diff --git a/README.html b/README.html
index c71d0ab..1b5f148 100644
--- a/README.html
+++ b/README.html
@@ -383,15 +383,15 @@ along with this program. If not, see http
-u ( t ) = { 1 , t > 0 1 2 , t = 0 0 , t < 0 Unit Step Function sgn ( t ) = { + 1 , t > 0 0 , t = 0 − 1 , t < 0 Signum Function sinc ( 2 W t ) = sin ( 2 π W t ) 2 π W t sinc Function rect ( t ) = Π ( t ) = { 1 , − 0.5 < t < 0.5 0 , ∣ t ∣ > 0.5 Rectangular/Gate Function tri ( t / T ) = { 1 − ∣ t ∣ T , ∣ t ∣ < T 0 , ∣ t ∣ > T = Π ( t / T ) ∗ Π ( t / T ) Triangle Function g ( t ) ∗ h ( t ) = ( g ∗ h ) ( t ) = ∫ ∞ ∞ g ( τ ) h ( t − τ ) d τ Convolution \begin{align*}
+u ( t ) = { 1 , t > 0 1 2 , t = 0 0 , t < 0 Unit Step Function sgn ( t ) = { + 1 , t > 0 0 , t = 0 − 1 , t < 0 Signum Function sinc ( 2 W t ) = sin ( 2 π W t ) 2 π W t sinc Function rect ( t ) = Π ( t ) = { 1 , − 0.5 < t < 0.5 0 , ∣ t ∣ > 0.5 Rectangular/Gate Function tri ( t / T ) = { 1 − ∣ t ∣ T , ∣ t ∣ < T 0 , ∣ t ∣ ≥ T = 1 T Π ( t / T ) ∗ Π ( t / T ) Triangle Function g ( t ) ∗ h ( t ) = ( g ∗ h ) ( t ) = ∫ ∞ ∞ g ( τ ) h ( t − τ ) d τ Convolution \begin{align*}
u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\
\text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
\text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\
\text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\
- \text{tri}(t/T) &= \begin{cases} 1 - \frac{|t|}{T}, & \lvert t\rvert < T \\ 0, & \lvert t \rvert > T \end{cases}=\Pi(t/T)*\Pi(t/T)&\text{Triangle Function}\\
+ \text{tri}(t/T) &= \begin{cases} 1 - \frac{|t|}{T}, & \lvert t\rvert < T \\ 0, & \lvert t \rvert \geq T \end{cases}=\frac{1}{T}\Pi(t/T)*\Pi(t/T)&\text{Triangle Function}\\
g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
\end{align*}
- u ( t ) sgn ( t ) sinc ( 2 W t ) rect ( t ) = Π ( t ) tri ( t / T ) g ( t ) ∗ h ( t ) = ( g ∗ h ) ( t ) = ⎩ ⎨ ⎧ 1 , 2 1 , 0 , t > 0 t = 0 t < 0 = ⎩ ⎨ ⎧ + 1 , 0 , − 1 , t > 0 t = 0 t < 0 = 2 πW t sin ( 2 πW t ) = { 1 , 0 , − 0.5 < t < 0.5 ∣ t ∣ > 0.5 = { 1 − T ∣ t ∣ , 0 , ∣ t ∣ < T ∣ t ∣ > T = Π ( t / T ) ∗ Π ( t / T ) = ∫ ∞ ∞ g ( τ ) h ( t − τ ) d τ Unit Step Function Signum Function sinc Function Rectangular/Gate Function Triangle Function Convolution
+ u ( t ) sgn ( t ) sinc ( 2 W t ) rect ( t ) = Π ( t ) tri ( t / T ) g ( t ) ∗ h ( t ) = ( g ∗ h ) ( t ) = ⎩ ⎨ ⎧ 1 , 2 1 , 0 , t > 0 t = 0 t < 0 = ⎩ ⎨ ⎧ + 1 , 0 , − 1 , t > 0 t = 0 t < 0 = 2 πW t sin ( 2 πW t ) = { 1 , 0 , − 0.5 < t < 0.5 ∣ t ∣ > 0.5 = { 1 − T ∣ t ∣ , 0 , ∣ t ∣ < T ∣ t ∣ ≥ T = T 1 Π ( t / T ) ∗ Π ( t / T ) = ∫ ∞ ∞ g ( τ ) h ( t − τ ) d τ Unit Step Function Signum Function sinc Function Rectangular/Gate Function Triangle Function Convolution
Required for some questions on sampling :
@@ -408,46 +408,46 @@ along with this program. If not, see http
C n = T s 1 ∫ T s x p ( t ) exp ( − j 2 π f s t ) d t = T s 1 X ( n f s ) (TODO: Check) x ( t − n T s ) is contained in the interval T s Shape functions
+
+Random processes examples
+Example: separate RV from expression X ( t ) = A cos ( 2 π f c t ) A ∼ N ( μ = 5 , σ 2 = 1 ) ⟹ E [ X ( t ) ] = E [ A cos ( 2 π f c t ) ] = E [ A ] cos ( 2 π f c t ) = 5 cos ( 2 π f c t ) Example: random phase X ( t ) = B cos ( 2 π f c t + θ ) θ ∼ U ( 0 , 2 π ) ⟹ E [ X ( t ) ] = E [ B cos ( 2 π f c t + θ ) ] = B ∫ 0 2 π 1 2 π ⏟ uniform cos ( 2 π f c t + θ ) d θ = 0 \begin{align*}
+ &\text{Example: separate RV from expression}\\
+ X(t) &= A\cos(2\pi f_c t)\quad A\thicksim \mathcal{N}(\mu=5,\sigma^2=1)\\
+ \implies E[X(t)] &= E[A\cos(2\pi f_c t)] = E[A]\cos(2\pi f_c t) = 5\cos(2\pi f_c t)\\
+ &\text{Example: random phase}\\
+ X(t) &= B\cos(2\pi f_c t+\theta)\quad \theta\thicksim \mathcal{U}(0,2\pi)\\
+ \implies E[X(t)] &= E[B\cos(2\pi f_c t+\theta)] = B\int_0^{2\pi}\underbrace{\frac{1}{2\pi}}_{\text{uniform}}\cos(2\pi f_c t+\theta)d\theta=0
+\end{align*}
+ X ( t ) ⟹ E [ X ( t )] X ( t ) ⟹ E [ X ( t )] Example: separate RV from expression = A cos ( 2 π f c t ) A ∼ N ( μ = 5 , σ 2 = 1 ) = E [ A cos ( 2 π f c t )] = E [ A ] cos ( 2 π f c t ) = 5 cos ( 2 π f c t ) Example: random phase = B cos ( 2 π f c t + θ ) θ ∼ U ( 0 , 2 π ) = E [ B cos ( 2 π f c t + θ )] = B ∫ 0 2 π uniform 2 π 1 cos ( 2 π f c t + θ ) d θ = 0
+Wide sense stationary (WSS)
+Two conditions for WSS:
-rect \text{rect} rect tri \text{tri} tri Constant mean
+Autocorrelation only dependent on time difference
-TODO: Add graphic.
