Jekyll doesn't use ``math `` blocks for rendering math

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Peter 2024-10-29 17:01:55 +08:00
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@ -69,7 +69,7 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
| $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ | | $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ |
| $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ | | $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ |
```math $$
\begin{align*} \begin{align*}
u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\ u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
\text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\ \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
@ -77,7 +77,7 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
\text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\ \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\ g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
\end{align*} \end{align*}
``` $$
### Fourier transform of continuous time periodic signal ### 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 $T_s$: Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
```math $$
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s} X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
``` $$
Calculate $C_n$ coefficient as follows from $x_p(t)$: Calculate $C_n$ coefficient as follows from $x_p(t)$:
<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ --> <!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
```math $$
\begin{align*} \begin{align*}
% C_n&=X_p(nf_s)\\ % C_n&=X_p(nf_s)\\
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
\end{align*} \end{align*}
``` $$
### $\text{rect}$ function ### $\text{rect}$ function
@ -109,16 +109,16 @@ Calculate $C_n$ coefficient as follows from $x_p(t)$:
### Bessel function ### Bessel function
```math $$
\begin{align*} \begin{align*}
\sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\ \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
J_n(\beta)&=(-1)^nJ_{-n}(\beta) J_n(\beta)&=(-1)^nJ_{-n}(\beta)
\end{align*} \end{align*}
``` $$
### White noise ### White noise
```math $$
\begin{align*} \begin{align*}
R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\ R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
G_w(f)&=\frac{N_0}{2}\\ G_w(f)&=\frac{N_0}{2}\\
@ -126,27 +126,27 @@ N_0&=kT\\
G_y(f)&=|H(f)|^2G_w(f)\\ G_y(f)&=|H(f)|^2G_w(f)\\
G_y(f)&=G(f)G_w(f)\\ G_y(f)&=G(f)G_w(f)\\
\end{align*} \end{align*}
``` $$
### WSS ### WSS
```math $$
\begin{align*} \begin{align*}
\mu_X(t) &= \mu_X\text{ Constant}\\ \mu_X(t) &= \mu_X\text{ Constant}\\
R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\ R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)] E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
\end{align*} \end{align*}
``` $$
### Ergodicity ### Ergodicity
```math $$
\begin{align*} \begin{align*}
\braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\ \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
\braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\ \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\
E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\ E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\
\end{align*} \end{align*}
``` $$
| Type | Normal | Mean square sense | | 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 ### Other identities
```math $$
\begin{align*} \begin{align*}
f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\ f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\
a(f*g) &= (af)*g \quad\text{Convolution associative}\\ a(f*g) &= (af)*g \quad\text{Convolution associative}\\
\sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right) \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right)
\end{align*} \end{align*}
``` $$
### Other trig ### Other trig
```math $$
\begin{align*} \begin{align*}
\cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\ \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\
e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\ e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\
@ -180,9 +180,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
\cos(A+\pi/2)&=-\sin(A)\\ \cos(A+\pi/2)&=-\sin(A)\\
\int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\ \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\
\end{align*} \end{align*}
``` $$
```math $$
\begin{align*} \begin{align*}
\cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\ \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\
\sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\ \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\
@ -190,9 +190,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
\cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\ \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\
\sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\ \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\
\end{align*} \end{align*}
``` $$
```math $$
\begin{align*} \begin{align*}
\cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
\cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
@ -201,7 +201,7 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
\cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
\cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
\end{align*} \end{align*}
``` $$
## IQ/Complex envelope ## 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 ### Convert complex envelope representation to time-domain representation of signal
```math $$
\begin{align*} \begin{align*}
g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\ g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\
&=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\ &=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\
@ -219,23 +219,23 @@ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\
g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\ g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\
g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\ g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\
\end{align*} \end{align*}
``` $$
### For transfer function ### For transfer function
```math $$
\begin{align*} \begin{align*}
h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
