<>Hilbert变换简介

x^(t)=H[x(t)]=x(t)∗1πt=1π∫−∞∞x(τ)t−τdτ \begin{array}{ll} \hat{x}(t)&
=H[x(t)]\\ &=x(t)*\frac{1}{\pi t}\\ &=\frac{1}{\pi
}\int^{\infty}_{-\infty}\frac{x(\tau)}{t-\tau}d\tau \end{array}x^(t)​=H[x(t)]=x(
t)∗πt1​=π1​∫−∞∞​t−τx(τ)​dτ​

x(t)=H−1[x(t)]=−1π∫−∞∞x^(τ)t−τdτ \begin{array}{ll} x(t)&=H^{-1}[x(t)]\\
&=-\frac{1}{\pi }\int^{\infty}_{-\infty}\frac{\hat{x}(\tau)}{t-\tau}d\tau
\end{array}x(t)​=H−1[x(t)]=−π1​∫−∞∞​t−τx^(τ)​dτ​

h(t)=1πth(t)=\frac{1}{\pi t}h(t)=πt1​。

H(ω)=−jsgn(ω)=∣H(ω)∣ejϕ(ω)H(\omega)=-jsgn(\omega)\\ =\lvert H(\omega)\rvert
e^{j\phi(\omega)}H(ω)=−jsgn(ω)=∣H(ω)∣ejϕ(ω)

H(ω)={−jω>00ω=0+jω<0H(\omega)= \left\{ \begin{array}{lcl} {-j}
&\omega>0\\ 0&\omega=0\\ {+j} &\omega<0 \end{array} \right.H(ω
)=⎩⎨⎧​−j0+j​ω>0ω=0ω<0​

sgn(ω)={1ω>00ω=0−1ω<0sgn(\omega)= \left\{ \begin{array}{ll} {1}
&\omega>0\\ 0&\omega=0\\ {-1} &\omega<0 \end{array} \right.sgn
(ω)=⎩⎨⎧​10−1​ω>0ω=0ω<0​

X^(ω)=X(ω)∗H(ω)=X(ω)∗(−jsgn(ω))=−jX(ω)∗sgn(ω) \begin{array}{ll}
\hat{X}(\omega)&=X(\omega)*H(\omega)\\ &=X(\omega)*(-jsgn(\omega))\\
&=-jX(\omega)*sgn(\omega)\\ \end{array}X^(ω)​=X(ω)∗H(ω)=X(ω)∗(−jsgn(ω))=−jX(
ω)∗sgn(ω)​

X^(ω)={−jX(ω)ω>00ω=0jX(ω)ω<0 \hat{X}(\omega)= \left\{ \begin{array}{ll}
-jX(\omega) &\omega>0\\ 0&\omega=0\\ jX(\omega) &\omega<0
\end{array} \right.X^(ω)=⎩⎨⎧​−jX(ω)0jX(ω)​ω>0ω=0ω<0​

X^(ω)={−jω∣ω∣X(ω)ω≠00ω=0 \hat{X}(\omega)= \left\{ \begin{array}{ll}
-j\frac{\omega}{|\omega|}X(\omega) &\omega\neq0\\ 0&\omega=0\\
\end{array} \right.X^(ω)={−j∣ω∣ω​X(ω)0​ω̸​=0ω=0​

X′(ω)=jωX(ω)X'(\omega)=j\omega X(\omega)X′(ω)=jωX(ω)

∣ω∣−1。

，而对于负频率成分，移动了pi/2pi/2pi/2。

Hilbert变换的性质有：

* It is anti-symmetric: H(−x)=−H(x)H(-x)=-H(x)H(−x)=−H(x)
* It suppresses the DC component: H(0)=0H(0)=0H(0)=0
* Its energy is equal to one for all nonzero frequencies: ∣H(x)∣=1∀x≠0
\begin{array}{ll} \lvert H(x)\rvert=1 & \forall x\ne0\end{array}∣H(x)∣=1​∀x̸
​=0​
* x(t)x(t)x(t)的Hilbert变换信号x^(t)\hat{x}(t)x^(t)与x(t)x(t)x(t)正交：∫ωx^(t)x(t)dt=0
\int_{\omega}\hat{x}(t)x(t)dt=0∫ω​x^(t)x(t)dt=0
<>Hilbert变换物理意义