+μ X ( t ) = μ X Constant \mu_X(t) = \mu_X\text{ Constant} μ X ( t ) = μ X Constant R X X ( t 1 , t 2 ) = R X ( t 1 − t 2 ) = R X ( τ ) R_{XX}(t_1,t_2)=R_X(t_1-t_2)=R_X(\tau) R XX ( t 1 , t 2 ) = R X ( t 1 − t 2 ) = R X ( τ )
+
+μ X ( t ) = E [ X ( t ) ] \mu_X(t)=E[X(t)] μ X ( t ) = E [ X ( t )] E [ X ( t 1 ) X ( t 2 ) ] = E [ X ( t ) X ( t + τ ) ] E[X(t_1)X(t_2)]=E[X(t)X(t+\tau)] E [ X ( t 1 ) X ( t 2 )] = E [ X ( t ) X ( t + τ )]
-Tri placeholder: For tri ( t / T ) = 1 − ∥ t ∥ / T \text{tri}(t/T)=1-\|t\|/T tri ( t / T ) = 1 − ∥ t ∥/ T x x x − T -T − T T T T y y y 1 1 1
-Bessel function
-∑ n ∈ Z J n 2 ( β ) = 1 J n ( β ) = ( − 1 ) n J − n ( β ) \begin{align*}
- \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\
- J_n(\beta)&=(-1)^nJ_{-n}(\beta)
-\end{align*}
- n ∈ Z ∑ J n 2 ( β ) J n ( β ) = 1 = ( − 1 ) n J − n ( β )
-
-White noise
-R W ( τ ) = N 0 2 δ ( τ ) = k T 2 δ ( τ ) = σ 2 δ ( τ ) G w ( f ) = N 0 2 N 0 = k T G y ( f ) = ∣ H ( f ) ∣ 2 G w ( f ) G y ( f ) = G ( f ) G w ( f ) \begin{align*}
-R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
-G_w(f)&=\frac{N_0}{2}\\
-N_0&=kT\\
-G_y(f)&=|H(f)|^2G_w(f)\\
-G_y(f)&=G(f)G_w(f)\\
-\end{align*}
- R W ( τ ) G w ( f ) N 0 G y ( f ) G y ( f ) = 2 N 0 δ ( τ ) = 2 k T δ ( τ ) = σ 2 δ ( τ ) = 2 N 0 = k T = ∣ H ( f ) ∣ 2 G w ( f ) = G ( f ) G w ( f )
-
-WSS
-μ X ( t ) = μ X Constant R X X ( t 1 , t 2 ) = R X ( t 1 − t 2 ) = R X ( τ ) E [ X ( t 1 ) X ( t 2 ) ] = E [ X ( t ) X ( t + τ ) ] \begin{align*}
- \mu_X(t) &= \mu_X\text{ Constant}\\
- R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\
- E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)]
-\end{align*}
- μ X ( t ) R XX ( t 1 , t 2 ) E [ X ( t 1 ) X ( t 2 )] = μ X Constant = R X ( t 1 − t 2 ) = R X ( τ ) = E [ X ( t ) X ( t + τ )]
-
Ergodicity
⟨ X ( t ) ⟩ T = 1 2 T ∫ − T T x ( t ) d t ⟨ X ( t + τ ) X ( t ) ⟩ T = 1 2 T ∫ − T T x ( t + τ ) x ( t ) d t E [ ⟨ X ( t ) ⟩ T ] = 1 2 T ∫ − T T x ( t ) d t = 1 2 T ∫ − T T m X d t = m X \begin{align*}
\braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\
@@ -467,17 +467,13 @@ G_y(f)&=G(f)G_w(f)\\
ergodic in mean
-lim T → ∞ ⟨ X ( t ) ⟩ T = m X ( t ) = m X \lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X T → ∞ lim ⟨ X ( t ) ⟩ T = m X ( t ) = m X
-lim T → ∞ VAR [ ⟨ X ( t ) ⟩ T ] = 0 \lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0 T → ∞ lim VAR [ ⟨ X ( t ) ⟩ T ] = 0
-lim T → ∞ ⟨ X ( t ) ⟩ T = m X ( t ) = m X \lim_{T\to\infty}\braket{X(t)}_T=m_X(t)=m_X lim T → ∞ ⟨ X ( t ) ⟩ T = m X ( t ) = m X lim T → ∞ VAR [ ⟨ X ( t ) ⟩ T ] = 0 \lim_{T\to\infty}\text{VAR}[\braket{X(t)}_T]=0 lim T → ∞ VAR [ ⟨ X ( t ) ⟩ T ] = 0
ergodic in autocorrelation function
-lim T → ∞ ⟨ X ( t + τ ) X ( t ) ⟩ T = R X ( τ ) \lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau) T → ∞ lim ⟨ X ( t + τ ) X ( t ) ⟩ T = R X ( τ )
-lim T → ∞ VAR [ ⟨ X ( t + τ ) X ( t ) ⟩ T ] = 0 \lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0 T → ∞ lim VAR [ ⟨ X ( t + τ ) X ( t ) ⟩ T ] = 0
-lim T → ∞ ⟨ X ( t + τ ) X ( t ) ⟩ T = R X ( τ ) \lim_{T\to\infty}\braket{X(t+\tau)X(t)}_T=R_X(\tau) lim T → ∞ ⟨ X ( t + τ ) X ( t ) ⟩ T = R X ( τ ) lim T → ∞ VAR [ ⟨ X ( t + τ ) X ( t ) ⟩ T ] = 0 \lim_{T\to\infty}\text{VAR}[\braket{X(t+\tau)X(t)}_T]=0 lim T → ∞ VAR [ ⟨ X ( t + τ ) X ( t ) ⟩ T ] = 0
@@ -556,24 +552,24 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
h ( t ) ⇒ h ~ ( t ) = h I ( t ) cos ( 2 π f c t ) − h Q ( t ) sin ( 2 π f c t ) = 2 Re [ h ~ ( t ) exp ( j 2 π f c t ) ] = h I ( t ) /2 + j h Q ( t ) /2 = A ( t ) /2 exp ( j ϕ ( t ))
AM
-CAM
-x ( t ) = A c cos ( 2 π f c t ) [ 1 + k a m ( t ) ] = A c cos ( 2 π f c t ) [ 1 + m a m ( t ) / A c ] , where m ( t ) = A m m ^ ( t ) and m ^ ( t ) is the normalized modulating signal m a = ∣ min t ( k a m ( t ) ) ∣ A c k a is the amplitude sensitivity ( volt − 1 ), m a is the modulation index. m a = A max − A min A max + A min (Symmetrical m ( t ) ) m a = k a A m (Symmetrical m ( t ) ) P c = A c 2 2 Carrier power P x = 1 4 m a 2 A c 2 η = Signal Power Total Power = P x P x + P c B T = 2 f m = 2 B \begin{align*}
