mirror of
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synced 2024-11-30 12:00:18 +08:00
Jekyll doesn't use ``math `` blocks for rendering math
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@ -69,7 +69,7 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
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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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```math
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$$
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\begin{align*}
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u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
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\text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
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@ -77,7 +77,7 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
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\text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
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g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
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\end{align*}
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```
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$$
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### Fourier transform of continuous time periodic signal
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@ -87,21 +87,21 @@ Required for some questions on **sampling**:
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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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```math
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$$
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X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
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```
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$$
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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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```math
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$$
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\begin{align*}
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% C_n&=X_p(nf_s)\\
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C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
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&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
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\end{align*}
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```
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$$
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### $\text{rect}$ function
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@ -109,16 +109,16 @@ Calculate $C_n$ coefficient as follows from $x_p(t)$:
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### Bessel function
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```math
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$$
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\begin{align*}
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\sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
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J_n(\beta)&=(-1)^nJ_{-n}(\beta)
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\end{align*}
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```
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$$
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### White noise
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```math
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$$
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\begin{align*}
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R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
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G_w(f)&=\frac{N_0}{2}\\
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@ -126,27 +126,27 @@ N_0&=kT\\
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G_y(f)&=|H(f)|^2G_w(f)\\
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G_y(f)&=G(f)G_w(f)\\
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\end{align*}
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```
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$$
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### WSS
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```math
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$$
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\begin{align*}
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\mu_X(t) &= \mu_X\text{ Constant}\\
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R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
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E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
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\end{align*}
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```
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$$
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### Ergodicity
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```math
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$$
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\begin{align*}
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\braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
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\braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\
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E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\
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\end{align*}
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```
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$$
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| Type | Normal | Mean square sense |
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| ----------------------------------- | ------------------------------------------------------- | ----------------------------------------------------------- |
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@ -157,17 +157,17 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
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### Other identities
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```math
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$$
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\begin{align*}
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f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\
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a(f*g) &= (af)*g \quad\text{Convolution associative}\\
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\sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right)
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\end{align*}
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```
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$$
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### Other trig
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```math
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$$
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\begin{align*}
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\cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\
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e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\
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@ -180,9 +180,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
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\cos(A+\pi/2)&=-\sin(A)\\
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\int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\
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\end{align*}
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```
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$$
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```math
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$$
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\begin{align*}
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\cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\
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\sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\
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@ -190,9 +190,9 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
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\cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\
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\sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\
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\end{align*}
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```
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$$
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```math
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$$
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\begin{align*}
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\cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\
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\cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\
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@ -201,7 +201,7 @@ Note: **A WSS random process needs to be both ergodic in mean and autocorrelatio
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\cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\
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\cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\
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\end{align*}
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```
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$$
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## IQ/Complex envelope
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@ -209,7 +209,7 @@ Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $
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### Convert complex envelope representation to time-domain representation of signal
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```math
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$$
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\begin{align*}
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g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\
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&=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\
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@ -219,23 +219,23 @@ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\
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g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\
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g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\
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\end{align*}
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```
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$$
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### For transfer function
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```math
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$$
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\begin{align*}
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h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
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&=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\
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\Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))}
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\end{align*}
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```
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$$
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## AM
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### CAM
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```math
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$$
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\begin{align*}
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m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
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m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
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@ -247,7 +247,7 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
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\eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
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B_T&=2f_m=2B
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\end{align*}
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```
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$$
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$B_T$: Signal bandwidth
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$B$: Bandwidth of modulating wave
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@ -256,16 +256,16 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
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### DSB-SC
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```math
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$$
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\begin{align*}
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x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
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B_T&=2f_m=2B
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\end{align*}
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```
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$$
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## FM/PM
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```math
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$$
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\begin{align*}
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s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\
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s(t) &= A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\
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@ -274,30 +274,30 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$
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\Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\
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D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}
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\end{align*}
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```
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$$
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### Bessel form and magnitude spectrum (single tone)
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```math
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$$
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\begin{align*}
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s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
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\end{align*}
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```
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$$
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### FM signal power
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```math
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$$
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\begin{align*}
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P_\text{av}&=\frac{ {A_c}^2}{2}\\
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P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
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\text{band\_index}&=0\implies f_c+0f_m\\
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\text{band\_index}&=1\implies f_c+1f_m,\dots\\
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\end{align*}
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```
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$$
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### Carson's rule to find $B$ (98% power bandwidth rule)
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```math
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$$
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\begin{align*}
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B &= 2Mf_m = 2(\beta + 1)f_m\\
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&= 2(\Delta f+f_m)\\
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@ -308,7 +308,7 @@ B &= \begin{cases}
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2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message}
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\end{cases}\\
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\end{align*}
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```
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$$
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#### $\Delta f$ of arbitrary modulating signal
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@ -316,7 +316,7 @@ Find instantaneous frequency $f_\text{FM}$.
