Butterworth approximation

The classical method of analog filters design is Butterworth approximation. The Butterworth

filters are also known as maximally flat filters. Squared magnitude response of a Butterworth

low-pass filter is defined as follows

where - radian frequency, - constant scaling frequency, -

order of the filter.

Some properties of the Butterworth filters are:

gain at DC is equal to 1;

has a maximum at

The first derivatives of (3.1) are equal to zero at . This is why Butterworth

filters are known as maximally flat filters.

Poles locations

Using property (1.26) expression (3.1) can be rearranged to the form

Given this expression can be written as follows

Function has poles and doesn't have any finite zeros. It is easy to see

that if is a pole of (3.2), then is also a pole of (3.2). In order to find the poles of

transfer function that satisfy (3.2), we have to select one pole from each pair

of the poles of expression (3.2). As it was mentioned before, the poles

of a valid filter have to have negative real parts.

The poles of (3.2) can be found as roots of equation

Observing that , where stands for the odd number, roots of (3.3)

can be obtained as solutions to the equation

The solution of (3.4) can be presented in the form

So, roots (3.5) are the poles of .

All poles lie on a circle of radius in the complex s-plane. Since the difference

has the same value for all roots, it can be concluded that the poles are

equally spaced on the circumference.

Fig 3.1 Pole locations of

the squared magnitude

response for the

Butterworth low-pass

filters with orders N=6

and N=5. Scaling

frequency .

It is easy to see that

for the

. Therefore, N roots of (3.5) have the negative real parts; these are the poles of

. The remaining roots are the poles of . Real parts in (3.5) are also never zero,

so poles never fall to the imaginary axis. For the Butterworth low-pass filters with odd

orders, two of the poles have zero imaginary parts, so they fall on the real axis in the s-

plane. For the filters with even orders, all imaginary and real parts are nonzero.

The poles of Butterworth filter lie on left half of the s-plane, and they can be given as

follows

Minimum order determination

The practical low-pass filter specification is determined by four parameters: .

The first step to design a filter with these parameters is to determine the minimum order

of the filter that meets this specification. The signal attenuation for the Butterworth

approximation can be expressed as follows

Applying (3.7) to the pass-band and stop-band edges results in the following system of

two equations

These equations can be rearranged as

Variables and must be obtained from system (3.9). The order of the filter is an

integer variable, and scaling frequency is the real variable. Parameters

are real and they are set in the specification (parameters were

replaced with ). In general, system of equations (3.9) doesn't have a precise

solution for those kinds of variables. But it can be solved if integer variable is replaced

with the real variable .

The precise solution of system (3.10) can be easily found as

(3.11) is a precise solution to the equations (3.10). It can be used to determine the

approach to the solution of the system (3.9). Minimum order of the Butterworth filter that

meets specification can be found as

where brackets [] stand for the nearest integer exceeding .

Since , the solution of the equations (3.9) will be exceeding the specification

requirements. In order to complete the Butterworth filter design, the scaling frequency

must be determined.

Cutoff frequency determination

Cutoff frequency for the ideal low-pass filters is determined as a frequency up to which

the signals pass well and the signals beyond it are rejected. Due to the transition band in

the practical filters, the definition of cutoff frequency is unclear. It is a common practice to

characterize practical filters with a natural cutoff frequency, which is determined as a

frequency at which the power gain is 1/2 of the power gain at the passband frequencies.

Since the power of the signal is proportional to the amplitude, the magnitude response at

the natural cutoff frequency is as follows

Converting gain to decibels results in

Therefore, the natural cutoff frequency can be determined as a frequency where the

signal loss through the filter is approximately 3 dB.

It is easy to see from (3.1), that for Butterworth filters, the natural cutoff frequency

, and it doesn't depend on the order of the filter.

The minimum required order for the Butterworth filter could be computed using (3.12).

The scaling frequency cannot be determined to precisely satisfy both of the edge

conditions (3.9). The scaling frequency can be determined in such a way as to satisfy the

system of inequalities

where the order of the filter is determined from (3.12). If these inequalities are

satisfied, then Butterworth filter meets or exceeds the specification requirements.

System of inequalities (3.14) has multiple solutions. Let us consider some practical

solutions.

Case 1. Specification requirements at the pass-band edge are met precisely.

