Parallel and ladder realization of a filter

Given:

$$G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}}$$

(a) Parallel scheme

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.

(b) Ladder scheme

Write the filter in positive powers of $z$:

$$G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}$$

In ladder programming, $G(z)$ is expanded as a continued fraction:

$$G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}$$

Each $1/(B_iz)$ is realized by a delay $z^{-1}/B_i$, and each $A_i$ by a constant.

Step 1: $A_0$.

$$G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}$$

So $A_0 = 2$.

Step 2: $B_1$.

$$\frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44}$$

So $B_1 = 1/1.4 = 0.7143$.

Step 3: $A_1$.

$$\frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12}$$

So $A_1 = 49/3 = 16.333$.

Step 4: $B_2$ and $A_2$.

$$\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05$$

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$.

$$G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}}$$

Ladder diagram:

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

Signal equations of the ladder ($o$ = take-off point, (S) = summer):

$$\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned}$$

With the values: $1/B_1 = 1.4$, $1/A_1 = 0.06122$, $1/B_2 = 28$, $1/A_2 = -0.05$. Only two delays are needed.