- 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], \\
+Conventional AM modulation (CAM)
+x ( t ) = A c cos ( 2 π f c t ) [ 1 + k a m ( t ) ] = A c cos ( 2 π f c t ) [ 1 + m a m ( t ) / A c ] CAM signal where m ( t ) = A m m ^ ( t ) and m ^ ( t ) is the normalized modulating signal m a = ∣ min t ( k a m ( t ) ) ∣ A c k a is the amplitude sensitivity ( volt − 1 ), m a is the modulation index. m a = A max − A min A max + A min (Symmetrical m ( t ) ) m a = k a A m (Symmetrical m ( t ) ) P c = A c 2 2 Carrier power P s = 1 4 m a 2 A c 2 Signal power, total of all 4 sideband power, single-tone case η = Signal Power Total Power = P s P s + P c = P s P x Power efficiency B T = 2 f m = 2 B \begin{align*}
+ 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]\quad\text{CAM signal}\\
&\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\
m_a &= \frac{|\min_t(k_a m(t))|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
m_a&=k_a A_m \quad\text{ (Symmetrical $m(t)$)}\\
P_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\
- P_x &=\frac{1}{4}{m_a}^2{A_c}^2\\
- \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\
- B_T&=2f_m=2B
+ P_s &=\frac{1}{4}{m_a}^2{A_c}^2\quad\text{Signal power, \textbf{total} of all 4 sideband power, \textbf{single-tone} case}\\
+ \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_s}{P_s+P_c}=\frac{P_s}{P_x}\quad\text{Power efficiency}\\
+ B_T&=2f_m=2B\\
\end{align*}
- x ( t ) m a m a m a P c P x η B T = A c cos ( 2 π f c t ) [ 1 + k a m ( t ) ] = A c cos ( 2 π f c t ) [ 1 + m a m ( t ) / A c ] , where m ( t ) = A m m ^ ( t ) and m ^ ( t ) is the normalized modulating signal = A c ∣ min t ( k a m ( t )) ∣ k a is the amplitude sensitivity ( volt − 1 ), m a is the modulation index. = A max + A min A max − A min (Symmetrical m ( t ) ) = k a A m (Symmetrical m ( t ) ) = 2 A c 2 Carrier power = 4 1 m a 2 A c 2 = Total Power Signal Power = P x + P c P x = 2 f m = 2 B
+ x ( t ) m a m a m a P c P s η B T = A c cos ( 2 π f c t ) [ 1 + k a m ( t ) ] = A c cos ( 2 π f c t ) [ 1 + m a m ( t ) / A c ] CAM signal where m ( t ) = A m m ^ ( t ) and m ^ ( t ) is the normalized modulating signal = A c ∣ min t ( k a m ( t )) ∣ k a is the amplitude sensitivity ( volt − 1 ), m a is the modulation index. = A max + A min A max − A min (Symmetrical m ( t ) ) = k a A m (Symmetrical m ( t ) ) = 2 A c 2 Carrier power = 4 1 m a 2 A c 2 Signal power, total of all 4 sideband power, single-tone case = Total Power Signal Power = P s + P c P s = P x P s Power efficiency = 2 f m = 2 B
B T B_T B T B B B
Overmodulation (resulting in phase reversals at crossing points): m a > 1 m_a>1 m a > 1
-DSB-SC
+Double sideband suppressed carrier (DSB-SC)
x DSB ( t ) = A c cos ( 2 π f c t ) m ( t ) B T = 2 f m = 2 B \begin{align*}
x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\
B_T&=2f_m=2B
@@ -593,58 +589,58 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
\end{align*}
s ( t ) s ( t ) s ( t ) f i ( t ) Δ f max Δ f max β D = A c cos [ 2 π f c t + k p m ( t ) ] Phase modulated (PM) = A c cos ( θ i ( t )) = A c cos [ 2 π f c t + 2 π k f ∫ − ∞ t m ( τ ) d τ ] Frequency modulated (FM) = A c cos [ 2 π f c t + β sin ( 2 π f m t ) ] FM single tone = 2 π 1 d t d θ i ( t ) = f c + k f m ( t ) = f c + Δ f max m ^ ( t ) Instantaneous frequency = t max ∣ f i ( t ) − f c ∣ = k f t max ∣ m ( t ) ∣ Maximum frequency deviation = k f A m Maximum frequency deviation (sinusoidal) = f m Δ f max Modulation index = W m Δ f max Deviation ratio, where W m is bandwidth of m ( t ) (Use FT)
-
-s ( t ) = A c cos [ 2 π f c t + β sin ( 2 π f m t ) ] ⇔ s ( t ) = A c ∑ n = − ∞ ∞ J n ( β ) cos [ 2 π ( f c + n f m ) t ] \begin{align*}
- s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
+Bessel function
