我正在开发一个小的测试程序，以确定向ADC添加噪声以获得过采样位是否可行。在我们开始之前，先讲一下一些理论知识。奈奎斯特采样定理表明，增加1位分辨率需要四个额外的采样，通常情况下，增加n位需要2^(n+1)个额外采样。我正在模拟一个完美的10位ADC，它从0..1023单调地返回一个值，并且对于0-2V的输入没有噪声。
为了获得更多的位数，必须添加随机分布的噪声（它不必实际上是随机的，但它必须看起来像白噪声）。我的问题是，尽管分辨率正在增加，但实际读数偏移了一些小的负值。这是一个输入1伏特时的样本输出（参考电压为2伏特，所以对于单调ADC，计数应该正好减半）：
事实上，无论我运行多少次模拟，最终结果总是约为0.9995，而不是1；最后一个值应该是134,217,728，而不是134,152,551，相差约65,771个单位，大约是额外18位分辨率的1/4（巧合吗？我除以4了）。我怀疑我的伪随机数生成器存在某种偏差，但我正在使用Python默认提供的Mersenne Twister。
为了获得更多的位数，必须添加随机分布的噪声（它不必实际上是随机的，但它必须看起来像白噪声）。我的问题是，尽管分辨率正在增加，但实际读数偏移了一些小的负值。这是一个输入1伏特时的样本输出（参考电压为2伏特，所以对于单调ADC，计数应该正好减半）：
10 bits: 512 volts: 1.0
11 bits: 1024 volts: 1.0
12 bits: 2046 volts: 0.9990234375
13 bits: 4093 volts: 0.999267578125
14 bits: 8189 volts: 0.999633789062
15 bits: 16375 volts: 0.999450683594
16 bits: 32753 volts: 0.999542236328
17 bits: 65509 volts: 0.999588012695
18 bits: 131013 volts: 0.999549865723
24 bits: 8384565 volts: 0.999518036842
28 bits: 134152551 volts: 0.999514393508
事实上，无论我运行多少次模拟，最终结果总是约为0.9995，而不是1；最后一个值应该是134,217,728，而不是134,152,551，相差约65,771个单位，大约是额外18位分辨率的1/4（巧合吗？我除以4了）。我怀疑我的伪随机数生成器存在某种偏差，但我正在使用Python默认提供的Mersenne Twister。
#!/usr/bin/python
#
# Demonstrates how oversampling/supersampling with noise can be used
# to improve the resolution of an ADC reading.
#
# Public domain.
#
import random, sys
volts = 1.000
reference = 2.000
noise = 0.01
adc_res = 10
def get_rand_bit():
return random.choice([-1, 1])
def volts_with_noise():
if get_rand_bit() == 1:
return volts + (noise * random.random() * get_rand_bit())
else:
return volts
def sample_adc(v):
# Sample ADC with adc_res bits on given voltage.
frac = v / reference
frac = max(min(frac, 1.0), 0.0) # clip voltage
return int(frac * (2 ** adc_res))
def adc_do_no_noise_sample():
return sample_adc(volts)
def adc_do_noise_sample(extra_bits_wanted):
# The number of extra samples required to gain n bits (according to
# Nyquist) is 2^(n+1). So for 1 extra bit, we need to sample 4 times.
samples = 2 ** (extra_bits_wanted + 1)
print "Sampling ", samples, " times for ", extra_bits_wanted, " extra bits."
# Sample the number of times and add the totals.
total = 0
for i in range(samples):
if i % 100000 == 99999:
print float(i * 100) / samples
sys.stdout.flush()
total += sample_adc(volts_with_noise())
# Divide by two (to cancel out the +1 in 2^(n+1)) and return the integer part.
return int(total / 2)
def convert_integer_to_volts(val, num_bits):
# Get a fraction.
frac = float(val) / (2 ** num_bits)
# Multiply by the reference.
return frac * reference
if __name__ == '__main__':
# First test: we want a 10 bit sample.
_10_bits = adc_do_no_noise_sample()
# Next, create additional samples.
_11_bits = adc_do_noise_sample(1)
_12_bits = adc_do_noise_sample(2)
_13_bits = adc_do_noise_sample(3)
_14_bits = adc_do_noise_sample(4)
_15_bits = adc_do_noise_sample(5)
_16_bits = adc_do_noise_sample(6)
_17_bits = adc_do_noise_sample(7)
_18_bits = adc_do_noise_sample(8)
_24_bits = adc_do_noise_sample(14)
_28_bits = adc_do_noise_sample(18)
# Print results both as integers and voltages.
print "10 bits: ", _10_bits, " volts: ", convert_integer_to_volts(_10_bits, 10)
print "11 bits: ", _11_bits, " volts: ", convert_integer_to_volts(_11_bits, 11)
print "12 bits: ", _12_bits, " volts: ", convert_integer_to_volts(_12_bits, 12)
print "13 bits: ", _13_bits, " volts: ", convert_integer_to_volts(_13_bits, 13)
print "14 bits: ", _14_bits, " volts: ", convert_integer_to_volts(_14_bits, 14)
print "15 bits: ", _15_bits, " volts: ", convert_integer_to_volts(_15_bits, 15)
print "16 bits: ", _16_bits, " volts: ", convert_integer_to_volts(_16_bits, 16)
print "17 bits: ", _17_bits, " volts: ", convert_integer_to_volts(_17_bits, 17)
print "18 bits: ", _18_bits, " volts: ", convert_integer_to_volts(_18_bits, 18)
print "24 bits: ", _24_bits, " volts: ", convert_integer_to_volts(_24_bits, 24)
print "28 bits: ", _28_bits, " volts: ", convert_integer_to_volts(_28_bits, 28)
我希望你能提出任何建议。我的计划是最终采用低成本微控制器来实现高分辨率ADC。速度相当重要，因此我可能会使用LFSR生成PRNG位，这不会像Mersenne twister一样好，但应该足够大多数用途，并且希望对此有足够的贡献。
volts_with_noise()实现方式让我感到有些奇怪：50%的采样将不会添加任何噪声。这是故意的吗？ - Andre Holzner