Butterworth filters

Filter design - sidney burrus Page 1 / 1

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 12

Three different orders of lowpass Butterworth analog filters:

n 1410

. 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,

\hat{G_{p}} 20 H ω -10 1 ω_{p} ω_{c} 2 n

and for the stopband in dB,

\hat{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 \hat{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 \hat{G_{s}} 10 1 10 \hat{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.

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Filter design - sidney burrus style. OpenStax CNX. May 07, 2009 Download for free at http://cnx.org/content/col10701/1.1

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Filter design - sidney burrus style' conversation and receive update notifications?

Ask

	15 AP 15 Autonomic Nervous System Essay By OpenStax Start Flashcards
	36 Biology 36 Sensory Systems MCQ By OpenStax Start Quiz
	2 AP Key Terms 02 Chemical Level of Organization By OpenStax Start Key Terms
	Anatomy & Physiology By OpenStax Read Online Course
	1 Physiotherapy Flashcards Set 1 By Rhodes Start Flashcards
	Spanish Lesson 4 By Anonymous User Start Quiz
	5 AP Key Terms 05 The Integumentary System By OpenStax Start Key Terms
	7 Biology 07 Cellular Respiration MCQ By OpenStax Start Quiz
	24 Biology 24 Fungi MCQ By OpenStax Start Quiz
	Evolutionary Biology Ecology By Olivia D'Ambrogio Start Quiz