python으로 소리 처리 ~ 부립엽 레벨로 펼쳐진 톱니파 ~

\begin{align}
c &=\frac{1}{T} \int_0 ^T f(t)  dt =\frac{1}{T} \int_0 ^T \frac{A}{T} t \,dt \\
  &=\frac{A}{T^2} \Big[t^2\Big]^T_0 \\
  &=\frac{A}{2}
\end{align}

\begin{align}
a_n &= \frac{2}{T} \int_0 ^T f(t) \sin n\omega t dt = \frac{2}{T} \int_0 ^T \frac{A}{T}t \sin n\omega t dt \\
&= \frac{2A}{T^2}\int_0^T t\sin n \omega t dt = \frac{2A}{T^2}\Big[ - \frac{t \cos n \omega t}{n \omega} + \frac{\sin n \omega t}{(n \omega)^2} \Big]^T_0 \\
  &= - \frac{A}{n \pi} \, (\because \omega = \frac{2 \pi}{T})
\end{align}

\begin{align}
b_n &= \frac{2}{T} \int_0 ^T f(t) \cos n\omega t dt = \frac{2}{T} \int_0 ^T \frac{A}{T}t \cos n\omega t dt \\
&= \frac{2A}{T^2}\int_0^T t\cos n \omega tdt = \frac{2A}{T^2}\Big[ \frac{t \sin n \omega t}{n \omega} + \frac{\cos n \omega t}{(n \omega)^2} \Big]^T_0 \\
  &= 0 \, (\because \omega = \frac{2 \pi}{T})
\end{align}

f(t) = -\frac{A}{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin n\omega t + \frac{A}{2}

도표로 톱날의 치파를 표시하다

def sawtooth_wave(N):
    dt = 0.01
    T = 2 * np.pi / 4.
    A = 1
    omega = 2 * np.pi / T
    t = np.arange(0, 2*np.pi, 0.01)
    y = np.zeros(len(t))
    for n in range(1,N):
        y += - A / (np.pi * n) * np.sin( n * 2 * np.pi / T * t)
    y += A / 2.
    plt.plot(t, y, label="n = {}".format(N))
    plt.yticks([0.5], ["A/2"])
    plt.xticks([T, 2*T, 3*T, 4*T], ["T", "2T", "3T", "4T"])
    plt.legend(loc="upper left")

톱날 파도

import numpy as np

# 定数
SR = 44100 # サンプリング周波数
f = 440.        # 周波数
A = 1             # 振幅
T = 5             # 音声の時間
N = 5000     # sin波の重ねる数

# 各時間
t = np.linspace(0, T, SR * T)[:-1] 

# ノコギリ波の計算
y = np.zeros(len(t))
for n in range(1,N):
    y += - A / (np.pi * n) * np.sin( n * 2 * np.pi * f  * t)
y += A / 2.

# 16bitで量子化
y2 = (y * np.iinfo(np.int16).max).astype(np.int16)

# ステレオに変換
y3 = np.array(np.c_[y2,y2])    

# 保存
import scipy.io.wavfile as siw
siw.write("sawtooth.wav", SR, y3)

총결산