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lol. lmao even.
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            Idiot's guide to ELEC4402 communication systems

            

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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.

Download PDF 📄

It is recommended to refer to use the PDF copy instead of whatever GitHub renders.

License and information

Notes are open-source and licensed under the GNU GPL-3.0. You must include the full-text of the license and follow its terms when using these notes or any diagrams in derivative works (but not when printing as notes)

Copyright (C) 2024 Peter Tanner


GPL copyright information

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

the Free Software Foundation, either version 3 of the License, or

(at your option) any later version.

This program is distributed in the hope that it will be useful,

but WITHOUT ANY WARRANTY; without even the implied warranty of

MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the

GNU General Public License for more details.

You should have received a copy of the GNU General Public License

along with this program. If not, see http://www.gnu.org/licenses/.


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  $erf(x)\text{erf}(x)$  table). So bring good notes.

Doing this unit after signal processing is a good idea.









https://www.petertanner.dev/posts/Idiots-guide-to-ELEC4402-Communications-Systems/

Notes are open-source and licensed under the GNU GPL-3.0. Suggest any corrections or changes on GitHub.



Fourier transform identities and properties


Time domain  $x (t)$ 

Frequency domain  $X (f)$ 

 $rect(tT)Π(tT)\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ 

 $\text{sinc}(fT)$ 

 $sinc(2Wt)\text{sinc}(2Wt)$ 

 $12Wrect(f2W)12WΠ(f2W)\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ 

 $exp⁡(−at)u(t),a>0\exp(-at)u(t),\quad a>0$ 

 $1a+j2πf\frac{1}{a + j2\pi f}$ 

 $exp⁡(−a∣t∣),a>0\exp(-a\lvert t \rvert),\quad a>0$ 

 $2aa2+(2πf)2\frac{2a}{a^2 + (2\pi f)^2}$ 

 $exp⁡(−πt2)\exp(-\pi t^2)$ 

 $exp⁡(−πf2)\exp(-\pi f^2)$ 

 $\frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T\quad\text{tri}(t/T)$ 

 $\text{sinc}^2(fT)$ 

 $δ(t)\delta(t)$ 

 $1$ 

 $1$ 

 $δ(f)\delta(f)$ 

 $δ(t−t0)\delta(t - t_0)$ 

 $exp⁡(−j2πft0)\exp(-j2\pi f t_0)$ 

 $exp⁡(j2πfct)\exp(j2\pi f_c t)$ 

 $δ(f−fc)\delta(f - f_c)$ 

 $cos⁡(2πfct)\cos(2\pi f_c t)$ 

 $12[δ(f−fc)+δ(f+fc)]\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$ 

 $cos⁡(2πfct+θ)\cos(2\pi f_c t+\theta)$ 

 $recv.\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}$ 

 $sin⁡(2πfct)\sin(2\pi f_c t)$ 

 $12j[δ(f−fc)−δ(f+fc)]\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$ 

 $sin⁡(2πfct+θ)\sin(2\pi f_c t+\theta)$ 

 $12j[δ(f−fc)exp⁡(jθ)−δ(f+fc)exp⁡(−jθ)]\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]$ 

 $sgn(t)\text{sgn}(t)$ 

 $1jπf\frac{1}{j\pi f}$ 

 $1πt\frac{1}{\pi t}$ 

 $\text{sgn}(f)$ 

 $u (t)$ 

 $12δ(f)+1j2πf\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ 

 $∑n=−∞∞δ(t−nT0)\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ 

 $1T0∑n=−∞∞δ(f−nT0)=f0∑n=−∞∞δ(f−nf0)\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ 

