希尔伯特变换

解析信号

Much of digital signal processing requires working with negative frequencies. Negative frequencies in practice do not mean anything and introducing such frequencies in digital signal processing analysis may be troublesome. It is easy to convert a signal that contains negative frequencies into one that does not. A converter that removes negative frequencies from an analytical signal is called a Hilbert transform.

解析信号(analytic signal):在数学和信号处理中,解析信号是没有负频率分量的复值函数。解析信号的实部和虚部是由希尔伯特变换相关联的实值函数。The analytic representation of a real-valued function is an analytic signal, comprising the original function and its Hilbert transform.

表示成解析信号的好处:This representation facilitates many mathematical manipulations. The basic idea is that the negative frequency components of the Fourier transform (or spectrum) of a real-valued function are superfluous (多余的), due to the Hermitian symmetry of such a spectrum. These negative frequency components can be discarded with no loss of information, provided one is willing to deal with a complex-valued function instead. That makes certain attributes of the function more accessible and facilitates the derivation of modulation and demodulation techniques, such as single-sideband.

解析信号到实际信号:As long as the manipulated function has no negative frequency components (that is, it is still analytic), the conversion from complex back to real is just a matter of discarding the imaginary part. The analytic representation is a generalization of the phasor (向量,矢量) concept: while the phasor is restricted to time-invariant amplitude, phase, and frequency, the analytic signal allows for time-variable parameters.

如果\(s(t)\)是一个实数函数,其傅里叶变换为\(S(f)\),则\(S(f)\)对于\(f=0\)轴具有Hermitian symmetry,即\(S(-f)=S(f)^*\)。定义函数$$ \begin{aligned} S_{\mathrm{a}}(f) & \triangleq \begin{cases}2 S(f), & \text { for } f>0, \\ S(f), & \text { for } f=0, \\ 0, & \text { for } f<0\end{cases} \\ & =\underbrace{2 \mathrm{u}(f)}_{1+\operatorname{sgn}(f)} S(f)=S(f)+\operatorname{sgn}(f) S(f) \end{aligned} $$其中\(u(f)\)为Heaviside step function,\(\operatorname{sgn}(f)\)为sign function。那么\(S_{\mathrm{a}}(f)\)只含有\(S(f)\)非负频率的分量。显然,由于\(S(f)\)的Hermitian symmetry特点,我们也可以通过\(S_{\mathrm{a}}(f)\)来表示\(S(f)\),二者的转换是reversible的。实数函数\(s(t)\)的解析信号,就是\(S_{\mathrm{a}}(f)\)的傅里叶逆变换。
其中\(\hat{s}(t) \triangleq \mathcal{H}[s(t)]\)为\(s(t)\)的希尔伯特变换;\(*\)是binary convolution operator;\(j\)是虚部单位。

Noting that \(s(t)=s(t) * \delta(t)\), this can also be expressed as a filtering operation that directly removes negative frequency components:$$ s_{\mathrm{a}}(t)=s(t) * \underbrace{\left[\delta(t)+j \frac{1}{\pi t}\right]}_{\mathcal{F}^{-1}\{2 u(f)\}} $$正如我们在之前傅里叶变换的笔记中谈到的那样,时域的卷积对应于频域的乘法,即上式频域的写法为\(S_{\mathrm{a}}(f)=S(f) 2 \mathrm{u}(f)\)。

负频率分量:由\(s_{\mathrm{a}}(t)=s(t)+j \hat{s}(t)\),易知\(s(t)=\operatorname{Re}\left[s_{\mathrm{a}}(t)\right]\),即恢复负频率分量可以通过丢掉解析信号虚部的方法来实现。同样对于\(s_{\mathrm{a}}^*(t)\) comprises only the negative frequency components. And therefore \(s(t)=\operatorname{Re}\left[s_{\mathrm{a}}^*(t)\right]\) restores the suppressed positive frequency components. Another viewpoint is that the imaginary component in either case is a term that subtracts frequency components from \( s(t)\). The \(\operatorname{Re}\) operator removes the subtraction, giving the appearance of adding new components.

Analytic signal的例子

例子-1,\(s(t)=\cos (\omega t)\), where \(\omega>0\)
\(s(t)\)的希尔伯特变换\(\hat{s}(t)=\displaystyle\frac{1}{\pi t} * s(t)\),写成对应频域的形式,右侧可以表示为\( \mathcal{F}(\displaystyle\frac{1}{\pi t})\mathcal{F}(s(t))=-i  \cdot \operatorname{sgn}(f)S(f)\),即对\(S(f)\)正频率的部分乘以\(-i\),对负频率的部分乘以\( i\)。假设某个正频率\(f\)对应的分量表示为\(a+ib\),那么乘以\(- i\)之后就变成了\(b-ia\),即顺时针旋转了90度相位。总之无论是从相位移动上去理解还是直接算傅里叶变换,我们都将得到如下结果:$$ \begin{aligned} \hat{s}(t) & =\cos \left(\omega t-\frac{\pi}{2}\right)=\sin (\omega t) \\ s_{\mathrm{a}}(t) & =s(t)+j \hat{s}(t)=\cos (\omega t)+j \sin (\omega t)=e^{j \omega t} \end{aligned} $$回忆一下由欧拉公式我们有\(\cos (\omega t)=\frac{1}{2}\left(e^{j \omega t}+e^{j(-\omega) t}\right)\),丢弃其中的负频率,然后让正频率翻倍,即得到上面的结果。

例子-2,Here we use Euler’s formula to identify and discard the negative frequency.$$ s(t)=\cos (\omega t+\theta)=\frac{1}{2}\left(e^{j(\omega t+\theta)}+e^{-j(\omega t+\theta)}\right) $$Then:$$ s_{\mathrm{a}}(t)= \begin{cases}e^{j(\omega t+\theta)}=e^{j|\omega| t} \cdot e^{j \theta}, & \text { if } \omega>0 \\ e^{-j(\omega t+\theta)}=e^{j|\omega| t} \cdot e^{-j \theta}, & \text { if } \omega<0\end{cases} $$

性质:待补充

希尔伯特变换

Relationship with the Fourier transform: \( \mathcal{F}(\mathrm{H}(u))(\omega)=-i \operatorname{sgn}(\omega) \cdot \mathcal{F}(u)(\omega)\)其中$$ -i \operatorname{sgn}(\omega)=\left\{\begin{array}{cc} i=e^{+\frac{i \pi}{2}}, & \text { for } \omega<0 \\ 0, & \text { for } \omega=0 \\ -i=e^{-\frac{i \pi}{2}}, & \text { for } \omega>0 \end{array}\right. $$

为什么需要希尔伯特变换:

参考资料:

联系希尔伯特一起写

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希尔伯特变换和瞬时频率问题–连载(二)

K-K关系

 

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