Idiot's guide to ELEC4402 communication systems

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. I accept pull requests for fixes or suggestions but the content must not be copyrighted under a non-GPL compatible license.

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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"
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.

Listing of examination papers on OneSearch

Communications Systems ELEC3302 Examination paper [2008 Supplementary]
Communications Systems ELEC4402 Examination paper [2014 Semester 2]
Communications Systems ELEC3302 Examination paper [2014 Semester 2]
Communications Systems ELEC3302 Examination paper [2008 Semester 1]
Digital Communications and Networking ENGT4301 Examination paper [2005 Supplementary]
Digital Communications and Networking ELEC4301 Examination paper [2009 Supplementary]

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.

Fourier transform identities and properties

Time domain $x(t)$	Frequency domain $X(f)$
$\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$	$T \text{sinc}(fT)$
$\text{sinc}(2Wt)$	$\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$
$\exp(-at)u(t),\quad a>0$	$\frac{1}{a + j2\pi f}$
$\exp(-a\lvert t \rvert),\quad a>0$	$\frac{2a}{a^2 + (2\pi f)^2}$
$\exp(-\pi t^2)$	$\exp(-\pi f^2)$
$1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T\quad\text{tri}(t/T)$	$T \text{sinc}^2(fT)$
$\delta(t)$	$1$
$1$	$\delta(f)$
$\delta(t - t_0)$	$\exp(-j2\pi f t_0)$
$\exp(j2\pi f_c t)$	$\delta(f - f_c)$
$\cos(2\pi f_c t)$	$\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$
$\cos(2\pi f_c t+\theta)$	$\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}$
$\sin(2\pi f_c t)$	$\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$
$\sin(2\pi f_c t+\theta)$	$\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]$
$\text{sgn}(t)$	$\frac{1}{j\pi f}$
$\frac{1}{\pi t}$	$-j \text{sgn}(f)$
$u(t)$	$\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$
$\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$	$\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$

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}\\ \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}\\ 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

Transform a continuous time-periodic signal

x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)

with period

T_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*}

Shape functions

Random processes examples

\begin{align*} &\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 \end{align*}

Wide sense stationary (WSS)

Ergodicity

Constant mean	Autocorrelation only dependent on time difference
$\mu_X(t) = \mu_X\text{ Constant}$	$R_{XX}(t_1,t_2)=R_X(t_1-t_2)=R_X(\tau)$
$\mu_X(t)=E[X(t)]$	$E[X(t_1)X(t_2)]=E[X(t)X(t+\tau)]$

\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*}

Conventional AM modulation (CAM)

\begin{align*} 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}\\ &\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\ 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 $m(t)$)}\\ P_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\ 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\\ \end{align*}

Double sideband suppressed carrier (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)}\\ 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}\\ 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)}\\ \end{align*}

Bessel function

\begin{align*} 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*}

Bessel form of FM signal

\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] \end{align*}

FM signal power

\begin{align*} 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\\ \end{align*}

Carson's rule to find

B

(98% power bandwidth rule)

\begin{align*} B &= 2(\beta + 1)f_m\\ B &= 2(\Delta f_\text{max}+f_m)\\ B &= 2(D+1)W_m\\ B &= \begin{cases} 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} \end{cases}\\\\ & D<1,\beta<1 \implies \text{Narrowband}\quad D>1,\beta>1\implies \text{Wideband} \end{align*}

Complex envelope of a FM signal

\begin{align*} 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))\\ 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*}

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)}\\ 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}\\ 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 $x(t)$}\\ \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}\\ \end{align*}

Noise performance

\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)&=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}\\ \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.

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.

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)

Transform a continuous time-periodic signal

x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)

with period

T_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*}

Nyquist criterion for zero-ISI

Do not transmit more than

2B

samples per second over a channel of

B

bandwidth.

Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to images/sampling.png)

Cannot add directly due to copyright! TODO: Make an open source replacement for this diagram Send a PR to GitHub.

Quantizer

\begin{align*} \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\quad\text{Quantizer step size $\Delta$}\\ \end{align*}

Quantization noise

\begin{align*} e &:= y-x\quad\text{Quantization error}\\ \mu_E &= E[E] = 0\quad\text{Zero mean}\\ 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} \end{align*}

Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to images/quantizer.png)

Cannot add directly due to copyright! TODO: Make an open source replacement for this diagram Send a PR to GitHub.

Line codes

\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*}

TODO: Someone please make plots of the PSD for all line code types in Mathematica or Python! Send a PR to GitHub.