&=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\ &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\
\Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))} \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))}
\end{align*} \end{align*}
``` $$
## AM ## AM
### CAM ### CAM
```math $$
\begin{align*} \begin{align*}
m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\ m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\ m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
@ -247,7 +247,7 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
\eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\ \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
B_T&=2f_m=2B B_T&=2f_m=2B
\end{align*} \end{align*}
``` $$
$B_T$: Signal bandwidth $B_T$: Signal bandwidth
$B$: Bandwidth of modulating wave $B$: Bandwidth of modulating wave
@ -256,16 +256,16 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
### DSB-SC ### DSB-SC
```math $$
\begin{align*} \begin{align*}
x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\ x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
B_T&=2f_m=2B B_T&=2f_m=2B
\end{align*} \end{align*}
``` $$
## FM/PM ## FM/PM
```math $$
\begin{align*} \begin{align*}
s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\ s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\ s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\
@ -274,30 +274,30 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
\Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\ \Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)} D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}
\end{align*} \end{align*}
``` $$
### Bessel form and magnitude spectrum (single tone) ### Bessel form and magnitude spectrum (single tone)
```math $$
\begin{align*} \begin{align*}
s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t] s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
\end{align*} \end{align*}
``` $$
### FM signal power ### FM signal power
```math $$
\begin{align*} \begin{align*}
P_\text{av}&=\frac{ {A_c}^2}{2}\\ P_\text{av}&=\frac{ {A_c}^2}{2}\\
P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\ P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
\text{band\_index}&=0\implies f_c+0f_m\\ \text{band\_index}&=0\implies f_c+0f_m\\
\text{band\_index}&=1\implies f_c+1f_m,\dots\\ \text{band\_index}&=1\implies f_c+1f_m,\dots\\
\end{align*} \end{align*}
``` $$
### Carson's rule to find $B$ (98% power bandwidth rule) ### Carson's rule to find $B$ (98% power bandwidth rule)
```math $$
\begin{align*} \begin{align*}
B &= 2Mf_m = 2(\beta + 1)f_m\\ B &= 2Mf_m = 2(\beta + 1)f_m\\
&= 2(\Delta f+f_m)\\ &= 2(\Delta f+f_m)\\
@ -308,7 +308,7 @@ B &= \begin{cases}
2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message} 2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
\end{cases}\\ \end{cases}\\
\end{align*} \end{align*}
``` $$
#### $\Delta f$ of arbitrary modulating signal #### $\Delta f$ of arbitrary modulating signal
@ -316,7 +316,7 @@ Find instantaneous frequency $f_\text{FM}$.
$M$: Number of **pairs** of significant sidebands $M$: Number of **pairs** of significant sidebands
```math $$
\begin{align*} \begin{align*}
s(t)&=A_c\cos(\theta_\text{FM}(t))\\ s(t)&=A_c\cos(\theta_\text{FM}(t))\\
f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\ f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\
@ -327,17 +327,17 @@ W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\
D &= \frac{\Delta f}{W_m}\\ D &= \frac{\Delta f}{W_m}\\
B_T &= 2(D+1)W_m B_T &= 2(D+1)W_m
\end{align*} \end{align*}
``` $$
### Complex envelope ### Complex envelope
```math $$
\begin{align*} \begin{align*}
s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\ s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\ s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
\tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t) \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
\end{align*} \end{align*}
``` $$
### Band ### Band
@ -347,7 +347,7 @@ B_T &= 2(D+1)W_m
## Power, energy and autocorrelation ## Power, energy and autocorrelation
```math $$
\begin{align*} \begin{align*}
G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_\text{WGN}(f)&=\frac{N_0}{2}\\
G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\ G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
@ -360,24 +360,24 @@ B_T &= 2(D+1)W_m
E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\ E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\
R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation} R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
\end{align*} \end{align*}
``` $$
## ##
## Noise performance ## Noise performance
```math $$
\begin{align*} \begin{align*}
\text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\
\text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\
\text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\ \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\
\text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio} \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio}
\end{align*} \end{align*}
``` $$
## Sampling ## Sampling
```math $$
\begin{align*} \begin{align*}
t&=nT_s\\ t&=nT_s\\
T_s&=\frac{1}{f_s}\\ T_s&=\frac{1}{f_s}\\
@ -385,7 +385,7 @@ B_T &= 2(D+1)W_m
X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\ X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\ B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\
\end{align*} \end{align*}
``` $$
### Procedure to reconstruct sampled signal ### 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 $T_s$: Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
```math $$
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s} X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
``` $$
Calculate $C_n$ coefficient as follows from $x_p(t)$: Calculate $C_n$ coefficient as follows from $x_p(t)$:
<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ --> <!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
```math $$
\begin{align*} \begin{align*}
% C_n&=X_p(nf_s)\\ % C_n&=X_p(nf_s)\\
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
\end{align*} \end{align*}
``` $$
<!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ --> <!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ -->
@ -440,15 +440,15 @@ Cannot add directly due to copyright!
