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 lowpass filter is defined as follows
where  radian frequency,  constant scaling frequency,  order of the filter.
Some properties of the Butterworth filters are:
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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 splane. 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 lowpass 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 lowpass filters with odd orders, two of the poles have zero imaginary parts, so they fall on the real axis in the splane. For the filters with even orders, all imaginary and real parts are nonzero.
The poles of Butterworth filter lie on left half of the splane, and they can be given as follows
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Cutoff frequency determination
Cutoff frequency for the ideal lowpass 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 passband 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 stopband 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 passband 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 passband edge, the scale frequency can be computed using (3.15), and the attenuation at the stopband will be exceeding the specification requirements.

Case2. Specification requirements at the stopband 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 passband 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 stopband 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 stopband edge, the scale frequency can be computed using (3.18), and attenuation at the passband 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 passband and stopband 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 passband are exceeded but requirements at the stopband are not met.
Fig 3.4
Requirements at the stopband are exceeded but requirements at the passband 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.
Matheonics Technology Inc, 2009
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:
Matheonics Technology Inc, 2009