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)\\
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*}
```
<!-- MATH END -->
```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) \\
\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*}
```
<!-- MATH END -->
```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) \\
\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*}
```
<!-- MATH END -->
## 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
```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)}]\\
& =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*}
```
<!-- MATH END -->
### 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*}
```
<!-- MATH END -->
## AM
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### Conventional AM modulation (CAM)
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```math
\begin{align*}
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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]\quad\text{CAM signal}\\
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& \text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\
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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.}\\
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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 $m(t)$)}\\
P_c & =\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\
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P_s & =\frac{1}{4}{m_a}^2{A_c}^2\quad\text{Signal power, \textbf{total} of all 4 sideband power, \textbf{single-tone} case}\\
\eta& =\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_s}{P_s+P_c}=\frac{P_s}{P_x}\quad\text{Power efficiency}\\
B_T& =2f_m=2B\\
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\end{align*}
```
<!-- MATH END -->
$B_T$: Signal bandwidth
$B$: Bandwidth of modulating wave
Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
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### Double sideband suppressed carrier (DSB-SC)
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```math
\begin{align*}
x_\text{DSB}(t) & = A_c \cos{(2\pi f_c t)} m(t)\\
B_T& =2f_m=2B
\end{align*}
```
<!-- MATH END -->
## 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(\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)=f_c+k_f m(t)=f_c+\Delta f_\text{max}\hat m(t)\quad\text{Instantaneous frequency}\\
\Delta f_\text{max}& =\max_t|f_i(t)-f_c|=k_f \max_t |m(t)|\quad\text{Maximum frequency deviation}\\
\Delta f_\text{max}& =k_f A_m\quad\text{Maximum frequency deviation (sinusoidal)}\\
\beta& =\frac{\Delta f_\text{max}}{f_m}\quad\text{Modulation index}\\
D& =\frac{\Delta f_\text{max}}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}\\
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\end{align*}
```
<!-- MATH END -->
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### Bessel function
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```math
\begin{align*}
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J_n(\beta)& =\begin{cases}
J_{-n}(\beta) & \text{$n$ is even}\\
-J_{-n}(\beta) & \text{$n$ is odd}
\end{cases}\\
1& =\sum_{n\in\mathbb{Z}}{J_n}^2(\beta)& \text{Conservation of power}\\
\end{align*}
```
<!-- MATH END -->
### Bessel form of FM signal
```math
\begin{align*}
s(t) & = A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\\
\Longleftrightarrow s(t) & = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
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\end{align*}
```
<!-- MATH END -->
### FM signal power
```math
\begin{align*}
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P_\text{av}& =\frac{ {A_c}^2}{2}& \text{Av. power of full signal}\\
P_\text{i}& =\frac{ {A_c}^2|{J_\text{i}}(\beta)|^2}{2}& \text{Av. power of band $i$}\\
i=0& \implies f_c+0f_m& \text{Middle band}\\
i=1& \implies f_c+1f_m& \text{1st sideband}\\
i=-1& \implies f_c-1f_m& \text{-1st sideband}\\
& \dots\\
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\end{align*}
```
<!-- MATH END -->
### Carson's rule to find $B$ (98% power bandwidth rule)
```math
\begin{align*}
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B & = 2(\beta + 1)f_m\\
B & = 2(\Delta f_\text{max}+f_m)\\
B & = 2(D+1)W_m\\
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B & = \begin{cases}
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2(\Delta f_\text{max}+f_m)=2(\Delta f_\text{max}+W_m) & \text{FM, sinusoidal message}\\
2(\Delta\phi_\text{max} + 1)f_m=2(\Delta \phi_\text{max}+1)W_m & \text{PM, sinusoidal message}
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\end{cases}\\\\
& D< 1 , \beta < 1 \implies \text { Narrowband } \quad D > 1,\beta>1\implies \text{Wideband}
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\end{align*}
```
<!-- MATH END -->
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### Complex envelope of a FM signal
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```math
\begin{align*}
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s(t)& =A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t))\\
\Longleftrightarrow \tilde{s}(t) & = A_c\exp(j\beta\sin(2\pi f_m t))\\
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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*}
```
<!-- MATH END -->
## 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)}\\
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)}\\
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 }\\
E_x& =\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\quad\text{Parseval's theorem}\\
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R_x(\tau) & = \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}\\
P_x & = R_x(0)\quad\text{Average power of WSS process $x(t)$}\\
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\end{align*}
```
<!-- MATH END -->
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### 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}\\
\end{align*}
```
<!-- MATH END -->
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## Noise performance
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<!-- Coherent detection system.