## Quantizer ## Quantizer
```math $$
\begin{align*} \begin{align*}
\Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\ \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\
\end{align*} \end{align*}
``` $$
### Quantization noise ### Quantization noise
```math $$
\begin{align*} \begin{align*}
e &:= y-x\quad\text{Quantization error}\\ e &:= y-x\quad\text{Quantization error}\\
\mu_E &= E[E] = 0\quad\text{Zero mean}\\ \mu_E &= E[E] = 0\quad\text{Zero mean}\\
@ -456,7 +456,7 @@ Cannot add directly due to copyright!
\text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\ \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\
\text{SQNR(dB)}&=10\log_{10}(\text{SQNR}) \text{SQNR(dB)}&=10\log_{10}(\text{SQNR})
\end{align*} \end{align*}
``` $$
### 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`) ### 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) ![binary_codes](/assets/img/2024-10-29-Idiots-guide-to-ELEC/Line_Codes.drawio.svg)
```math $$
\begin{align*} \begin{align*}
R_b&\rightarrow\text{Bit rate}\\ R_b&\rightarrow\text{Bit rate}\\
D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\ D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\
@ -482,7 +482,7 @@ Cannot add directly due to copyright!
G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\ G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\
G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b
\end{align*} \end{align*}
``` $$
## Modulation and basis functions ## Modulation and basis functions
@ -492,27 +492,27 @@ Cannot add directly due to copyright!
#### Basis functions #### Basis functions
```math $$
\begin{align*} \begin{align*}
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
\end{align*} \end{align*}
``` $$
#### Symbol mapping #### Symbol mapping
```math $$
b_n:\{1,0\}\to a_n:\{1,0\} b_n:\{1,0\}\to a_n:\{1,0\}
``` $$
#### 2 possible waveforms #### 2 possible waveforms
```math $$
\begin{align*} \begin{align*}
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\ s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\
s_1(t)&=0\\ s_1(t)&=0\\
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$} &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$}
\end{align*} \end{align*}
``` $$
Distance is $d=\sqrt{2E_b}$ Distance is $d=\sqrt{2E_b}$
@ -520,27 +520,27 @@ Distance is $d=\sqrt{2E_b}$
#### Basis functions #### Basis functions
```math $$
\begin{align*} \begin{align*}
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
\end{align*} \end{align*}
``` $$
#### Symbol mapping #### Symbol mapping
```math $$
b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\} b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}
``` $$
#### 2 possible waveforms #### 2 possible waveforms
```math $$
\begin{align*} \begin{align*}
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\ s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\
s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\ s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$} &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$}
\end{align*} \end{align*}
``` $$
Distance is $d=2\sqrt{E_b}$ Distance is $d=2\sqrt{E_b}$
@ -548,45 +548,45 @@ Distance is $d=2\sqrt{E_b}$
#### Basis functions #### Basis functions
```math $$
\begin{align*} \begin{align*}
T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\ T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\
\varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\ \varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\
\varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\ \varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\
\end{align*} \end{align*}
``` $$
### 4 possible waveforms ### 4 possible waveforms
```math $$
\begin{align*} \begin{align*}
s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\ s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\
s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\ s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\
s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\ s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\
s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\ s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\
\end{align*} \end{align*}
``` $$
Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows: Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows:
```math $$
\begin{align*} \begin{align*}
s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
&=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
&=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
\end{align*} \end{align*}
``` $$
#### Signal #### Signal
```math $$
\begin{align*} \begin{align*}
\text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\ \text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\
I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\ I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\
Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\ Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\
x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)] x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)]
