These notes should provide a printable summary of all the formulas and procedures required to solve questions in the exam and tests. This unit allows you to bring infinite physical notes (except books borrowed from the UWA library) to all tests and the final exam. You can’t rely on what material they provide in the test/exam, it is very minimal to say the least. Hope this helps.

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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.

Printable notes begins on next page (in PDF)

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

Required for some questions on sampling:

Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:

\[X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}\]

Calculate $C_n$ coefficient as follows from $x_p(t)$:

\[\begin{align*} C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*}\]

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)

Two conditions for WSS:

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

Ergodicity

\[\begin{align*} \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\ \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\ E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\ \end{align*}\]

Type	Normal	Mean square sense
ergodic in mean	$\lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X$	$\lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0$
ergodic in autocorrelation function	$\lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau)$	$\lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0$

Note: A WSS random process needs to be both ergodic in mean and autocorrelation to be considered an ergodic process

Other identities

\[\begin{align*} f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\ a(f*g) &= (af)*g \quad\text{Convolution associative}\\ \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right) \end{align*}\]

Other trig

\[\begin{align*} \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\ e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\ e^{-j\alpha}+e^{j\alpha}&=2 \cos(\alpha)\\ \cos(-A)&=\cos(A)\\ \sin(-A)&=-\sin(A)\\ \sin(A+\pi/2)&=\cos(A)\\ \sin(A-\pi/2)&=-\cos(A)\\ \cos(A-\pi/2)&=\sin(A)\\ \cos(A+\pi/2)&=-\sin(A)\\ \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\ \end{align*}\] \[\begin{align*} \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\ \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\ \cos(A)\cos(B) &= \frac{1}{2} (\cos (A-B)+\cos (A+B)) \\ \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\ \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\ \end{align*}\] \[\begin{align*} \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\ \sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\ \sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\ \end{align*}\]

IQ/Complex envelope

Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $e^{j\theta}$ form.

Convert complex envelope representation to time-domain representation of signal

\[\begin{align*} g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\ &=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\ &=A(t)\cos(2\pi f_c t+\phi(t))\\ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\ \phi(t)&\quad\text{Phase}\\ g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\ g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\ \end{align*}\]

For transfer function

\[\begin{align*} h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\ \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))} \end{align*}\]

AM

Conventional AM modulation (CAM)

\[\begin{align*} x(t)&=A_c\cos(2\pi f_c t)\left[1+k_a m(t)\right]=A_c\cos(2\pi f_c t)\left[1+m_a m(t)/A_c\right]\quad\text{CAM signal}\\ &\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\ m_a &= \frac{|\min_t(k_a m(t))|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\ m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\ m_a&=k_a A_m \quad\text{ (Symmetrical $m(t)$)}\\ P_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\ P_s &=\frac{1}{4}{m_a}^2{A_c}^2\quad\text{Signal power, \textbf{total} of all 4 sideband power, \textbf{single-tone} case}\\ \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_s}{P_s+P_c}=\frac{P_s}{P_x}\quad\text{Power efficiency}\\ B_T&=2f_m=2B\\ \end{align*}\]

$B_T$: Signal bandwidth $B$: Bandwidth of modulating wave

Overmodulation (resulting in phase reversals at crossing points): $m_a>1$

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 and magnitude spectrum (single tone)

\[\begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\\ \Longleftrightarrow s(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t] \end{align*}\]

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)

\[\begin{align*} B &= 2(\beta + 1)f_m\\ B &= 2(\Delta f_\text{max}+f_m)\\ B &= 2(D+1)W_m\\ B &= \begin{cases} 2(\Delta f_\text{max}+f_m)=2(\Delta f_\text{max}+W_m) & \text{FM, sinusoidal message}\\ 2(\Delta\phi_\text{max} + 1)f_m=2(\Delta \phi_\text{max}+1)W_m & \text{PM, sinusoidal message} \end{cases}\\ \end{align*}\]

Complex envelope of a FM signal

\[\begin{align*} s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t))\\ \Longleftrightarrow \tilde{s}(t) &= A_c\exp(j\beta\sin(2\pi f_m t))\\ s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\ \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t) \end{align*}\]

