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2.1 Butterworth filters

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Describes the design of analog lowpass Butterworth filters.

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

Magnitude of butterworth filter transfer function

H ω 1 1 ω ω c 2 n
where n is the order of the filter and ω c is the cutoff frequency . The cutoff frequency is the frequency where the magnitude experiences a 3 dB dropoff(where H ω 1 2 ).

Three different orders of lowpass Butterworth analog filters: n 1 4 10 . As n increases, the filter more closely approximates an ideal brickwall lowpassresponse.

The important aspects of are that it does not ripple in the passband or stopband as otherfilters tend to, and that the larger n , the sharper the cutoff (the smaller the transition band ).

This transfer function is often seen in its normalized form of

Magnitude of normalized transfer function for lowpass butterworth filter

H ω 1 1 ω 2 n

Butterworth filters give transfer functions ( H ω and H s ) that are rational functions . They also have only poles , resulting in a transfer function of the form

1 s s 1 s s 2 ⋯ s s n
and a pole-zero plot of

Poles of a 10th-order ( n 5 ) lowpass Butterworth filter.

Note that the poles lie along a circle in the s-plane.

Designing a butterworth filter

Designing a Butterworth filter is a trivial task. Since we know that the filter contains only poles, we know that we canwrite it as

H s 1 s n a n - 1 s n 1 ⋯ a 1 s 1
From this, we may look up the a i from a table (like the one below) for any desired n . We can also find them in Matlab by using the buttap command. The real challenge of designing a Butterworth filter comes withfiguring out the optimal characteristics for the given application.

n a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9
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 0 ω 10 and a stopband not to exceed 0.0316 (-30 dB gain) for ω 20 .

The first step is to determine n . To do this, we must solve for n using the passband and stopband criteria. We begin by finding theequation for the gain in the passband in dB,

G p ^ 20 H ω -10 1 ω p ω c 2 n
and for the stopband in dB,
G s ^ 20 H ω -10 1 ω s ω c 2 n
these equations can also take the form
ω x ω c 2 n 10 G x ^ 10 1
In this form, we may divide the passband equation by the stopband equation to get rid of the ω c . From there, we can solve for n to get
n 10 G s ^ 10 1 10 G p ^ 10 1 2 ω s ω p
By plugging in, we find n 5.9569 . However, since n must be an integer, we round this up to n 6

The next step is to find ω c . We can do this by substituting n 6 into the equations for the passband and stopband and solving for ω c . This yields ω c 11.1919 for the passband equation and ω c 11.2478 for the stopband equation. The difference in these solutions is a result of n 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 ω c 11.2478 .

Now, we can find the normalized transfer function. Since we know this to be a sixth-order Butterworth, we candetermine from the table that

H s 1 s 6 3.863703 s 5 7.464102 s 4 9.141620 s 3 7.464102 s 2 3.863703 s 1

Finally, we can determine the final transfer function.

H s 1 s 11.2478 6 3.863703 s 11.2478 5 7.464102 s 11.2478 4 9.141620 s 11.2478 3 7.464102 s 11.2478 2 3.863703 s 11.2478 1
Rather than multiplying this out and factoring, we will leave it in this form for readability, since the numberscan get quite large otherwise.
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Source:  OpenStax, Communications b : filters and transmission lines. OpenStax CNX. Nov 30, 2012 Download for free at http://cnx.org/content/col11169/1.2
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