+J n ( β ) = { J − n ( β ) n is even − J − n ( β ) n is odd 1 = ∑ n ∈ Z J n 2 ( β ) Conservation of power \begin{align*}
+ J_n(\beta)&=\begin{cases}
+ J_{-n}(\beta) & \text{$n$ is even}\\
+ -J_{-n}(\beta) & \text{$n$ is odd}
+ \end{cases}\\
+ 1&=\sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&\text{Conservation of power}\\
\end{align*}
- s ( t ) = A c cos [ 2 π f c t + β sin ( 2 π f m t ) ] ⇔ s ( t ) = A c n = − ∞ ∑ ∞ J n ( β ) cos [ 2 π ( f c + n f m ) t ]
+ J n ( β ) 1 = { J − n ( β ) − J − n ( β ) n is even n is odd = n ∈ Z ∑ J n 2 ( β ) Conservation of power
+
+
+s ( t ) = A c cos [ 2 π f c t + β sin ( 2 π f m t ) ] ⟺ s ( t ) = A c ∑ n = − ∞ ∞ J n ( β ) cos [ 2 π ( f c + n f m ) t ] \begin{align*}
+ s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\\
+ \Longleftrightarrow s(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t]
+\end{align*}
+ s ( t ) ⟺ s ( t ) = A c cos [ 2 π f c t + β sin ( 2 π f m t ) ] = A c n = − ∞ ∑ ∞ J n ( β ) cos [ 2 π ( f c + n f m ) t ]
FM signal power
-P av = A c 2 2 P band_index = A c 2 J band_index 2 ( β ) 2 band_index = 0 ⟹ f c + 0 f m band_index = 1 ⟹ f c + 1 f m , … \begin{align*}
- P_\text{av}&=\frac{ {A_c}^2}{2}\\
- P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\
- \text{band\_index}&=0\implies f_c+0f_m\\
- \text{band\_index}&=1\implies f_c+1f_m,\dots\\
+P av = A c 2 2 Av. power of full signal P i = A c 2 ∣ J i ( β ) ∣ 2 2 Av. power of band i i = 0 ⟹ f c + 0 f m Middle band i = 1 ⟹ f c + 1 f m 1st sideband i = − 1 ⟹ f c − 1 f m -1st sideband … \begin{align*}
+ P_\text{av}&=\frac{ {A_c}^2}{2}&\text{Av. power of full signal}\\
+ P_\text{i}&=\frac{ {A_c}^2|{J_\text{i}}(\beta)|^2}{2}&\text{Av. power of band $i$}\\
+ i=0&\implies f_c+0f_m&\text{Middle band}\\
+ i=1&\implies f_c+1f_m&\text{1st sideband}\\
+ i=-1&\implies f_c-1f_m&\text{-1st sideband}\\
+ &\dots\\
\end{align*}
- P av P band_index band_index band_index = 2 A c 2 = 2 A c 2 J band_index 2 ( β ) = 0 ⟹ f c + 0 f m = 1 ⟹ f c + 1 f m , …
+ P av P i i = 0 i = 1 i = − 1 = 2 A c 2 = 2 A c 2 ∣ J i ( β ) ∣ 2 ⟹ f c + 0 f m ⟹ f c + 1 f m ⟹ f c − 1 f m … Av. power of full signal Av. power of band i Middle band 1st sideband -1st sideband
Carson's rule to find B B B
-B = 2 M f m = 2 ( β + 1 ) f m = 2 ( Δ f max + f m ) = 2 ( D + 1 ) W m B = { 2 ( Δ f max + f m ) = 2 ( Δ f max + W m ) FM, sinusoidal message 2 ( Δ ϕ max + 1 ) f m = 2 ( Δ ϕ max + 1 ) W m PM, sinusoidal message \begin{align*}
-B &= 2Mf_m = 2(\beta + 1)f_m\\
- &= 2(\Delta f_\text{max}+f_m)\\
- &= 2(D+1)W_m\\
+B = 2 ( β + 1 ) f m B = 2 ( Δ f max + f m ) B = 2 ( D + 1 ) W m B = { 2 ( Δ f max + f m ) = 2 ( Δ f max + W m ) FM, sinusoidal message 2 ( Δ ϕ max + 1 ) f m = 2 ( Δ ϕ max + 1 ) W m PM, sinusoidal message D < 1 , β < 1 ⟹ Narrowband D > 1 , β > 1 ⟹ Wideband \begin{align*}
+B &= 2(\beta + 1)f_m\\
+B &= 2(\Delta f_\text{max}+f_m)\\
+B &= 2(D+1)W_m\\
B &= \begin{cases}
2(\Delta f_\text{max}+f_m)=2(\Delta f_\text{max}+W_m) & \text{FM, sinusoidal message}\\
2(\Delta\phi_\text{max} + 1)f_m=2(\Delta \phi_\text{max}+1)W_m & \text{PM, sinusoidal message}
-\end{cases}\\
+\end{cases}\\\\
+ & D<1,\beta<1 \implies \text{Narrowband}\quad D>1,\beta>1\implies \text{Wideband}
\end{align*}
- B B = 2 M f m = 2 ( β + 1 ) f m = 2 ( Δ f max + f m ) = 2 ( D + 1 ) W m = { 2 ( Δ f max + f m ) = 2 ( Δ f max + W m ) 2 ( Δ ϕ max + 1 ) f m = 2 ( Δ ϕ max + 1 ) W m FM, sinusoidal message PM, sinusoidal message
+ B B B B = 2 ( β + 1 ) f m = 2 ( Δ f max + f m ) = 2 ( D + 1 ) W m = { 2 ( Δ f max + f m ) = 2 ( Δ f max + W m ) 2 ( Δ ϕ max + 1 ) f m = 2 ( Δ ϕ max + 1 ) W m FM, sinusoidal message PM, sinusoidal message D < 1 , β < 1 ⟹ Narrowband D > 1 , β > 1 ⟹ Wideband
-Complex envelope
-s ( t ) = A c cos ( 2 π f c t + β sin ( 2 π f m t ) ) ⇔ s ~ ( t ) = A c exp ( j β sin ( 2 π f m t ) ) s ( t ) = Re [ s ~ ( t ) exp ( j 2 π f c t ) ] s ~ ( t ) = A c ∑ n = − ∞ ∞ J n ( β ) exp ( j 2 π f m t ) \begin{align*}
- s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\
