如何在没有相位跳跃的情况下连接正弦波

我需要创建一个生成给定频率的正弦波并使用pyaudio(阻塞模式)播放它们的 python脚本,我还需要能够在运行时更改此频率,调制它并使用pyqtgraph绘制它.现在我有一个生成数据块的线程,我的’连接’那些正弦的方法是得到fft然后计算角度(numpy.angle),将它存储在一个变量中并用它作为相位偏移到下一个大块,但我没有得到我预期的结果,也许我错过了什么或混合起来.

import matplotlib.pyplot as plt
import numpy as np
import pyaudio

#-----------------------
CHUNK = 1024
RATE =  44100
CHANNELS = 2
FORMAT = pyaudio.paFloat32
#-----------------------

samples = int(CHUNK)
t = np.arange(samples) / RATE
con = 0


def generate_sine(a: float = 0.5,freq: float = 440.0):

    global con

    sine = a * np.sin(2.0 * np.pi * freq * t + con)

    # get the angle of the wave

    phase = np.angle(np.fft.fft(sine))

    # update ref var to generate subsequent sines
    # begining where the last ended

    con = phase[-1]

    return sine


def play_sine(data):

    pa = pyaudio.PyAudio()

    stream = pa.open(format=FORMAT,channels=CHANNELS,rate=RATE,input=False,output=True,frames_per_buffer=CHUNK)

    stream.write(np.array(data).astype(np.float32).tostring())

    stream.close()

if __name__ == '__main__':

    f = 80

    chunks = generate_sine(freq=f)

    for i in range(0,4):

        chunks = np.concatenate((chunks,generate_sine(freq=f)))

    #for i in range(0,10):

    #play_sine(chunks)

    plt.plot(chunks)

    plt.show()

演示情节

您可以在链接的图像中看到x = 1024,x = 2048周围存在不连续性,依此类推.

a * sin(2πf * t + con)

con = 2.0 * np.pi * f * CHUNK/RATE + con

con = math.fmod(2.0 * np.pi * f * CHUNK/RATE + con,2.0 * np.pi)

a * sin(2π * (f * t + con))

con = math.modf(f * CHUNK/RATE + con)[0]

2πf?*0/RATE+con?,2πf?*1/RATE+con?,...,2πf?*1022/RATE+con?,2πf?*1023/RATE+con?

2πf?*0/RATE + con? = 2πf?*1024/RATE + con?

0      + con? = 2πf?*1024/RATE + con?

con? = 2πf?*1024/RATE + con?

con = 2.0 * np.pi * f * CHUNK/RATE + con