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$\mathrm{x\left(t\right)}$ is obtained by the convolution of $\mathrm{x\left(t\right)}$ and (1/Ï€t),i.e.,,

$$\mathrm{\hat{x}(t)\:=\:x(t)\:\cdot\:\left( \frac{1}{\pi t} \right)}$$

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 $\mathrm{x(t)}$, which is in time domain. The Hilbert transform of $\mathrm{x(t)}$, i.e., $\mathrm{\hat{x}(t)}$ is obtained by the convolution of $\mathrm{x(t)}$ and.$\mathrm{\left( \frac{1}{\pi t} \right)}$ Hence, the function $\mathrm{\hat{x}(t)}$ 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 $\mathrm{\hat{x}(t)}$ is given by,

$$\mathrm{\hat{X}(\omega)\:=\: -j\: sgn (\omega)\:X (\omega)}$$

$$\mathrm{\because\:\left | -j\:sgn (\omega) \right|\:=\:1}$$

Therefore,

$$\mathrm{\left |\hat{X}(\omega) \right|\:=\:\left |X(\omega)\right |}$$

This proves that the Hilbert transform does not change the magnitude spectrum of a signal, i.e., $\mathrm{x(t)}$ and $\mathrm{\hat{x}(t)}$ have the same magnitude spectrum.

Also, the function $\mathrm{x(t)}$ and $\mathrm{\hat{x}(t)}$ have the same energy density function and same autocorrelation function. If the function $\mathrm{x(t)}$ is band limited, then its Hilbert transform $\mathrm{\hat{x}(t)}$ is also band limited.

Property 3

A signal $\mathrm{x(t)}$ and its Hilbert transform $\mathrm{\hat{x}(t)}$ are orthogonal to each other.

Proof

In order to prove the Orthogonality between $\mathrm{x(t)}$ and $\mathrm{\hat{x}(t)}$, we have to show that,

$$\mathrm{\int_{-\infty }^{\infty }x\left(t\right)\:\hat{x}\left(t\right)\:dt\:=\:\mathrm{0}}$$

Now, according to Rayleigh's energy theorem, we get,

$$\mathrm{\int_{-\infty }^{\infty }x(t)\:\hat{x}(t)\:dt\:=\:\int_{-\infty }^{\infty }x(t)\:\hat{x}^{*}(t)\:dt}$$

$$\mathrm{\Rightarrow\: \int_{-\infty }^{\infty }x(t)\:\hat{x}(t)\:dt\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty }X(\omega)\hat{X}^{*}(\omega)\:d\omega}$$

$$\mathrm{\Rightarrow\: \int_{-\infty }^{\infty }x(t)\:\hat{x}(t)\:dt\:=\:\frac{1}{2\pi }\int_{-\infty }^{\infty }X\left(\omega\right)\left |j\: sgn\left(\omega\right)\hat{X}^{*}\left(\omega\right) \right |\: d\omega}$$

$$\mathrm{\Rightarrow\: \int_{-\infty}^{\infty }x(t)\:\hat{x}(t)\:dt\:=\:\frac{j}{2\pi }\int_{-\infty}^{\infty}\: sgn(\omega) \:| X(\omega)|^{2}\:d\omega}$$

As the function $\mathrm{sgn(\omega)}$ is an odd function and the function $\mathrm{| X(\omega)|^{2}}$ is an even function. Therefore, the integral on the RHS is zero, i.e.,

$$\mathrm{\int_{-\infty}^{\infty}\:x(t)\:\hat{x}(t)\:dt\:=\:0}$$

Hence, this proves that $\mathrm{x(t)}$ and $\hat{x}(t)$ are orthogonal to each other over the interval $(-\infty)$ to $(\infty)$.

Property 4

If the Hilbert transform of $\mathrm{x(t)}$ is $\mathrm{\hat{x}(t)}$, then the Hilbert transform of $\hat{x}(t)$ is $\mathrm{[-x(t)]}$.

Proof

The Hilbert transform of a signal $\mathrm{x(t)}$ is equivalent to passing the signal $\mathrm{x(t)}$ through a device which is having a transfer function equal to $\mathrm{[ -j\:sgn(\omega)]}$. Therefore, a double Hilbert transform of $\mathrm{x(t)}$ is equivalent to passing $\mathrm{x(t)}$ through a cascade of such devices. Hence, the overall transfer function of such cascaded system is,

$$\mathrm{[-j\: sgn(\omega)]^{2} \:=\: -1 [ \mathrm{sgn}(\omega)]^{2} \:=\: -1.(1) \:=\: -1;\:\:for\:all \:frequencies}$$

Hence, the resulting output is $\mathrm{[-x(t)]}$, i.e., the Hilbert transform of $\mathrm{\hat{x}(t)}$ is $\mathrm{ [-x(t)]}$.