Example: Second-Order Butterworth Lowpass
In the second-order case, we have, for the analog prototype,
where, from Eq.
, so that
To convert this to digital form, we apply the bilinear transform
)), where, as discussed in §I.3
, we set
to obtain a digital cut-off frequency at
second. For example, choosing
(a cut off at
one-fourth the sampling rate
), we get
and the digital filter transfer function
Note that the numerator is
, as predicted earlier.
As a check, we can verify that the dc
gain is 1:
It is also immediately verified that
, that there
is a (double) notch at half the sampling
In the analog prototype,
the cut-off frequency is
), the amplitude response
. Since we mapped the cut-off frequency precisely
under the bilinear transform, we expect the digital filter
precisely this gain.
The digital frequency response
at one-fourth the sampling rate is
Note from Eq.
) 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
contributing +45 degrees, and two poles
, 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.
plane, it is not as easy to use the pole-zero diagram
to calculate the phase at
, but using Eq.
and exact agreement with
)] is verified.
A related example appears in §9.2.4
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