Hilbert Transform | Spectral Audio Signal Processing (original) (raw)

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_:

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

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

$\displaystyle y = h \ast x. % \qquad \hbox{(Hilbert transform of <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">x</span></span></span></span>)}. $ (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

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:

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.

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