BUTTERWORTH ANALOG FILTER DESIGN

1. INTRODUCTION

Butterworth filters are characterized by a flat pass and attenuation bands, as shown in Fig. 1. The transfer function for Butterworth low-pass filters (see previous blog) is given by

                                                │H(ω) │ = 1/[1+ (ω/ω_c)²ⁿ]^1/2    (1)

It can be seen that in this equation, only the first and last terms of the binomial expression [(1+(ω/ω_c)² ]ⁿ are present; all the intermediate terms between the first term ‘1’ and the last term ‘ω/ω_c’, are absent as they do not contribute to the activities of the filter. Since only the first and the last terms are present in the equation, the filter-transfer function is called an approximation.

We now describe the general procedure for designing a Butterworth low-pass filter. It may be noted that all filters are designed at first as a low-pass filter. Thus, we design the high-pass, band-pass, and band-reject filters as low-pass filters. Then they are converted into the desired filter by frequency-transformation techniques.

Filters are designed based on requirements known as specifications. The important specifications of a filter are:

1.      Pass-band gain

2.      Pass-band width

3.      Attenuation starting frequency

4.      Attenuation required                             

These points are illustrated in Fig. 1.

2. DESIGN STEPS WHEN H(w) IS A NUMBER

The filter specifications can be given in number form or in the unit of decibels. We first consider the design when the specs are given in numbers. The following example will illustrate the entire design procedure.

2.1 SPECIFICATIONS

Design a Butterworth low-pass filter to meet the following specifications:

            1. Pass-band gain required                      :                       H₁(w)

            2. Pass-band frequency limit                     :                       w₁

                3. Attenuation required                               :                       H₂(w)

            4. Attenuation-band starting frequency   :                       w₂

2.2       Design Steps

Step 1: Determination of the order n and cut-off frequency ω_c of the filter

Consider Eq. (1), which gives the transfer function of an n^th-order Butterworth low-pass filter. The two unknowns in this equation are n, the order of the filter and ω_c, its cut-off frequency. Using the values given in the specifications, we can determine the values of n and ω_c from Eq. (1). For this, first we find that at w = w₁, we have

                             │H₁(ω) │ = 1/[1+ (ω₁/ω_c)²ⁿ]^1/2    (2)

                           

Similarly, at w = w₂, we obtain

                            │H₂(ω) │ = 1/[1+ (ω₂/ω_c)²ⁿ]^1/2    (3)

                                   

Equations (2) and (3) are used for obtaining the values of n and ω_c.

Step 2: Determination of the Poles of H(s)

The poles of the Butterworth polynomial lie on a circle of radius ω_c, as shown in Fig. 2. To obtain the poles of the Butterworth filter, the following conditions are used:

1.  Number of Butterworth poles =  2n, where n is the order of the filter.                                                                                   

2.  Angle between poles q = 360^º/2n

3.  If n is even, the first pole will be located at q/2, where q is measured with respect to the +x axis.

4.   The second pole will be at (q/2) + q, the third pole will be at (q/2) + 2q,  and so on. 

5.   If n is odd, the first pole will be at q and the remaining poles will be at q +q, q +2q, and so on.

6.  Valid poles are those that lie in the left-half of the s plane, i.e., in  between 90° and 270°

To illustrate the above points, let us choose n = 2.Using the condition given above, we get

1.The number of Butterworth poles = 2n = 4                                                                               

2.  Angle between any two poles q = 360º/4 = 90°

3. Since n =2 is even, the first pole will be at 90°/2 = 45°

4. Other poles are at 135° ( = 45° + 90°), 225° (=45°+ 180°), and 315°(= 45°+270°), respectively.

5. The valid poles are those that are located at 135° and 225°, respectively

From the above results, we find the locations of the valid poles, with ω_c = 1, as:

          Location of pole B₁ = cos135º + jsin135º = ‒0.707+j0.707                                     

           Location of conjugate pole B₂ = cos135º-j sin135º= ‒0.707‒j0.707

These poles are located on the periphery of the circle of radius ω_c = 1, as shown in Fig. 3.

            It may be noted that if ω_c is not equal to unity, then the respective terms must be multiplied by the value of ω_c. We then find that

                                    B₁ is at ω_c(‒0.707+j0.707) 

                         

                                    B₂ is at ω_c (‒0.707‒j0.707)

             To illustrate the above points, let us assume that ω_c = 1.1. Then we observe that the poles will be located, respectively, at

   1.1 × (cos 135º+j sin 135º)  =(‒0.777+j0.777)

                                                                                          

                            1.1 × (cos 135º-j sin 135º) =(‒0.777‒j0.777)

                                                                                                     

Step 3: Transfer Gain H(s) expressed in the Laplace (s) Domain

When the location of the poles, the transfer function of a second-order filter can be obtained by using the general expression for the transfer function of a second-order filter in the s domain given by:

where (–a+jb) and (–a–jb) are the poles, and d is the damping factor. We now substitute the values of the valid poles from the above to get

Using the transfer function given in (5) or (6), we can design the filter, as illustrated in the numerical example given in the next blog.