Parallel realization of a second-order filter

In parallel programming, $G(z)$ is expanded into partial fractions in $z^{-1}$, and each term is realized as a separate branch.

Step 1: Factor the denominator

$$1 + 0.4z^{-1} - 0.12z^{-2} = (1 + 0.6z^{-1})(1 - 0.2z^{-1})$$

The poles are $z = -0.6$ and $z = 0.2$.

Step 2: Partial fractions

The numerator and denominator have the same degree in $z^{-1}$, so

$$G(z) = C_0 + \frac{C_1}{1 - 0.2z^{-1}} + \frac{C_2}{1 + 0.6z^{-1}}$$

Let $w = z^{-1}$.

$$\begin{aligned} C_0 &= \lim_{w\to\infty}\frac{2 + 2.2w + 0.2w^2}{1 + 0.4w - 0.12w^2} = \frac{0.2}{-0.12} = -1.6667 \\ C_1 &= \frac{2 + 2.2w + 0.2w^2}{1 + 0.6w}\Big|_{w=5} = \frac{2 + 11 + 5}{4} = 4.5 \\ C_2 &= \frac{2 + 2.2w + 0.2w^2}{1 - 0.2w}\Big|_{w=-5/3} = \frac{2 - 3.6667 + 0.5556}{1.3333} = -0.8333 \end{aligned}$$ $$G(z) = -1.6667 + \frac{4.5}{1 - 0.2z^{-1}} - \frac{0.8333}{1 + 0.6z^{-1}}$$

Check at $z^{-1} = 0$: $-1.6667 + 4.5 - 0.8333 = 2$. ✔

Step 3: Difference equations

• $w_1(k) = x(k) + 0.2w_1(k-1)$
• $w_2(k) = x(k) - 0.6w_2(k-1)$
• $y(k) = -1.6667x(k) + 4.5w_1(k) - 0.8333w_2(k)$

Step 4: Block diagram

Overall structure (three branches added):

        +-------------[-1.6667]-------------+
        |                                   v
x(k)-o--+-->[ 4.5/(1-0.2z^-1) ]----------->(S)--> y(k)
        |                                   ^
        +-->[ -0.8333/(1+0.6z^-1) ]---------+

Branch 1:

x(k)-->(S)------------o--[1]-------->(S)->[4.5]-> y1(k)
        ^             |
        |          [z^-1]
        |             | w1(k-1)
        +<[0.2]-------o

Branch 2:

x(k)-->(S)------------o--[1]-------->(S)->[-0.8333]-> y2(k)
        ^             |
        |          [z^-1]
        |             | w2(k-1)
        +<[-0.6]------o

Two delays are used, one per pole.