<>解析过程概念

x~=x(t)+jx^(t)\widetilde{x}=x(t)+j\hat{x}(t)x=x(t)+jx^(t)

RX^X(τ)=−R^X(τ)RXX^(τ)=R^X(τ)RX(τ)=2[RX(τ)+jR^X(τ)]
R_{\hat{X}X}(\tau)=-\hat{R}_{X}(\tau)\\ R_{X\hat{X}}(\tau)=\hat{R}_{X}(\tau)\\
R_{X}(\tau)=2[R_{X}(\tau)+j\hat{R}_{X}(\tau)]RX^X​(τ)=−R^X​(τ)RXX^​(τ)=R^X​(τ)RX
​(τ)=2[RX​(τ)+jR^X​(τ)]

X~(ω)=X(ω)2u(ω)={2X(ω)ω>0X(0)ω=00ω<0
\widetilde{X}(\omega)=X(\omega)2u(\omega)=\\ \{ \begin{array}{lcl} {2X(\omega)}
&\omega>0\\ {X(0)} &\omega=0\\ {0} &\omega<0 \end{array}X(ω)=X
(ω)2u(ω)={2X(ω)X(0)0​ω>0ω=0ω<0​

<>欧拉公式（Euler‘s formula）的启发

eix=cos(x)+isin(x)e^{ix} = cos(x)+isin(x)eix=cos(x)+isin(x)

cos(ω0t)cos(\omega_0t)cos(ω0​t) π[δ(ω+ω0)+δ(ω−ω0)]
\pi[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]π[δ(ω+ω0​)+δ(ω−ω0​)]
sin(ω0t)sin(\omega_0t)sin(ω0​t) jπ[δ(ω+ω0)−δ(ω−ω0)]
j\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]jπ[δ(ω+ω0​)−δ(ω−ω0​)]

。只有正频率，且是两倍。虽然时域上是复数，但是在频域只有正分量，实际上是一种简化。

<>Hilbert解调原理

​t+ϕ(t))

a(t)=[1+\sum^{M}_{m=1}x_mcos(2\pi f_mt+\gamma_m)]a(t)=[1+∑m=1M​xm​cos(2πfm​t+γm​
)]，fmf_mfm​为调幅信号a(t)a(t)a(t)的频率分量，γm\gamma_mγm​为fmf_mfm​的各初相角。因此a(t)a(t)a(t)

z(t)=x(t)+x^(t)=A(t)ejΦ(t)z(t)=x(t)+\hat{x}(t)=A(t)e^{j\Phi(t)}z(t)=x(t)+x^(t)=
A(t)ejΦ(t)

f_mt+\gamma_m)]A(t)=a(t)=[1+∑m=1M​xm​cos(2πfm​t+γm​)]，Φ(t)=2πfst+ϕ(t)
\Phi(t)=2\pi f_st+\phi(t)Φ(t)=2πfs​t+ϕ(t)。

a(t)=A(t)=x2(t)+x^2(t)a(t)=A(t)=\sqrt{x^2(t)+\hat{x}^2(t)}a(t)=A(t)=x2(t)+x^2(t
)​
ϕ(t)=Φ(t)−2πfst=actan(x(t)x^(t))−2πfst\phi(t)=\Phi(t)-2\pi
f_st=actan(\frac{x(t)}{\hat{x}(t)})-2\pi f_stϕ(t)=Φ(t)−2πfs​t=actan(x^(t)x(t)​)−
2πfs​t

f(t)=\frac{1}{2\pi}\frac{d\phi(t)}{dt}=\frac{1}{2\pi}\frac{d\Phi(t)}{dt}-f_sf(t)
=2π1​dtdϕ(t)​=2π1​dtdΦ(t)​−fs​

<>希尔伯特变换的意义

= sqrt(x^2(t) +
Hilbert(x(t))^2)，而瞬时相位就是虚部（Hilbert变换后的）和实部（原始信号）在某一时间点的比值的arctan，瞬时频率就是它的导数。