\end{align*} \end{align*}
``` $$
### Example of waveform ### 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: Find transfer function $h(t)$ of matched filter and apply to an input:
```math $$
\begin{align*} \begin{align*}
h(t)&=s_1(T-t)-s_2(T-t)\\ h(t)&=s_1(T-t)-s_2(T-t)\\
h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\ h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\
s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\ s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\
n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output} n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output}
\end{align*} \end{align*}
``` $$
### 2. Bit error rate ### 2. Bit error rate
Bit error rate (BER) from matched filter outputs and filter output noise Bit error rate (BER) from matched filter outputs and filter output noise
```math $$
\begin{align*} \begin{align*}
% H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right) % H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right)
@ -660,7 +660,7 @@ Bit error rate (BER) from matched filter outputs and filter output noise
\text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\ \text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\
\text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\ \text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\
\end{align*} \end{align*}
``` $$
<div style="page-break-after: always;"></div> <div style="page-break-after: always;"></div>
@ -745,7 +745,7 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
### Receiver output shit ### Receiver output shit
```math $$
\begin{align*} \begin{align*}
r_o(t)&=\begin{cases} r_o(t)&=\begin{cases}
s_{o1}(t)+n_o(t) & \text{code 1}\\ s_{o1}(t)+n_o(t) & \text{code 1}\\
@ -753,14 +753,14 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
\end{cases}\\ \end{cases}\\
n&: \text{AWGN with }\sigma_o^2\\ n&: \text{AWGN with }\sigma_o^2\\
\end{align*} \end{align*}
``` $$
<!-- <!--
```math $$
\begin{align*} \begin{align*}
G_x(f) G_x(f)
\end{align*} \end{align*}
``` --> $$ -->
## ISI, channel model ## ISI, channel model
@ -770,22 +770,22 @@ TODO:
### Nomenclature ### Nomenclature
```math $$
\begin{align*} \begin{align*}
D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\ D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\
T&\rightarrow\text{Symbol Duration}\\ T&\rightarrow\text{Symbol Duration}\\
M&\rightarrow\text{Symbol set size}\\ M&\rightarrow\text{Symbol set size}\\
W&\rightarrow\text{Bandwidth}\\ W&\rightarrow\text{Bandwidth}\\
\end{align*} \end{align*}
``` $$
### Raised cosine (RC) pulse ### Raised cosine (RC) pulse
![Raised cosine pulse](/assets/img/2024-10-29-Idiots-guide-to-ELEC/RC.drawio.svg) ![Raised cosine pulse](/assets/img/2024-10-29-Idiots-guide-to-ELEC/RC.drawio.svg)
```math $$
0\leq\alpha\leq1 0\leq\alpha\leq1
``` $$
⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method. ⚠ 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 $M$ #### Symbol set size $M$
```math $$
\begin{align*} \begin{align*}
D\text{ symbol/s}&=\frac{2W\text{ Hz}}{1+\alpha}\\ D\text{ symbol/s}&=\frac{2W\text{ Hz}}{1+\alpha}\\
R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\ R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\
M\text{ symbol/set}&=2^k\\ M\text{ symbol/set}&=2^k\\
E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\ E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
\end{align*} \end{align*}
``` $$
### Nyquist stuff ### Nyquist stuff
#### TODO: Condition for 0 ISI #### TODO: Condition for 0 ISI
```math $$
P_r(kT)=\begin{cases} P_r(kT)=\begin{cases}
1 & k=0\\ 1 & k=0\\
0 & k\neq0 0 & k\neq0
\end{cases} \end{cases}
``` $$
#### Other #### Other
```math $$
\begin{align*} \begin{align*}
\text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\ \text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\
\alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
T&=1/D T&=1/D
\end{align*} \end{align*}
``` $$
### Table of bandpass signalling and BER ### 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 ### Entropy for discrete random variables
```math $$
\begin{align*} \begin{align*}
H(x) &\geq 0\\ H(x) &\geq 0\\
H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\ H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\
@ -878,7 +878,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems
H(x|y) &= H(x,y)-H(y)\\ H(x|y) &= H(x,y)-H(y)\\
H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\ H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\
\end{align*} \end{align*}
``` $$
Entropy is **maximized** when all have an equal probability. 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 TODO: Cut out if not required
```math $$
\begin{align*} \begin{align*}
h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx
\end{align*} \end{align*}
``` $$
### Mutual information ### Mutual information
Amount of entropy decrease of $x$ after observation by $y$. Amount of entropy decrease of $x$ after observation by $y$.
