Hilbert transform properties; distortionless transmission

Hilbert transform

The Hilbert transform $\hat x(t)$ of a signal $x(t)$ is obtained by shifting the phase of every positive-frequency component by $-90°$ and every negative-frequency component by $+90°$, without changing amplitudes.

Time domain: it is the convolution of $x(t)$ with $\frac{1}{\pi t}$:

$$\hat x(t)=x(t)*\frac{1}{\pi t}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau$$

Frequency response: since $\frac{1}{\pi t}\leftrightarrow-j\,\text{sgn}(f)$,

$$H(f)=-j\,\text{sgn}(f)=\begin{cases}-j=1\angle-90°, & f>0\\ +j=1\angle+90°, & f<0\end{cases},\qquad \hat X(f)=-j\,\text{sgn}(f)X(f)$$

  |H(f)|              angle H(f)
 1 +-------+------    +90 ----+
   |       |                  |
 --+-------+------> f  -------+-------> f
           0                  |
                              +---- -90

Inverse: $x(t)=-\hat{\hat x}(t)$, i.e. $x(t)=-\frac{1}{\pi}\int\frac{\hat x(\tau)}{t-\tau}d\tau$.

Properties of the Hilbert transform

1. $x(t)$ and $\hat x(t)$ have the same amplitude spectrum, hence the same ESD/PSD, energy and power.
2. Applying it twice gives the negative: $\hat{\hat x}(t)=-x(t)$.
3. $x(t)$ and $\hat x(t)$ are orthogonal: $\int_{-\infty}^{\infty}x(t)\hat x(t)dt=0$.
4. $\widehat{\cos\omega t}=\sin\omega t$ and $\widehat{\sin\omega t}=-\cos\omega t$.
5. For a low-pass $m(t)$ and carrier $f_c$ greater than its bandwidth: $\widehat{m(t)\cos\omega_ct}=m(t)\sin\omega_ct$.
6. It is a linear operation, and an even signal gives an odd Hilbert transform (and vice versa).

Use: SSB generation $s(t)=m\cos\omega_ct\mp\hat m\sin\omega_ct$, pre-envelope and band-pass signal representation.

Distortionless transmission (line)

A transmission system or line is distortionless if the output is an exact replica of the input except for a constant scale factor and a constant delay:

$$y(t)=K\,x(t-t_d)\ \Rightarrow\ H(f)=K\,e^{-j2\pi ft_d}$$

Frequency response requirements:

• Amplitude: $\lvert H(f)\rvert=K$ constant for all frequencies in the signal band; otherwise amplitude distortion.
• Phase: $\theta(f)=-2\pi ft_d$, a straight line through the origin with slope $-2\pi t_d$; i.e. constant group delay $\tau_g=-\frac{1}{2\pi}\frac{d\theta}{df}=t_d$. Otherwise phase (delay) distortion.

  |H(f)|                theta(f)
 K +--------------        |\
   |                 -----+-\------> f
   +-------------> f      |  \ slope -2*pi*td

For a physical transmission line with parameters $R, L, G, C$ per unit length, this is achieved when the Heaviside condition $\frac{R}{L}=\frac{G}{C}$ holds. Then the attenuation constant $\alpha=\sqrt{RG}$ and the velocity $v=\frac{1}{\sqrt{LC}}$ are both independent of frequency, so all frequency components are attenuated and delayed equally. Telephone lines achieve this approximately by loading coils that increase $L$.