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1172 lines
54 KiB
Markdown
1172 lines
54 KiB
Markdown
# Idiot's guide to ELEC4402 communication systems
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## License and information
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Notes are open-source and licensed under the GNU GPL-3.0. **You must include the [full-text of the license](/COPYING.txt) and follow its terms when using these notes or any diagrams in derivative works** (but not when printing as notes)
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Copyright (C) 2024 Peter Tanner
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<details>
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<summary>GPL copyright information</summary>
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Copyright (C) 2024 Peter Tanner
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program. If not, see <http://www.gnu.org/licenses/>.
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</details>
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[Access a web-copy of the notes for printing](/README.html)
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I accept pull requests or suggestions but the content must not be copyrighted under a non-GPL compatible license.
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## Fourier transform identities
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| **Time Function** | **Fourier Transform** |
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| --------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
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| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \, \text{sinc}(fT)$ |
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| $\text{sinc}(2Wt)$ | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ |
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| $\exp(-at)u(t), \, a>0$ | $\frac{1}{a + j2\pi f}$ |
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| $\exp(-a\lvert t \rvert), \, a>0$ | $\frac{2a}{a^2 + (2\pi f)^2}$ |
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| $\exp(-\pi t^2)$ | $\exp(-\pi f^2)$ |
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| $\begin{cases} 1 - \frac{\lvert t \rvert}{T}, & \lvert t \rvert < T \\ 0, & \lvert t \rvert \geq T \end{cases}$ | $T \, \text{sinc}^2(fT)$ |
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| $\delta(t)$ | $1$ |
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| $1$ | $\delta(f)$ |
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| $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ |
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| $g(t-a)$ | $\exp(-j2\pi fa)G(f)\quad\text{shift property}$ |
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| $g(bt)$ | $\frac{G(f/b)}{\|b\|}\quad\text{scaling property}$ |
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| $g(bt-a)$ | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)\quad\text{shift \& scale}$ |
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| $\frac{d}{dt}g(t)$ | $j2\pi fG(f)\quad\text{differentiation property}$ |
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| $G(t)$ | $g(-f)\quad\text{duality property}$ |
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| $g(t)h(t)$ | $G(f)*H(f)$ |
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| $g(t)*h(t)$ | $G(f)H(f)$ |
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| $\exp(j2\pi f_c t)$ | $\delta(f - f_c)$ |
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| $\cos(2\pi f_c t)$ | $\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$ |
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| $\sin(2\pi f_c t)$ | $\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$ |
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| $\text{sgn}(t)$ | $\frac{1}{j\pi f}$ |
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| $\frac{1}{\pi t}$ | $-j \, \text{sgn}(f)$ |
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| $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ |
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| $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ |
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| **Function Name** | **Formula** |
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| -------------------- | ------------------------------------------------------------------------------------------------------- |
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| Unit Step Function | $u(t) = \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}$ |
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| Signum Function | $\text{sgn}(t) = \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}$ |
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| sinc Function | $\text{sinc}(2Wt) = \frac{\sin(2\pi W t)}{2\pi W t}$ |
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| Rectangular Function | $\text{rect}(t) = \Pi(t) = \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}$ |
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| Convolution | $g(t)*h(t)=(g*h)(t)=\int_\infty^\infty g(\tau)h(t-\tau)d\tau$ |
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### Fourier transform of continuous time periodic signal
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Required for some questions on **sampling**:
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<!-- Transform a continuous time-periodic signal $x(t)=\sum_{n=-\infty}^\infty x_p(t-nT_s)$ with period $T_s$: -->
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Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
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$$
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X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
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$$
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Calculate $C_n$ coefficient as follows from $x_p(t)$:
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<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
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$$
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\begin{align*}
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% C_n&=X_p(nf_s)\\
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C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
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&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
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\end{align*}
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$$
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### $\text{rect}$ function
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### Bessel function
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$$
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\begin{align*}
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\sum_{n\in\Z}{J_n}^2(\beta)&=1\\
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J_n(\beta)&=(-1)^nJ_{-n}(\beta)
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\end{align*}
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$$
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### White noise
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$$
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\begin{align*}
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R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
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G_w(f)&=\frac{N_0}{2}\\
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N_0&=kT\\
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G_y(f)&=|H(f)|^2G_w(f)\\
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G_y(f)&=G(f)G_w(f)\\
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\end{align*}
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$$
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### WSS
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$$
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\begin{align*}
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\mu_X(t) &= \mu_X\text{ Constant}\\
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R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
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E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
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\end{align*}
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$$
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### Ergodicity
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$$
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\begin{align*}
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\braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
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\braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\
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E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\
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\end{align*}
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$$
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| Type | Normal | Mean square sense |
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| ----------------------------------- | ------------------------------------------------------- | ----------------------------------------------------------- |