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$M$: Number of **pairs** of significant sidebands
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```math
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$$
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\begin{align*}
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s(t)&=A_c\cos(\theta_\text{FM}(t))\\
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f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\
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@ -327,17 +327,17 @@ W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\
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D &= \frac{\Delta f}{W_m}\\
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B_T &= 2(D+1)W_m
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\end{align*}
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```
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$$
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### Complex envelope
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```math
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$$
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\begin{align*}
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s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
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s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
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\tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
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\end{align*}
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```
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$$
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### Band
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@ -347,7 +347,7 @@ B_T &= 2(D+1)W_m
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## Power, energy and autocorrelation
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```math
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$$
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\begin{align*}
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G_\text{WGN}(f)&=\frac{N_0}{2}\\
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G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
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@ -360,24 +360,24 @@ B_T &= 2(D+1)W_m
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E&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\
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R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
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\end{align*}
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```
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$$
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##
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## Noise performance
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```math
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$$
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\begin{align*}
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\text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\
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\text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\
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\text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\
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\text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio}
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\end{align*}
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```
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$$
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## Sampling
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```math
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$$
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\begin{align*}
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t&=nT_s\\
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T_s&=\frac{1}{f_s}\\
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@ -385,7 +385,7 @@ B_T &= 2(D+1)W_m
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X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\
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B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\
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\end{align*}
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```
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$$
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### Procedure to reconstruct sampled signal
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@ -407,21 +407,21 @@ Required for some questions on **sampling**:
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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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```math
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$$
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X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}
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```
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$$
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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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```math
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$$
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\begin{align*}
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% C_n&=X_p(nf_s)\\
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C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\
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&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
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\end{align*}
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```
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$$
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<!-- Reconstruct from $\bar{X_s}(f)$ within the range $[-f_s/2,f_s/2]$ -->
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@ -440,15 +440,15 @@ Cannot add directly due to copyright!
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## Quantizer
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```math
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$$
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\begin{align*}
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\Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\
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\end{align*}
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```
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$$
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### Quantization noise
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```math
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$$
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\begin{align*}
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e &:= y-x\quad\text{Quantization error}\\
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\mu_E &= E[E] = 0\quad\text{Zero mean}\\
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@ -456,7 +456,7 @@ Cannot add directly due to copyright!
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\text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\
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\text{SQNR(dB)}&=10\log_{10}(\text{SQNR})
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\end{align*}
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```
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$$
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### 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`)
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@ -468,7 +468,7 @@ Cannot add directly due to copyright!
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```math
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$$
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\begin{align*}
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R_b&\rightarrow\text{Bit rate}\\
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D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\
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|
@ -482,7 +482,7 @@ Cannot add directly due to copyright!
|
|||
G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\
|
||||
G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
## Modulation and basis functions
|
||||
|
||||
|
|
@ -492,27 +492,27 @@ Cannot add directly due to copyright!