Band

Narrowband	Wideband
$D<1,\beta<1$	$D>1,\beta>1$

Power, energy and autocorrelation

\[\begin{align*} G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\ G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\ G_x(f)&=\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\text{ (PSD)}\\ G_x(f)&=\mathfrak{F}[R_x(\tau)]\text{ (WSS)}\\ P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df\quad\text{For zero mean}\\ P_x&={\sigma_x}^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\quad\text{For zero mean}\\ P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\ E_x&=\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\quad\text{Parseval's theorem}\\ R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}\\ P_x &= R_x(0)\quad\text{Average power of WSS process $x(t)$}\\ \end{align*}\]

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

Use formualas from previous section, Power, energy and autocorrelation.
Use these formulas in particular:

\[\begin{align*} G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_x(f)&=|H(f)|^2G_w(f)&\text{Note the square in $|H(f)|^2$}\\ P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df&\text{Often perform graphical integration}\\ \end{align*}\] \[\begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\ \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio} \end{align*}\]

Sampling

\[\begin{align*} t&=nT_s\\ T_s&=\frac{1}{f_s}\\ x_s(t)&=x(t)\delta_s(t)=x(t)\sum_{n\in\mathbb{Z}}\delta(t-nT_s)=\sum_{n\in\mathbb{Z}}x(nT_s)\delta(t-nT_s)\\ X_s(f)&=f_s X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=f_s X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\ \implies X_s(f)&=\sum_{n\in\mathbb{Z}}f_s X\left(f-n f_s\right)\quad\text{Sampling (FT)}\\ B&>\frac{1}{2}f_s\implies 2B>f_s\rightarrow\text{Aliasing}\\ \end{align*}\]

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.

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 $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:

\[X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}\]

Calculate $C_n$ coefficient as follows from $x_p(t)$:

\[\begin{align*} C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*}\]

Nyquist criterion for zero-ISI

Do not transmit more than $2B$ samples per second over a channel of $B$ bandwidth.

\[\text{Nyquist rate} = 2B\quad\text{Nyquist interval}=\frac{1}{2B}\]

Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to assets/img/2024-10-29-Idiots-guide-to-ELEC/sampling.png)

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 assets/img/2024-10-29-Idiots-guide-to-ELEC/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

\[b_n:\{1,0\}\to a_n:\{1,0\}\]

2 possible waveforms

\[\begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\ s_1(t)&=0\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$} \end{align*}\]

Distance is $d=\sqrt{2E_b}$

BPSK

Basis functions

\[\begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*}\]

Symbol mapping

\[b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}\]

2 possible waveforms

\[\begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\ s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$} \end{align*}\]

Distance is $d=2\sqrt{E_b}$

QPSK ($M=4$ PSK)

Basis functions

\[\begin{align*} T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\ \varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\ \varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\ \end{align*}\]

4 possible waveforms

\[\begin{align*} s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\ s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\ s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\ s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\ \end{align*}\]

Note on energy per symbol: Since $

s_i(t)

=A_c$, have to normalize distance as follows:

\[\begin{align*} s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ \end{align*}\]

Signal

\[\begin{align*} \text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\ I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\ Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\ x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)] \end{align*}\]

Example of waveform

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)"]

Remember that $T=2T_b$

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

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 $x=T/2$.

\[\begin{align*} h(t)&=s_1(T-t)-s_2(T-t)\\ h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\ s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\ n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output} \end{align*}\]

2. Bit error rate of matched filter

Bit error rate (BER) from matched filter outputs and filter output noise

\[\begin{align*} Q(x)&=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Leftrightarrow\text{erf}\left(\frac{x}{\sqrt{2}}\right)=1-2Q(x)\\ E_b&=d^2=\int_{-\infty}^\infty|s_1(t)-s_2(t)|^2dt\quad\text{Energy per bit/Distance}\\ T&=1/R_b\quad\text{$R_b$: Bitrate}\\ E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\ P(\text{W})&=10^{\frac{P(\text{dB})}{10}}\\ P_\text{RX}(W)&=P_\text{TX}(W)\cdot10^{\frac{P_\text{loss}(\text{dB})}{10}}\quad \text{$P_\text{loss}$ is expressed with negative sign e.g. "-130 dB"}\\ \text{BER}_\text{MatchedFilter}&=Q\left(\sqrt{\frac{d^2}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{2N_0}}\right)\\ \text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\ \text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\ \end{align*}\]

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.