Time domain  $x (t)$ 

Frequency domain  $X (f)$ 

Property

 $g (t - a)$ 

 $exp⁡(−j2πfa)G(f)\exp(-j2\pi fa)G(f)$ 

Time shifting

 $exp⁡(−j2πfct)g(t)\exp(-j2\pi f_c t)g(t)$ 

 $G(f-f_c)$ 

Frequency shifting

 $g (b t)$ 

 $G(f/b)∣b∣\frac{G(f/b)}{|b|}$ 

Time scaling

 $g (b t - a)$ 

 $1∣b∣exp⁡(−j2πa(f/b))⋅G(f/b)\frac{1}{|b|}\exp(-j2\pi a(f/b))\cdot G(f/b)$ 

Time scaling and shifting

 $ddtg(t)\frac{d}{dt}g(t)$ 

 $j2πfG(f)j2\pi fG(f)\quad$ 

Differentiation wrt time

 $t g (t)$ 

 $12πddfG(f)\frac{1}{2\pi}\frac{d}{df}G(f)\quad$ 

Differentiation wrt frequency

 $g^*(t)$ 

 $G^*(-f)$ 

Conjugate functions

 $G (t)$ 

 $g (- f)$ 

Duality

 $∫−∞tg(τ)dτ\int_{-\infty}^t g(\tau)d\tau$ 

 $1j2πfG(f)+G(0)2δ(f)\frac{1}{j2\pi f}G(f)+\frac{G(0)}{2}\delta(f)$ 

Integration wrt time

 $g (t) h (t)$ 

 $G (f) * H (f)$ 

Time multiplication

 $g (t) * h (t)$ 

 $G (f) H (f)$ 

Time convolution

 $a g (t) + bh (t)$ 

 $a G (f) + b H (f)$ 

Linearity  $a, b$  constants

 $∫−∞∞x(t)y∗(t)dt\int_{-\infty}^\infty x(t)y^*(t)dt$ 

 $∫−∞∞X(f)Y∗(f)df\int_{-\infty}^\infty X(f)Y^*(f)df$ 

Parseval's theorem

 $Ex=∫−∞∞∣x(t)∣2dtE_x=\int_{-\infty}^\infty |x(t)|^2dt$ 

 $Ex=∫−∞∞∣X(f)∣2dfE_x=\int_{-\infty}^\infty |X(f)|^2df$ 

Parseval's theorem

Description

Property

 $g(0)=∫−∞∞G(f)dfg(0)=\int_{-\infty}^\infty G(f)df$ 

Area under  $G (f)$ 

 $G(0)=∫−∞∞G(t)dtG(0)=\int_{-\infty}^\infty G(t)dt$ 

Area under  $g (t)$ 

 $Functiong(t)∗h(t)=(g∗h)(t)=∫∞∞g(τ)h(t−τ)dτConvolution\begin{align*} u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\ \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\ \text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\ \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\ \text{tri}(t/T) &= \begin{cases} 1 - \frac{|t|}{T}, & \lvert t\rvert < T \\ 0, & \lvert t \rvert \geq T \end{cases}=\frac{1}{T}\Pi(t/T)*\Pi(t/T)&\text{Triangle Function}\\ g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\ \end{align*}$ 

Fourier transform of continuous time periodic signal

Required for some questions on sampling:


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

 $Xp(f)=∑n=−∞∞Cnδ(f−nfs)fs=1TsX_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}$ 

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


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

Shape functions



Random processes examples

 $E[X(t)]=E[Bcos⁡(2πfct+θ)]=B∫02π12π⏟uniformcos⁡(2πfct+θ)dθ=0\begin{align*} &\text{Example: separate RV from expression}\\ X(t) &= A\cos(2\pi f_c t)\quad A\thicksim \mathcal{N}(\mu=5,\sigma^2=1)\\ \implies E[X(t)] &= E[A\cos(2\pi f_c t)] = E[A]\cos(2\pi f_c t) = 5\cos(2\pi f_c t)\\ &\text{Example: random phase}\\ X(t) &= B\cos(2\pi f_c t+\theta)\quad \theta\thicksim \mathcal{U}(0,2\pi)\\ \implies E[X(t)] &= E[B\cos(2\pi f_c t+\theta)] = B\int_0^{2\pi}\underbrace{\frac{1}{2\pi}}_{\text{uniform}}\cos(2\pi f_c t+\theta)d\theta=0 \end{align*}$ 

Wide sense stationary (WSS)

Two conditions for WSS:


Constant mean

Autocorrelation only dependent on time difference

 $Constant\mu_X(t) = \mu_X\text{ Constant}$ 

 $RXX(t1,t2)=RX(t1−t2)=RX(τ)R_{XX}(t_1,t_2)=R_X(t_1-t_2)=R_X(\tau)$ 

 $μX(t)=E[X(t)]\mu_X(t)=E[X(t)]$ 

 $E[X(t1)X(t2)]=E[X(t)X(t+τ)]E[X(t_1)X(t_2)]=E[X(t)X(t+\tau)]$ 

Ergodicity

 $⟨X(t)⟩T=12T∫−TTx(t)dt⟨X(t+τ)X(t)⟩T=12T∫−TTx(t+τ)x(t)dtE[⟨X(t)⟩T]=12T∫−TTx(t)dt=12T∫−TTmXdt=mX\begin{align*} \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\ \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\ E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\ \end{align*}$ 

Type

Normal

Mean square sense

ergodic in mean

 $lim⁡T→∞⟨X(t)⟩T=mX(t)=mX\lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X$ 

 $lim⁡T→∞VAR[⟨X(t)⟩T]=0\lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0$ 

ergodic in autocorrelation function

 $lim⁡T→∞⟨X(t+τ)X(t)⟩T=RX(τ)\lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau)$ 

 $lim⁡T→∞VAR[⟨X(t+τ)X(t)⟩T]=0\lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0$ 

