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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

Time Function	Fourier Transform
`\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)`	`T \text{sinc}(fT)`
`\text{sinc}(2Wt)`	`\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)`
`\exp(-at)u(t),\quad a>0`	`\frac{1}{a + j2\pi f}`
`\exp(-a\lvert t \rvert),\quad a>0`	`\frac{2a}{a^2 + (2\pi f)^2}`
`\exp(-\pi t^2)`	`\exp(-\pi f^2)`
`1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T`	`T \text{sinc}^2(fT)`
`\delta(t)`	`1`
`1`	`\delta(f)`
`\delta(t - t_0)`	`\exp(-j2\pi f t_0)`
`g(t-a)`	`\exp(-j2\pi fa)G(f)\quad\text{shift property}`
`g(bt)`	`\frac{G(f/b)}{\\|b\\|}\quad\text{scaling property}`
`g(bt-a)`	`\frac{1}{\\|b\\|}\exp(-j2\pi a(f/b))\cdot G(f/b)\quad\text{shift and scale}`
`\frac{d}{dt}g(t)`	`j2\pi fG(f)\quad\text{differentiation property}`
`G(t)`	`g(-f)\quad\text{duality property}`
`g(t)h(t)`	`G(f)*H(f)`
`g(t)*h(t)`	`G(f)H(f)`
`\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)]`
`\sin(2\pi f_c t)`	`\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]`
`\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)`


\begin{align*}
    u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
    \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
    \text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\
    \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
    g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
\end{align*}

Fourier transform of continuous time periodic signal

Required for some questions on sampling:

Transform a continuous time-periodic signal x_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*}

`\text{rect}` function

Bessel function


\begin{align*}
    \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
    J_n(\beta)&=(-1)^nJ_{-n}(\beta)
\end{align*}

White noise


\begin{align*}
R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
G_w(f)&=\frac{N_0}{2}\\
N_0&=kT\\
G_y(f)&=|H(f)|^2G_w(f)\\
G_y(f)&=G(f)G_w(f)\\
\end{align*}

WSS


\begin{align*}
    \mu_X(t) &= \mu_X\text{ Constant}\\
    R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
    E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
\end{align*}

Ergodicity


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

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

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

Other identities


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

Other trig


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


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


\begin{align*}
    \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
    \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
    \sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\
    \sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
    \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
    \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
\end{align*}

IQ/Complex envelope

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

Convert complex envelope representation to time-domain representation of signal


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

For transfer function


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

AM

CAM


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

B_T: Signal bandwidth B: Bandwidth of modulating wave

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

DSB-SC


\begin{align*}
    x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
    B_T&=2f_m=2B
\end{align*}

FM/PM


\begin{align*}
    s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
    s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^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}\\
    \beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\
    \Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
    D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}
\end{align*}

Bessel form and magnitude spectrum (single tone)


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

FM signal power


\begin{align*}
    P_\text{av}&=\frac{ {A_c}^2}{2}\\
    P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
    \text{band\_index}&=0\implies f_c+0f_m\\
    \text{band\_index}&=1\implies f_c+1f_m,\dots\\
\end{align*}

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


\begin{align*}
B &= 2Mf_m = 2(\beta + 1)f_m\\
    &= 2(\Delta f+f_m)\\
    &= 2(k_f A_m+f_m)\\
    &= 2(D+1)W_m\\
B &= \begin{cases}
    2(\Delta f+f_m) & \text{FM, sinusoidal message}\\
    2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
\end{cases}\\
\end{align*}

`\Delta f` of arbitrary modulating signal

Find instantaneous frequency f_\text{FM}.

M: Number of pairs of significant sidebands


\begin{align*}
s(t)&=A_c\cos(\theta_\text{FM}(t))\\
f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\
A_m &= \max_t|m(t)|\\
\Delta f &= \max_t(f_\text{FM}(t)) - f_c\\
W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\
\text{Example: }&\text{sinc}(At+t)+2\cos(2\pi t)=\frac{\sin(2\pi((At+t)/2))}{\pi(At+t)}+2\cos(2\pi t)\to W_m=\max\left(\frac{A+1}{2},1\right)\\
D &= \frac{\Delta f}{W_m}\\
B_T &= 2(D+1)W_m
\end{align*}

Complex envelope


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

Band

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

Power, energy and autocorrelation


\begin{align*}
    G_\text{WGN}(f)&=\frac{N_0}{2}\\
    G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
    G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\
    G_x(f)&=\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\text{ (PSD)}\\
    G_x(f)&=\mathfrak{F}[R_x(\tau)]\text{ (WSS)}\\
    P&=\sigma^2=\int_\mathbb{R}G_x(f)df\\
    P&=\sigma^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\\
    P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\
    E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\
    R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
\end{align*}

Noise performance


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

Sampling


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

Procedure to reconstruct sampled signal

Analog signal x'(t) which can be reconstructed from a sampled signal x_s(t): Put x_s(t) through LPF with maximum frequency of f_s/2 and minimum frequency of -f_s/2. Anything outside of the BPF will be attenuated, therefore n which results in frequencies outside the BPF will evaluate to 0 and can be ignored.