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| ergodic in mean | $$\lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X$$ | $$\lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0$$ |
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| ergodic in autocorrelation function | $$\lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau)$$ | $$\lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0$$ |
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**A WSS random process needs to be both ergodic in mean and autocorrelation to be considered an ergodic process**
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### Other identities
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$$
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\begin{align*}
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f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\
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a(f*g) &= (af)*g \quad\text{Convolution associative}\\
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\sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right)
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\end{align*}
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$$
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### Other trig
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$$
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\begin{align*}
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\cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\
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e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\
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e^{-j\alpha}+e^{j\alpha}&=2 \cos(\alpha)\\
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\cos(-A)&=\cos(A)\\
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\sin(-A)&=-\sin(A)\\
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\sin(A+\pi/2)&=\cos(A)\\
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\sin(A-\pi/2)&=-\cos(A)\\
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\cos(A-\pi/2)&=\sin(A)\\
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\cos(A+\pi/2)&=-\sin(A)\\
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\int_{x\in\R}\text{sinc}(A x) &= \frac{1}{|A|}\\
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\end{align*}
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$$
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$$
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\begin{align*}
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\cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\
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\sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\
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\cos(A)\cos(B) &= \frac{1}{2} (\cos (A-B)+\cos (A+B)) \\
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\cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\
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\sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\
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\end{align*}
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$$
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$$
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\begin{align*}
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\cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
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\cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
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\sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\
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\sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
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\cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
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\cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
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\end{align*}
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$$
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## IQ/Complex envelope
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Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $e^{j\theta}$ form.
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### Convert complex envelope representation to time-domain representation of signal
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$$
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\begin{align*}
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g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\
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&=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\
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&=A(t)\cos(2\pi f_c t+\phi(t))\\
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A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\
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\phi(t)&\quad\text{Phase}\\
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g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\
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g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\
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\end{align*}
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$$
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### For transfer function
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$$
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\begin{align*}
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h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
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&=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\
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\Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))}
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\end{align*}
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$$
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## AM
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### CAM
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$$
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\begin{align*}
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m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
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m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
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m_a&=k_a A_m \quad\text{ (Symmetrical $m(t)$)}\\
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x(t)&=A_c\cos(2\pi f_c t)\left[1+k_a m(t)\right]=A_c\cos(2\pi f_c t)\left[1+m_a m(t)/A_c\right], \\
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&\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\
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P_c &=\frac{{A_c}^2}{2}\quad\text{Carrier power}\\
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P_x &=\frac{1}{4}{m_a}^2{A_c}^2\\
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\eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
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B_T&=2f_m=2B
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\end{align*}
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$$
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$B_T$: Signal bandwidth
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$B$: Bandwidth of modulating wave
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Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
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### DSB-SC
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$$
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\begin{align*}
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x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
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B_T&=2f_m=2B
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\end{align*}
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$$
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## FM/PM
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$$
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\begin{align*}
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s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
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s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\
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s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\quad\text{FM single tone}\\
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\beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\
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\Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
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D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}
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\end{align*}
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$$
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### Bessel form and magnitude spectrum (single tone)
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$$
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\begin{align*}
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s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
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\end{align*}
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$$
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<!-- ### Bessel repr.
