希尔伯特变换

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.

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$$S_{\mathrm{a}}(f)=S(f) 2 \mathrm{u}(f)$$。

Analytic signal的例子

$$s(t)$$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。

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

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