|
|||
|
||||
#### Basis functions
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Symbol mapping
|
||||
|
||||
```math
|
||||
$$
|
||||
b_n:\{1,0\}\to a_n:\{1,0\}
|
||||
```
|
||||
$$
|
||||
|
||||
#### 2 possible waveforms
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\
|
||||
s_1(t)&=0\\
|
||||
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
Distance is $d=\sqrt{2E_b}$
|
||||
|
||||
|
|
@ -520,27 +520,27 @@ Distance is $d=\sqrt{2E_b}$
|
|||
|
||||
#### Basis functions
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Symbol mapping
|
||||
|
||||
```math
|
||||
$$
|
||||
b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}
|
||||
```
|
||||
$$
|
||||
|
||||
#### 2 possible waveforms
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\
|
||||
s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\
|
||||
&\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
Distance is $d=2\sqrt{E_b}$
|
||||
|
||||
|
|
@ -548,45 +548,45 @@ Distance is $d=2\sqrt{E_b}$
|
|||
|
||||
#### Basis functions
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\
|
||||
\varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\
|
||||
\varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### 4 possible waveforms
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\
|
||||
s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\
|
||||
s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\
|
||||
s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows:
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
|
||||
&=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
|
||||
&=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Signal
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\
|
||||
I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\
|
||||
Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\
|
||||
x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)]
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Example of waveform
|
||||
|
||||
|
|
@ -632,20 +632,20 @@ Remember that $T=2T_b$
|
|||
|
||||
Find transfer function $h(t)$ of matched filter and apply to an input:
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
h(t)&=s_1(T-t)-s_2(T-t)\\
|
||||
h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\
|
||||
s_{on}(t)&=h(t)*s_n(t)=\int_\infty^\infty h(\tau)s_n(t-\tau)d\tau\quad\text{Filter output}\\
|
||||
n_o(t)&=h(t)*n(t)\quad\text{Noise at filter output}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### 2. Bit error rate
|
||||
|
||||
Bit error rate (BER) from matched filter outputs and filter output noise
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
% H_\text{opt}(f)&=\max_{H(f)}\left(\frac{s_{o1}-s_{o2}}{2\sigma_o}\right)
|
||||
|
||||
|
|
@ -660,7 +660,7 @@ Bit error rate (BER) from matched filter outputs and filter output noise
|
|||
\text{BER}_\text{unipolarNRZ|BASK}&=Q\left(\sqrt{\frac{d^2}{N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)\\
|
||||
\text{BER}_\text{polarNRZ|BPSK}&=Q\left(\sqrt{\frac{2d^2}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
<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
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
r_o(t)&=\begin{cases}
|
||||
s_{o1}(t)+n_o(t) & \text{code 1}\\
|
||||
|
|
@ -753,14 +753,14 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
|
|||
\end{cases}\\
|
||||
n&: \text{AWGN with }\sigma_o^2\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
<!--
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
G_x(f)
|
||||
\end{align*}
|
||||
``` -->
|
||||
$$ -->
|
||||
|
||||
## ISI, channel model
|
||||
|
||||
|
|
@ -770,22 +770,22 @@ TODO:
|
|||
|
||||
### Nomenclature
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\
|
||||
T&\rightarrow\text{Symbol Duration}\\
|
||||
M&\rightarrow\text{Symbol set size}\\
|
||||
W&\rightarrow\text{Bandwidth}\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Raised cosine (RC) pulse
|
||||
|
||||
|
||||
|
||||
```math
|
||||
$$
|
||||
0\leq\alpha\leq1
|
||||
```
|
||||
$$
|
||||
|
||||
⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method.
|
||||
|
||||
|
|
@ -802,35 +802,35 @@ To solve this type of question:
|
|||
|
||||
#### Symbol set size $M$
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
D\text{ symbol/s}&=\frac{2W\text{ Hz}}{1+\alpha}\\
|
||||
R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\
|
||||
M\text{ symbol/set}&=2^k\\
|
||||
E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Nyquist stuff
|
||||
|
||||
#### TODO: Condition for 0 ISI
|
||||
|
||||
```math
|
||||
$$
|
||||
P_r(kT)=\begin{cases}
|
||||
1 & k=0\\
|
||||
0 & k\neq0
|
||||
\end{cases}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Other
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\
|
||||
\alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\
|
||||
T&=1/D
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Table of bandpass signalling and BER
|
||||
|
||||
|
|
@ -866,7 +866,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems
|
|||
|
||||
### Entropy for discrete random variables
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
H(x) &\geq 0\\
|
||||
H(x) &= -\sum_{x_i\in A_x} p_X(x_i) \log_2(p_X(x_i))\\
|
||||
|
|
@ -878,7 +878,7 @@ Adapted from table 11.4 M F Mesiya - Contemporary Communication Systems
|
|||
H(x|y) &= H(x,y)-H(y)\\
|
||||
H(x,y) &= H(x) + H(y|x) = H(y) + H(x|y)\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
Entropy is **maximized** when all have an equal probability.
