Parallel programming of a digital filter

In parallel programming, $G(z)$ is expanded into partial fractions in $z^{-1}$. Each term is realized separately (first- or second-order), and the outputs are added.

Step 1: Write in powers of $z^{-1}$

$$\frac{Y(z)}{X(z)} = \frac{4(1-z^{-1})(1+1.2z^{-1}+z^{-2})}{(1+0.1z^{-1})(1-0.3z^{-1}+0.8z^{-2})}$$

The numerator and denominator have equal degree. So there is a constant term, plus a first-order term for the real pole and a second-order term for the complex pair, $z = 0.15 \pm j0.882$.

Step 2: Partial fractions (let $w = z^{-1}$)

$$G = C_0 + \frac{C_1}{1+0.1w} + \frac{D_0 + D_1w}{1 - 0.3w + 0.8w^2}$$

• $C_0 = \lim_{w\to\infty} G = \dfrac{4(-1)(1)}{(0.1)(0.8)} = -50$
• $C_1 = \dfrac{4(1-w)(1+1.2w+w^2)}{1-0.3w+0.8w^2}\Big|_{w=-10} = \dfrac{4(11)(89)}{84} = 46.619$
• $D_0$ and $D_1$: at $w = 0$, $4 = -50 + 46.619 + D_0$, so $D_0 = 7.381$. At $w = 1$ (that is, $z = 1$, where $G = 0$): $0 = -50 + \dfrac{46.619}{1.1} + \dfrac{7.381 + D_1}{1.5}$, so $D_1 = 4.0476$.

$$\frac{Y(z)}{X(z)} = -50 + \frac{46.619}{1 + 0.1z^{-1}} + \frac{7.381 + 4.0476z^{-1}}{1 - 0.3z^{-1} + 0.8z^{-2}}$$

Check: At $z = 2$, both forms give the same value. At $z \to \infty$: $-50 + 46.619 + 7.381 = 4$. ✔

Step 3: Difference equations of the branches

• $y_1(k) = -50x(k)$
• $w_2(k) = x(k) - 0.1w_2(k-1)$, and $y_2(k) = 46.619\,w_2(k)$
• $w_3(k) = x(k) + 0.3w_3(k-1) - 0.8w_3(k-2)$, and $y_3(k) = 7.381w_3(k) + 4.0476w_3(k-1)$
• $y(k) = y_1(k) + y_2(k) + y_3(k)$

Step 4: Block diagram

Overall structure:

        +-------------[-50]-------------+ y1
        |                               v
x(k)-o--+-->[ 46.619/(1+0.1z^-1) ]--->(S)--> y(k)
        |                       y2      ^
        +-->[ second-order branch ]-----+ y3

Branch 2:

x(k)-->(S)------------o--[1]-------->(S)->[46.619]-> y2(k)
        ^             |
        |          [z^-1]
        |             | w2(k-1)
        +<[-0.1]------o

Branch 3:

x(k)-->(S)------------o--[7.381]---->(S)--> y3(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w3(k-1)       |
       (S)<[0.3]------o--[4.0476]-----+
        ^             |
        |          [z^-1]
        |             | w3(k-2)
        +<[-0.8]------o

Total delays: $1 + 2 = 3$, the minimum for a third-order controller. A coefficient error in one branch affects only that branch's pole(s).