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The Butterworth filter is a filter that can be constructed out of passive R, L, C circuits. The magnitude of the transferfunction for this filter is
The important aspects of are that it does not ripple in the passband or stopband as otherfilters tend to, and that the larger , the sharper the cutoff (the smaller the transition band ).
This transfer function is often seen in its normalized form of
Butterworth filters give transfer functions ( and ) that are rational functions . They also have only poles , resulting in a transfer function of the form
Note that the poles lie along a circle in the s-plane.
Designing a Butterworth filter is a trivial task. Since we know that the filter contains only poles, we know that we canwrite it as
buttap
command. The
real challenge of designing a Butterworth filter comes withfiguring out the optimal characteristics for the given
application.
n | |||||||||
---|---|---|---|---|---|---|---|---|---|
2 | 1.414214 | ||||||||
3 | 2.000000 | 2.000000 | |||||||
4 | 2.613126 | 3.414214 | 2.613126 | ||||||
5 | 3.236068 | 5.236068 | 5.236068 | 3.236068 | |||||
6 | 3.863703 | 7.464102 | 9.141620 | 7.464102 | 3.863703 | ||||
7 | 4.493959 | 10.097835 | 14.591794 | 14.591794 | 10.097835 | 4.493959 | |||
8 | 5.125831 | 13.137071 | 21.846151 | 25.688356 | 21.846151 | 13.137071 | 5.125831 | ||
9 | 5.758770 | 16.581719 | 31.163437 | 41.986386 | 41.986386 | 31.163437 | 16.581719 | 5.758770 | |
10 | 6.392453 | 20.431729 | 42.802061 | 64.882396 | 74.233429 | 64.882396 | 42.802061 | 20.431729 | 6.392453 |
Design a Butterworth filter with a passband gain between 1 and 0.891 (-1 dB gain) for and a stopband not to exceed 0.0316 (-30 dB gain) for .
The first step is to determine . To do this, we must solve for using the passband and stopband criteria. We begin by finding theequation for the gain in the passband in dB,
The next step is to find . We can do this by substituting into the equations for the passband and stopband and solving for . This yields for the passband equation and for the stopband equation. The difference in these solutions is a result of needing to be an integer. If we choose the solution from the passband equation, the passband will meet itsrequirements exactly, and the stopband will surpass its requirements. If we choose the solution from the stopbandequation instead, the stopband requirements will be met exactly, while we will exceed the passband requirements.Therefore, we may choose either value or any value in between. For this example, we will choose .
Now, we can find the normalized transfer function. Since we know this to be a sixth-order Butterworth, we candetermine from the table that
Finally, we can determine the final transfer function.
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