In this case, the first inequality in (3.14) should be replaced with equality, and scale

frequency can be found as follows

Inserting scale frequency (3.15) to (3.7), the attenuation at the stop-band edge can be

computed

This attenuation must satisfy the second inequality in (3.14). To demonstrate it, the

following equations for the precise solution and for the solution of the system that

precisely meets the specification requirements at the pass-band edge can be considered.

Since and , it is easy to see from (3.17) that the second inequality in

(3.14) is satisfied.

For the system that precisely meets the specification requirements at the pass-band

edge, the scale frequency can be computed using (3.15), and the attenuation at the stop-

band will be exceeding the specification requirements.

Case2. Specification requirements at the stop-band edge are met precisely.

In this case, the second inequality in (3.14) should be replaced with equality. The scale

frequency can be expressed as follows

Attenuation at the pass-band edge

where scale frequency is expressed by (3.18).

Equations for the precise solution and for the solution of the system that precisely meets

the specification requirements at the stop-band edge are

Since and , it is easy to see from (3.20) that , and the first

inequality in (3.14) is satisfied. For the system that precisely meets the specification

requirements at the stop-band edge, the scale frequency can be computed using (3.18),

and attenuation at the pass-band will be exceeding the specification requirements.

Case3. Specification requirements on both edges are exceeded.

At some conditions, the scale frequency which was determined for the precise

solution (3.11) can be used to satisfy the system of inequalities (3.14). Attenuation for the

precise solution of the system is given by

Attenuation of the Butterworth filter with order and scaling frequency is as

follows

Parameters in these expressions are determined by (3.11).

Fig 3.2 displays the precise solution (3.21) and solution (3.22) for the case when

.

Fig. 3.2 Case .

It is clear that in this case the specification requirements are exceeded on both pass-band

and stop-band edges.

Fig 3.3 and Fig 3.4 display solutions (3.21) and (3.22) for the cases when and

. In both cases one of the edge requirements is not met.

Fig 3.3

Requirements at the pass-band are exceeded but requirements at the stop-band are not

met.

Fig 3.4

Requirements at the stop-band are exceeded but requirements at the pass-band are not

met.

Consequently, the scaling frequency can be used to meet or exceed specification

requirements on both edges only in the case when . For most

practical filters, these conditions are met. If there is the need to design a filter with

different conditions, it is recommended to use expressions (3.15) or (3.18) to compute

scaling frequency for the Butterworth filters.

How to obtain transfer function of the low-pass Butterworth filters

The transfer function of the low-pass Butterworth filter can be expressed as follows

where are poles of transfer function.

As it was mentioned before, all poles of the filter with even order occur in complex

conjugate pairs. The product of every pair of complex conjugate poles

can be expressed using formulas (3.5) for Butterworth poles

Therefore, the transfer function of the low-pass Butterworth filter with even order can

be expressed as follows

In the case when the order is an odd number, the Butterworth filters have a single real

pole equal to and complex conjugate pairs. Therefore, the transfer

function of the Butterworth low-pass filter with an odd order can be expressed as follows

Note that the normalized form of transfer functions can be obtained by setting in

(3.25) and (3.26).

As it was shown earlier in (1.16), the transfer function of a serial system is determined as

a product of transfer functions of the linked blocks. Therefore, Butterworth filters of any

order can be presented in the form of serially linked blocks of the first and second

orders:

When the order of the filter is even, the transfer function of the Butterworth low-pass

filter can be submitted in the form

When the order of the filter is odd, the transfer function of the Butterworth low-pass

filter can be submitted in the form

Design samples using FAZA

Sample 1:

Apass = 0.5 db

Astop = 3.0 db

Fpass = 0.5 rad/sec

Fstop = 1.5 rad/sec

Design results: min order = 1; Transfer function:

Sample 2:

Apass = 0.5 db

Astop = 10.0 db

Fpass = 0.5 rad/sec

Fstop = 1.5 rad/sec

Design results: min order = 2; Transfer function:

Sample 3:

Apass = 0.5 db

Astop = 15.0 db

Fpass = 0.5 rad/sec

Fstop = 1.5 rad/sec

Design results: min order = 3; Transfer function:

Sample 4:

Apass = 0.5 db

Astop = 20.0 db

Fpass = 0.5 rad/sec

Fstop = 1.5 rad/sec

Design results: min order = 4; Transfer function:

Sample 5:

Apass = 0.5 db

Astop = 30.0 db

Fpass = 0.5 rad/sec

Fstop = 1.5 rad/sec

Design results: min order = 5; Transfer function: