压摆率测量

Question

我必须测量信号中的压摆率，如下图所示。我需要灰色箭头标记的部分的转换速率。

此刻，我用hann窗口使信号平滑，以消除最终的噪音并使峰值变平。然后我搜索（右起）30％和70％的点并计算这两点之间的转换率。 但我的问题是，平滑后信号会变平。因此，计算的转换速率不会高到应有的水平。如果我减少平滑，那么峰值（你可以看到图像中的右侧）变得更高，最终在错误的位置找到30％的点。

是否有更好/更安全的方法来找到所需的压摆率？

Answer 1

如果你知道你的信号转换的值和你的噪音不是太大，你可以简单地计算30％的所有交叉点和70％的所有交叉点之间的时间差，并保持最小的一个：

import numpy as np
import matplotlib.pyplot as plt

s100, s0 = 5, 0

signal = np.concatenate((np.ones((25,)) * s100,
                         s100 + (np.random.rand(25) - 0.5) * (s100-s0),
                         np.linspace(s100, s0, 25),
                         s0 + (np.random.rand(25) - 0.5) * (s100-s0),
                         np.ones((25,)) * s0))


# Interpolate to find crossings with 30% and 70% of signal
# The general linear interpolation formula between (x0, y0) and (x1, y1) is:
# y = y0 + (x-x0) * (y1-y0) / (x1-x0)
# to find the x at which the crossing with y happens:
# x = x0 + (y-y0) * (x1-x0) / (y1-y0)
# Because we are using indices as time, x1-x0 == 1, and if the crossing
# happens within the interval, then 0 <= x <= 1.
# The following code is just a vectorized version of the above
delta_s = np.diff(signal)
t30 = (s0 + (s100-s0)*.3 - signal[:-1]) / delta_s
idx30 = np.where((t30 > 0) & (t30 < 1))[0]
t30 = idx30 + t30[idx30]
t70 = (s0 + (s100-s0)*.7 - signal[:-1]) / delta_s
idx70 = np.where((t70 > 0) & (t70 < 1))[0]
t70 = idx70 + t70[idx70]

# compute all possible transition times, keep the smallest
idx = np.unravel_index(np.argmin(t30[:, None] - t70),
                       (len(t30), len(t70),))

print t30[idx[0]] - t70[idx[1]]
# 9.6

plt. plot(signal)
plt.plot(t30, [s0 + (s100-s0)*.3]*len(t30), 'go')
plt.plot(t30[idx[0]], [s0 + (s100-s0)*.3], 'o', mec='g', mfc='None', ms=10)
plt.plot(t70, [s0 + (s100-s0)*.7]*len(t70), 'ro')
plt.plot(t70[idx[1]], [s0 + (s100-s0)*.7], 'o', mec='r', mfc='None', ms=10 )
plt.show()