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

Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems

$\text{erf}(x)$ function

\[Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}(\frac{x}{\sqrt{2}})\]

$x$	$\text{erf}(x)$	$x$	$\text{erf}(x)$	$x$	$\text{erf}(x)$
$0.00$	$0.00000$	$0.75$	$0.71116$	$1.50$	$0.96611$
$0.05$	$0.05637$	$0.80$	$0.74210$	$1.55$	$0.97162$
$0.10$	$0.11246$	$0.85$	$0.77067$	$1.60$	$0.97635$
$0.15$	$0.16800$	$0.90$	$0.79691$	$1.65$	$0.98038$
$0.20$	$0.22270$	$0.95$	$0.82089$	$1.70$	$0.98379$
$0.25$	$0.27633$	$1.00$	$0.84270$	$1.75$	$0.98667$
$0.30$	$0.32863$	$1.05$	$0.86244$	$1.80$	$0.98909$
$0.35$	$0.37938$	$1.10$	$0.88021$	$1.85$	$0.99111$
$0.40$	$0.42839$	$1.15$	$0.89612$	$1.90$	$0.99279$
$0.45$	$0.47548$	$1.20$	$0.91031$	$1.95$	$0.99418$
$0.50$	$0.52050$	$1.25$	$0.92290$	$2.00$	$0.99532$
$0.55$	$0.56332$	$1.30$	$0.93401$	$2.50$	$0.99959$
$0.60$	$0.60386$	$1.35$	$0.94376$	$3.00$	$0.99998$
$0.65$	$0.64203$	$1.40$	$0.95229$	$3.30$	$0.999998$**
$0.70$	$0.67780$	$1.45$	$0.95970$

**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

Using identity.

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

Receiver output shit

\[\begin{align*} r_o(t)&=\begin{cases} s_{o1}(t)+n_o(t) & \text{code 1}\\ s_{o2}(t)+n_o(t) & \text{code 0}\\ \end{cases}\\ n&: \text{AWGN with }\sigma_o^2\\ \end{align*}\]

ISI, channel model

Raised cosine (RC) pulse

\[0\leq\alpha\leq1\]

⚠ NOTE might not be safe to assume $T’=T$, if you can solve the question without $T$ then use that method.

Nyquist criterion for zero ISI

\[\begin{align*} D &> 2W\quad\text{Use $W$ from table below depending on modulation scheme.}\\ B_\text{Nyquist} &= \frac{W}{1+\alpha}\\ \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ \end{align*}\]

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

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
$W=B_\text{\color{green}abs-abs}$	$W=B_\text{\color{green}abs}$
$W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$	$W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$
$D=\frac{W\text{ symbol/s}}{1+\alpha}$	$D=\frac{2W\text{ symbol/s}}{1+\alpha}$

\[\begin{align*} R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\ M\text{ symbol/set}&=2^k\\ T\text{ s/symbol}&=1/(D\text{ symbol/s})\\ E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\ \end{align*}\]