Modulation and basis functions

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

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*}

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

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*}

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

Matched filter

1. Filter function


$b_n$
$I(t)$ (Odd, 1st bits)
$Q(t)$ (Even, 2nd bits)

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*}

2. Bit error rate of matched filter

\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*}

Value tables for

\text{erf}(x)

and

Q(x)

Q(x)

function

You should use

\text{erf}

function table instead in exams using the identity

Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)

. Use this for validation.

$x$	$Q(x)$	$x$	$Q(x)$	$x$	$Q(x)$	$x$	$Q(x)$
$0.00$	$0.5$	$2.30$	$0.010724$	$4.55$	$2.6823 \times 10^{-6}$	$6.80$	$5.231 \times 10^{-12}$
$0.05$	$0.48006$	$2.35$	$0.0093867$	$4.60$	$2.1125 \times 10^{-6}$	$6.85$	$3.6925 \times 10^{-12}$
$0.10$	$0.46017$	$2.40$	$0.0081975$	$4.65$	$1.6597 \times 10^{-6}$	$6.90$	$2.6001 \times 10^{-12}$
$0.15$	$0.44038$	$2.45$	$0.0071428$	$4.70$	$1.3008 \times 10^{-6}$	$6.95$	$1.8264 \times 10^{-12}$
$0.20$	$0.42074$	$2.50$	$0.0062097$	$4.75$	$1.0171 \times 10^{-6}$	$7.00$	$1.2798 \times 10^{-12}$
$0.25$	$0.40129$	$2.55$	$0.0053861$	$4.80$	$7.9333 \times 10^{-7}$	$7.05$	$8.9459 \times 10^{-13}$
$0.30$	$0.38209$	$2.60$	$0.0046612$	$4.85$	$6.1731 \times 10^{-7}$	$7.10$	$6.2378 \times 10^{-13}$
$0.35$	$0.36317$	$2.65$	$0.0040246$	$4.90$	$4.7918 \times 10^{-7}$	$7.15$	$4.3389 \times 10^{-13}$
$0.40$	$0.34458$	$2.70$	$0.003467$	$4.95$	$3.7107 \times 10^{-7}$	$7.20$	$3.0106 \times 10^{-13}$
$0.45$	$0.32636$	$2.75$	$0.0029798$	$5.00$	$2.8665 \times 10^{-7}$	$7.25$	$2.0839 \times 10^{-13}$
$0.50$	$0.30854$	$2.80$	$0.0025551$	$5.05$	$2.2091 \times 10^{-7}$	$7.30$	$1.4388 \times 10^{-13}$
$0.55$	$0.29116$	$2.85$	$0.002186$	$5.10$	$1.6983 \times 10^{-7}$	$7.35$	$9.9103 \times 10^{-14}$
$0.60$	$0.27425$	$2.90$	$0.0018658$	$5.15$	$1.3024 \times 10^{-7}$	$7.40$	$6.8092 \times 10^{-14}$
$0.65$	$0.25785$	$2.95$	$0.0015889$	$5.20$	$9.9644 \times 10^{-8}$	$7.45$	$4.667 \times 10^{-14}$
$0.70$	$0.24196$	$3.00$	$0.0013499$	$5.25$	$7.605 \times 10^{-8}$	$7.50$	$3.1909 \times 10^{-14}$
$0.75$	$0.22663$	$3.05$	$0.0011442$	$5.30$	$5.7901 \times 10^{-8}$	$7.55$	$2.1763 \times 10^{-14}$
$0.80$	$0.21186$	$3.10$	$0.0009676$	$5.35$	$4.3977 \times 10^{-8}$	$7.60$	$1.4807 \times 10^{-14}$
$0.85$	$0.19766$	$3.15$	$0.00081635$	$5.40$	$3.332 \times 10^{-8}$	$7.65$	$1.0049 \times 10^{-14}$
$0.90$	$0.18406$	$3.20$	$0.00068714$	$5.45$	$2.5185 \times 10^{-8}$	$7.70$	$6.8033 \times 10^{-15}$
$0.95$	$0.17106$	$3.25$	$0.00057703$	$5.50$	$1.899 \times 10^{-8}$	$7.75$	$4.5946 \times 10^{-15}$