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$$s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m) t)$$ -->
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### FM signal power
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$$
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\begin{align*}
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P_\text{av}&=\frac{{A_c}^2}{2}\\
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P_\text{band\_index}&=\frac{{A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
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\text{band\_index}&=0\implies f_c+0f_m\\
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\text{band\_index}&=1\implies f_c+1f_m,\dots\\
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\end{align*}
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$$
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### Carson's rule to find $B$ (98% power bandwidth rule)
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$$
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\begin{align*}
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B &= 2Mf_m = 2(\beta + 1)f_m\\
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&= 2(\Delta f+f_m)\\
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&= 2(k_f A_m+f_m)\\
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&= 2(D+1)W_m\\
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B &= \begin{cases}
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2(\Delta f+f_m) & \text{FM, sinusoidal message}\\
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2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
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\end{cases}\\
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\end{align*}
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$$
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#### $\Delta f$ of arbitrary modulating signal
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Find instantaneous frequency $f_\text{FM}$.
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$M$: Number of **pairs** of significant sidebands
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$$
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\begin{align*}
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s(t)&=A_c\cos(\theta_\text{FM}(t))\\
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f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\
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A_m &= \max_t|m(t)|\\
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\Delta f &= \max_t(f_\text{FM}(t)) - f_c\\
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W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\
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\text{Example: }&\text{sinc}(At+t)+2\cos(2\pi t)=\frac{\sin(2\pi((At+t)/2))}{\pi(At+t)}+2\cos(2\pi t)\to W_m=\max\left(\frac{A+1}{2},1\right)\\
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D &= \frac{\Delta f}{W_m}\\
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B_T &= 2(D+1)W_m
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\end{align*}
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$$
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### Complex envelope
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$$
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\begin{align*}
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s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
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s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
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\tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
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\end{align*}
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$$
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### Band
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| Narrowband | Wideband |
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| ------------- | ------------- |
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| $D<1,\beta<1$ | $D>1,\beta>1$ |
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## Power, energy and autocorrelation
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$$
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\begin{align*}
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G_\text{WGN}(f)&=\frac{N_0}{2}\\
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G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
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G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\
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G_x(f)&=\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\text{ (PSD)}\\
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G_x(f)&=\mathfrak{F}[R_x(\tau)]\text{ (WSS)}\\
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P&=\sigma^2=\int_\mathbb{R}G_x(f)df\\
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P&=\sigma^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\\
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P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\
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E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\
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R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
