Hilbert Transform

The Hilbert transform $y(t)$ of a real, continuous-time signal $x(t)$ may be expressed as the convolution of $x$ with the Hilbert transform kernel:

$$h(t) \isdefs \frac{1}{\pi t},\qquad t\in(-\infty,\infty) \protect$$ (5.17)

That is, the Hilbert transform of $x$ is given by

$$y = h \ast x.$$ (5.18)

Thus, the Hilbert transform is a non-causal linear time-invariant filter.

The complex analytic signal $x_a(t)$ corresponding to the real signal $x(t)$ is then given by

$$x_a(t) \isdef x(t) + j y(t)$$

$$= \frac{1}{\pi}\displaystyle\int_0^{\infty} X(\omega)e^{j\omega t}\, d\omega$$ (5.19)

To show this last equality (note the lower limit of 0 instead of the usual $-\infty$ ), it is easiest to apply (4.16) in the frequency domain:

$$X_a(\omega) \isdef X(\omega) + j Y(\omega)$$

$$\isdef (X_++X_-) + j (-j X_+ + j X_-)$$ (5.20)

$$= 2X_+(\omega)$$ (5.21)

Thus, the negative-frequency components of $X$ are canceled, while the positive-frequency components are doubled. This occurs because, as discussed above, the Hilbert transform is an allpass filter that provides a $90$ degree phase shift at all negative frequencies, and a $-90$ degree phase shift at all positive frequencies, as indicated in (4.16). The use of the Hilbert transform to create an analytic signal from a real signal is one of its main applications. However, as the preceding sections make clear, a Hilbert transform in practice is far from ideal because it must be made finite-duration in some way.