$1.00$	$0.15866$	$3.30$	$0.00048342$	$5.55$	$1.4283 \times 10^{-8}$	$7.80$	$3.0954 \times 10^{-15}$
$1.05$	$0.14686$	$3.35$	$0.00040406$	$5.60$	$1.0718 \times 10^{-8}$	$7.85$	$2.0802 \times 10^{-15}$
$1.10$	$0.13567$	$3.40$	$0.00033693$	$5.65$	$8.0224 \times 10^{-9}$	$7.90$	$1.3945 \times 10^{-15}$
$1.15$	$0.12507$	$3.45$	$0.00028029$	$5.70$	$5.9904 \times 10^{-3}$	$7.95$	$9.3256 \times 10^{-16}$
$1.20$	$0.11507$	$3.50$	$0.00023263$	$5.75$	$4.4622 \times 10^{-9}$	$8.00$	$6.221 \times 10^{-16}$
$1.25$	$0.10565$	$3.55$	$0.00019262$	$5.80$	$3.3157 \times 10^{-9}$	$8.05$	$4.1397 \times 10^{-16}$
$1.30$	$0.0968$	$3.60$	$0.00015911$	$5.85$	$2.4579 \times 10^{-9}$	$8.10$	$2.748 \times 10^{-16}$
$1.35$	$0.088508$	$3.65$	$0.00013112$	$5.90$	$1.8175 \times 10^{-9}$	$8.15$	$1.8196 \times 10^{-16}$
$1.40$	$0.080757$	$3.70$	$0.0001078$	$5.95$	$1.3407 \times 10^{-9}$	$8.20$	$1.2019 \times 10^{-16}$
$1.45$	$0.073529$	$3.75$	$8.8417 \times 10^{-5}$	$6.00$	$9.8659 \times 10^{-10}$	$8.25$	$7.9197 \times 10^{-17}$
$1.50$	$0.066807$	$3.80$	$7.2348 \times 10^{-5}$	$6.05$	$7.2423 \times 10^{-10}$	$8.30$	$5.2056 \times 10^{-17}$
$1.55$	$0.060571$	$3.85$	$5.9059 \times 10^{-5}$	$6.10$	$5.3034 \times 10^{-10}$	$8.35$	$3.4131 \times 10^{-17}$
$1.60$	$0.054799$	$3.90$	$4.8096 \times 10^{-5}$	$6.15$	$3.8741 \times 10^{-10}$	$8.40$	$2.2324 \times 10^{-17}$
$1.65$	$0.049471$	$3.95$	$3.9076 \times 10^{-5}$	$6.20$	$2.8232 \times 10^{-10}$	$8.45$	$1.4565 \times 10^{-17}$
$1.70$	$0.044565$	$4.00$	$3.1671 \times 10^{-5}$	$6.25$	$2.0523 \times 10^{-10}$	$8.50$	$9.4795 \times 10^{-18}$
$1.75$	$0.040059$	$4.05$	$2.5609 \times 10^{-5}$	$6.30$	$1.4882 \times 10^{-10}$	$8.55$	$6.1544 \times 10^{-18}$
$1.80$	$0.03593$	$4.10$	$2.0658 \times 10^{-5}$	$6.35$	$1.0766 \times 10^{-10}$	$8.60$	$3.9858 \times 10^{-18}$
$1.85$	$0.032157$	$4.15$	$1.6624 \times 10^{-5}$	$6.40$	$7.7688 \times 10^{-11}$	$8.65$	$2.575 \times 10^{-18}$
$1.90$	$0.028717$	$4.20$	$1.3346 \times 10^{-5}$	$6.45$	$5.5925 \times 10^{-11}$	$8.70$	$1.6594 \times 10^{-18}$
$1.95$	$0.025588$	$4.25$	$1.0689 \times 10^{-5}$	$6.50$	$4.016 \times 10^{-11}$	$8.75$	$1.0668 \times 10^{-18}$
$2.00$	$0.02275$	$4.30$	$8.5399 \times 10^{-6}$	$6.55$	$2.8769 \times 10^{-11}$	$8.80$	$6.8408 \times 10^{-19}$
$2.05$	$0.020182$	$4.35$	$6.8069 \times 10^{-6}$	$6.60$	$2.0558 \times 10^{-11}$	$8.85$	$4.376 \times 10^{-19}$
$2.10$	$0.017864$	$4.40$	$5.4125 \times 10^{-6}$	$6.65$	$1.4655 \times 10^{-11}$	$8.90$	$2.7923 \times 10^{-19}$
$2.15$	$0.015778$	$4.45$	$4.2935 \times 10^{-6}$	$6.70$	$1.0421 \times 10^{-11}$	$8.95$	$1.7774 \times 10^{-19}$
$2.20$	$0.013903$	$4.50$	$3.3977 \times 10^{-6}$	$6.75$	$7.3923 \times 10^{-12}$	$9.00$	$1.1286 \times 10^{-19}$
$2.25$	$0.012224$

\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.