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

Other identities

 $associative∑x=−∞∞(f(xa)δ(ω−xb))=f(ωab)\begin{align*} f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\ a(f*g) &= (af)*g \quad\text{Convolution associative}\\ \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right) \end{align*}$ 

Other trig

 $cos⁡2θ=2cos⁡2θ−1⇔cos⁡2θ+12=cos⁡2θe−jα−ejα=−2jsin⁡(α)e−jα+ejα=2cos⁡(α)cos⁡(−A)=cos⁡(A)sin⁡(−A)=−sin⁡(A)sin⁡(A+π/2)=cos⁡(A)sin⁡(A−π/2)=−cos⁡(A)cos⁡(A−π/2)=sin⁡(A)cos⁡(A+π/2)=−sin⁡(A)∫x∈Rsinc(Ax)=1∣A∣\begin{align*} \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\ e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\ e^{-j\alpha}+e^{j\alpha}&=2 \cos(\alpha)\\ \cos(-A)&=\cos(A)\\ \sin(-A)&=-\sin(A)\\ \sin(A+\pi/2)&=\cos(A)\\ \sin(A-\pi/2)&=-\cos(A)\\ \cos(A-\pi/2)&=\sin(A)\\ \cos(A+\pi/2)&=-\sin(A)\\ \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\ \end{align*}$ 

 $cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)cos⁡(A)cos⁡(B)=12(cos⁡(A−B)+cos⁡(A+B))cos⁡(A)sin⁡(B)=12(sin⁡(A+B)−sin⁡(A−B))sin⁡(A)sin⁡(B)=12(cos⁡(A−B)−cos⁡(A+B))\begin{align*} \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\ \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\ \cos(A)\cos(B) &= \frac{1}{2} (\cos (A-B)+\cos (A+B)) \\ \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\ \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\ \end{align*}$ 

 $cos⁡(A)+cos⁡(B)=2cos⁡(A2−B2)cos⁡(A2+B2)cos⁡(A)−cos⁡(B)=−2sin⁡(A2−B2)sin⁡(A2+B2)sin⁡(A)+sin⁡(B)=2sin⁡(A2+B2)cos⁡(A2−B2)sin⁡(A)−sin⁡(B)=2sin⁡(A2−B2)cos⁡(A2+B2)cos⁡(A)+sin⁡(B)=−2sin⁡(A2−B2−π4)sin⁡(A2+B2+π4)cos⁡(A)−sin⁡(B)=−2sin⁡(A2+B2−π4)sin⁡(A2−B2+π4)\begin{align*} \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\ \sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\ \sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\ \end{align*}$ 

IQ/Complex envelope

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

Convert complex envelope representation to time-domain representation of signal

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

For transfer function

 $h(t)=hI(t)cos⁡(2πfct)−hQ(t)sin⁡(2πfct)=2Re[h~(t)exp⁡(j2πfct)]⇒h~(t)=hI(t)/2+jhQ(t)/2=A(t)/2exp⁡(jϕ(t))\begin{align*} h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\ \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))} \end{align*}$ 

AM

Conventional AM modulation (CAM)

 $m(t)=Amm^(t) and m^(t) is the normalized modulating signalma=∣min⁡t(kam(t))∣Acka is the amplitude sensitivity (volt−1), ma is the modulation index.ma=Amax−AminAmax+Amin (Symmetrical m(t))ma=kaAm (Symmetrical m(t))Pc=Ac22Carrier powerPs=14ma2Ac2Signal power, total of all 4 sideband power, single-tone caseη=Signal PowerTotal Power=PsPs+Pc=PsPxPower efficiencyBT=2fm=2B\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)

 $xDSB(t)=Accos⁡(2πfct)m(t)BT=2fm=2B\begin{align*} x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\ B_T&=2f_m=2B \end{align*}$ 

FM/PM

 $tonefi(t)=12πddtθi(t)=fc+kfm(t)=fc+Δfmaxm^(t)Instantaneous frequencyΔfmax=max⁡t∣fi(t)−fc∣=kfmax⁡t∣m(t)∣Maximum frequency deviationΔfmax=kfAmMaximum frequency deviation (sinusoidal)β=ΔfmaxfmModulation indexD=ΔfmaxWmDeviation ratio, where Wm is bandwidth of m(t) (Use FT)\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

 $power\begin{align*} J_n(\beta)&=\begin{cases} J_{-n}(\beta) & \text{$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

 $s(t)=Accos⁡[2πfct+βsin⁡(2πfmt)]⟺s(t)=Ac∑n=−∞∞Jn(β)cos⁡[2π(fc+nfm)t]\begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\\ \Longleftrightarrow s(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t] \end{align*}$ 

FM signal power

 $sideband…\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)