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\end{align*}
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$$
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##
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## Noise performance
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$$
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\begin{align*}
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\text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\
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\text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\
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\text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\
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\text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio}
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\end{align*}
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$$
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## Sampling
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$$
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\begin{align*}
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t&=nT_s\\
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|
T_s&=\frac{1}{f_s}\\
|
|
x_s(t)&=x(t)\delta_s(t)=x(t)\sum_{n\in\mathbb{Z}}\delta(t-nT_s)=\sum_{n\in\mathbb{Z}}x(nT_s)\delta(t-nT_s)\\
|
|
X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
|
|
B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\
|
|
\end{align*}
|
|
$$
|
|
|
|
### 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 $x(t)=\sum_{n=-\infty}^\infty x_p(t-nT_s)$ with period $T_s$: -->
|
|
|
|
Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:
|
|
|
|
$$
|
|
X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
|
|
$$
|
|
|
|
Calculate $C_n$ coefficient as follows from $x_p(t)$:
|
|
|
|
<!-- Remember $X_p(f)\leftrightarrow x_p(t)$ and **NOT** $\color{red}X_p(f)\leftrightarrow x_p(t-nT_s)$ -->
|
|
|
|
$$
|
|
\begin{align*}
|
|
% C_n&=X_p(nf_s)\\
|
|
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
|
|
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
|
|
\end{align*}
|
|
$$
|
|
|
|
<!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ -->
|
|
|
|
### 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`)
|
|
|
|
|
|
|
|
|
|
## 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`)
|
|
|
|
|
|
|
|
|
|
## 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
|
|
|
|
$$b_n:\{1,0\}\to a_n:\{1,\color{lime}-1\color{white}\}$$
|
|
|
|
#### 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
|
|
|
|
<details>
|
|
<summary>Code</summary>
|
|
<pre><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)"]
|
|
</code></pre>
|
|
|
|
</details>
|
|
|
|
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
|
|
|
|
$$
|
|
\begin{align*}
|
|
% H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right)
|
|
|
|
% \text{BER}_\text{bin}&=p Q\left(\frac{s_{o1}-V_T}{\sigma_o}\right)+(1-p)Q\left(\frac{V_T-s_{o2}}{\sigma_o}\right)\text{, $p\rightarrow$Probability $s_1(t)$ sent, $V_T\rightarrow$Threshold voltage}
|
|
Q(x)&=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Leftrightarrow\text{erf}\left(\frac{x}{\sqrt{2}}\right)=1-2Q(x)\\
|
|
E_b&=d^2=\int_{-\infty}^\infty|s_1(t)-s_2(t)|^2dt\quad\text{Energy per bit/Distance}\\
|
|
T&=1/R_b\quad\text{$R_b$: Bitrate}\\
|
|
E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
|
|
P(\text{W})&=10^{\frac{P(\text{dB})}{10}}\\
|
|
P_\text{RX}(W)&=P_\text{TX}(W)\cdot10^{\frac{P_\text{loss}(\text{dB})}{10}}\quad \text{$P_\text{loss}$ is expressed with negative sign e.g. "-130 dB"}\\
|
|
\text{BER}_\text{MatchedFilter}&=Q\left(\sqrt{\frac{d^2}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{2N_0}}\right)\\
|
|
\text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\
|
|
\text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\
|
|
\end{align*}
|
|
$$
|
|
|
|
<div style="page-break-after: always;"></div>
|
|
|
|
## Value tables for $\text{erf}(x)$ and $Q(x)$
|
|
|
|
### $\text{erf}(x)$ function
|
|
|
|
| $x$ | $\text{erf}(x)$ | $x$ | $\text{erf}(x)$ | $x$ | $\text{erf}(x)$ |
|
|
| ------ | --------------- | ------ | --------------- | ------ | --------------- |
|
|
| $0.00$ | $0.00000$ | $0.75$ | $0.71116$ | $1.50$ | $0.96611$ |
|
|
| $0.05$ | $0.05637$ | $0.80$ | $0.74210$ | $1.55$ | $0.97162$ |
|
|
| $0.10$ | $0.11246$ | $0.85$ | $0.77067$ | $1.60$ | $0.97635$ |
|
|
| $0.15$ | $0.16800$ | $0.90$ | $0.79691$ | $1.65$ | $0.98038$ |
|
|
| $0.20$ | $0.22270$ | $0.95$ | $0.82089$ | $1.70$ | $0.98379$ |
|
|
| $0.25$ | $0.27633$ | $1.00$ | $0.84270$ | $1.75$ | $0.98667$ |
|
|
| $0.30$ | $0.32863$ | $1.05$ | $0.86244$ | $1.80$ | $0.98909$ |
|
|
| $0.35$ | $0.37938$ | $1.10$ | $0.88021$ | $1.85$ | $0.99111$ |
|
|
| $0.40$ | $0.42839$ | $1.15$ | $0.89612$ | $1.90$ | $0.99279$ |
|
|
| $0.45$ | $0.47548$ | $1.20$ | $0.91031$ | $1.95$ | $0.99418$ |
|
|
| $0.50$ | $0.52050$ | $1.25$ | $0.92290$ | $2.00$ | $0.99532$ |
|
|
| $0.55$ | $0.56332$ | $1.30$ | $0.93401$ | $2.50$ | $0.99959$ |
|
|
| $0.60$ | $0.60386$ | $1.35$ | $0.94376$ | $3.00$ | $0.99998$ |
|
|
| $0.65$ | $0.64203$ | $1.40$ | $0.95229$ | $3.30$ | $0.999998$\*\* |
|
|
| $0.70$ | $0.67780$ | $1.45$ | $0.95970$ | | |
|
|
|
|
### $Q(x)$ 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
|
|
|
|
\*\*The value of $\text{erf}(3.30)$ should be $\approx0.999997$ instead, but this value is quoted in the formula table.
