Ladder programming of a third-order transfer function

Ladder programming expands $X(z) = \dfrac{z^3 + 4z^2 + 10z + 7}{z^3 + 3z^2 + 11z + 18}$ as a continued fraction:

$$G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}$$

Step 1: $A_0 = 1$

$$X(z) = 1 + \frac{z^2 - z - 11}{z^3 + 3z^2 + 11z + 18}$$

Step 2: $B_1$

$$\frac{z^3 + 3z^2 + 11z + 18}{z^2 - z - 11} = z + \frac{4z^2 + 22z + 18}{z^2 - z - 11}$$

So $B_1 = 1$.

Step 3: $A_1$

$$\frac{z^2 - z - 11}{4z^2 + 22z + 18} = 0.25 + \frac{-6.5z - 15.5}{4z^2 + 22z + 18}$$

So $A_1 = 0.25$.

Step 4: $B_2$

$$\frac{4z^2 + 22z + 18}{-6.5z - 15.5} = -0.6154z + \frac{12.4615z + 18}{-6.5z - 15.5}$$

So $B_2 = -0.6154$.

Step 5: $A_2$

$$\frac{-6.5z - 15.5}{12.4615z + 18} = -0.5216 + \frac{-6.1111}{12.4615z + 18}$$

So $A_2 = -0.5216$.

Step 6: $B_3$ and $A_3$

$$\frac{12.4615z + 18}{-6.1111} = -2.0392z - 2.9455$$

So $B_3 = -2.0392$ and $A_3 = -1/2.9455 = -0.3395$.

Exact values: $B_2 = -8/13$, $A_2 = -169/324$, $B_3 = -1458/715$, $A_3 = -55/162$.

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
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+

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

$$\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\ Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\ Y &= A_0X + Y_1 \end{aligned}$$

Gains in the diagram: $1/B_1 = 1$, $1/A_1 = 4$, $1/B_2 = -1.625$, $1/A_2 = -1.917$, $1/B_3 = -0.4904$, $1/A_3 = -2.9455$.

Substituting back reproduces $X(z)$ exactly. Three delays are used.