+Complex envelope of a FM signal
+s ( t ) = A c cos ( 2 π f c t + β sin ( 2 π f m t ) ) ⟺ s ~ ( t ) = A c exp ( j β sin ( 2 π f m t ) ) s ( t ) = Re [ s ~ ( t ) exp ( j 2 π f c t ) ] s ~ ( t ) = A c ∑ n = − ∞ ∞ J n ( β ) exp ( j 2 π f m t ) \begin{align*}
+ s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t))\\
+ \Longleftrightarrow \tilde{s}(t) &= A_c\exp(j\beta\sin(2\pi f_m t))\\
s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\
\tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t)
\end{align*}
- s ( t ) s ( t ) s ~ ( t ) = A c cos ( 2 π f c t + β sin ( 2 π f m t )) ⇔ s ~ ( t ) = A c exp ( j β sin ( 2 π f m t )) = Re [ s ~ ( t ) exp ( j 2 π f c t ) ] = A c n = − ∞ ∑ ∞ J n ( β ) exp ( j 2 π f m t )
+ s ( t ) ⟺ s ~ ( t ) s ( t ) s ~ ( t ) = A c cos ( 2 π f c t + β sin ( 2 π f m t )) = A c exp ( j β sin ( 2 π f m t )) = Re [ s ~ ( t ) exp ( j 2 π f c t ) ] = A c n = − ∞ ∑ ∞ J n ( β ) exp ( j 2 π f m t )
-Band
-
-
-
-Narrowband
-Wideband
-
-
-
-
-D < 1 , β < 1 D<1,\beta<1 D < 1 , β < 1 D > 1 , β > 1 D>1,\beta>1 D > 1 , β > 1
-
-
Power, energy and autocorrelation
-G WGN ( f ) = N 0 2 G x ( f ) = ∣ H ( f ) ∣ 2 G w ( f ) (PSD) G x ( f ) = G ( f ) G w ( f ) (PSD) G x ( f ) = lim T → ∞ ∣ X T ( f ) ∣ 2 T (PSD) G x ( f ) = F [ R x ( τ ) ] (WSS) P x = σ x 2 = ∫ R G x ( f ) d f For zero mean P x = σ x 2 = lim t → ∞ 1 T ∫ − T / 2 T / 2 ∣ x ( t ) ∣ 2 d t For zero mean P [ A cos ( 2 π f t + ϕ ) ] = A 2 2 Power of sinusoid E x = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ − ∞ ∞ ∣ X ( f ) ∣ 2 d f Parseval’s theorem R x ( τ ) = F ( G x ( f ) ) PSD to Autocorrelation \begin{align*}
+G WGN ( f ) = N 0 2 G x ( f ) = ∣ H ( f ) ∣ 2 G w ( f ) (PSD) G x ( f ) = G ( f ) G w ( f ) (PSD) G x ( f ) = lim T → ∞ ∣ X T ( f ) ∣ 2 T (PSD) G x ( f ) = F [ R x ( τ ) ] (WSS) P x = σ x 2 = ∫ R G x ( f ) d f For zero mean P x = σ x 2 = lim t → ∞ 1 T ∫ − T / 2 T / 2 ∣ x ( t ) ∣ 2 d t For zero mean P [ A cos ( 2 π f t + ϕ ) ] = A 2 2 Power of sinusoid E x = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ − ∞ ∞ ∣ X ( f ) ∣ 2 d f Parseval’s theorem R x ( τ ) = F ( G x ( f ) ) PSD to Autocorrelation P x = R x ( 0 ) Average power of WSS process x ( t ) \begin{align*}
G_\text{WGN}(f)&=\frac{N_0}{2}\\
G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\
G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\
@@ -654,21 +650,29 @@ B &= \begin{cases}
P_x&={\sigma_x}^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\quad\text{For zero mean}\\
P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\
E_x&=\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\quad\text{Parseval's theorem}\\
- R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}
+ R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation}\\
+ P_x &= R_x(0)\quad\text{Average power of WSS process $x(t)$}\\
\end{align*}
- G WGN ( f ) G x ( f ) G x ( f ) G x ( f ) G x ( f ) P x P x P [ A cos ( 2 π f t + ϕ )] E x R x ( τ ) = 2 N 0 = ∣ H ( f ) ∣ 2 G w ( f ) (PSD) = G ( f ) G w ( f ) (PSD) = T → ∞ lim T ∣ X T ( f ) ∣ 2 (PSD) = F [ R x ( τ )] (WSS) = σ x 2 = ∫ R G x ( f ) df For zero mean = σ x 2 = t → ∞ lim T 1 ∫ − T /2 T /2 ∣ x ( t ) ∣ 2 d t For zero mean = 2 A 2 Power of sinusoid = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ − ∞ ∞ ∣ X ( f ) ∣ 2 df Parseval’s theorem = F ( G x ( f )) PSD to Autocorrelation
+ G WGN ( f ) G x ( f ) G x ( f ) G x ( f ) G x ( f ) P x P x P [ A cos ( 2 π f t + ϕ )] E x R x ( τ ) P x = 2 N 0 = ∣ H ( f ) ∣ 2 G w ( f ) (PSD) = G ( f ) G w ( f ) (PSD) = T → ∞ lim T ∣ X T ( f ) ∣ 2 (PSD) = F [ R x ( τ )] (WSS) = σ x 2 = ∫ R G x ( f ) df For zero mean = σ x 2 = t → ∞ lim T 1 ∫ − T /2 T /2 ∣ x ( t ) ∣ 2 d t For zero mean = 2 A 2 Power of sinusoid = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ − ∞ ∞ ∣ X ( f ) ∣ 2 df Parseval’s theorem = F ( G x ( f )) PSD to Autocorrelation = R x ( 0 ) Average power of WSS process x ( t )
-
-
-
-Coherent detection system.
