Filtering and Windowing the Ideal
Hilbert-Transform Impulse Response

Let $h_i(t)$ denote the convolution kernel of the continuous-time Hilbert transform from (4.17) above:

$$h_i(t) = \frac{1}{\pi t} \protect$$ (5.22)

Convolving a real signal $x(t)$ with this kernel produces the imaginary part of the corresponding analytic signal. The way the ``window method'' for digital filter design is classically done is to simply sample the ideal impulse response to obtain $h_i(nT)$ and then window it to give $\hat{h}_i(nT)=w(n)h_i(nT)$ . However, we know from above (e.g., §4.5.2) that we need to provide transition bands in order to obtain a reasonable design. A single-sideband filter needs a transition band between dc and $\Omega_w$ , or higher, where $\Omega_w$ denotes the main-lobe width (in rad/sample) of the window $w$ we choose, and a second transition band is needed between $\pi-\Omega_w$ and $\pi$ .

Note that we cannot allow a time-domain sample at time 0 in (4.22) because it would be infinity. Instead, time 0 should be taken to lie between two samples, thereby introducing a small non-integer advance or delay. We'll choose a half-sample delay. As a result, we'll need to delay the real-part filter by half a sample as well when we make a complete single-sideband filter.

The matlab below illustrates the design of an FIR Hilbert-transform filter by the window method using a Kaiser window. For a more practical illustration, the sampling-rate assumed is set to $f_s=22,050$ Hz instead of being normalized to 1 as usual. The Kaiser-window $\beta$ parameter is set to $8$ , which normally gives ``pretty good'' audio performance (cf. Fig.3.28). From Fig.3.28, we see that we can expect a stop-band attenuation better than $50$ dB. The choice of $\beta$ , in setting the time-bandwidth product of the Kaiser window, determines both the stop-band rejection and the transition bandwidths required by our FIR frequency response.

  M = 257;    % window length = FIR filter length (Window Method)
  fs = 22050; % sampling rate assumed (Hz)
  f1 = 530;   % lower pass-band limit = transition bandwidth (Hz)
  beta = 8;   % beta for Kaiser window for decent side-lobe rejection

Recall that, for a rectangular window, our minimum transition bandwidth would be $2f_s/M \approx 172$ Hz, and for a Hamming window, $4f_s/M \approx 343$ Hz. In this example, using a Kaiser window with $\beta=8$ ( $\alpha=\beta/\pi\approx 2.5$ ), the main-lobe width is on the order of $5f_s/M \approx 430$ Hz, so we expect transition bandwidths of this width. The choice $f_1=530$ above should therefore be sufficient, but not ``tight''.^5.8 For each doubling of the filter length (or each halving of the sampling rate), we may cut $f_1$ in half.