```math $$
\begin{align*} \begin{align*}
I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\ I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\
\end{align*} \end{align*}
``` $$
### Channel model ### Channel model
Vertical, $x$: input\ Vertical, $x$: input\
Horizontal, $y$: output Horizontal, $y$: output
```math $$
\mathbf{P}=\left[\begin{matrix} \mathbf{P}=\left[\begin{matrix}
p_{11} & p_{12} &\dots & p_{1N}\\ p_{11} & p_{12} &\dots & p_{1N}\\
p_{21} & p_{22} &\dots & p_{2N}\\ p_{21} & p_{22} &\dots & p_{2N}\\
\vdots & \vdots &\ddots & \vdots\\ \vdots & \vdots &\ddots & \vdots\\
p_{M1} & p_{M2} &\dots & p_{MN}\\ p_{M1} & p_{M2} &\dots & p_{MN}\\
\end{matrix}\right] \end{matrix}\right]
``` $$
```math $$
\begin{array}{c|cccc} \begin{array}{c|cccc}
P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline
x_1 & p_{11} & p_{12} & \dots & p_{1N} \\ x_1 & p_{11} & p_{12} & \dots & p_{1N} \\
@ -924,7 +924,7 @@ Horizontal, $y$: output
\vdots & \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \\
x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\ x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\
\end{array} \end{array}
``` $$
Input has probability distribution $p_X(a_i)=P(X=a_i)$ 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 $p_Y(b_j)=P(y=b_j)$ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
```math $$
\begin{align*} \begin{align*}
p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\ p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\
&= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\ &= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\
[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P} [\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}
\end{align*} \end{align*}
``` $$
#### Fast procedure to calculate $I(y;x)$ #### Fast procedure to calculate $I(y;x)$
```math $$
\begin{align*} \begin{align*}
&\text{1. Find }H(x)\\ &\text{1. Find }H(x)\\
&\text{2. Find }[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] = [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}\\ &\text{2. Find }[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] = [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}\\
@ -951,7 +951,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
&\text{5. Find }H(x|y)=H(x,y)-H(y)\\ &\text{5. Find }H(x|y)=H(x,y)-H(y)\\
&\text{6. Find }I(y;x)=H(x)-H(x|y)\\ &\text{6. Find }I(y;x)=H(x)-H(x|y)\\
\end{align*} \end{align*}
``` $$
### Channel types ### Channel types
@ -962,7 +962,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
#### Channel capacity of weakly symmetric channel #### Channel capacity of weakly symmetric channel
```math $$
\begin{align*} \begin{align*}
C &\to\text{Channel capacity (bits/channels used)}\\ C &\to\text{Channel capacity (bits/channels used)}\\
N &\to\text{Output alphabet size}\\ N &\to\text{Output alphabet size}\\
@ -970,30 +970,30 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\ C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\
R &< C \text{ for error-free transmission} R &< C \text{ for error-free transmission}
\end{align*} \end{align*}
``` $$
#### Channel capacity of an AWGN channel #### Channel capacity of an AWGN channel
```math $$
y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2) y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
``` $$
```math $$
C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right) C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)
``` $$
#### Channel capacity of a bandwidth AWGN channel #### Channel capacity of a bandwidth AWGN channel
Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition. Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition.
```math $$
\begin{align*} \begin{align*}
P_s&\to\text{Bandwidth limited average power}\\ P_s&\to\text{Bandwidth limited average power}\\
y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\ y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\
C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\ C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\
C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W) C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W)
\end{align*} \end{align*}
``` $$
## Channel code ## 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 $\mathbf{G}$. Each generator vector is a binary string of size $n$. There are $k$ generator vectors in $\mathbf{G}$.