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```math
\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*}
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``` -->
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<!-- MATH END -->
Use formualas from previous section, [Power, energy and autocorrelation ](#power-energy-and-autocorrelation ).\
Use these formulas in particular:
```math
\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*}
```
<!-- MATH END -->
```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*}
```
<!-- MATH END -->
## Sampling
```math
\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)\\
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X_s(f)& =f_s X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=f_s X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
\implies X_s(f)& =\sum_{n\in\mathbb{Z}}f_s X\left(f-n f_s\right)\quad\text{Sampling (FT)}\\
B& >\frac{1}{2}f_s\implies 2B>f_s\rightarrow\text{Aliasing}\\
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\end{align*}
```
<!-- MATH END -->
### 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 $T_s$:
```math
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
```
<!-- MATH END -->
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& =\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*}
```
<!-- MATH END -->
<!-- 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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```math
\text{Nyquist rate} = 2B\quad\text{Nyquist interval}=\frac{1}{2B}
```
<!-- MATH END -->
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### Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to `images/sampling.png`)
Cannot add directly due to copyright!
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**TODO: Make an open source replacement for this diagram [Send a PR to GitHub ](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/issues/new ).**
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< img src = "copyrighted_images/sampling.png" alt = "sampling" class = "copyrighted" >
## Quantizer
```math
\begin{align*}
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\Delta & = \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\quad\text{Quantizer step size $\Delta$}\\
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\end{align*}
```
<!-- MATH END -->
### Quantization noise
```math
\begin{align*}
e & := y-x\quad\text{Quantization error}\\
\mu_E & = E[E] = 0\quad\text{Zero mean}\\
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P_E& ={\sigma_E}^2=\frac{\Delta^2}{12}=2^{-2m}V^2/3\quad\text{Uniformly distributed error}\\
\text{SQNR}& =\frac{\text{Signal power}}{\text{Quantization noise power}}=\frac{P_x}{P_E}\\
\text{SQNR(dB)}& =10\log_{10}(\text{SQNR})\\
m\to m+A\text{ bits}& \implies \text{newSQNR(dB)}=\text{SQNR(dB)}+6A\text{ dB}
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\end{align*}
```
<!-- MATH END -->
### Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to `images/quantizer.png`)
Cannot add directly due to copyright!
2024-11-05 02:18:47 +08:00
**TODO: Make an open source replacement for this diagram [Send a PR to GitHub ](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/issues/new ).**
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< img src = "copyrighted_images/quantizer.png" alt = "quantizer" class = "copyrighted" >
## Line codes
```math
\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*}
```
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**TODO: Someone please make plots of the PSD for all line code types in Mathematica or Python! [Send a PR to GitHub ](https://github.com/peter-tanner/IDIOTS-GUIDE-TO-ELEC4402-communication-systems/issues/new ).**
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<!-- MATH END -->
## Modulation and basis functions
### BASK
#### 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*}
```
<!-- MATH END -->
#### Symbol mapping
```math
b_n:\{1,0\}\to a_n:\{1,0\}
```
<!-- MATH END -->
#### 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*}
```
<!-- MATH END -->
Distance is $d=\sqrt{2E_b}$
### BPSK
#### 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*}
```
<!-- MATH END -->
#### Symbol mapping
```math
b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}
```
<!-- MATH END -->
#### 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*}
```
<!-- MATH END -->
Distance is $d=2\sqrt{E_b}$
### QPSK ($M=4$ PSK)
#### 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*}
```
<!-- MATH END -->
### 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*}
```
<!-- MATH END -->
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*}
```
<!-- MATH END -->
#### 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*}
```
<!-- MATH END -->
### 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 $T=2T_b$
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| | |
| ----------------------- | ---------------------------------- |
| $b_n$ |  |
| $I(t)$ (Odd, 1st bits) |  |
| $Q(t)$ (Even, 2nd bits) |  |
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<!-- ADJUST ACCORDING TO PDF OUTPUT -->
< div style = "page-break-after: always;" > < / div >
## Matched filter
### 1. Filter function
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 $x=T/2$.
```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*}
```
<!-- MATH END -->
### 2. Bit error rate of matched filter
Bit error rate (BER) from matched filter outputs and filter output noise
```math
\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& =P_\text{av}T=P_\text{av}/R_b\quad\text{Energy per bit}\\
P_\text{av}& =E_b/T=E_bR_b\quad\text{Average power}\\
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*}
```
<!-- MATH END -->
< div style = "page-break-after: always;" > < / div >
## Value tables for $\text{erf}(x)$ and $Q(x)$
### $Q(x)$ function
\*\*The value of $\text{erf}(3.30)$ should be $\approx0.999997$ instead, but this value is quoted in the formula table.
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### $Q(x)$ fast reference
Using identity.