|
|
|
|
### Receiver output shit
|
|
|
|
$$
|
|
\begin{align*}
|
|
r_o(t)&=\begin{cases}
|
|
s_{o1}(t)+n_o(t) & \text{code 1}\\
|
|
s_{o2}(t)+n_o(t) & \text{code 0}\\
|
|
\end{cases}\\
|
|
n&: \text{AWGN with }\sigma_o^2\\
|
|
\end{align*}
|
|
$$
|
|
|
|
<!--
|
|
$$
|
|
\begin{align*}
|
|
G_x(f)
|
|
\end{align*}
|
|
$$ -->
|
|
|
|
## 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:
|
|
|
|
1. Use the formula for $D$ below
|
|
2. Consult the BER table below to get the BER which relates the noise of the channel $N_0$ to $E_b$ and to $R_b$.
|
|
|
|
<!-- $$
|
|
V_\text{RC}(f)=\begin{cases}
|
|
T & 0\le|f|\le(1-\alpha)/2T\\
|
|
\frac{T}{2}\left(1+\cos\left[\right]\right)
|
|
\end{cases}
|
|
$$ -->
|
|
|
|
| Linear modulation ($M$-PSK, $M$-QAM) | NRZ unipolar encoding |
|
|
| --------------------------------------------------- | -------------------------------------------------- |
|
|
| $W=B_\text{\color{lime}abs-abs}$ | $W=B_\text{\color{lime}abs}$ |
|
|
| $W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ | $W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ |
|
|
| $D=\frac{W\text{ symbol/s}}{1+\alpha}$ | $D=\frac{2W\text{ symbol/s}}{1+\alpha}$ |
|
|
|
|
#### 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
|
|
|
|
#### Condition for 0 ISI
|
|
|
|
$$
|
|
P_r(kT)=\begin{cases}
|
|
1 & k=0\\
|
|
0 & k\neq0
|
|
\end{cases}
|
|
$$
|
|
|
|
#### Other
|
|
|
|
$$
|
|
\begin{align*}
|
|
\text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\
|
|
\alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
|
|
T&=1/D
|
|
\end{align*}
|
|
$$
|
|
|
|
<div style="page-break-after: always;"></div>
|
|
|
|
### 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**: $x\neq\bold{0}$, 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:
|
|
|
|
1. Zero vector must be present at least once
|
|
2. 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 $\mathbf{G}$.
|
|
|
|
$$
|
|
\begin{align*}
|
|
\mathbf{g}_i&=[\begin{matrix}
|
|
g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1}
|
|
\end{matrix}]\\
|
|
\color{darkgray}\mathbf{g}_0&\color{darkgray}=[1010]\quad\text{Example for $n=4$}\\
|
|
\mathbf{G}&=\left[\begin{matrix}
|
|
\mathbf{g}_0\\
|
|
\mathbf{g}_1\\
|
|
\vdots\\
|
|
\mathbf{g}_{k-1}\\
|
|
\end{matrix}\right]=\left[\begin{matrix}
|
|
g_{0,0}& \dots & g_{0,n-2} & g_{0,n-1}\\
|
|
g_{1,0}& \dots & g_{1,n-2} & g_{1,n-1}\\
|
|
\vdots & \ddots & \vdots & \vdots\\
|
|
g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\
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|
\end{matrix}\right]
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|
\end{align*}
|
|
$$
|
|
|
|
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
|
|
|
|
1. From the number of codewords, find $k=\log_2(N)$
|
|
2. Partition codewords into $k$ information bits and remaining bits into $n-k$ parity bits. The information bits should be a simple counter (?).
|
|
3. Express parity bits as a linear combination of information bits
|
|
4. 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 $x_1,x_2$.
|
|
|
|
$$
|
|
\begin{align*}
|
|
\begin{align*}
|
|
x_3 &= x_1\oplus x_2\\
|
|
x_4 &= x_1\oplus x_2\\
|
|
x_5 &= x_2\\
|
|
\end{align*}
|
|
\implies\textbf{P}&=
|
|
\begin{array}{c|ccc}
|
|
& x_1 & x_2 \\
|
|
\hline
|
|
x_3&1&1&\\
|
|
x_4&1&1&\\
|
|
x_5&0&1&\\
|
|
\end{array}\\
|
|
\textbf{H}&=\left[
|
|
\begin{array}{c|c}
|
|
\begin{matrix}
|
|
1&1\\
|
|
1&1\\
|
|
0&1\\
|
|
\end{matrix}
|
|
&
|
|
\begin{matrix}
|
|
1 & 0 & 0\\
|
|
0 & 1 & 0\\
|
|
0 & 0 & 1\\
|
|
\end{matrix}\end{array}\right]
|
|
\end{align*}
|
|
$$
|
|
|
|
#### 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
|