Primer on Hilbert Transform Theory

We need a Hilbert-transform filter ${\cal H}$ to compute the imaginary part $y(t)$ of the analytic signal $x_a(t)$ given its real part $x(t)$ . That is,

$$x_a(t) \eqsp x(t) + j\,y(t)$$ (5.14)

where $y = {\cal H}\{x\}$ . In the frequency domain, we have

$$X_a(\omega) \eqsp X(\omega) + j\,Y(\omega) \eqsp \left[1+j\,H(\omega)\right]\,X(\omega)$$ (5.15)

where $H(\omega)$ denotes the frequency response of the Hilbert transform ${\cal H}$ . Since by definition we have $X_a(\omega)=0$ for $\omega<0$ , we must have $j\,H(\omega)=-1$ for $\omega<0$ , so that $H(\omega)=j$ for negative frequencies (an allpass response with phase-shift $+90$ degrees). To pass the positive-frequency components unchanged, we would most naturally define $H(\omega)=0$ for $\omega>0$ . However, conventionally, the positive-frequency Hilbert-transform frequency response is defined more symmetrically as $H(\omega)=-j$ for $\omega>0$ , which gives $j\,H(\omega)=+1$ and $\mbox{re\ensuremath{\left\{X_a(\omega)\right\}}}=2X(\omega)$ , i.e., the positive-frequency components of $X$ are multiplied by $2$ .

In view of the foregoing, the frequency response of the ideal Hilbert-transform filter may be defined as follows:

$$H(\omega) \isdefs \left\{\begin{array}{ll} \quad\! j, & \omega<0 \\ [5pt] \quad\!0, & \omega=0 \\ [5pt] -j, & \omega>0 \\ \end{array} \right. \protect$$ (5.16)

Note that the point at $\omega = 0$ can be defined arbitrarily since the inverse-Fourier transform integral is not affected by a single finite point (being a ``set of measure zero'').

The ideal filter impulse response $h(n)$ is obtained by finding the inverse Fourier transform of (4.16). For discrete time, we may take the inverse DTFT of (4.16) to obtain the ideal discrete-time Hilbert-transform impulse response, as pursued in Problem 10. We will work with the usual continuous-time limit $h(t)=1/(\pi t)$ in the next section.