-y ( t ) = m ( t ) cos ( 2 π f c t + θ ) = m ( t ) cos 2 ( 2 π f c t + θ ) After IF mixing. = m ( t ) 1 2 ( 1 + cos ( 4 f c t + 2 θ ) ) = 1 2 m ( t ) + 1 2 cos ( 4 f c t + 2 θ ) ⟹ S u ( f ) = ( 1 2 ) 2 S u ( f ) After LPF. \begin{align*}
- y(t) &= m(t)\cos(2\pi f_c t+\theta) = m(t)\cos^2(2\pi f_c t+\theta)\text{ After IF mixing.}\\
- &= m(t)\frac{1}{2}(1+\cos(4 f_c t+2\theta))\\
- &= \frac{1}{2}m(t)+\frac{1}{2}\cos(4 f_c t+2\theta)\\
- \implies S_u(f) &= \left(\frac{1}{2}\right)^2S_u(f)\quad\text{After LPF.}
+White noise
+R W ( τ ) = N 0 2 δ ( τ ) = k T 2 δ ( τ ) = σ 2 δ ( τ ) G w ( f ) = N 0 2 \begin{align*}
+R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\
+G_w(f)&=\frac{N_0}{2}\\
\end{align*}
- y ( t ) ⟹ S u ( f ) = m ( t ) cos ( 2 π f c t + θ ) = m ( t ) cos 2 ( 2 π f c t + θ ) After IF mixing. = m ( t ) 2 1 ( 1 + cos ( 4 f c t + 2 θ )) = 2 1 m ( t ) + 2 1 cos ( 4 f c t + 2 θ ) = ( 2 1 ) 2 S u ( f ) After LPF.
+ R W ( τ ) G w ( f ) = 2 N 0 δ ( τ ) = 2 k T δ ( τ ) = σ 2 δ ( τ ) = 2 N 0
+
+
+
Use formualas from previous section, Power, energy and autocorrelation .
@@ -727,27 +731,30 @@ Use these formulas in particular:
-Cannot add directly due to copyright!
+Cannot add directly due to copyright!
+TODO: Make an open source replacement for this diagram Send a PR to GitHub .
Quantizer
-Δ = x Max − x Min 2 k for k -bit quantizer (V/lsb) \begin{align*}
- \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\
+Δ = x Max − x Min 2 k for k -bit quantizer (V/lsb) Quantizer step size Δ \begin{align*}
+ \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\quad\text{Quantizer step size $\Delta$}\\
\end{align*}
- Δ = 2 k x Max − x Min for k -bit quantizer (V/lsb)
+ Δ = 2 k x Max − x Min for k -bit quantizer (V/lsb) Quantizer step size Δ
Quantization noise
-e : = y − x Quantization error μ E = E [ E ] = 0 Zero mean σ E 2 = E [ E 2 ] − 0 2 = ∫ − Δ / 2 Δ / 2 e 2 × ( 1 Δ ) d e Where E ∼ 1 / Δ uniform over ( − Δ / 2 , Δ / 2 ) SQNR = Signal power Quantization noise SQNR(dB) = 10 log 10 ( SQNR ) \begin{align*}
+e : = y − x Quantization error μ E = E [ E ] = 0 Zero mean P E = σ E 2 = Δ 2 12 = 2 − 2 m V 2 / 3 Uniformly distributed error SQNR = Signal power Quantization noise power = P x P E SQNR(dB) = 10 log 10 ( SQNR ) m → m + A bits ⟹ newSQNR(dB) = SQNR(dB) + 6 A dB \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})
+ P_E&={\sigma_E}^2=\frac{\Delta^2}{12}=2^{-2m}V^2/3\quad\text{Uniformly distributed error}\\
+ \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise power}}=\frac{P_x}{P_E}\\
+ \text{SQNR(dB)}&=10\log_{10}(\text{SQNR})\\
+ m\to m+A\text{ bits}&\implies \text{newSQNR(dB)}=\text{SQNR(dB)}+6A\text{ dB}
\end{align*}
- e μ E σ E 2 SQNR SQNR(dB) := y − x Quantization error = E [ E ] = 0 Zero mean = E [ E 2 ] − 0 2 = ∫ − Δ/2 Δ/2 e 2 × ( Δ 1 ) d e Where E ∼ 1/Δ uniform over ( − Δ/2 , Δ/2 ) = Quantization noise Signal power = 10 log 10 ( SQNR )
+ e μ E P E SQNR SQNR(dB) m → m + A bits := y − x Quantization error = E [ E ] = 0 Zero mean = σ E 2 = 12 Δ 2 = 2 − 2 m V 2 /3 Uniformly distributed error = Quantization noise power Signal power = P E P x = 10 log 10 ( SQNR ) ⟹ newSQNR(dB) = SQNR(dB) + 6 A dB
-Cannot add directly due to copyright!
+Cannot add directly due to copyright!
+TODO: Make an open source replacement for this diagram Send a PR to GitHub .
Line codes
@@ -766,6 +773,7 @@ Use these formulas in particular:
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*}
R b D A V ( f ) V rectangle ( f ) G MunipolarNRZ ( f ) G MpolarNRZ ( f ) G unipolarNRZ ( f ) G polarNRZ ( f ) G unipolarNRZ ( f ) G unipolarRZ ( f ) → Bit rate → Symbol rate | R d | 1/ T b → m a → Pulse shape = T sinc ( f T × DutyCycle ) = 12 ( M 2 − 1 ) A 2 D ∣ V ( f ) ∣ 2 + 4 ( M − 1 ) 2 ( D A ) 2 l = − ∞ ∑ ∞ ∣ V ( l D ) ∣ 2 δ ( f − l D ) = 3 ( M 2 − 1 ) A 2 D ∣ V ( f ) ∣ 2 = 4 R b A 2 ( sinc 2 ( R b f ) + R b δ ( f ) ) , NB 0 = R b = R b A 2 sinc 2 ( R b f ) = 4 R b A 2 ( sinc 2 ( R b f ) + R b δ ( f ) ) = 16 A 2 ( l = − ∞ ∑ ∞ δ ( f − T b l ) ∣ sinc ( duty × l ) ∣ 2 + T b ∣ sinc ( duty × f T b ) ∣ 2 ) , NB 0 = 2 R b TODO: Someone please make plots of the PSD for all line code types in Mathematica or Python! Send a PR to GitHub .
