In the second-order case, we have, for the analog prototype,
where, from Eq.(I.2), , so that
(from Eq.(I.9)), where, as discussed in §I.3, we set
to obtain a digital cut-off frequency at radians per second. For example, choosing (a cut off at one-fourth the sampling rate), we get
and the digital filter transfer function is
(I.4) | |||
(I.5) | |||
(I.6) | |||
(I.7) |
It is also immediately verified that , i.e., that there is a (double) notch at half the sampling rate.
In the analog prototype, the cut-off frequency is rad/sec, where, from Eq.(I.1), the amplitude response is . Since we mapped the cut-off frequency precisely under the bilinear transform, we expect the digital filter to have precisely this gain. The digital frequency response at one-fourth the sampling rate is
Note from Eq.(I.8) that the phase at cut-off is exactly -90 degrees in the digital filter. This can be verified against the pole-zero diagram in the plane, which has two zeros at , each contributing +45 degrees, and two poles at , each contributing -90 degrees. Thus, the calculated phase-response at the cut-off frequency agrees with what we expect from the digital pole-zero diagram.
In the plane, it is not as easy to use the pole-zero diagram to calculate the phase at , but using Eq.(I.3), we quickly obtain
and exact agreement with [Eq.(I.8)] is verified.
A related example appears in §9.2.4.