diff --git a/_posts/2024-10-29-Idiots-guide-to-ELEC4402-Communications-Systems.md b/_posts/2024-10-29-Idiots-guide-to-ELEC4402-Communications-Systems.md index 5575c6c..0c582e2 100644 --- a/_posts/2024-10-29-Idiots-guide-to-ELEC4402-Communications-Systems.md +++ b/_posts/2024-10-29-Idiots-guide-to-ELEC4402-Communications-Systems.md @@ -69,7 +69,7 @@ along with this program. If not, see . | $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ | | $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ | -```math +$$ \begin{align*} u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\ \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\ @@ -77,7 +77,7 @@ along with this program. If not, see . \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\ g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\ \end{align*} -``` +$$ ### Fourier transform of continuous time periodic signal @@ -87,21 +87,21 @@ 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 $T_s$: -```math +$$ X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s} -``` +$$ Calculate $C_n$ coefficient as follows from $x_p(t)$: -```math +$$ \begin{align*} % C_n&=X_p(nf_s)\\ C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*} -``` +$$ ### $\text{rect}$ function @@ -109,16 +109,16 @@ Calculate $C_n$ coefficient as follows from $x_p(t)$: ### Bessel function -```math +$$ \begin{align*} \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\ J_n(\beta)&=(-1)^nJ_{-n}(\beta) \end{align*} -``` +$$ ### White noise -```math +$$ \begin{align*} R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\ G_w(f)&=\frac{N_0}{2}\\ @@ -126,27 +126,27 @@ N_0&=kT\\ G_y(f)&=|H(f)|^2G_w(f)\\ G_y(f)&=G(f)G_w(f)\\ \end{align*} -``` +$$ ### WSS -```math +$$ \begin{align*} \mu_X(t) &= \mu_X\text{ Constant}\\ R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\ E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)] \end{align*} -``` +$$ ### Ergodicity -```math +$$ \begin{align*} \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\ \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\ E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\ \end{align*} -``` +$$ | Type | Normal | Mean square sense | | ----------------------------------- | ------------------------------------------------------- | ----------------------------------------------------------- | @@ -157,17 +157,17 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio ### Other identities -```math +$$ \begin{align*} f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\ a(f*g) &= (af)*g \quad\text{Convolution associative}\\ \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right) \end{align*} -``` +$$ ### Other trig -```math +$$ \begin{align*} \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\ e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\ @@ -180,9 +180,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio \cos(A+\pi/2)&=-\sin(A)\\ \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\ \end{align*} -``` +$$ -```math +$$ \begin{align*} \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\ \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\ @@ -190,9 +190,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\ \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\ \end{align*} -``` +$$ -```math +$$ \begin{align*} \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\ @@ -201,7 +201,7 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\ \end{align*} -``` +$$ ## IQ/Complex envelope @@ -209,7 +209,7 @@ Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $ ### Convert complex envelope representation to time-domain representation of signal -```math +$$ \begin{align*} g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\ &=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\ @@ -219,23 +219,23 @@ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\ g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\ g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\ \end{align*} -``` +$$ ### For transfer function -```math +$$ \begin{align*} h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\ \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))} \end{align*} -``` +$$ ## AM ### CAM -```math +$$ \begin{align*} m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\ m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\ @@ -247,7 +247,7 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\ B_T&=2f_m=2B \end{align*} -``` +$$ $B_T$: Signal bandwidth $B$: Bandwidth of modulating wave @@ -256,16 +256,16 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$ ### DSB-SC -```math +$$ \begin{align*} x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\ B_T&=2f_m=2B \end{align*} -``` +$$ ## FM/PM -```math +$$ \begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\ s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\ @@ -274,30 +274,30 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$ \Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\ D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)} \end{align*} -``` +$$ ### Bessel form and magnitude spectrum (single tone) -```math +$$ \begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t] \end{align*} -``` +$$ ### FM signal power -```math +$$ \begin{align*} P_\text{av}&=\frac{ {A_c}^2}{2}\\ P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\ \text{band\_index}&=0\implies f_c+0f_m\\ \text{band\_index}&=1\implies f_c+1f_m,\dots\\ \end{align*} -``` +$$ ### Carson's rule to find $B$ (98% power bandwidth rule) -```math +$$ \begin{align*} B &= 2Mf_m = 2(\beta + 1)f_m\\ &= 2(\Delta f+f_m)\\ @@ -308,7 +308,7 @@ B &= \begin{cases} 2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message} \end{cases}\\ \end{align*} -``` +$$ #### $\Delta