```math $$
\begin{align*} \begin{align*}
\mathbf{g}_i&=[\begin{matrix} \mathbf{g}_i&=[\begin{matrix}
g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1} g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1}
@ -1040,11 +1040,11 @@ Each generator vector is a binary string of size $n$. There are $k$ generator ve
g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\ g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\
\end{matrix}\right] \end{matrix}\right]
\end{align*} \end{align*}
``` $$
A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$: A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$:
```math $$
\begin{align*} \begin{align*}
\mathbf{m}&=[\begin{matrix} \mathbf{m}&=[\begin{matrix}
m_{0}& \dots & m_{n-2} & m_{k-1} m_{0}& \dots & m_{n-2} & m_{k-1}
@ -1052,13 +1052,13 @@ A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codew
\color{darkgray}\mathbf{m}&\color{darkgray}=[101001]\quad\text{Example for $k=6$}\\ \color{darkgray}\mathbf{m}&\color{darkgray}=[101001]\quad\text{Example for $k=6$}\\
\mathbf{x} &= \mathbf{m}\mathbf{G}=m_0\mathbf{g}_0+m_1\mathbf{g}_1+\dots+m_{k-1}\mathbf{g}_{k-1} \mathbf{x} &= \mathbf{m}\mathbf{G}=m_0\mathbf{g}_0+m_1\mathbf{g}_1+\dots+m_{k-1}\mathbf{g}_{k-1}
\end{align*} \end{align*}
``` $$
### Systemic linear block code ### Systemic linear block code
Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits. Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits.
```math $$
\begin{align*} \begin{align*}
\mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[ \mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[
\begin{array}{c|c} \begin{array}{c|c}
@ -1081,13 +1081,13 @@ Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits afte
\mathbf{x} &= \mathbf{m}\mathbf{G}= \mathbf{m} \begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\begin{array}{c|c}[\mathbf{mI}_k & \mathbf{mP}]\end{array}=\begin{array}{c|c}[\mathbf{m} & \mathbf{b}]\end{array}\\ \mathbf{x} &= \mathbf{m}\mathbf{G}= \mathbf{m} \begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\begin{array}{c|c}[\mathbf{mI}_k & \mathbf{mP}]\end{array}=\begin{array}{c|c}[\mathbf{m} & \mathbf{b}]\end{array}\\
\mathbf{b} &= \mathbf{m}\mathbf{P}\quad\text{Parity bits of $\mathbf{x}$} \mathbf{b} &= \mathbf{m}\mathbf{P}\quad\text{Parity bits of $\mathbf{x}$}
\end{align*} \end{align*}
``` $$
#### Parity check matrix $\mathbf{H}$ #### Parity check matrix $\mathbf{H}$
Transpose $\mathbf{P}$ for the parity check matrix Transpose $\mathbf{P}$ for the parity check matrix
```math $$
\begin{align*} \begin{align*}
\mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\ \mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\
&=\left[ &=\left[
@ -1114,7 +1114,7 @@ Transpose $\mathbf{P}$ for the parity check matrix
\end{matrix}\end{array}\right]\\ \end{matrix}\end{array}\right]\\
\mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid} \mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid}
\end{align*} \end{align*}
``` $$
#### Procedure to find parity check matrix from list of codewords #### Procedure to find parity check matrix from list of codewords
@ -1125,7 +1125,7 @@ Transpose $\mathbf{P}$ for the parity check matrix
Example: Example:
```math $$
\begin{array}{cccc} \begin{array}{cccc}
x_1 & x_2 & x_3 & x_4 & x_5 \\\hline x_1 & x_2 & x_3 & x_4 & x_5 \\\hline
\color{magenta}1&\color{magenta}0&1&1&0\\ \color{magenta}1&\color{magenta}0&1&1&0\\
@ -1133,11 +1133,11 @@ Example:
\color{magenta}0&\color{magenta}0&0&0&0\\ \color{magenta}0&\color{magenta}0&0&0&0\\
\color{magenta}1&\color{magenta}1&0&0&1\\ \color{magenta}1&\color{magenta}1&0&0&1\\
\end{array} \end{array}
``` $$
Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$. Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
```math $$
\begin{align*} \begin{align*}
\begin{aligned} \begin{aligned}
x_3 &= x_1\oplus x_2\\ x_3 &= x_1\oplus x_2\\
@ -1165,7 +1165,7 @@ Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
0 & 0 & 1\\ 0 & 0 & 1\\
\end{matrix}\end{array}\right] \end{matrix}\end{array}\right]
\end{align*} \end{align*}
``` $$
#### Error detection and correction #### Error detection and correction