Q(x)

fast reference

Receiver output shit

$x$	$Q(x)$
$\sqrt{2}$	$0.07865$
$2\sqrt{2}$	$0.00234$

\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*}

ISI, channel model

Raised cosine (RC) pulse

⚠ 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

\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*}

Bandwidth

W

and bit error rate of modulation schemes

Linear modulation	Half
BPSK, QPSK, $M$ -PSK, $M$ -QAM, ASK, FSK	$M$ -PAM, PAM
RZ unipolar, Manchester	NRZ Unipolar, NRZ Polar, Bipolar RZ
$W=B_\text{\color{green}abs-abs}$	$W=B_\text{\color{green}abs}$
$W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$	$W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$
$D=\frac{W\text{ symbol/s}}{1+\alpha}$	$D=\frac{2W\text{ symbol/s}}{1+\alpha}$

\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*}

Table of bandpass signalling and BER

PSD of modulated signals

Symbol error probability

Information theory

Stats

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	$2M R_b / \log_2 M$			$\frac{M - 1}{2 \log_2 M} 0.5\exp({-(\log_2 M) E_b / 2N_0})$

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	$\color{red}\frac{\|V(f)\|^2}{2}\sum_{l=-\infty}^\infty R(l)\exp(-j2\pi l f T)\quad\text{What??}$

\begin{align*} P(A|B) &= \frac{P(B|A)P(A)}{P(B)} = \frac{P(A,B)}{P(B)}\\ \end{align*}

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*}

Transition probability diagram

\begin{align*} 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] \end{align*}

Mutual information

Channel model

Vertical,

x

: input
Horizontal,

y

: output
Remember

\mathbf{P}

is a matrix where each element is

P(y_j|x_i)

\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}

\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 $p_{XY}(a_i,b_i)=P(b_i|a_i)P(a_i)$}\\ &\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(x;y)=H(x)-H(x|y)\\ \end{align*}

\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]

Channel types

Channel capacity of weakly symmetric channel

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

\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*}

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}

)

Channel capacity of an AWGN channel

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)\\ 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

Linear block code

Code generation


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, $d_\text{min}=w_\text{min}$

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}

\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

\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*}

Idiot's guide to ELEC4402 communication systems

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Fourier transform identities and properties

Fourier transform of continuous time periodic signal

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Wide sense stationary (WSS)

Ergodicity

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Other trig

IQ/Complex envelope

Convert complex envelope representation to time-domain representation of signal

For transfer function

AM

Conventional AM modulation (CAM)

Double sideband suppressed carrier (DSB-SC)

FM/PM

Bessel function

Bessel form of FM signal

FM signal power

Carson's rule to find BBB (98% power bandwidth rule)

Complex envelope of a FM signal

Power, energy and autocorrelation

White noise

Noise performance

Sampling

Procedure to reconstruct sampled signal

Fourier transform of continuous time periodic signal (1)

Nyquist criterion for zero-ISI

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Quantizer

Quantization noise

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Line codes

Modulation and basis functions

BASK

Basis functions

Symbol mapping

2 possible waveforms

BPSK

Basis functions

Symbol mapping

2 possible waveforms

QPSK (M=4M=4M=4 PSK)

Basis functions

4 possible waveforms

Signal

Example of waveform

Matched filter

1. Filter function

2. Bit error rate of matched filter

Value tables for erf(x)\text{erf}(x)erf(x) and Q(x)Q(x)Q(x)

Q(x)Q(x)Q(x) function

erf(x)\text{erf}(x)erf(x) function

Q(x)Q(x)Q(x) fast reference

Receiver output shit

ISI, channel model

Raised cosine (RC) pulse

Nyquist criterion for zero ISI

Nomenclature

Bandwidth WWW and bit error rate of modulation schemes

Table of bandpass signalling and BER

PSD of modulated signals

Symbol error probability

Information theory

Stats

Entropy for discrete random variables

Transition probability diagram

Mutual information

Channel model

Fast procedure to calculate I(y;x)I(y;x)I(y;x)

Channel types

Channel capacity of weakly symmetric channel

Channel capacity of an AWGN channel

Channel capacity of a bandwidth limited AWGN channel

Shannon limit

Carson's rule to find $B$ (98% power bandwidth rule)

Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to `images/sampling.png`)

Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to `images/quantizer.png`)

QPSK ( $M=4$ PSK)

Value tables for $\text{erf}(x)$ and $Q(x)$

$Q(x)$ function

$\text{erf}(x)$ function

$Q(x)$ fast reference

Bandwidth $W$ and bit error rate of modulation schemes

Fast procedure to calculate $I(y;x)$

Parity check matrix $\mathbf{H}$