f$ of arbitrary modulating signal @@ -316,7 +316,7 @@ Find instantaneous frequency $f_\text{FM}$. $M$: Number of **pairs** of significant sidebands -```math +$$ \begin{align*} s(t)&=A_c\cos(\theta_\text{FM}(t))\\ f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\ @@ -327,17 +327,17 @@ W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\ D &= \frac{\Delta f}{W_m}\\ B_T &= 2(D+1)W_m \end{align*} -``` +$$ ### Complex envelope -```math +$$ \begin{align*} s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\ s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\ \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t) \end{align*} -``` +$$ ### Band @@ -347,7 +347,7 @@ B_T &= 2(D+1)W_m ## Power, energy and autocorrelation -```math +$$ \begin{align*} G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\ @@ -360,24 +360,24 @@ B_T &= 2(D+1)W_m E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\ R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation} \end{align*} -``` +$$ ## ## Noise performance -```math +$$ \begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\ \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio} \end{align*} -``` +$$ ## Sampling -```math +$$ \begin{align*} t&=nT_s\\ T_s&=\frac{1}{f_s}\\ @@ -385,7 +385,7 @@ B_T &= 2(D+1)W_m X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\ B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\ \end{align*} -``` +$$ ### Procedure to reconstruct sampled signal @@ -407,21 +407,21 @@ 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 $T_s$: -```math +$$ X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s} -``` +$$ Calculate $C_n$ coefficient as follows from $x_p(t)$: -```math +$$ \begin{align*} % C_n&=X_p(nf_s)\\ C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*} -``` +$$ @@ -440,15 +440,15 @@ Cannot add directly due to copyright! ## Quantizer -```math +$$ \begin{align*} \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\ \end{align*} -``` +$$ ### Quantization noise -```math +$$ \begin{align*} e &:= y-x\quad\text{Quantization error}\\ \mu_E &= E[E] = 0\quad\text{Zero mean}\\ @@ -456,7 +456,7 @@ Cannot add directly due to copyright! \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\ \text{SQNR(dB)}&=10\log_{10}(\text{SQNR}) \end{align*} -``` +$$ ### 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`) @@ -468,7 +468,7 @@ Cannot add directly due to copyright! ![binary_codes](/assets/img/2024-10-29-Idiots-guide-to-ELEC/Line_Codes.drawio.svg) -```math +$$ \begin{align*} R_b&\rightarrow\text{Bit rate}\\ D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\ @@ -482,7 +482,7 @@ Cannot add directly due to copyright! G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\ G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b \end{align*} -``` +$$ ## Modulation and basis functions @@ -492,27 +492,27 @@ Cannot add directly due to copyright! #### Basis functions -```math +$$ \begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*} -``` +$$ #### Symbol mapping -```math +$$ b_n:\{1,0\}\to a_n:\{1,0\} -``` +$$ #### 2 possible waveforms -```math +$$ \begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\ s_1(t)&=0\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$} \end{align*} -``` +$$ Distance is $d=\sqrt{2E_b}$ @@ -520,27 +520,27 @@ Distance is $d=\sqrt{2E_b}$ #### Basis functions -```math +$$ \begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*} -``` +$$ #### Symbol mapping -```math +$$ b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\} -``` +$$ #### 2 possible waveforms -```math +$$ \begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\ s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$} \end{align*} -``` +$$ Distance is $d=2\sqrt{E_b}$ @@ -548,45 +548,45 @@ Distance is $d=2\sqrt{E_b}$ #### Basis functions -```math +$$ \begin{align*} T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\ \varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\ \varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\ \end{align*} -``` +$$ ### 4 possible waveforms -```math +$$ \begin{align*} s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\ s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\ s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\ s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\ \end{align*} -``` +$$ Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows: -```math +$$ \begin{align*} s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ \end{align*} -``` +$$ #### Signal -```math +$$ \begin{align*} \text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\ I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\ Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\ x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)] \end{align*} -``` +$$ ### Example of waveform @@ -632,20 +632,20 @@ Remember that $T=2T_b$ Find transfer function $h(t)$ of matched filter and apply to an input: -```math +$$ \begin{align*} h(t)&=s_1(T-t)-s_2(T-t)\\ h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\ s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\ n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output} \end{align*} -``` +$$ ### 2. Bit error rate Bit error rate (BER) from matched filter outputs and filter output noise -```math +$$ \begin{align*} % H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right) @@ -660,7 +660,7 @@ Bit error rate (BER) from matched filter outputs and filter output noise \text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\ \text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\ \end{align*} -``` +$$