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

Complex envelope of a FM signal

 $s(t)=Accos⁡(2πfct+βsin⁡(2πfmt))⟺s~(t)=Acexp⁡(jβsin⁡(2πfmt))s(t)=Re[s~(t)exp⁡(j2πfct)]s~(t)=Ac∑n=−∞∞Jn(β)exp⁡(j2πfmt)\begin{align*} s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t))\\ \Longleftrightarrow \tilde{s}(t) &= A_c\exp(j\beta\sin(2\pi f_m t))\\ s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\ \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t) \end{align*}$ 

Power, energy and autocorrelation

 $x(t)\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

 $RW(τ)=N02δ(τ)=kT2δ(τ)=σ2δ(τ)Gw(f)=N02\begin{align*} R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\ G_w(f)&=\frac{N_0}{2}\\ \end{align*}$ 

Noise performance


Use formualas from previous section, Power, energy and autocorrelation.


Use these formulas in particular:

 $integration\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*}$ 

 $ratio\begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\ \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio} \end{align*}$ 

Sampling

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

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:  $LPF∈[−2500,2500]f_s=5000\implies \text{LPF}\in[-2500,2500]$ 

Then iterate for  $n=0,1,−1,2,−2,…n=0,1,-1,2,-2,\dots$  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  $Xˉs(f)\bar X_s(f)$  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  $xp(t)=∑n=−∞∞x(t−nTs)x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$  with period  $T_s$ :

 $Xp(f)=∑n=−∞∞Cnδ(f−nfs)fs=1TsX_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}$ 

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


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

Nyquist criterion for zero-ISI

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

 $interval=12B\text{Nyquist rate} = 2B\quad\text{Nyquist interval}=\frac{1}{2B}$ 



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

Cannot add directly due to copyright!

TODO: Make an open source replacement for this diagram Send a PR to GitHub.





Quantizer

 $Δ\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

 $dB\begin{align*} e &:= y-x\quad\text{Quantization error}\\ \mu_E &= E[E] = 0\quad\text{Zero mean}\\ P_E&={\sigma_E}^2=\frac{\Delta^2}{12}=2^{-2m}V^2/3\quad\text{Uniformly distributed error}\\ \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise power}}=\frac{P_x}{P_E}\\ \text{SQNR(dB)}&=10\log_{10}(\text{SQNR})\\ m\to m+A\text{ bits}&\implies \text{newSQNR(dB)}=\text{SQNR(dB)}+6A\text{ dB} \end{align*}$ 

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



 $shapeVrectangle(f)=Tsinc(fT×DutyCycle)GMunipolarNRZ(f)=(M2−1)A2D12∣V(f)∣2+(M−1)24(DA)2∑l=−∞∞∣V(lD)∣2δ(f−lD)GMpolarNRZ(f)=(M2−1)A2D3∣V(f)∣2GunipolarNRZ(f)=A24Rb(sinc2(fRb)+Rbδ(f)),NB0=RbGpolarNRZ(f)=A2Rbsinc2(fRb)GunipolarNRZ(f)=A24Rb(sinc2(fRb)+Rbδ(f))GunipolarRZ(f)=A216(∑l=−∞∞δ(f−lTb)∣sinc(duty×l)∣2+Tb∣sinc(duty×fTb)∣2),NB0=2Rb\begin{align*} R_b&\rightarrow\text{Bit rate}\\ D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\ A&\rightarrow m_a\\ V(f)&\rightarrow\text{Pulse shape}\\ V_\text{rectangle}(f)&=T\text{sinc}(fT\times\text{DutyCycle})\\ G_\text{MunipolarNRZ}(f)&=\frac{(M^2-1)A^2D}{12}|V(f)|^2+\frac{(M-1)^2}{4}(DA)^2\sum_{l=-\infty}^{\infty}|V(lD)|^2\delta(f-lD)\\ G_\text{MpolarNRZ}(f)&=\frac{(M^2-1)A^2D}{3}|V(f)|^2\\ G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right), \text{NB}_0=R_b\\ G_\text{polarNRZ}(f)&=\frac{A^2}{R_b}\text{sinc}^2\left(\frac{f}{R_b}\right)\\ G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\ G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b \end{align*}$ 

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

 $φ1(t)=2Tbcos⁡(2πfct)0≤t≤Tb\begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*}$ 

Symbol mapping

 $bn:{1,0}→an:{1,0}b_n:\{1,0\}\to a_n:\{1,0\}$ 

2 possible waveforms

 $Eb=Eaverage=12(Ac22×Tb+0)=Ac24Tb\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  $d=2Ebd=\sqrt{2E_b}$ 

BPSK

Basis functions

 $φ1(t)=2Tbcos⁡(2πfct)0≤t≤Tb\begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*}$ 

Symbol mapping

 $bn:{1,0}→an:{1,−1}b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}$ 

2 possible waveforms

 $Eb=Eaverage=12(Ac22×Tb+Ac22×Tb)=Ac22Tb\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  $d=2Ebd=2\sqrt{E_b}$ 