Modulation and basis functions
@@ -1246,6 +1254,18 @@ h400000v40h-400000z"/>
Value tables for erf ( x ) \text{erf}(x) erf ( x ) Q ( x ) Q(x) Q ( x )
Q ( x ) Q(x) Q ( x ) You should use erf \text{erf} erf Q ( x ) = 1 2 − 1 2 erf ( x 2 ) Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right) Q ( x ) = 2 1 − 2 1 erf ( 2 x )
@@ -1724,6 +1744,19 @@ h400000v40h-400000z"/>
Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems
erf ( x ) \text{erf}(x) erf ( x ) Q ( x ) = 1 2 − 1 2 erf ( x 2 ) Q(x)=\frac{1}{2}-\frac{1}{2}\text{erf}(\frac{x}{\sqrt{2}})
+ Q ( x ) = 2 1 − 2 1 erf ( 2 x )
@@ -1859,6 +1892,48 @@ h400000v40h-400000z"/>
**The value of erf ( 3.30 ) \text{erf}(3.30) erf ( 3.30 ) ≈ 0.999997 \approx0.999997 ≈ 0.999997
+Q ( x ) Q(x) Q ( x ) Using identity.
+
+
+
+x x x Q ( x ) Q(x) Q ( x )
+
+
+
+2 \sqrt{2} 2 0.07865 0.07865 0.07865
+
+2 2 2\sqrt{2} 2 2 0.00234 0.00234 0.00234
+
+
Receiver output shit
r o ( t ) = { s o 1 ( t ) + n o ( t ) code 1 s o 2 ( t ) + n o ( t ) code 0 n : AWGN with σ o 2 \begin{align*}
@@ -2165,6 +2240,12 @@ M1001 80h400000v40h-400000z"/>
+Stats
+P ( A ∣ B ) = P ( B ∣ A ) P ( A ) P ( B ) = P ( A , B ) P ( B ) \begin{align*}
+ P(A|B) &= \frac{P(B|A)P(A)}{P(B)} = \frac{P(A,B)}{P(B)}\\
+\end{align*}
+ P ( A ∣ B ) = P ( B ) P ( B ∣ A ) P ( A ) = P ( B ) P ( A , B )
+
Entropy for discrete random variables
H ( x ) ≥ 0 H ( x ) = − ∑ x i ∈ A x p X ( x i ) log 2 ( p X ( x i ) ) H ( x , y ) = − ∑ x i ∈ A x ∑ y i ∈ A y p X Y ( x i , y i ) log 2 ( p X Y ( x i , y i ) ) Joint entropy H ( x , y ) = H ( x ) + H ( y ) Joint entropy if x and y independent H ( x ∣ y = y j ) = − ∑ x i ∈ A x p X ( x i ∣ y = y j ) log 2 ( p X ( x i ∣ y = y j ) ) Conditional entropy H ( x ∣ y ) = − ∑ y j ∈ A y p Y ( y j ) H ( x ∣ y = y j ) Average conditional entropy, equivocation H ( x ∣ y ) = − ∑ x i ∈ A x ∑ y i ∈ 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 ) \begin{align*}
H(x) &\geq 0\\
@@ -2180,13 +2261,19 @@ M1001 80h400000v40h-400000z"/>
H ( x ) H ( x ) H ( x , y ) H ( x , y ) H ( x ∣ y = y j ) H ( x ∣ y ) H ( x ∣ y ) H ( x ∣ y ) H ( x , y ) ≥ 0 = − x i ∈ A x ∑ p X ( x i ) log 2 ( p X ( x i )) = − x i ∈ A x ∑ y i ∈ A y ∑ p X Y ( x i , y i ) log 2 ( p X Y ( x i , y i )) Joint entropy = H ( x ) + H ( y ) Joint entropy if x and y independent = − x i ∈ A x ∑ p X ( x i ∣ y = y j ) log 2 ( p X ( x i ∣ y = y j )) Conditional entropy = − y j ∈ A y ∑ p Y ( y j ) H ( x ∣ y = y j ) Average conditional entropy, equivocation = − x i ∈ A x ∑ y i ∈ A y ∑ p X ( x i , y j ) log 2 ( p X ( x i ∣ y = y j )) = H ( x , y ) − H ( y ) = H ( x ) + H ( y ∣ x ) = H ( y ) + H ( x ∣ y )
Entropy is maximized when all have an equal probability.
-Differential entropy for continuous random variables
-TODO: Cut out if not required
-h ( x ) = − ∫ R f X ( x ) log 2 ( f X ( x ) ) d x \begin{align*}
- h(x) &= -\int_\mathbb{R}f_X(x)\log_2(f_X(x))dx
+Transition probability diagram
+Example for binary erasure channel where X X X Y Y Y
+
+Equivalent to:
+P [ Y = 0 ∣ X = 0 ] = 1 − p P [ Y = e ∣ X = 0 ] = p P [ Y = 1 ∣ X = 1 ] = 1 − p P [ Y = e ∣ X = 1 ] = p P [ X = 0 ∣ Y = 0 ] = 0 Note the direction P [ Y = 0 ] = P [ Y = 0 ∣ X = 0 ] P [ X = 0 ] \begin{align*}
+ P[Y=0|X=0] &= 1-p\\
+ P[Y=e|X=0] &= p\\
+ P[Y=1|X=1] &= 1-p\\
+ P[Y=e|X=1] &= p\\
+ P[X=0|Y=0] &= 0\quad\text{Note the direction}\\
+ P[Y=0] &= P[Y=0|X=0] P[X=0]
\end{align*}
- h ( x ) = − ∫ R f X ( x ) log 2 ( f X ( x )) d x
-
+ P [ Y = 0∣ X = 0 ] P [ Y = e ∣ X = 0 ] P [ Y = 1∣ X = 1 ] P [ Y = e ∣ X = 1 ] P [ X = 0∣ Y = 0 ] P [ Y = 0 ] = 1 − p = p = 1 − p = p = 0 Note the direction = P [ Y = 0∣ X = 0 ] P [ X = 0 ]
Amount of entropy decrease of x x x y y y
@@ -2197,7 +2284,8 @@ M1001 80h400000v40h-400000z"/>
Channel model
Vertical, x x x y y y
+Horizontal, y y y P \mathbf{P} P P ( y j ∣ x i ) P(y_j|x_i) P ( y j ∣ x i ) P = [ p 11 p 12 … p 1 N p 21 p 22 … p 2 N ⋮ ⋮ ⋱ ⋮ p M 1 p M 2 … p M N ] \mathbf{P}=\left[\begin{matrix}
p_{11} & p_{12} &\dots & p_{1N}\\
