Example: Second-Order Butterworth Lowpass Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

#### Example: Second-Order Butterworth Lowpass

In the second-order case, we have, for the analog prototype, where, from Eq. (G.2), , so that (G.3)

To convert this to digital form, we apply the bilinear transform (from Eq. (G.9)), where, as discussed in §G.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   (G.4)  (G.5)  (G.6)  (G.7)

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 , 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. (G.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 (G.8)

and dB as expected.

Note from Eq. (G.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. (G.3), we quickly obtain and exact agreement with (Eq. (G.8)) is verified.

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