QPSK ( $M = 4$  PSK)

Basis functions

 $Tbφ1(t)=2Tcos⁡(2πfct)0≤t≤Tφ2(t)=2Tsin⁡(2πfct)0≤t≤T\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

 $s1(t)=Es/2[φ1(t)+φ2(t)]s2(t)=Es/2[φ1(t)−φ2(t)]s3(t)=Es/2[−φ1(t)+φ2(t)]s4(t)=Es/2[−φ1(t)−φ2(t)]\begin{align*} s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\ s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\ s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\ s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\ \end{align*}$ 

Note on energy per symbol: Since  $s_i(t)|=A_c$ , have to normalize distance as follows:

 $si(t)=AcT/2/2×[α1iφ1(t)+α2iφ2(t)]=TAc2/4[α1iφ1(t)+α2iφ2(t)]=Es/2[α1iφ1(t)+α2iφ2(t)]\begin{align*} s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ \end{align*}$ 

Signal

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

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  $x = T /2$ .

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

2. Bit error rate of matched filter

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

 $dB"BERMatchedFilter=Q(d22N0)=Q(Eb2N0)BERunipolarNRZ|BASK=Q(d2N0)=Q(EbN0)BERpolarNRZ|BPSK=Q(2d2N0)=Q(2EbN0)\begin{align*} Q(x)&=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Leftrightarrow\text{erf}\left(\frac{x}{\sqrt{2}}\right)=1-2Q(x)\\ E_b&=d^2=\int_{-\infty}^\infty|s_1(t)-s_2(t)|^2dt\quad\text{Energy per bit/Distance}\\ T&=1/R_b\quad\text{$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  $erf(x)\text{erf}(x)$  and  $Q (x)$ 





 $Q (x)$  function

You should use  $erf\text{erf}$  function table instead in exams using the identity  $Q(x)=12−12erf(x2)Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)$ . 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



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

 $Q(x)=12−12erf(x2)Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}(\frac{x}{\sqrt{2}})$ 

 $x$ 

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

 $x$ 

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

 $x$ 

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

 $0.00$ 

 $0.00000$ 

 $0.75$ 

 $0.71116$ 

 $1.50$ 

 $0.96611$ 

 $0.05$ 

 $0.05637$ 

 $0.80$ 

 $0.74210$ 

 $1.55$ 

 $0.97162$ 

 $0.10$ 

 $0.11246$ 

 $0.85$ 

 $0.77067$ 

 $1.60$ 

 $0.97635$ 

 $0.15$ 

 $0.16800$ 

 $0.90$ 

 $0.79691$ 

 $1.65$ 

 $0.98038$ 

 $0.20$ 

 $0.22270$ 

 $0.95$ 

 $0.82089$ 

 $1.70$ 

 $0.98379$ 

 $0.25$ 

 $0.27633$ 

 $1.00$ 

 $0.84270$ 

 $1.75$ 

 $0.98667$ 

 $0.30$ 

 $0.32863$ 

 $1.05$ 

 $0.86244$ 

 $1.80$ 

 $0.98909$ 

 $0.35$ 

 $0.37938$ 

 $1.10$ 

 $0.88021$ 

 $1.85$ 

 $0.99111$ 

 $0.40$ 

 $0.42839$ 

 $1.15$ 

 $0.89612$ 

 $1.90$ 

 $0.99279$ 

 $0.45$ 

 $0.47548$ 

 $1.20$ 

 $0.91031$ 

 $1.95$ 

 $0.99418$ 

 $0.50$ 

 $0.52050$ 

 $1.25$ 

 $0.92290$ 

 $2.00$ 

 $0.99532$ 

 $0.55$ 

 $0.56332$ 

 $1.30$ 

 $0.93401$ 

 $2.50$ 

 $0.99959$ 

 $0.60$ 

 $0.60386$ 

 $1.35$ 

 $0.94376$ 

 $3.00$ 

 $0.99998$ 

 $0.65$ 

 $0.64203$ 

 $1.40$ 

 $0.95229$ 

 $3.30$ 

 $0.999998$ **

 $0.70$ 

 $0.67780$ 

 $1.45$ 

 $0.95970$ 

**The value of  $erf(3.30)\text{erf}(3.30)$  should be  $≈0.999997\approx0.999997$  instead, but this value is quoted in the formula table.



 $Q (x)$  fast reference

Using identity.