|
||||
|
||||
|
|
@ -886,37 +886,37 @@ Entropy is **maximized** when all have an equal probability.
|
|||
|
||||
TODO: Cut out if not required
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Mutual information
|
||||
|
||||
Amount of entropy decrease of $x$ after observation by $y$.
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
I(x;y) &= H(x)-H(x|y)=H(y)-H(y|x)\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Channel model
|
||||
|
||||
Vertical, $x$: input\
|
||||
Horizontal, $y$: output
|
||||
|
||||
```math
|
||||
$$
|
||||
\mathbf{P}=\left[\begin{matrix}
|
||||
p_{11} & p_{12} &\dots & p_{1N}\\
|
||||
p_{21} & p_{22} &\dots & p_{2N}\\
|
||||
\vdots & \vdots &\ddots & \vdots\\
|
||||
p_{M1} & p_{M2} &\dots & p_{MN}\\
|
||||
\end{matrix}\right]
|
||||
```
|
||||
$$
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{array}{c|cccc}
|
||||
P(y_j|x_i)& y_1 & y_2 & \dots & y_N \\\hline
|
||||
x_1 & p_{11} & p_{12} & \dots & p_{1N} \\
|
||||
|
|
@ -924,7 +924,7 @@ Horizontal, $y$: output
|
|||
\vdots & \vdots & \vdots & \ddots & \vdots \\
|
||||
x_M & p_{M1} & p_{M2} & \dots & p_{MN} \\
|
||||
\end{array}
|
||||
```
|
||||
$$
|
||||
|
||||
Input has probability distribution $p_X(a_i)=P(X=a_i)$
|
||||
|
||||
|
|
@ -932,17 +932,17 @@ Channel maps alphabet $`\{a_1,\dots,a_M\} \to \{b_1,\dots,b_N\}`$
|
|||
|
||||
Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
p_Y(b_j) &= \sum_{i=1}^{M}P[x=a_i,y=b_j]\quad 1\leq j\leq N \\
|
||||
&= \sum_{i=1}^{M}P[X=a_i]P[Y=b_j|X=a_i]\\
|
||||
[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] &= [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Fast procedure to calculate $I(y;x)$
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
&\text{1. Find }H(x)\\
|
||||
&\text{2. Find }[\begin{matrix}p_Y(b_0)&p_Y(b_1)&\dots&p_Y(b_j)\end{matrix}] = [\begin{matrix}p_X(a_0)&p_X(a_1)&\dots&p_X(a_i)\end{matrix}]\times\mathbf{P}\\
|
||||
|
|
@ -951,7 +951,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
|
|||
&\text{5. Find }H(x|y)=H(x,y)-H(y)\\
|
||||
&\text{6. Find }I(y;x)=H(x)-H(x|y)\\
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Channel types
|
||||
|
||||
|
|
@ -962,7 +962,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
|
|||
|
||||
#### Channel capacity of weakly symmetric channel
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
C &\to\text{Channel capacity (bits/channels used)}\\
|
||||
N &\to\text{Output alphabet size}\\
|
||||
|
|
@ -970,30 +970,30 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$
|
|||
C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\
|
||||
R &< C \text{ for error-free transmission}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Channel capacity of an AWGN channel
|
||||
|
||||
```math
|
||||
$$
|
||||
y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
|
||||
```
|
||||
$$
|
||||
|
||||
```math
|
||||
$$
|
||||
C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)
|
||||
```
|
||||
$$
|
||||
|
||||
#### Channel capacity of a bandwidth AWGN channel
|
||||
|
||||
Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition.
