BUTTERWORTH IIR FILTERS-IV STEP-BY-STEP DESIGN PROCEDURE

EDITOR: B. SOMANATHAN NAIR

1.    INTRODUCTION

In this blog, we explain the step-by-step design procedure of designing a Butterworth filter in accordance with the given specifications. As stated in a previous blog, the step-by-step design of Butterworth digital filters begins with the design of its analog counterpart. From the analog design steps, we find expression for H(s). From H(s) we find H(z) either by direct conversion or with the help of a suitable transformation. Then using H(z) we construct the desired filter structure.

            For the design of digital filters, we use the same set of equations that we have used for the design of analog filters, with some modifications incorporated in them. We know that digital signals are obtained from analog signals by means of analog-to-digital conversion. For A/D conversion, analog signals are to be sampled by means of sampling pulses. Hence, sampling frequency must be included into the specifications.

To incorporate sampling, we modify the specifications suitably, as illustrated in the example given below.

Example 1: Design a Butterworth digital low-pass filter for the following specifications:

1.      Pass-band gain required                            :           0.89 

2.      Frequency  up to which pass-

band gain must remain more or

less steady, f₁                                          :           25 Hz.

3.  Amount of attenuation required                 :           0.215

4.   Frequency from which

     the attenuation must start,  f₂                 :           75 Hz

      5.  Sampling frequency, f_s                                :           300 Hz

            In the specifications given above, the only modification we have made is to add the sampling-frequency term f_S. This is the frequency with which analog-to-digital conversion is done. According to sampling theorem, we must have

                                                                 f_s  ³ 2 f_m (1)                                                                                     

where f_m is the maximum frequency in the analog input signal to be processed. In our example, we have been given f_s = 300 Hz. This value is much higher than the required minimum criterion given in (1). If it is not specified in a given design problem, then we can assume a suitable value for f_s based on (1). We now proceed with our design.

STEP 1: NORMALIZATION OF FREQUENCIES

Before starting the actual procedure for the digital filter design, it is quite common to normalize the given frequencies for the convenience of using small values of w’s. Similar techniques may be used in analog designs also. It may be noted that if normalized frequencies are initially used, then to get the final results denormalization must be done.

            To normalize a given frequency, we use the equation

                                                           

ω_n = 2πf/f_s  (2)

                                                                                                                         

where ω_n = normalized angular frequency, f is either f₁ or f₂, and f_s = sampling frequency. Note that the normalized frequency is a mere number and has no units attached to it.

 In certain cases, frequencies will be specified in terms of angular frequencies (i.e. in terms of ω’s). In such cases, we still proceed in the same way as given above for normalizing the frequencies. This is because, by definition, angular frequency

             ω = 2πf (3)

Then

                                                       ω₁ = 2πf₁ (4)

                                                             ω₂ = 2πf₂ (5)                                                                                

Now, using Eq. (10.65), we get the normalized frequencies

                                        ω₁ = ω_n1 = 2πf₁/f_s = 2πx25/300 = π/6 (6)

                         

                   ω₂ = ω_n2 = 2πf₂/f_s= 2πx75/300 = π/2 (7)     

                                                              

STEP 2: DETERMINATION OF n AND w_C

Given

                                  H₁(ω) = 0.89

We substitute w₁ = p/6 in the Butterworth polynomial to yield

                                                (0.89)² = 1/[1+(π/6ω_c)²ⁿ]   (8)

                                                                   

Similarly, at w₂ = p/2 in the Butterworth polynomial, we have

                                                (0.215)² = 1/[1+(π/2ω_c)²ⁿ]   (9)

Solving (8) and (9) simultaneously, we get n = 1.98. Choosing the next higher integer, we find the order of the filter

                                         n = 2    (10)                                                                  

To find ω_c, we use Eq. (10.74). Substituting for n = 2 in (9) and solving, we get    

                                                            ω_c = 0.737 (11)

                                                                                                                   

STEPS 2 AND 3: DETERMINATION POLES AND VALID POLES

We have already seen that the poles of a second-order filter lie at 90^o with respect to each other on the circle of radius ω_c, and that the valid poles are

            B₁ = ω_c(‒0.707+j0.707) = 0.737(‒0.707+j0.707) = ‒0.521+j0.521 (12)

The second pole B₂ is the complex conjugate of B₁, and therefore

                                 B₂ = ‒0.521‒j0.521  (13)                                       

STEP 4: FINDING THE EXPRESSION FOR H(S) IN THE ANALOG DOMAIN     

By using the poles, we can now write the expression for H(s) as:

                        H(s) = ω_c²/(s² + 2δω_cs+ ω_c²) = ω_c²/(s+a+jb)(s+a‒jb)

                                = 0.737²/(s+0.521+j0.521)(s+0.521‒j0.521)

                                = 0.543/(s+0.521)²+0.521²)   (14)

Taking the inverse Laplace transform of (14), we obtain

                             h(t) = 1.042e^‒0.521t sin (0.521t)  (15)

By taking the z transform of h(t), the transfer function of desired digital filter will be obtained.