DIGITAL INFINITE-IMPULSE RESPONSE (IIR) FILTERS-III

EDITOR: B. SOMANATHAN NAIR

Example 1: Obtain the digital filter-transfer function given, given

                                                       H(s) = 1/(s + a) (s + b)    (1)

                                                        

Solution: Equation (1) can be written in partial-fraction form as

                             H(s) = [A/(s + a)] + [B/(s + b)]  (2)                                         

To find A and B, rewrite (2) in the form

         H(s) = [A(s + b) + B(s + a)]/(s + a) (s + b)    (3)

Comparing (1) and (3) and equating the coefficients of s, we get

                                                            A + B  = 0   (4)

Hence

                                                                        A = ‒B (5)

Similarly, equating constants yields

               

                                                            Ab + aB = 1  (6)

                                                           

Substituting for B from (5) into (6), we find

                                                            A = 1/(b ‒ a)  (7)

And

                                                            B = 1/(a ‒ b)   (8)

Using the values of A and B, we can express (2) as

               H(s) = 1/(b ‒ a) (s + a) + 1/(a ‒ b)(s + b)]  (9)                                                          

Taking the inverse Laplace transform of (9), we get

                                                                               

                        h(t) = e^‒at/(b ‒ a) ‒ e^‒bt/(b ‒ a)   (10)

The z transform of (10) is given by

                          H(z) =[1/(b ‒ a)(1 ‒ e^‒az ^‒1)] ‒ [1/(b ‒ a)(1 ‒ e^‒bz ^‒1)]   (11)

Let us now assume that a = 0.3 and b = 0.5. Then (11) gives

                                                                                               

  H(z) =[5/(1 ‒ 0.741z ^‒1)] ‒ [5/(1 ‒ 0.607z ^‒1)]  

                        = H­₁(z) + H­₂(z)  (12)                                

where

                                             H­₁(z) = 5/(1 ‒ 0.741z ^‒1)  (13)

and

         H­₂(z) = ­‒5/(1 ‒ 0.607z ^‒1)  (14)

Equation (13) may be expanded as

                                        H­₁(z) = Y₁(z)/ X₁(z) = 5/(1 ‒ 0.741z ^‒1)  (15)

From which we get

                                                Y₁(z) (1 ‒ 0.741z ^‒1) = 5 X₁(z)  (16)

Taking inverse z transform of (16) and rearrangement yields

                       

y₁(n) = 0.741y₁(n ‒ 1) + 5x₁(n)  (17)

In a similar manner, we obtain

                                    y₂(n) = 0.607y₂(n ‒ 1) ‒ 5x₂(n) (18)

Equations (17) and (18) can be implemented as shown in Figs. 1 and 2. Figure 1 shows the block schematics of parallel form of implementation in general and Fig. 2 shows the actual implementation.