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Hilbert transform: For any artitrary time series, X(t), we can always have its Hibert Transform, Y(t), as
$$ Y(t) = \frac{1}{\pi}P.V.\int_{-\infty}^{\infty}\frac{X(t')}{t-t'}dt' $$
With this definition, $X(t)$ and $Y(t)$ form the complex conjugate pair, so we can have an analytic signal, $Z(t)$, as
$$ Z(t) = X(t) + i Y(t)=a(t)e^{i\theta(t)} $$
in which
$$ a(t)=[X^2(t)+Y^2(t)]^{1/2},\ \ \theta(t)=\arctan\left(\frac{Y(t)}{X(t)}\right) $$
Then the instantaneous frequency is defined as
$$ \omega=\frac{d\theta(t)}{dt} $$
Intrinsic mode function (IMF) for which the instantaneous frequency can be defined everywhere.
定义瞬时频率的条件:
本征模函数与瞬时频率的关系?
Two conditions:
Knowing the well-behaved Hilbert transforms of the IMF components is only the starting point.
这就是一个两重的迭代过程
完备性根据如下公式可以得到保证:
$$ X(t) = \sum\limits_{i=1}^{n}c_i + r_n $$
正交性:
Orthogonality is a requirement only for linear decomposition systems; it would not make phhysical sense for a nonlinear decomposition as in EMD.
Fortunately, in most cases encountered, the leakage is small.
首先对原始数据进行经验模态分解,得到IMF的各个分量,然后对IMF的各个分量进行Hilbert变换,并计算每个分量的瞬时频率,则可得到原始数据的一个如下展开:
$$ X(t)=\sum\limits_{j=1}^{n}a_j(t)exp\left(i\int \omega_j(t)dt\right) $$
这里我们不考虑最后的那个残量,因为它或者是一个单调的趋势函数,或者是一个常数。
对比Fourier变化:
$$ X(t)=\sum\limits_{j=1}^{\infty}a_je^{i\omega_j t} $$
The IMF represents a generalized Fourier expansion. The variable amplitude and the instantaneous frequency have not only greatly improved the efficiency of the expansion, but also enabled the expansion to accomodate non-stationary data.
根据上面的展开,可以得到振幅的时频分布:$H(\omega,t)$。如果振幅与能量密度有关系,一般振幅的平方可以用来表示Hilbert能量谱。
The marginal spectrum:
$$ h(\omega) = \int_0^T H(\omega,t)dt $$
The instantaneous energy:
$$ IE(t)=\int_{\omega}H^2(\omega,t)d\omega $$
mean marginal spectrum:
$$ n(\omega) = \frac{1}{T}h(\omega) $$
the degree of stationarity:
$$ DS(\omega) = \frac{1}{T}\int_0^T\left(1-\frac{H(\omega,t)}{n(\omega)}\right)^2dt $$
把平稳性定义成频率的函数是合理的,因为对于某些频率来说可能是非平稳的,但是对于其他的频率分量来说却是平稳的。
A modified statistical stationary:
$$ DSS(\omega, \Delta T)=\frac{1}{T}\int_0^{T}\left(1-\frac{\overline{H(\omega,t)}}{n(\omega)}\right)^2dt $$
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