Table of bandpass signalling and BER

Binary Bandpass Signaling	$B_\text{null-null}$ (Hz)	$B_\text{abs-abs}\color{red}=2B_\text{abs}$ (Hz)	BER with Coherent Detection	BER with Noncoherent Detection
ASK, unipolar NRZ	$2R_b$	$R_b (1 + \alpha)$	$Q\left( \sqrt{E_b / N_0} \right)$	$0.5\exp(-E_b / (2N_0))$
BPSK	$2R_b$	$R_b (1 + \alpha)$	$Q\left( \sqrt{2E_b / N_0} \right)$	Requires coherent detection
Sunde’s FSK	$3R_b$		$Q\left( \sqrt{E_b / N_0} \right)$	$0.5\exp(-E_b / (2N_0))$
DBPSK, $M$-ary Bandpass Signaling	$2R_b$	$R_b (1 + \alpha)$		$0.5\exp(-E_b / N_0)$
QPSK/OQPSK ($M=4$, PSK)	$R_b$	$\frac{R_b (1 + \alpha)}{2}$	$Q\left( \sqrt{2E_b / N_0} \right)$	Requires coherent detection
MSK	$1.5R_b$	$\frac{3R_b (1 + \alpha)}{4}$	$Q\left( \sqrt{2E_b / N_0} \right)$	Requires coherent detection
$M$-PSK ($M > 4$)	$2R_b / \log_2 M$	$\frac{R_b (1 + \alpha)}{\log_2 M}$	$\frac{2}{\log_2 M} Q\left( \sqrt{2 \log_2 M \sin^2 \left( \pi / M \right) E_b / N_0} \right)$	Requires coherent detection
$M$-DPSK ($M > 4$)	$2R_b / \log_2 M$	$\frac{R_b (1 + \alpha)}{2 \log_2 M}$		$\frac{2}{\log_2 M} Q\left( \sqrt{4 \log_2 M \sin^2 \left( \pi / (2M) \right) E_b / N_0} \right)$
$M$-QAM (Square constellation)	$2R_b / \log_2 M$	$\frac{R_b (1 + \alpha)}{\log_2 M}$	$\frac{4}{\log_2 M} \left( 1 - \frac{1}{\sqrt{M}} \right) Q\left( \sqrt{\frac{3 \log_2 M}{M - 1} E_b / N_0} \right)$	Requires coherent detection
$M$-FSK Coherent	$\frac{(M + 3) R_b}{2 \log_2 M}$		$\frac{M - 1}{\log_2 M} Q\left( \sqrt{(\log_2 M) E_b / N_0} \right)$
Noncoherent	$2M R_b / \log_2 M$			$\frac{M - 1}{2 \log_2 M} 0.5\exp({-(\log_2 M) E_b / 2N_0})$

Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems

PSD of modulated signals

Modulation	$G_x(f)$
Quadrature	$\color{red}\frac{ {A_c}^2}{4}[G_I(f-f_c)+G_I(f+f_c)+G_Q(f-f_c)+G_Q(f+f_c)]$
Linear	$\color{red}\frac{|V(f)|^2}{2}\sum_{l=-\infty}^\infty R(l)\exp(-j2\pi l f T)\quad\text{What??}$

Symbol error probability

• Minimum distance between any two point
• Different from bit error since a symbol can contain multiple bits

Information theory

Stats

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

Entropy for discrete random variables

\[\begin{align*} H(x) &\geq 0\\ H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\ H(x,y) &= -\sum_{x_i\in A_x}\sum_{y_i\in A_y} p_{XY}(x_i,y_i)\log_2(p_{XY}(x_i,y_i)) \quad\text{Joint entropy}\\ H(x,y) &= H(x)+H(y) \quad\text{Joint entropy if $x$ and $y$ independent}\\ H(x|y=y_j) &= -\sum_{x_i\in A_x} p_X(x_i|y=y_j) \log_2(p_X(x_i|y=y_j)) \quad\text{Conditional entropy}\\ H(x|y) &= -\sum_{y_j\in A_y} p_Y(y_j) H(x|y=y_j) \quad\text{Average conditional entropy, equivocation}\\ H(x|y) &= -\sum_{x_i\in A_x}\sum_{y_i\in A_y} p_X(x_i,y_j) \log_2(p_X(x_i|y=y_j))\\ H(x|y) &= H(x,y)-H(y)\\ H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\ \end{align*}\]

Entropy is maximized when all have an equal probability.

Transition probability diagram

Example for binary erasure channel where $X$ is input and $Y$ is output:

Equivalent to:

\[\begin{align*} P[Y=0|X=0] &= 1-p\\ P[Y=e|X=0] &= p\\ P[Y=1|X=1] &= 1-p\\ P[Y=e|X=1] &= p\\ P[X=0|Y=0] &= 0\quad\text{Note the direction}\\ P[Y=0] &= P[Y=0|X=0] P[X=0] \end{align*}\]

Differential entropy for continuous random variables

TODO: Cut out if not required

\[\begin{align*} h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx \end{align*}\]

Mutual information

Amount of entropy decrease of $x$ after observation by $y$.

\[\begin{align*} I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\ \end{align*}\]

Channel model

Vertical, $x$: input
Horizontal, $y$: output
Remember $\mathbf{P}$ is a matrix where each element is $P(y_j|x_i)$

\[\mathbf{P}=\left[\begin{matrix} p_{11} & p_{12} &\dots & p_{1N}\\ p_{21} & p_{22} &\dots & p_{2N}\\ \vdots & \vdots &\ddots & \vdots\\ p_{M1} & p_{M2} &\dots & p_{MN}\\ \end{matrix}\right]\] \[\begin{array}{c|cccc} P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline x_1 & p_{11} & p_{12} & \dots & p_{1N} \\ x_2 & p_{21} & p_{22} & \dots & p_{2N} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\ \end{array}\]

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 $p_Y(b_j)=P(y=b_j)$

\[\begin{align*} p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\ &= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\ [\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P} \end{align*}\]

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

\[\begin{align*} &\text{1. Find }H(x)\\ &\text{2. Find }[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] = [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}\\ &\text{3. Multiply each row in $\textbf{P}$ by $p_X(a_i)$ since $p_{XY}(a_i,b_i)=P(b_i|a_i)P(a_i)$}\\ &\text{4. Find $H(x,y)$ using each element from (3.)}\\ &\text{5. Find }H(x|y)=H(x,y)-H(y)\\ &\text{6. Find }I(x;y)=H(x)-H(x|y)\\ \end{align*}\]

Example of step 3:

\[\mathbf{P_{XY}}=\left[\begin{matrix} P(y_1|x_1) P(x_1) & P(y_2|x_1) P(x_1) & \dots\\ P(y_1|x_2) P(x_2) & P(y_2|x_2) P(x_2) & \dots\\ \vdots & \vdots &\ddots \end{matrix}\right]\]

Channel types

Type	Definition
Symmetric channel	Every row is a permutation of every other row, Every column is a permutation of every other column. $\text{Symmetric}\implies\text{Weakly symmetric}$
Weakly symmetric	Every row is a permutation of every other row, Every column has the same sum

Channel capacity of weakly symmetric channel

\[\begin{align*} C &\to\text{Channel capacity (bits/channels used)}\\ N &\to\text{Output alphabet size}\\ \mathbf{p} &\to\text{Probability vector, any row of the transition matrix}\\ C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\ R_b &< C \text{ for error-free transmission} \end{align*}\]

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

\[y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)\] \[C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)\]

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

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 $\mathbf{G}$.

\[\begin{align*} \mathbf{g}_i&=[\begin{matrix} g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1} \end{matrix}]\\ \color{darkgray}\mathbf{g}_0&\color{darkgray}=[1010]\quad\text{Example for $n=4$}\\ \mathbf{G}&=\left[\begin{matrix} \mathbf{g}_0\\ \mathbf{g}_1\\ \vdots\\ \mathbf{g}_{k-1}\\ \end{matrix}\right]=\left[\begin{matrix} g_{0,0}& \dots & g_{0,n-2} & g_{0,n-1}\\ g_{1,0}& \dots & g_{1,n-2} & g_{1,n-1}\\ \vdots & \ddots & \vdots & \vdots\\ g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\ \end{matrix}\right] \end{align*}\]

A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$:

\[\begin{align*} \mathbf{m}&=[\begin{matrix} m_{0}& \dots & m_{n-2} & m_{k-1} \end{matrix}]\\ \color{darkgray}\mathbf{m}&\color{darkgray}=[101001]\quad\text{Example for $k=6$}\\ \mathbf{x} &= \mathbf{m}\mathbf{G}=m_0\mathbf{g}_0+m_1\mathbf{g}_1+\dots+m_{k-1}\mathbf{g}_{k-1} \end{align*}\]