p_{21} & p_{22} &\dots & p_{2N}\\
@@ -2228,16 +2316,25 @@ M347 1759 V0 H263 V1759 v1800 v1759 h84z"/> p Y ( b j ) [ p Y ( b 0 ) p Y ( b 1 ) … p Y ( b j ) ] = i = 1 ∑ M P [ x = a i , y = b j ] 1 ≤ j ≤ N = i = 1 ∑ M P [ X = a i ] P [ Y = b j ∣ X = a i ] = [ p X ( a 0 ) p X ( a 1 ) … p X ( a i ) ] × P
Fast procedure to calculate I ( y ; x ) I(y;x) I ( y ; x )
-1. Find H ( x ) 2. Find [ p Y ( b 0 ) p Y ( b 1 ) … p Y ( b j ) ] = [ p X ( a 0 ) p X ( a 1 ) … p X ( a i ) ] × P 3. Multiply each row in P by p X ( a i ) since p X Y ( x i , y i ) = P ( y i ∣ x i ) P ( x i ) 4. Find H ( x , y ) using each element from (3.) 5. Find H ( x ∣ y ) = H ( x , y ) − H ( y ) 6. Find I ( y ; x ) = H ( x ) − H ( x ∣ y ) \begin{align*}
+1. Find H ( x ) 2. Find [ p Y ( b 0 ) p Y ( b 1 ) … p Y ( b j ) ] = [ p X ( a 0 ) p X ( a 1 ) … p X ( a i ) ] × P 3. Multiply each row in P by p X ( a i ) since p X Y ( a i , b i ) = P ( b i ∣ a i ) P ( a i ) 4. Find H ( x , y ) using each element from (3.) 5. Find H ( x ∣ y ) = H ( x , y ) − H ( y ) 6. Find I ( x ; y ) = H ( x ) − H ( x ∣ y ) \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{3. Multiply each row in $\textbf{P}$ by $p_X(a_i)$ since $p_{XY}(a_i,b_i)=P(b_i|a_i)P(a_i)$}\\
&\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)\\
+ &\text{6. Find }I(x;y)=H(x)-H(x|y)\\
\end{align*}
- 1. Find H ( x ) 2. Find [ p Y ( b 0 ) p Y ( b 1 ) … p Y ( b j ) ] = [ p X ( a 0 ) p X ( a 1 ) … p X ( a i ) ] × P 3. Multiply each row in P by p X ( a i ) since p X Y ( x i , y i ) = P ( y i ∣ x i ) P ( x i ) 4. Find H ( x , y ) using each element from (3.) 5. Find H ( x ∣ y ) = H ( x , y ) − H ( y ) 6. Find I ( y ; x ) = H ( x ) − H ( x ∣ y )
+ 1. Find H ( x ) 2. Find [ p Y ( b 0 ) p Y ( b 1 ) … p Y ( b j ) ] = [ p X ( a 0 ) p X ( a 1 ) … p X ( a i ) ] × P 3. Multiply each row in P by p X ( a i ) since p X Y ( a i , b i ) = P ( b i ∣ a i ) P ( a i ) 4. Find H ( x , y ) using each element from (3.) 5. Find H ( x ∣ y ) = H ( x , y ) − H ( y ) 6. Find I ( x ; y ) = H ( x ) − H ( x ∣ y )
+Example of step 3 :
+P X Y = [ P ( y 1 ∣ x 1 ) P ( x 1 ) P ( y 2 ∣ x 1 ) P ( x 1 ) … P ( y 1 ∣ x 2 ) P ( x 2 ) P ( y 2 ∣ x 2 ) P ( x 2 ) … ⋮ ⋮ ⋱ ] \mathbf{P_{XY}}=\left[\begin{matrix}
+ P(y_1|x_1) P(x_1) & P(y_2|x_1) P(x_1) & \dots\\
+ P(y_1|x_2) P(x_2) & P(y_2|x_2) P(x_2) & \dots\\
+ \vdots & \vdots &\ddots
+\end{matrix}\right]
+ P XY = P ( y 1 ∣ x 1 ) P ( x 1 ) P ( y 1 ∣ x 2 ) P ( x 2 ) ⋮ P ( y 2 ∣ x 1 ) P ( x 1 ) P ( y 2 ∣ x 2 ) P ( x 2 ) ⋮ … … ⋱
Channel types
@@ -2266,6 +2363,7 @@ M347 1759 V0 H263 V1759 v1800 v1759 h84z"/>C N p C R b → Channel capacity (bits/channels used) → Output alphabet size → Probability vector, any row of the transition matrix = log 2 ( N ) − H ( p ) Capacity for weakly symmetric and symmetric channels < C for error-free transmission Note that the channel capacity is realized when the channel inputs are uniformly distributed (i.e. P ( x 1 ) = P ( x 2 ) = ⋯ = P ( x N ) = 1 N P(x_1)=P(x_2)=\dots=P(x_N)=\frac{1}{N} P ( x 1 ) = P ( x 2 ) = ⋯ = P ( x N ) = N 1
Channel capacity of an AWGN channel
y i = x i + n i n i ∼ N ( 0 , N 0 / 2 ) y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2)
@@ -2486,7 +2584,14 @@ M347 1759 V0 H263 V1759 v0 v1759 h84z"/> h ~ ( t ) \tilde{h}(t) h ~ ( t )
Convolutions: do not forget width when using graphical method
-todo: add more items to check
+2 W 2W 2 W Scale sampled spectrum by f s f_s f s
+2 f c 2f_c 2 f c Square transfer function for PSD G y ( f ) = ∣ H ( f ) ∣ 2 G x ( f ) G_y(f)=|H(f)|^\mathbf{2}G_x(f) G y ( f ) = ∣ H ( f ) ∣ 2 G x ( f )
+Square besselJ function for FM power ∣ J n ( β ) ∣ 2 |J_n(\beta)|^\mathbf{2} ∣ J n ( β ) ∣ 2
+Bandwidth: only consider positive frequencies (so the bandwidth of an AM signal will be the range from the lowest to greatest sideband frequency. For a rectangular function, it will be from 0 to W).
+TODO: add more items to check
+TODO: add some graphics for these checklist items