 $x$ 

 $Q (x)$ 

 $2\sqrt{2}$ 

 $0.07865$ 

 $222\sqrt{2}$ 

 $0.00234$ 



Receiver output shit

 $σo2\begin{align*} r_o(t)&=\begin{cases} s_{o1}(t)+n_o(t) & \text{code 1}\\ s_{o2}(t)+n_o(t) & \text{code 0}\\ \end{cases}\\ n&: \text{AWGN with }\sigma_o^2\\ \end{align*}$ 

ISI, channel model

Raised cosine (RC) pulse



 $0≤α≤10\leq\alpha\leq1$ 

⚠ 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


 $BWBNyquist=Babs−BNyquistBNyquist\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

 $sizeW→Bandwidth\begin{align*} D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\ T&\rightarrow\text{Symbol Duration}\\ M&\rightarrow\text{Symbol set size}\\ W&\rightarrow\text{Bandwidth}\\ \end{align*}$ 

Bandwidth  $W$  and bit error rate of modulation schemes

To solve this type of question:



Use the formula for  $D$  below

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

 $W=Babs-absW=B_\text{\color{green}abs-abs}$ 

 $W=BabsW=B_\text{\color{green}abs}$ 

 $W=Babs-abs=1+αT=(1+α)DW=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ 

 $W=Babs=1+α2T=(1+α)D/2W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ 

 $symbol/s1+αD=\frac{W\text{ symbol/s}}{1+\alpha}$ 

 $symbol/s1+αD=\frac{2W\text{ symbol/s}}{1+\alpha}$ 

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

Table of bandpass signalling and BER


Binary Bandpass Signaling

 $Bnull-nullB_\text{null-null}$  (Hz)

 $Babs-abs=2BabsB_\text{abs-abs}\color{red}=2B_\text{abs}$  (Hz)

BER with Coherent Detection

BER with Noncoherent Detection

ASK, unipolar NRZ

 $2R_b$ 

 $Rb(1+α)R_b (1 + \alpha)$ 

 $Q(Eb/N0)Q\left( \sqrt{E_b / N_0} \right)$ 

 $0.5\exp(-E_b / (2N_0))$ 

BPSK

 $2R_b$ 

 $Rb(1+α)R_b (1 + \alpha)$ 

 $Q(2Eb/N0)Q\left( \sqrt{2E_b / N_0} \right)$ 

Requires coherent detection

Sunde's FSK

 $3R_b$ 

 $Q(Eb/N0)Q\left( \sqrt{E_b / N_0} \right)$ 

 $0.5\exp(-E_b / (2N_0))$ 

DBPSK,  $M$ -ary Bandpass Signaling

 $2R_b$ 

 $Rb(1+α)R_b (1 + \alpha)$ 

 $0.5\exp(-E_b / N_0)$ 

QPSK/OQPSK ( $M = 4$ , PSK)

 $R_b$ 

 $Rb(1+α)2\frac{R_b (1 + \alpha)}{2}$ 

 $Q(2Eb/N0)Q\left( \sqrt{2E_b / N_0} \right)$ 

Requires coherent detection

MSK

 $1.5R_b$ 

 $3Rb(1+α)4\frac{3R_b (1 + \alpha)}{4}$ 

 $Q(2Eb/N0)Q\left( \sqrt{2E_b / N_0} \right)$ 

Requires coherent detection

 $M$ -PSK ( $M > 4$ )

 $2R_b / \log_2 M$ 

 $Rb(1+α)log⁡2M\frac{R_b (1 + \alpha)}{\log_2 M}$ 

 $2log⁡2MQ(2log⁡2Msin⁡2(π/M)Eb/N0)\frac{2}{\log_2 M} Q\left( \sqrt{2 \log_2 M \sin^2 \left( \pi / M \right) E_b / N_0} \right)$ 

Requires coherent detection

 $M$ -DPSK ( $M > 4$ )

 $2R_b / \log_2 M$ 

 $Rb(1+α)2log⁡2M\frac{R_b (1 + \alpha)}{2 \log_2 M}$ 

 $2log⁡2MQ(4log⁡2Msin⁡2(π/(2M))Eb/N0)\frac{2}{\log_2 M} Q\left( \sqrt{4 \log_2 M \sin^2 \left( \pi / (2M) \right) E_b / N_0} \right)$ 

 $M$ -QAM (Square constellation)

 $2R_b / \log_2 M$ 

 $Rb(1+α)log⁡2M\frac{R_b (1 + \alpha)}{\log_2 M}$ 

 $4log⁡2M(1−1M)Q(3log⁡2MM−1Eb/N0)\frac{4}{\log_2 M} \left( 1 - \frac{1}{\sqrt{M}} \right) Q\left( \sqrt{\frac{3 \log_2 M}{M - 1} E_b / N_0} \right)$ 

Requires coherent detection

 $M$ -FSK Coherent

 $(M+3)Rb2log⁡2M\frac{(M + 3) R_b}{2 \log_2 M}$ 

 $M−1log⁡2MQ((log⁡2M)Eb/N0)\frac{M - 1}{\log_2 M} Q\left( \sqrt{(\log_2 M) E_b / N_0} \right)$ 