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
P_s&\to\text{Bandwidth limited average power}\\
|
||||
y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\
|
||||
C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\
|
||||
C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W)
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
## Channel code
|
||||
|
||||
|
|
@ -1022,7 +1022,7 @@ For a linear block code, $d_\text{min}=w_\text{min}$
|
|||
|
||||
Each generator vector is a binary string of size $n$. There are $k$ generator vectors in $\mathbf{G}$.
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\mathbf{g}_i&=[\begin{matrix}
|
||||
g_{i,0}& \dots & g_{i,n-2} & g_{i,n-1}
|
||||
|
|
@ -1040,11 +1040,11 @@ Each generator vector is a binary string of size $n$. There are $k$ generator ve
|
|||
g_{k-1,0}& \dots & g_{k-1,n-2} & g_{k-1,n-1}\\
|
||||
\end{matrix}\right]
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codewords in $\mathbf{G}$:
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\mathbf{m}&=[\begin{matrix}
|
||||
m_{0}& \dots & m_{n-2} & m_{k-1}
|
||||
|
|
@ -1052,13 +1052,13 @@ A message block $\mathbf{m}$ is coded as $\mathbf{x}$ using the generation codew
|
|||
\color{darkgray}\mathbf{m}&\color{darkgray}=[101001]\quad\text{Example for $k=6$}\\
|
||||
\mathbf{x} &= \mathbf{m}\mathbf{G}=m_0\mathbf{g}_0+m_1\mathbf{g}_1+\dots+m_{k-1}\mathbf{g}_{k-1}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
### Systemic linear block code
|
||||
|
||||
Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits after the message bits.
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\mathbf{G}&=\begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\left[
|
||||
\begin{array}{c|c}
|
||||
|
|
@ -1081,13 +1081,13 @@ Contains $k$ message bits (Copy $\mathbf{m}$ as-is) and $(n-k)$ parity bits afte
|
|||
\mathbf{x} &= \mathbf{m}\mathbf{G}= \mathbf{m} \begin{array}{c|c}[\mathbf{I}_k & \mathbf{P}]\end{array}=\begin{array}{c|c}[\mathbf{mI}_k & \mathbf{mP}]\end{array}=\begin{array}{c|c}[\mathbf{m} & \mathbf{b}]\end{array}\\
|
||||
\mathbf{b} &= \mathbf{m}\mathbf{P}\quad\text{Parity bits of $\mathbf{x}$}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Parity check matrix $\mathbf{H}$
|
||||
|
||||
Transpose $\mathbf{P}$ for the parity check matrix
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\mathbf{H}&=\begin{array}{c|c}[\mathbf{P}^\text{T} & \mathbf{I}_{n-k}]\end{array}\\
|
||||
&=\left[
|
||||
|
|
@ -1114,7 +1114,7 @@ Transpose $\mathbf{P}$ for the parity check matrix
|
|||
\end{matrix}\end{array}\right]\\
|
||||
\mathbf{xH}^\text{T}&=\mathbf{0}\implies\text{Codeword is valid}
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Procedure to find parity check matrix from list of codewords
|
||||
|
||||
|
|
@ -1125,7 +1125,7 @@ Transpose $\mathbf{P}$ for the parity check matrix
|
|||
|
||||
Example:
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{array}{cccc}
|
||||
x_1 & x_2 & x_3 & x_4 & x_5 \\\hline
|
||||
\color{magenta}1&\color{magenta}0&1&1&0\\
|
||||
|
|
@ -1133,11 +1133,11 @@ Example:
|
|||
\color{magenta}0&\color{magenta}0&0&0&0\\
|
||||
\color{magenta}1&\color{magenta}1&0&0&1\\
|
||||
\end{array}
|
||||
```
|
||||
$$
|
||||
|
||||
Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
|
||||
|
||||
```math
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{aligned}
|
||||
x_3 &= x_1\oplus x_2\\
|
||||
|
|
@ -1165,7 +1165,7 @@ Set $x_1,x_2$ as information bits. Express $x_3,x_4,x_5$ in terms of $x_1,x_2$.
|
|||
0 & 0 & 1\\
|
||||
\end{matrix}\end{array}\right]
|
||||
\end{align*}
|
||||
```
|
||||
$$
|
||||
|
||||
#### Error detection and correction
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue
Block a user