# 为什么有负频率，什么是相位谱 —— 关于傅立叶变换的随笔

$$c(t) = \sum_{k=-\infty}^{+\infty} c_k e^{ik\omega t}$$

$$c_k = \frac{1}{T}\int_0^Tc(t)e^{-ik\omega t}dt$$

### 关于正、负频率系数之间的关系

$$c_k = \frac{1}{T}\int_0^Tc(t)[cos(-k\omega t) + i sin(-k\omega t)] dt = \frac{1}{T}\int_0^Tc(t)[cos(k\omega t) - i sin(k\omega t)] dt$$

$$c_{-k} = \frac{1}{T}\int_0^Tc(t)[cos(k\omega t) + i sin(k\omega t)] dt$$

### 考察 $$c(t)$$ 展开式中的求和

$$c(t) = \sum_{k=-\infty}^{+\infty} c_k e^{ik\omega t} = c_0 + \sum_{k = 1}^{+\infty}(c_k e^{ik\omega t} + c_{-k} e^{-ik\omega t})$$

$$c_k e^{ik\omega t} + c_{-k} e^{-ik\omega t} = (a + i b)(cos(k\omega t) + i sin(k\omega t)) + (a - i b)(cos(-k\omega t) + i sin(-k\omega t))$$

$$c_k e^{ik\omega t} + c_{-k} e^{-ik\omega t} = 2[a cos(\theta) - b sin(\theta) ]$$

$$c(t) = c_0 + 2 \sum_{k = 1}^{+\infty}( a_k cos(\theta_k) - b_k sin(\theta_k) )$$

### 相位谱

$$c(t) = c_0 + 2 \sum_{k = 1}^{+\infty}( a_k cos(\theta_k) - b_k sin(\theta_k) )$$

$$c(t) = c_0 + 2 \sum_{k = 1}^{+\infty}r_k( \frac{a_k}{r_k} cos(\theta_k) - \frac{b_k}{r_k} sin(\theta_k) ) = c_0 + 2 \sum_{k = 1}^{+\infty}r_k cos(\theta_k + \phi_k)$$

posted @ 2015-04-08 22:08  Bruce Bi  阅读(2715)  评论(0编辑  收藏  举报