Systemic linear block code

Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits.

\[\begin{align*} \mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[ \begin{array}{c|c} \begin{matrix} 1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0& 0 & \dots & 1\\ \end{matrix} & \begin{matrix} p_{0,0}& \dots & p_{0,n-2} & p_{0,n-1}\\ p_{1,0}& \dots & p_{1,n-2} & p_{1,n-1}\\ \vdots & \ddots & \vdots & \vdots\\ p_{k-1,0}& \dots & p_{k-1,n-2} & p_{k-1,n-1}\\ \end{matrix}\end{array}\right]\\ \mathbf{m}&=[\begin{matrix} m_{0}& \dots & m_{n-2} & m_{k-1} \end{matrix}]\\ \mathbf{x} &= \mathbf{m}\mathbf{G}= \mathbf{m} \begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\begin{array}{c|c}[\mathbf{mI}_k & \mathbf{mP}]\end{array}=\begin{array}{c|c}[\mathbf{m} & \mathbf{b}]\end{array}\\ \mathbf{b} &= \mathbf{m}\mathbf{P}\quad\text{Parity bits of $\mathbf{x}$} \end{align*}\]

Parity check matrix $\mathbf{H}$

Transpose $\mathbf{P}$ for the parity check matrix

\[\begin{align*} \mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\ &=\left[ \begin{array}{c|c} \begin{matrix} {\textbf{p}_0}^\text{T} & {\textbf{p}_1}^\text{T} & \dots & {\textbf{p}_{k-1}}^\text{T} \end{matrix} & \mathbf{I}_{n-k}\end{array}\right]\\ &=\left[ \begin{array}{c|c} \begin{matrix} p_{0,0}& \dots & p_{0,k-2} & p_{0,k-1}\\ p_{1,0}& \dots & p_{1,k-2} & p_{1,k-1}\\ \vdots & \ddots & \vdots & \vdots\\ p_{n-1,0}& \dots & p_{n-1,k-2} & p_{n-1,k-1}\\ \end{matrix} & \begin{matrix} 1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0& 0 & \dots & 1\\ \end{matrix}\end{array}\right]\\ \mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid} \end{align*}\]

Procedure to find parity check matrix from list of codewords

1. From the number of codewords, find $k=\log_2(N)$
2. Partition codewords into $k$ information bits and remaining bits into $n-k$ parity bits. The information bits should be a simple counter (?).
3. Express parity bits as a linear combination of information bits
4. Put coefficients into $\textbf{P}$ matrix and find $\textbf{H}$

Example:

\[\begin{array}{cccc} x_1 & x_2 & x_3 & x_4 & x_5 \\\hline \color{magenta}1&\color{magenta}0&1&1&0\\ \color{magenta}0&\color{magenta}1&1&1&1\\ \color{magenta}0&\color{magenta}0&0&0&0\\ \color{magenta}1&\color{magenta}1&0&0&1\\ \end{array}\]

Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.

\[\begin{align*} \begin{aligned} x_3 &= x_1\oplus x_2\\ x_4 &= x_1\oplus x_2\\ x_5 &= x_2\\ \end{aligned} \implies\textbf{P}&= \begin{array}{c|cc} & x_1 & x_2 \\\hline x_3&1&1&\\ x_4&1&1&\\ x_5&0&1&\\ \end{array}\\ \textbf{H}&=\left[ \begin{array}{c|c} \begin{matrix} 1&1\\ 1&1\\ 0&1\\ \end{matrix} & \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix}\end{array}\right] \end{align*}\]

Error detection and correction

Detection of $s$ errors: $d_\text{min}\geq s+1$

Correction of $u$ errors: $d_\text{min}\geq 2u+1$

CHECKLIST

• Transfer function in complex envelope form $\tilde{h}(t)$ should be divided by two.
• Convolutions: do not forget width when using graphical method
• $2W$ 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