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

Then iterate for n=0,1,-1,2,-2,\dots until the first iteration where the result is 0 since all terms are eliminated by the LPF.

TODO: Add example

Then add all terms and transform \bar X_s(f) back to time domain to get x_s(t)

Fourier transform of continuous time periodic signal (1)

Required for some questions on sampling:

Transform a continuous time-periodic signal x_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.

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!

Quantizer


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

Quantization noise


\begin{align*}
    e &:= y-x\quad\text{Quantization error}\\
    \mu_E &= E[E] = 0\quad\text{Zero mean}\\
    {\sigma_E}^2&=E[E^2]-0^2=\int_{-\Delta/2}^{\Delta/2}e^2\times\left(\frac{1}{\Delta}\right) de\quad\text{Where $E\thicksim 1/\Delta$ uniform over $(-\Delta/2,\Delta/2)$}\\
    \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\
    \text{SQNR(dB)}&=10\log_{10}(\text{SQNR})
\end{align*}

Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to `assets/img/2024-10-29-Idiots-guide-to-ELEC/quantizer.png`)

Cannot add directly due to copyright!

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*}

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:


\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

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

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

Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems

`\text{erf}(x)` function

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

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

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

Nyquist criterion for zero ISI

TODO:

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*}

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.

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 ($M$-PSK, $M$-QAM)	NRZ unipolar encoding
`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}`

Symbol set size `M`


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

Nyquist stuff

TODO: Condition for 0 ISI


P_r(kT)=\begin{cases}
    1 & k=0\\
    0 & k\neq0
\end{cases}

Other


\begin{align*}
    \text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\
    \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
    T&=1/D
\end{align*}

Table of bandpass signalling and BER

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

Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems

PSD of modulated signals

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

Symbol error probability

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

Information theory

Entropy for discrete random variables


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

Entropy is maximized when all have an equal probability.

Differential entropy for continuous random variables

TODO: Cut out if not required


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

Mutual information

Amount of entropy decrease of x after observation by y.


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

Channel model

Vertical, x: input
Horizontal, y: output


\mathbf{P}=\left[\begin{matrix}
    p_{11} & p_{12} &\dots & p_{1N}\\
    p_{21} & p_{22} &\dots & p_{2N}\\
    \vdots & \vdots &\ddots & \vdots\\
    p_{M1} & p_{M2} &\dots & p_{MN}\\
\end{matrix}\right]


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

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

Channel maps alphabet \{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}

Output has probabiltiy distribution p_Y(b_j)=P(y=b_j)


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

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


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

Channel types

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

Channel capacity of weakly symmetric channel


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

Channel capacity of an AWGN channel


y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)


C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)

Channel capacity of a bandwidth AWGN channel

Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition.


\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})\quad\text{SNR}=P_s/(N_0 W)
\end{align*}

Channel code


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:

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.

For a linear block code, d_\text{min}=w_\text{min}

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

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 \textbf{P} matrix and find \textbf{H}

Example:


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

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


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

Error detection and correction

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

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

CHECKLIST

Transfer function in complex envelope form \tilde{h}(t) should be divided by two.
Convolutions: do not forget width when using graphical method
todo: add more items to check

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Fourier transform identities

Fourier transform of continuous time periodic signal

\text{rect} function

Bessel function

White noise

WSS

Ergodicity

Other identities

Other trig

IQ/Complex envelope

Convert complex envelope representation to time-domain representation of signal

For transfer function

AM

CAM

DSB-SC

FM/PM

Bessel form and magnitude spectrum (single tone)

FM signal power

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

\Delta f of arbitrary modulating signal

Complex envelope

Band

Power, energy and autocorrelation

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 assets/img/2024-10-29-Idiots-guide-to-ELEC/sampling.png)

Quantizer

Quantization noise

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)

Line codes

Modulation and basis functions

BASK

Basis functions

Symbol mapping

2 possible waveforms

BPSK

Basis functions

Symbol mapping

2 possible waveforms

QPSK (M=4 PSK)

Basis functions

4 possible waveforms

Signal

Example of waveform

Matched filter

1. Filter function

2. Bit error rate

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

Q(x) function

\text{erf}(x) function

Receiver output shit

ISI, channel model

Nyquist criterion for zero ISI

Nomenclature

Raised cosine (RC) pulse

Symbol set size M

Nyquist stuff

TODO: Condition for 0 ISI

Other

Table of bandpass signalling and BER

PSD of modulated signals

Symbol error probability

Information theory

Entropy for discrete random variables

Differential entropy for continuous random variables

Mutual information

Channel model

Fast procedure to calculate I(y;x)

Channel types

Channel capacity of weakly symmetric channel

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`\text{rect}` function

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

`\Delta f` of arbitrary modulating signal

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`)

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`)

QPSK (`M=4` PSK)

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

`Q(x)` function

`\text{erf}(x)` function

Symbol set size `M`

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

Parity check matrix `\mathbf{H}`