@@ -745,7 +745,7 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems ### Receiver output shit -```math +$$ \begin{align*} r_o(t)&=\begin{cases} s_{o1}(t)+n_o(t) & \text{code 1}\\ @@ -753,14 +753,14 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems \end{cases}\\ n&: \text{AWGN with }\sigma_o^2\\ \end{align*} -``` +$$ +$$ --> ## ISI, channel model @@ -770,22 +770,22 @@ TODO: ### Nomenclature -```math +$$ \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*} -``` +$$ ### Raised cosine (RC) pulse ![Raised cosine pulse](/assets/img/2024-10-29-Idiots-guide-to-ELEC/RC.drawio.svg) -```math +$$ 0\leq\alpha\leq1 -``` +$$ ⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method. @@ -802,35 +802,35 @@ To solve this type of question: #### Symbol set size $M$ -```math +$$ \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*} -``` +$$ ### Nyquist stuff #### TODO: Condition for 0 ISI -```math +$$ P_r(kT)=\begin{cases} 1 & k=0\\ 0 & k\neq0 \end{cases} -``` +$$ #### Other -```math +$$ \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*} -``` +$$ ### Table of bandpass signalling and BER @@ -866,7 +866,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems ### Entropy for discrete random variables -```math +$$ \begin{align*} H(x) &\geq 0\\ H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\ @@ -878,7 +878,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems H(x|y) &= H(x,y)-H(y)\\ H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\ \end{align*} -``` +$$ Entropy is **maximized** when all have an equal probability. @@ -886,37 +886,37 @@ Entropy is **maximized** when all have an equal probability. TODO: Cut out if not required -```math +$$ \begin{align*} h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx \end{align*} -``` +$$ ### Mutual information Amount of entropy decrease of $x$ after observation by $y$. -```math +$$ \begin{align*} I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\ \end{align*} -``` +$$ ### Channel model Vertical, $x$: input\ Horizontal, $y$: output -```math +$$ \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] -``` +$$ -```math +$$ \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} \\ @@ -924,7 +924,7 @@ Horizontal, $y$: output \vdots & \vdots & \vdots & \ddots & \vdots \\ x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\ \end{array} -``` +$$ Input has probability distribution $p_X(a_i)=P(X=a_i)$ @@ -932,17 +932,17 @@ Channel maps alphabet $`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$ -```math +$$ \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*} -``` +$$ #### Fast procedure to calculate $I(y;x)$ -```math +$$ \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}\\ @@ -951,7 +951,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$ &\text{5. Find }H(x|y)=H(x,y)-H(y)\\ &\text{6. Find }I(y;x)=H(x)-H(x|y)\\ \end{align*} -``` +$$ ### Channel types @@ -962,7 +962,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$ #### Channel capacity of weakly symmetric channel -```math +$$ \begin{align*} C &\to\text{Channel capacity (bits/channels used)}\\ N &\to\text{Output alphabet size}\\ @@ -970,30 +970,30 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$ 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*} -``` +$$ #### Channel capacity of an AWGN channel -```math +$$ y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2) -``` +$$ -```math +$$ C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right) -``` +$$ #### Channel capacity of a bandwidth AWGN channel Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition. -```math +$$ \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*} -``` +$$ ## Channel code @@ -1022,7 +1022,7 @@ For a linear block code, $d_\text{min}=w_\text{min}$ Each generator vector is a binary string of size $n$. There are $k$ generator vectors in $\mathbf{G}$. -```math +$$ \begin{align*} \mathbf{g}_i&=[\begin{matrix} g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1} @@ -1040,11 +1040,11 @@ Each generator vector is a binary string of size $n$. There are $k$ generator ve g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\ \end{matrix}\right] \end{align*} -``` +$$ A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$: -```math +$$ \begin{align*} \mathbf{m}&=[\begin{matrix} m_{0}& \dots & m_{n-2} & m_{k-1} @@ -1052,13 +1052,13 @@ A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codew \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*} -``` +$$ ### Systemic linear block code Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits. -```math +$$ \begin{align*} \mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[ \begin{array}{c|c} @@ -1081,13 +1081,13 @@ Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits afte \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*} -``` +$$ #### Parity check matrix $\mathbf{H}$ Transpose $\mathbf{P}$ for the parity check matrix -```math +$$ \begin{align*} \mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\ &=\left[ @@ -1114,7 +1114,7 @@ Transpose $\mathbf{P}$ for the parity check matrix \end{matrix}\end{array}\right]\\ \mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid} \end{align*} -``` +$$ #### Procedure to find parity check matrix from list of codewords @@ -1125,7 +1125,7 @@ Transpose $\mathbf{P}$ for the parity check matrix Example: -```math +$$ \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 & x_5 \\\hline \color{magenta}1&\color{magenta}0&1&1&0\\ @@ -1133,11 +1133,11 @@ Example: \color{magenta}0&\color{magenta}0&0&0&0\\ \color{magenta}1&\color{magenta}1&0&0&1\\ \end{array} -``` +$$ Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$. -```math +$$ \begin{align*} \begin{aligned} x_3 &= x_1\oplus x_2\\ @@ -1165,7 +1165,7 @@ Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$. 0 & 0 & 1\\ \end{matrix}\end{array}\right] \end{align*} -``` +$$ #### Error detection and correction