$\Large \displaystyle \delta(t)\Leftrightarrow 1$

$\Large \displaystyle 1\Leftrightarrow 2\pi \delta(\omega)$

$\Large \displaystyle \delta(t-t_0)\Leftrightarrow e^{-j\omega t_0}$

$\Large \displaystyle e^{j\omega_0 t}\Leftrightarrow 2\pi \delta(\omega-\omega_0)$

$\Large \displaystyle e^{-j\omega_0 t}\Leftrightarrow 2\pi \delta(\omega+\omega_0)$

$\Large \displaystyle \cos(\omega_0 t) = \frac{1}{2}(e^{j\omega_0 t} + e^{-j\omega_0 t})$

$\Large \displaystyle \cos(\omega_0 t)\Leftrightarrow \pi(\delta(\omega - \omega_0)+\delta(\omega + \omega_0))$

$\Large \displaystyle \sin(\omega_0 t) = \frac{1}{2j}(e^{j\omega_0 t} - e^{-j\omega_0 t})$

$\Large \displaystyle \sin(\omega_0 t)\Leftrightarrow j\pi(\delta(\omega + \omega_0)-\delta(\omega - \omega_0))$

$\Large \displaystyle U(t)\Leftrightarrow \pi\delta(\omega)+\frac{1}{j\omega}$

$\Large \displaystyle \int_{-\infty}^{\infty}f_1(\tau)f_2(t-\tau)d\tau\Leftrightarrow F_1(\omega)F_2(\omega)$

$\Large \displaystyle f_1(t)f_2(t)\Leftrightarrow \frac{1}{2\pi}\int_{-\infty}^{\infty}F_1(\omega)F_2(\omega-y)dy$

$\Large \displaystyle f(t+T)=f(t)$

$\Large \displaystyle f(t)=\sum_{n=-\infty}^{\infty}\alpha_n e^{jn\omega_0 t}\ \ \ \omega_0=\frac{2\pi}{T}$

$\Large \displaystyle \alpha_n=\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-jn\omega_0 t}dt$

$\Large \displaystyle F(\omega)=2\pi\sum_{n=-\infty}^{\infty}\alpha_n\delta(\omega-n\omega_0)$

$\Large \displaystyle \sum_{-\infty}^{\infty}f_0(t+nT)=\frac{1}{T}\sum_{-\infty}^{\infty}e^{jn\omega_0 t}F_0(n\omega_0)$

$\Large \displaystyle \sum_{-\infty}^{\infty}\phi(t+nT)=\frac{1}{T}\sum_{-\infty}^{\infty}e^{jn\omega_0 t}\Phi(n\omega_0)\ \ \ \omega_0=\frac{2\pi}{T}$

$\Large \displaystyle f^*(t)=T\sum_{-\infty}^{\infty}f(nT)\delta(t-nT)$

$\Large \displaystyle F^*(\omega)=T\sum_{-\infty}^{\infty}f(nT)e^{-jnT\omega}$

$\Large \displaystyle F^*(\omega)=\sum_{-\infty}^{\infty}F(\omega+\frac{2\pi n}{T})$

$\Large \displaystyle F(\omega)=0\ \ \ |\omega|\geq \omega_c$

$\Large \displaystyle f_n=f(n\frac{\pi}{\omega_c})$

$\Large \displaystyle f(t)=\sum_{n=-\infty}^{\infty}f_n\frac{sin(\omega_c t-n\pi)}{\omega_c t-n\pi}$

$\Large \displaystyle f(t)=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}F(\omega)e^{j\omega t}d\omega$

$\Large \displaystyle f_n=f(n\frac{\pi}{\omega_c})=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}F(\omega)e^{j\omega n\frac{\pi}{\omega_c}}d\omega$

$\Large \displaystyle F(\omega)=\sum_{n=-\infty}^{\infty}A_n e^{-jnT\omega}=\sum_{n=-\infty}^{\infty}A_n e^{-jn\pi\frac{\omega}{\omega_c}}\ \ \ -\omega_c\lt \omega \lt \omega_c$

$\Large \displaystyle A_n=\frac{1}{2\omega_c}\int_{-\omega_c}^{\omega_c}F(\omega)e^{jn\pi\frac{\omega}{\omega_c}}d\omega=\frac{\pi}{\omega_c}f_n$

$\Large \displaystyle a(b+c)=ab+ac$

$\Large \displaystyle a\sum_{i=1}^n b_i=\sum_{i=1}^n ab_i$

$\Large \displaystyle \boxed{A}\longrightarrow\boxed{B}\longrightarrow\boxed{C}\longrightarrow\boxed{D}$

$\Large \displaystyle P(A,B,C,D)=P(D|C)P(C|B)P(B|A)P(A)$

$\Large \displaystyle P(C|D)=\frac{P(C,D)}{P(D)}=\frac{1}{\alpha}\sum_B \sum_A P(A,B,C,D)$

$\Large \displaystyle =\frac{1}{\alpha}\sum_B \sum_A (P(D|C)P(C|B)P(B|A)P(A))$

$\Large \displaystyle P(B|D)=\frac{P(B,D)}{P(D)}=\frac{1}{\beta}\sum_B \sum_A P(A,B,C,D)$

• bit1: 1=人の動きを検知、0=人の動きを不検知

• 動体を検出してからの時間（秒）

• bit6: 明るい
• bit7: 暗い