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3 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
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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
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The poles of (3.2) can be found as roots of equation
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All
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It is easy to see that
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The poles of Butterworth filter lie on left half of the s-plane, and they can be given as follows
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Minimum order determination
The practical low-pass filter specification is determined by four parameters:
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Since
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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
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It is easy to see from (3.1), that for Butterworth filters, the natural cutoff frequency
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The minimum required order for the Butterworth filter could be computed using (3.12). The scaling frequency
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System of inequalities (3.14) has multiple solutions. Let us consider some practical solutions.
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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.
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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.
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Case3. Specification requirements on both edges are exceeded.
At some conditions, the scale frequencywhich 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 whenand
. 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 frequencycan 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
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As it was mentioned before, all poles of the filter with even order
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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
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Design samples using FAZA
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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:
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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:
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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:
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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:
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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:
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