Noncoherent

 $2M R_b / \log_2 M$ 

 $M−12log⁡2M0.5exp⁡(−(log⁡2M)Eb/2N0)\frac{M - 1}{2 \log_2 M} 0.5\exp({-(\log_2 M) E_b / 2N_0})$ 

Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems

PSD of modulated signals


Modulation

 $G_x(f)$ 

Quadrature

 $Ac24[GI(f−fc)+GI(f+fc)+GQ(f−fc)+GQ(f+fc)]\color{red}\frac{ {A_c}^2}{4}[G_I(f-f_c)+G_I(f+f_c)+G_Q(f-f_c)+G_Q(f+f_c)]$ 

Linear

 $∣V(f)∣22∑l=−∞∞R(l)exp⁡(−j2πlfT)What??\color{red}\frac{|V(f)|^2}{2}\sum_{l=-\infty}^\infty R(l)\exp(-j2\pi l f T)\quad\text{What??}$ 

Symbol error probability



Minimum distance between any two point

Different from bit error since a symbol can contain multiple bits





Information theory

Stats

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

Entropy for discrete random variables

 $equivocationH(x∣y)=−∑xi∈Ax∑yi∈AypX(xi,yj)log⁡2(pX(xi∣y=yj))H(x∣y)=H(x,y)−H(y)H(x,y)=H(x)+H(y∣x)=H(y)+H(x∣y)\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:

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

Mutual information



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

 $I(x;y)=H(x)−H(x∣y)=H(y)−H(y∣x)\begin{align*} I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\ \end{align*}$ 

Channel model

Vertical,  $x$ : input


Horizontal,  $y$ : output


Remember  $P\mathbf{P}$  is a matrix where each element is  $P(y_j|x_i)$ 

 $P=[p11p12…p1Np21p22…p2N⋮⋮⋱⋮pM1pM2…pMN]\mathbf{P}=\left[\begin{matrix} p_{11} & p_{12} &\dots & p_{1N}\\ p_{21} & p_{22} &\dots & p_{2N}\\ \vdots & \vdots &\ddots & \vdots\\ p_{M1} & p_{M2} &\dots & p_{MN}\\ \end{matrix}\right]$ 

 $P(yj∣xi)y1y2…yNx1p11p12…p1Nx2p21p22…p2N⋮⋮⋮⋱⋮xMpM1pM2…pMN\begin{array}{c|cccc} P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline x_1 & p_{11} & p_{12} & \dots & p_{1N} \\ x_2 & p_{21} & p_{22} & \dots & p_{2N} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\ \end{array}$ 

Input has probability distribution  $p_X(a_i)=P(X=a_i)$ 

Channel maps alphabet  $‘{a1,…,aM}→{b1,…,bN}‘`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$ 

Output has probability distribution  $p_Y(b_j)=P(y=b_j)$ 

 $pY(bj)=∑i=1MP[x=ai,y=bj]1≤j≤N=∑i=1MP[X=ai]P[Y=bj∣X=ai][pY(b0)pY(b1)…pY(bj)]=[pX(a0)pX(a1)…pX(ai)]×P\begin{align*} p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\ &= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\ [\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P} \end{align*}$ 

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



 $I(x;y)=H(x)−H(x∣y)\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:

 $PXY=[P(y1∣x1)P(x1)P(y2∣x1)P(x1)…P(y1∣x2)P(x2)P(y2∣x2)P(x2)…⋮⋮⋱]\mathbf{P_{XY}}=\left[\begin{matrix} P(y_1|x_1) P(x_1) & P(y_2|x_1) P(x_1) & \dots\\ P(y_1|x_2) P(x_2) & P(y_2|x_2) P(x_2) & \dots\\ \vdots & \vdots &\ddots \end{matrix}\right]$ 

Channel types


Type

Definition

Symmetric channel

Every row is a permutation of every other row, Every column is a permutation of every other column.  $symmetric\text{Symmetric}\implies\text{Weakly symmetric}$ 

Weakly symmetric

Every row is a permutation of every other row, Every column has the same sum

Channel capacity of weakly symmetric channel

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

Note that the channel capacity is realized when the channel inputs are uniformly distributed (i.e.  $P(x1)=P(x2)=⋯=P(xN)=1NP(x_1)=P(x_2)=\dots=P(x_N)=\frac{1}{N}$ )


Channel capacity of an AWGN channel

 $yi=xi+nini∼N(0,N0/2)y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)$ 

 $C=12log⁡2(1+PavN0/2)C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)$ 

Channel capacity of a bandwidth limited AWGN channel

 $poweryi=bandpassW(xi)+nini∼N(0,N0/2)C=Wlog⁡2(1+PsN0W)C=Wlog⁡2(1+SNR)SNR=Ps/(N0W)\begin{align*} P_s&\to\text{Bandwidth limited average power}\\ y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\ C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\ C&=W\log_2(1+\text{SNR})\\ \text{SNR}&=P_s/(N_0 W) \end{align*}$ 

