Ladder realization of a second-order filter

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

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

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

Write the filter in positive powers of $z$:

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

Step 1: Constant term $A_0$

$$G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.12)}{z^2+0.4z-0.12} = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}$$

So $A_0 = 2$.

Step 2: $B_1$

$$\frac{z^2+0.4z-0.12}{1.4z+0.44} = \frac{1}{1.4}z + \frac{0.0857z - 0.12}{1.4z+0.44}$$

So $B_1 = 1/1.4 = 0.7143$. The remainder is $0.4z - 0.3143z = 0.0857z$ (exactly $3z/35$).

Step 3: $A_1$

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

So $A_1 = 49/3 = 16.333$.

Step 4: $B_2$ and $A_2$

$$\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05 = B_2z + \frac{1}{A_2}$$

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

$$G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z + \cfrac{1}{-20}}}}$$

Ladder block 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 (these define the ladder diagram; $o$ marks a take-off point):

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

With the values: $1/B_1 = 1.4$, $1/A_1 = 0.0612$, $1/B_2 = 28$, $1/A_2 = -0.05$.

Simulating these equations gives the same impulse response as the original difference equation $y(k) = -0.4y(k-1) + 0.12y(k-2) + 2x(k) + 2.2x(k-1) + 0.2x(k-2)$. The realization needs only 2 delays.