| $x$ | $Q(x)$ |
| ----------- | --------- |
| $\sqrt{2}$ | $0.07865$ |
| $2\sqrt{2}$ | $0.00234$ |
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< div style = "page-break-after: always;" > < / div >
### Receiver output shit
```math
\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*}
```
<!-- MATH END -->
<!--
```math
\begin{align*}
G_x(f)
\end{align*}
``` -->
## ISI, channel model
### Raised cosine (RC) pulse
```math
0\leq\alpha\leq1
```
<!-- MATH END -->
⚠ 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
<!-- P_r(kT)= \begin{cases}
1 & k=0\\
0 & k\neq0
\end{cases} -->
```math
\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*}
```
<!-- MATH END -->
### 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*}
```
<!-- MATH END -->
### 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 $R_b$.
| Linear modulation | Half |
| --------------------------------------------------- | -------------------------------------------------- |
| BPSK, QPSK, $M$-PSK, $M$-QAM, ASK, FSK | $M$-PAM, PAM |
| RZ unipolar, Manchester | NRZ Unipolar, NRZ Polar, Bipolar RZ |
| $W=B_\text{\color{green}abs-abs}$ | $W=B_\text{\color{green}abs}$ |
| $W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ | $W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ |
| $D=\frac{W\text{ symbol/s}}{1+\alpha}$ | $D=\frac{2W\text{ symbol/s}}{1+\alpha}$ |
```math
\begin{align*}
R_b\text{ bit/s}& =(D\text{ symbol/s})\times(k\text{ bit/symbol})\\
M\text{ symbol/set}& =2^k\\
T\text{ s/symbol}& =1/(D\text{ symbol/s})\\
E_b& =PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
\end{align*}
```
<!-- MATH END -->
### 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
<!-- ADJUST ACCORDING TO PDF OUTPUT -->
< div style = "page-break-after: always;" > < / div >
## Information theory
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### Stats
```math
\begin{align*}
P(A|B) & = \frac{P(B|A)P(A)}{P(B)} = \frac{P(A,B)}{P(B)}\\
\end{align*}
```
<!-- MATH END -->
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### 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))\\
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*}
```
<!-- MATH END -->
Entropy is **maximized** when all have an equal probability.
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### Transition probability diagram
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Example for binary erasure channel where $X$ is input and $Y$ is output:
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Equivalent to:
$$
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\begin{align*}
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P[Y=0|X=0] & = 1-p\\
P[Y=e|X=0] & = p\\
P[Y=1|X=1] & = 1-p\\
P[Y=e|X=1] & = p\\
P[X=0|Y=0] & = 0\quad\text{Note the direction}\\
P[Y=0] & = P[Y=0|X=0] P[X=0]
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\end{align*}
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$$
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### 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*}
```
<!-- MATH END -->
### Channel model
Vertical, $x$: input\
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Horizontal, $y$: output\
Remember $\mathbf{P}$ is a matrix where each element is $P(y_j|x_i)$
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```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 END -->
```math
\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}
```
<!-- MATH END -->
Input has probability distribution $p_X(a_i)=P(X=a_i)$
Channel maps alphabet $`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$
Output has probability 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*}
```
<!-- MATH END -->
#### 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}\\
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& \text{3. Multiply each row in $\textbf{P}$ by $p_X(a_i)$ since $p_{XY}(a_i,b_i)=P(b_i|a_i)P(a_i)$}\\
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& \text{4. Find $H(x,y)$ using each element from (3.)}\\
& \text{5. Find }H(x|y)=H(x,y)-H(y)\\
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& \text{6. Find }I(x;y)=H(x)-H(x|y)\\
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\end{align*}
```
<!-- MATH END -->
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Example of step **3** :
$$
\mathbf{P_{XY}}=\left[\begin{matrix}
P(y_1|x_1) P(x_1) & P(y_2|x_1) P(x_1) & \dots\\
P(y_1|x_2) P(x_2) & P(y_2|x_2) P(x_2) & \dots\\
\vdots & \vdots & \ddots
\end{matrix}\right]
$$
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### 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
```math
\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}\\
R_b & < C \text { for error-free transmission }
\end{align*}
```
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**Note that the channel capacity is realized when the channel inputs are uniformly distributed** (i.e. $P(x_1)=P(x_2)=\dots=P(x_N)=\frac{1}{N}$)
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<!-- MATH END -->
#### Channel capacity of an AWGN channel
```math
y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
```
<!-- MATH END -->
```math
C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)
```
<!-- MATH END -->
#### Channel capacity of a bandwidth limited AWGN channel
```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})\\
\text{SNR}& =P_s/(N_0 W)
\end{align*}
```
<!-- MATH END -->
#### Shannon limit
```math
\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*}
```
<!-- MATH END -->
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## Channel code
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.
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 $\mathbf{G}$.
```math
\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*}
```
<!-- MATH END -->
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}
\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*}
```
<!-- MATH END -->
### 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}
\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*}
```
<!-- MATH END -->
#### 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[
\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*}
```
<!-- MATH END -->
#### 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:
```math
\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}
```
<!-- MATH END -->
Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
```math
\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*}
```
<!-- MATH END -->
#### 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
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- $2W$ for rectangle functions
- Scale sampled spectrum by $f_s$
- $2f_c$ for spectrum after IF mixing.
- Square transfer function for PSD $G_y(f)=|H(f)|^\mathbf{2}G_x(f)$
- Square besselJ function for FM power $|J_n(\beta)|^\mathbf{2}$
- Bandwidth: only consider positive frequencies (so the bandwidth of an AM signal will be the range from the lowest to greatest sideband frequency. For a rectangular function, it will be from 0 to W).
- TODO: add more items to check
- TODO: add some graphics for these checklist items
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