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Properties of Hilbert Transform
Hilbert Transform
When the phase angles of all the positive frequency spectral components of a signal are shifted by (-90°) and the phase angles of all the negative frequency spectral components are shifted by (+90°), then the resulting function of time is called the Hilbert transform of the signal.
The Hilbert transform of a signal$\mathit{x\left(t\right)}$ is obtained by the convolution of $\mathit{x\left(t\right)}$ and (1/πt),i.e.,,
$$\mathrm{\mathit{\hat{x}\left(t\right)=x\left(t\right)*\left ( \frac{\mathrm{1}}{\mathit{\pi t}} \right )}}$$
Properties of Hilbert Transform
The statement and proofs of the properties of the Hilbert transform are given as follows −
Property 1
The Hilbert transform does not change the domain of a signal.
Proof
Let a signal $\mathit{x\left(t\right)}$, which is in time domain. The Hilbert transform of $\mathit{x\left(t\right)}$, i.e., $\mathit{\hat{x}\left(t\right)}$ is obtained by the convolution of $\mathit{x\left(t\right)}$ and.$\mathit{\left ( \frac{\mathrm{1}}{\pi t} \right )}$ Hence, the function $\mathit{\hat{x}\left(t\right)}$ is also in time domain. Therefore, it proves that the Hilbert transform does not change the domain of a signal.
Property 2
The Hilbert transform does not change the magnitude spectrum of a signal.
Proof
The Fourier transform of $\mathit{\hat{x}\left(t\right)}$ is given by,
$$\mathrm{\mathrm{\mathit{\hat{X}\left(\omega \right)}\mathrm{=}\mathit{-j}\:\mathrm{sgn}\left(\omega \right)\:X\left(\omega\right)}}$$
$$\mathrm{\mathit{\because\left | -j\:\mathrm{sgn}\left(\omega\right) \right |=\mathrm{1}}}$$
Therefore,
$$\mathrm{\mathit{\left | \hat{X}\left(\omega\right) \right |=\left | X\left(\omega\right)\right |}}$$
This proves that the Hilbert transform does not change the magnitude spectrum of a signal, i.e., $\mathit{x\left(t\right)}$ and $\mathit{\hat{x}\left(t\right)}$ have the same magnitude spectrum.
Also, the function $\mathit{x\left(t\right)}$ and $\mathit{\hat{x}\left(t\right)}$ have the same energy density function and same autocorrelation function. If the function $\mathit{x\left(t\right)}$ is band limited, then its Hilbert transform $\mathit{\hat{x}\left(t\right)}$ is also band limited.
Property 3
A signal $\mathit{x\left(t\right)}$ and its Hilbert transform $\mathit{\hat{x}\left(t\right)}$ are orthogonal to each other.
Proof
In order to prove the Orthogonality between $\mathit{x\left(t\right)}$ and $\mathit{\hat{x}\left(t\right)}$, we have to show that,
$$\mathrm{\mathit{\int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\mathrm{0}}}$$
Now, according to Rayleigh's energy theorem, we get,
$$\mathrm{\mathit{\int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}^{*}\left(t\right)\:dt}}$$
$$\mathrm{\mathit{\Rightarrow \int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X\left(\omega\right)\hat{X}^{*}\left(\omega\right)\:d\omega} }$$
$$\mathrm{\mathit{\Rightarrow \int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X\left(\omega\right)\left |j\: \mathrm{sgn}\left(\omega\right)\hat{X}^{*}\left(\omega\right) \right |\: d\omega }}$$
$$\mathrm{\mathit{\Rightarrow \int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\frac{j}{\mathrm{2}\pi }\int_{-\infty }^{\infty }\mathrm{sgn}\left(\omega\right)\left | X\left(\omega \right)\right |^{\mathrm{2}}\:d\omega } }$$
As the function $\mathit{\mathrm{\mathrm{sgn}}\left(\omega \right ) }$ is an odd function and the function $\mathit{\left | X\left(\omega\right) \right |^{\mathrm{2}}}$ is an even function. Therefore, the integral on the RHS is zero, i.e.,
$$\mathrm{\mathit{\int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt=\mathrm{0}}}$$
Hence, this proves that $\mathit{x\left(t\right)}$ and $\hat{x}(t)$ are orthogonal to each other over the interval $(-\infty)$ to $(\infty)$.
Property 4
If the Hilbert transform of $\mathit{x\left(t\right)}$ is $\mathit{\hat{x}\left(t\right)}$, then the Hilbert transform of $\hat{x}(t)$ is $\mathit{\left [-x(t)\right ]}$.
Proof
The Hilbert transform of a signal $\mathit{x\left(t\right)}$ is equivalent to passing the signal $\mathit{x\left(t\right)}$ through a device which is having a transfer function equal to $\mathit{\left [ -j\:\mathrm{sgn}\left(\omega\right) \right ]}$. Therefore, a double Hilbert transform of $\mathit{x\left(t\right)}$ is equivalent to passing $\mathit{x\left(t\right)}$ through a cascade of such devices. Hence, the overall transfer function of such cascaded system is,
$$\mathrm{\mathit{\left [ -j\:\mathrm{sgn}\left(\omega\right) \right ]^{\mathrm{2}}=-\mathrm{1}\left [ \mathrm{sgn}\left(\omega\right) \right ]^{\mathrm{2}}=-\mathrm{1}.\left(\mathrm{1}\right)\mathrm{=}-\mathrm{1};\:\:\mathrm{for} \:\mathrm{all} \:\mathrm{frequencies}}}$$
Hence, the resulting output is $\mathit{\left [-x(t)\right ]}$, i.e., the Hilbert transform of $\mathit{\hat{x}\left(t\right)}$ is $\mathit{\left [-x(t)\right ]}$.
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