Idiot's guide to ELEC4402 communication systems

This unit allows you to bring infinite physical notes (except books borrowed from the UWA library) to all tests and the final exam. You can't rely on what material they provide in the test/exam, it is very minimal to say the least. Hope this helps.

If you have issues or suggestions, raise them on GitHub. I accept pull requests for fixes or suggestions but the content must not be copyrighted under a non-GPL compatible license.

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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

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.

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

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

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

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

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

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 images/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 images/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

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

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

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

$Q(x)$ function