#### Example: Second-Order Butterworth Lowpass

In the second-order case, we have, for the analog prototype,To convert this to digital form, we apply the bilinear transform

(I.4) | |||

(I.5) | |||

(I.6) | |||

(I.7) |

Note that the numerator is , as predicted earlier. As a check, we can verify that the dc gain is 1:

*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

and dB as expected. 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

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Butterworth Lowpass Poles and Zeros