Question

我正在使用Python中的频谱包，使用multitaper方法计算60个磁力计局部磁北组分读数的样本的功率谱密度（PSD）（参见http://pyspectrum.readthedocs.io/en/latest/_modules/spectrum/mtm.html），每分钟采样，测量在nanotesla。

我的目标是仅在1.5-6mHz频段观看PSD，因为这是我研究中唯一重要的频段。

使用自适应方法运行函数（默认NFFT = 256，NW = 1.4，默认k）我得到以下图

我的理解是y轴的单位（nT）^ 2 / Hz很好，但是如何在x轴上得到感兴趣的特定mHz频段？我最初认为x轴是以Hz为单位，但对于我选择的任何NFFT值，图表看起来都相同，但只是从0跨越到x轴的NFFT值。

有人知道我如何寻找1.5 - 6 mHz范围内的功率（或者我是否完全采用了错误的方式？

谢谢！

如果它有帮助，那么函数正在执行的代码（对于它引用的其他函数，请参阅我上面发布的链接，我刚刚发布了这个特定函数，因为它可以更深入地了解正在发生的事情（使用np.FFT.FFT例如）：

def pmtm(x, NW=None, k=None, NFFT=None, e=None, v=None, method='adapt', show=False):
"""Multitapering spectral estimation

:param array x: the data
:param float NW: The time half bandwidth parameter (typical values are
    2.5,3,3.5,4). Must be provided otherwise the tapering windows and
    eigen values (outputs of dpss) must be provided
:param int k: uses the first k Slepian sequences. If *k* is not provided,
    *k* is set to *NW*2*.
:param NW:
:param e: the window concentrations (eigenvalues)
:param v: the matrix containing the tapering windows
:param str method: set how the eigenvalues are used. Must be
    in ['unity', 'adapt', 'eigen']
:param bool show: plot results
:return: Sk (complex), weights, eigenvalues

Usually in spectral estimation the mean to reduce bias is to use tapering window.
In order to reduce variance we need to average different spectrum. The problem
is that we have only one set of data. Thus we need to decompose a set into several
segments. Such method are well-known: simple daniell's periodogram, Welch's method
and so on. The drawback of such methods is a loss of resolution since the segments
used to compute the spectrum are smaller than the data set.
The interest of multitapering method is to keep a good resolution while reducing
bias and variance.

How does it work? First we compute different simple periodogram with the whole data
set (to keep good resolution) but each periodgram is computed with a different
tapering windows. Then, we average all these spectrum. To avoid redundancy and bias
due to the tapers mtm use special tapers.

.. plot::
    :width: 80%
    :include-source:

    from spectrum import data_cosine, dpss, pmtm

    data = data_cosine(N=2048, A=0.1, sampling=1024, freq=200)
    # If you already have the DPSS windows
    [tapers, eigen] = dpss(2048, 2.5, 4)
    res = pmtm(data, e=eigen, v=tapers, show=False)
    # You do not need to compute the DPSS before end
    res = pmtm(data, NW=2.5, show=False)
    res = pmtm(data, NW=2.5, k=4, show=True)


.. versionchanged:: 0.6.2

    APN modified method to return each Sk as complex values, the eigenvalues
    and the weights

"""
assert method in ['adapt','eigen','unity']

N = len(x)

# if dpss not provided, compute them
if e is None and v is None:
    if NW is not None:
        [tapers, eigenvalues] = dpss(N, NW, k=k)
    else:
        raise ValueError("NW must be provided (e.g. 2.5, 3, 3.5, 4")
elif e is not None and v is not None:
    eigenvalues = e[:]
    tapers = v[:]
else:
    raise ValueError("if e provided, v must be provided as well and viceversa.")
nwin = len(eigenvalues) # length of the eigen values vector to be used later

# set the NFFT
if NFFT==None:
    NFFT = max(256, 2**nextpow2(N))

Sk_complex = np.fft.fft(np.multiply(tapers.transpose(), x), NFFT)
Sk = abs(Sk_complex)**2

# si nfft smaller thqn N, cut otherwise add wero.
# compute
if method in ['eigen', 'unity']:
    if method == 'unity':
        weights = np.ones((nwin, 1))
    elif method == 'eigen':
        # The S_k spectrum can be weighted by the eigenvalues, as in Park et al.
        weights = np.array([_x/float(i+1) for i,_x in enumerate(eigenvalues)])
        weights = weights.reshape(nwin,1)

elif method == 'adapt':
    # This version uses the equations from [2] (P&W pp 368-370).

    # Wrap the data modulo nfft if N > nfft
    sig2 = np.dot(x, x) / float(N)
    Sk = abs(np.fft.fft(np.multiply(tapers.transpose(), x), NFFT))**2
    Sk = Sk.transpose()
    S = (Sk[:,0] + Sk[:,1]) / 2    # Initial spectrum estimate
    S = S.reshape(NFFT, 1)
    Stemp = np.zeros((NFFT,1))
    S1 = np.zeros((NFFT,1))
    # Set tolerance for acceptance of spectral estimate:
    tol = 0.0005 * sig2 / float(NFFT)
    i = 0
    a = sig2 * (1 - eigenvalues)

    # converges very quickly but for safety; set i<100
    while sum(np.abs(S-S1))/NFFT > tol and i<100:
        i = i + 1
        # calculate weights
        b1 = np.multiply(S, np.ones((1,nwin)))
        b2 = np.multiply(S,eigenvalues.transpose()) + np.ones((NFFT,1))*a.transpose()
        b = b1/b2

        # calculate new spectral estimate
        wk=(b**2)*(np.ones((NFFT,1))*eigenvalues.transpose())
        S1 = sum(wk.transpose()*Sk.transpose())/ sum(wk.transpose())
        S1 = S1.reshape(NFFT, 1)
        Stemp = S1
        S1 = S
        S = Stemp   # swap S and S1
    weights=wk

if show is True:
    from pylab import semilogy
    if method == "adapt":
        Sk = np.mean(Sk * weights, axis=1)
    else:
        Sk = np.mean(Sk * weights, axis=0)
    semilogy(Sk)

return Sk_complex, weights, eigenvalues

Answer 1

如果正确采样，以1/60 Hz采样的离散时间信号表示从 -1/120 Hz 到 1/120 Hz 的频率：https://en.wikipedia.org/wiki/Nyquist_frequency

如果样本都是实值，那么负频率和正频率分量是相同的。

离散时间信号的频率是圆形/模块化的，即在您的信号中，频率 f 和 f + 1/60 Hz 是相同的。

在PSD结果中，x轴在此范围内运行，从 0 Hz 开始，一直到 1/120 Hz ，这与 -1 / 120Hz ，并再次继续 0 Hz 之前。

bin x 表示的频率为 f = x * NFFT / 60 Hz ，您可以忽略所有 x＆gt; = NFFT / 2

如果您只对特定乐队感兴趣，那么您可以选择所需的值并丢弃其余的值，选择 NFFT 来确定分辨率。

理解Python中multitaper方法的频率轴

1 个答案: