2019-03-21

2019-03-22 本文已影响0人快乐的大脚aaa

时域信号与频域信号
时域表示：
- 频域表示： $\pi [\delta(f+f_c)+\delta(f-f_c)]$
- 矩形脉冲
  - $rect{\frac{t}{T}}$
  - $T\cdot sinc(fT)$
时移
- $x(t) = \int_{-\infty}^{\infty}X(f)e^{j2\pi ft}df$
- $x(t-t_0) = \int_{-\infty}^{\infty}X(f)e^{j2\pi f(t-t_0)}df$
- $= \int_{-\infty}^{\infty}X(f)e^{-j 2 \pi f t_0} \cdot e^{j 2\pi ft}df$
- $\mathcal{F}[x(t-t_0)] = X(f)e^{-j2\pi ft_0}$
频移
- $x(t) = \int_{-\infty}^{\infty}X(f)e^{j2\pi ft}df$
- 两边乘以频率为 $f_0$ 的复单频信号 $e^{j2\pi f_0t}$
- $x(t)e^{j2\pi f_0t} = \int_{-\infty}^{\infty}X(f)e^{j2\pi ft}e^{j2\pi f_0t}df$
- $= \int_{-\infty}^{\infty}X(f)e^{j2\pi (f+f_0)t}df$
- $= \int_{-\infty}^{\infty}X(f-f_0)e^{j2\pi ft}df$
- $\mathcal{F}[x(t)e^{j2\pi f_0 t}] = X(f-f_0)$
eg:的傅氏变换为,
- $cos(2\pi f_0 t+\theta)= \frac{1}{2}\{e^{j\theta} \cdot e^{j2\pi f_0t} + e^{-j\theta} \cdot e^{-j2\pi f_0t} \}$
- $s(t) = \frac{A}{2}m(t)e^{j\theta} \cdot e^{j2\pi f_0t} + \frac{A}{2}m(t) e^{-j\theta} \cdot e^{-j2\pi f_0t}$
- $\mathcal{F}[s(t)] = \frac{A}{2}e^{j\theta} M(f-f_0)+\frac{A}{2}e^{-j\theta} M(f+f_0)$
共轭
- $\mathcal{F}[x(t)] = X(f)$
- $\mathcal{F}[x^*(t)] =\int_{-\infty}^{\infty}x^*(t)e^{-j2\pi ft} dt$
- $=\int_{-\infty}^{\infty}[x(t)e^{j2\pi ft}dt]^*$
- $=\int_{-\infty}^{\infty}[x(t)e^{-j2\pi (-f)t}dt]^*$
- $= [X(-f)]^* = X^*(-f)$
- 实信号满足共轭对称性，
  - 因此其频谱满足共轭对称性 $X(f) = X^*(-f)$
时域镜像，频域也镜像
微分
- $x(t) = \int_{-\infty}^{\infty}X(f)e^{j2\pi f t}df$
- 两边对t微分
  - $x^{'}(t) = \int_{-\infty}^{\infty}(j2\pi f)X(f)e^{j2\pi f t}df$
  - $\mathcal{F}[x^{'}(t)] = (j2\pi f)X(f)$
- $\mathcal{F^{-1}}[X^{'}(f)] = (-j2\pi t)x(t)$
eg:的频谱是
- $H^{'}(f) = -j\cdot 2\delta(f)$
- $\mathcal{F^{-1}}[-j2\delta(f)] = -2j$
- $\mathcal{F^{-1}}[H^{'}(f)] = (-j2\pi t)h(t)$
- $h(t) = \frac{F^{-1} [-j2\delta(f)]}{-j2\pi t} = \frac{1}{\pi t}$

2019-03-21

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