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Fix modulation index, add tri

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Peter 2024-11-03 00:43:57 +08:00
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@ -82,13 +82,13 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
## Fourier transform identities and properties ## Fourier transform identities and properties
| Time domain $x(t)$ | Frequency domain $X(f)$ | | Time domain $x(t)$ | Frequency domain $X(f)$ |
| --------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- | | ---------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- |
| $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \text{sinc}(fT)$ | | $\text{rect}\left(\frac{t}{T}\right)\quad\Pi\left(\frac{t}{T}\right)$ | $T \text{sinc}(fT)$ |
| $\text{sinc}(2Wt)$ | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ | | $\text{sinc}(2Wt)$ | $\frac{1}{2W}\text{rect}\left(\frac{f}{2W}\right)\quad\frac{1}{2W}\Pi\left(\frac{f}{2W}\right)$ |
| $\exp(-at)u(t),\quad a>0$ | $\frac{1}{a + j2\pi f}$ | | $\exp(-at)u(t),\quad a>0$ | $\frac{1}{a + j2\pi f}$ |
| $\exp(-a\lvert t \rvert),\quad a>0$ | $\frac{2a}{a^2 + (2\pi f)^2}$ | | $\exp(-a\lvert t \rvert),\quad a>0$ | $\frac{2a}{a^2 + (2\pi f)^2}$ |
| $\exp(-\pi t^2)$ | $\exp(-\pi f^2)$ | | $\exp(-\pi t^2)$ | $\exp(-\pi f^2)$ |
| $1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T$ | $T \text{sinc}^2(fT)$ | | $1 - \frac{\lvert t \rvert}{T},\quad\lvert t \rvert < T\quad\text{tri}(t/T)$ | $T \text{sinc}^2(fT)$ |
| $\delta(t)$ | $1$ | | $\delta(t)$ | $1$ |
| $1$ | $\delta(f)$ | | $1$ | $\delta(f)$ |
| $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ | | $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ |
@ -130,6 +130,7 @@ along with this program. If not, see <http://www.gnu.org/licenses/>.
\text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\ \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\
\text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc 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{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}\\
g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\ g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\
\end{align*} \end{align*}
``` ```
@ -156,16 +157,19 @@ Calculate $C_n$ coefficient as follows from $x_p(t)$:
```math ```math
\begin{align*} \begin{align*}
C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
&=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$}
\end{align*} \end{align*}
``` ```
<!-- MATH END --> <!-- MATH END -->
### $\text{rect}$ function ### Shape functions
![rect](images/rect.drawio.svg) | $\text{rect}$ function | $\text{tri}$ function |
| ------------------------------- | --------------------- |
| ![rect](images/rect.drawio.svg) | TODO: Add graphic. |
Tri placeholder: For $\text{tri}(t/T)=1-\|t\|/T$, Intersects $x$ axis at $-T$ and $T$ and $y$ axis at $1$.
### Bessel function ### Bessel function
@ -319,7 +323,7 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\
\begin{align*} \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], \\ 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], \\
&\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating 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{|\min_t(k_a m(t))|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\
m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\ m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\
m_a&=k_a A_m \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_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\

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