Ladder realization (denominator −0.1z⁻²)

In positive powers of $z$:

$$G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.1z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.1}$$

Ladder programming expands $G(z)$ as:

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

Step 1: $A_0$

$$G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.1)}{z^2+0.4z-0.1} = 2 + \frac{1.4z + 0.4}{z^2 + 0.4z - 0.1}$$

Step 2: $B_1$

$$\frac{z^2 + 0.4z - 0.1}{1.4z + 0.4} = 0.7143z + \frac{0.1143z - 0.1}{1.4z + 0.4}$$

The remainder is $0.4z - 0.7143(0.4)z = 0.1143z$ (exactly $4z/35$). So $B_1 = 0.7143$.

Step 3: $A_1$

$$\frac{1.4z + 0.4}{0.1143z - 0.1} = 12.25 + \frac{0.4 + 12.25(0.1)}{0.1143z - 0.1} = 12.25 + \frac{1.625}{0.1143z - 0.1}$$

So $A_1 = 12.25$.

Step 4: $B_2$ and $A_2$

$$\frac{0.1143z - 0.1}{1.625} = 0.07033z - 0.06154 = B_2z + \frac{1}{A_2}$$

So $B_2 = 0.07033$ and $A_2 = -16.25$.

$$G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{12.25 + \cfrac{1}{0.07033z + \cfrac{1}{-16.25}}}}$$

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

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

Numerical gains: $1/B_1 = 1.4$, $1/A_1 = 0.08163$, $1/B_2 = 14.22$, $1/A_2 = -0.06154$.

Substituting back reproduces $G(z)$ exactly, so the realization is correct. It uses two delay elements.