Shannon limit

 $limited\begin{align*} R_b &< C\\ \implies R_b &< W\log_2\left(1+\frac{P_s}{N_0 W}\right)\quad\text{For bandwidth limited AWGN channel}\\ \frac{E_b}{N_0} &> \frac{2^\eta-1}{\eta}\quad\text{SNR per bit required for error-free transmission}\\ \eta &= \frac{R_b}{W}\quad\text{Spectral efficiency (bit/(s-Hz))}\\ \eta &\gg 1\quad\text{Bandwidth limited}\\ \eta &\ll 1\quad\text{Power limited} \end{align*}$ 



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

 $dH(x1,x2)=wH(x1⊕x2)d_H(x_1,x_2)=w_H(x_1\oplus x_2)$ 

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

 $dmind_\text{min}$ 

IMPORTANT:  $x≠0x\neq\textbf{0}$ , excludes weight of all-zero codeword. For a linear block code,  $dmin=wmind_\text{min}=w_\text{min}$ 

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:



Zero vector must be present at least once

The XOR of any codeword pair in the code must result in a codeword that is already present in the code table.



 $dmin=wmind_\text{min}=w_\text{min}$  (Implied by (1) and (2).)



Code generation

Each generator vector is a binary string of size  $n$ . There are  $k$  generator vectors in  $G\mathbf{G}$ .

 $n=4G=[g0g1⋮gk−1]=[g0,0…g0,n−2g0,n−1g1,0…g1,n−2g1,n−1⋮⋱⋮⋮gk−1,0…gk−1,n−2gk−1,n−1]\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  $m\mathbf{m}$  is coded as  $x\mathbf{x}$  using the generation codewords in  $G\mathbf{G}$ :

 $k=6x=mG=m0g0+m1g1+⋯+mk−1gk−1\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  $m\mathbf{m}$  as-is) and  $(n - k)$  parity bits after the message bits.

 $x\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  $H\mathbf{H}$ 



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

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

Procedure to find parity check matrix from list of codewords



From the number of codewords, find  $k=\log_2(N)$ 



Partition codewords into  $k$  information bits and remaining bits into  $n - k$  parity bits. The information bits should be a simple counter (?).

Express parity bits as a linear combination of information bits

Put coefficients into  $P\textbf{P}$  matrix and find  $H\textbf{H}$ 





Example:

 $x1x2x3x4x510110011110000011001\begin{array}{cccc} x_1 & x_2 & x_3 & x_4 & x_5 \\\hline \color{magenta}1&\color{magenta}0&1&1&0\\ \color{magenta}0&\color{magenta}1&1&1&1\\ \color{magenta}0&\color{magenta}0&0&0&0\\ \color{magenta}1&\color{magenta}1&0&0&1\\ \end{array}$ 

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

 $P=x1x2x311x411x501H=[111101100010001]\begin{align*} \begin{aligned} x_3 &= x_1\oplus x_2\\ x_4 &= x_1\oplus x_2\\ x_5 &= x_2\\ \end{aligned} \implies\textbf{P}&= \begin{array}{c|cc} & x_1 & x_2 \\\hline x_3&1&1&\\ x_4&1&1&\\ x_5&0&1&\\ \end{array}\\ \textbf{H}&=\left[ \begin{array}{c|c} \begin{matrix} 1&1\\ 1&1\\ 0&1\\ \end{matrix} & \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix}\end{array}\right] \end{align*}$ 

Error detection and correction

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

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

CHECKLIST



Transfer function in complex envelope form  $h~(t)\tilde{h}(t)$  should be divided by two.

Convolutions: do not forget width when using graphical method



 $2 W$  for rectangle functions

Scale sampled spectrum by  $f_s$ 





 $2f_c$  for spectrum after IF mixing.

Square transfer function for PSD  $Gy(f)=∣H(f)∣2Gx(f)G_y(f)=|H(f)|^\mathbf{2}G_x(f)$ 



Square besselJ function for FM power  $∣Jn(β)∣2|J_n(\beta)|^\mathbf{2}$ 



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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Idiot's guide to ELEC4402 communication systems

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Tests

Fourier transform identities and properties

Fourier transform of continuous time periodic signal

Shape functions

Random processes examples

Wide sense stationary (WSS)

Ergodicity

Other identities

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

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

Quantizer

Quantization noise

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

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

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 $erf(x)\text{erf}(x)$ and $Q (x)$

$Q (x)$ function

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

$Q (x)$ fast reference

Bandwidth $W$ and bit error rate of modulation schemes

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

Parity check matrix $H\mathbf{H}$
