Python 和波形样本宽度

Question

I'm playing around with the wave library and I'm having trouble saving a sine to a wave file whose resolution is more than 8 bits. My program works fine for 8 bits resolution (WIDTH = 1 byte in my program and comment out one line). If I use WIDTH = 2, the program works, but the output wave file does not look like a sine.

I hope I can get some help!

import math
import wave
import struct

BITRATE = 44100
FREQUENCY = 440
LENGTH = 1      # in seconds
WIDTH = 2       # in bytes. 1->8 bits, 2->16 bits

def GenerateSine(NumberOfFrames):
  wave = ''
  for i in xrange(NumberOfFrames):
    value = math.sin(i * 2 * math.pi * FREQUENCY / BITRATE)
    value += 1
    value /= 2
    value *= (256 ** WIDTH) - 1
    value = int(round(value))
    #wave += struct.pack("<B", value)       # this is good for WIDTH = 1
    wave += struct.pack("<H", value)        # i would think this would work for WIDTH = 2
  return wave

WaveData = GenerateSine(int(BITRATE * LENGTH))

wf = wave.open("output.wave", 'w')
wf.setnchannels(1)
wf.setsampwidth(WIDTH)
wf.setframerate(BITRATE)
wf.writeframes(WaveData)
wf.close()

Answer 1

Got it to work! I think the problem was that for WIDTH > 1, I must use a signed integer instead of unsigned.

Here's something that works for WIDTH = 2 or WIDTH = 4. It won't work with WIDTH = 1 or 3, but that's fine for what I want to do.

import math
import wave
import struct

BITRATE = 44100
FREQUENCY = 440
LENGTH = 1
WIDTH = 2

def GenerateSine(NumberOfFrames):
  wave = ''
  for i in xrange(NumberOfFrames):
    value = math.sin(i * 2 * math.pi * FREQUENCY / BITRATE)
    value *= 128
    value *= (256 ** (WIDTH - 1)) - 1
    value = int(round(value))
    if WIDTH == 2:
      wave += struct.pack("<h", value)
    elif WIDTH == 4:
      wave += struct.pack("<l", value)
  return wave

WaveData = GenerateSine(int(BITRATE * LENGTH))

wf = wave.open("output.wave", 'w')
wf.setnchannels(1)
wf.setsampwidth(WIDTH)
wf.setframerate(BITRATE)
wf.writeframes(WaveData)
wf.close()